dr. robert a. durham, phd, pe dr. marcus o. durham, phd, pe … nutshell master.pdf · dr. robert...
TRANSCRIPT
ELECTRICAL ENGINEERING in a Nutshell
Dr. Robert A. Durham, PhD, PE
Dr. Marcus O. Durham, PhD, PE
A Quick Reference
The book is designed with four purposes:
A Study Guide to provide assistance with preparation for
professional examinations.
A Reference for investigating those occasional problems outside your
regular specialty.
A Curriculum for professional classes for maintaining your license.
A Synopsis for university students in electrical engineering and other
technical disciplines.
Electrical Engineering
in a Nutshell
A Quick Reference
Contact:
THEWAY Corp.
P.O. Box 33124
Tulsa, OK 74153
www.ThewayCorp.com
Cover Design:
Cover photo:
Printed in United States of America
First printing: March 2006
Library of Congress Control Number
ISBN: 978-0-9719324-7-?
Copyright 2006 by Robert A. Durham
All rights reserved under International Copyright Law. Contents and/or cover may not be
reproduced in whole or in part in any form without the express written consent of the Publisher.
Contents Chapter 1 - Circuit Elements and Analysis ................................................................................................. 1-1
1.1 Introduction ............................................................................................................................... 1-1
1.2 Circuit Element Description & Construction .............................................................................. 1-3
1.2.1 Resistor ..................................................................................................................................... 1-3
1.2.2 Inductor ..................................................................................................................................... 1-3
1.3 Circuit Basics .............................................................................................................................. 1-5
1.3.1 Element Manipulation .............................................................................................................. 1-5
1.3.2 Element Combinations .............................................................................................................. 1-6
1.3.3 Sources ...................................................................................................................................... 1-7
1.3.4 Circuit Laws ............................................................................................................................... 1-8
1.4 Circuit Analysis ........................................................................................................................... 1-9
1.4.1 Loop Current Method ............................................................................................................. 1-10
1.4.2 Node Voltage Method ............................................................................................................. 1-12
1.4.3 Equivalent Circuit Method ...................................................................................................... 1-13
1.5 Single Source and Impedance .................................................................................................. 1-14
1.6 Steady State AC Analysis .......................................................................................................... 1-18
1.7 Electric Fields – Electrostatics .................................................................................................. 1-20
1.8 Magnetic Fields ........................................................................................................................ 1-21
1.9 Maxwell ................................................................................................................................... 1-24
Chapter 1 Problems ............................................................................................................................. 1-25
Chapter 2 Waveforms ................................................................................................................................ 2-1
2.1 Introduction ..................................................................................................................................... 2-1
2.2 Waveforms .................................................................................................................................... 2-1
2.3 Transients ......................................................................................................................................... 2-2
2.3.1 First Order Transients ............................................................................................................... 2-2
2.3.2 RL Circuits .................................................................................................................................. 2-4
2.3.3 RC Circuits ..................................................................................................................................... 2-5
2.4 LaPlace ............................................................................................................................................. 2-6
2.5 LaPlace Operational Rules ................................................................................................................ 2-8
2.5.1 Partial Fraction Expansion ......................................................................................................... 2-9
2.5.2 Alternate Approach ................................................................................................................. 2-10
2.6 Fourier Series ................................................................................................................................. 2-11
2.7 Signals - Modulation ...................................................................................................................... 2-15
2.7.1 Modulation Types ................................................................................................................... 2-15
2.7.2 Amplitude Modulation (AM) ................................................................................................... 2-17
2.7.3 Angle Modulation ................................................................................................................... 2-19
2.7.4 Frequency Modulation (FM) ................................................................................................... 2-20
2.7.5 Phase Modulation ................................................................................................................... 2-21
2.7.6 Sampled Messages .................................................................................................................. 2-22
2.7.7 Digital - Pulse Modulation ....................................................................................................... 2-22
2.8 Signal transmission ........................................................................................................................ 2-23
2.8.1 dBm ......................................................................................................................................... 2-23
2.8.2 Noise ....................................................................................................................................... 2-23
2.8.3 Propagation - Transmit ........................................................................................................... 2-25
2.8.4 Reflections ............................................................................................................................... 2-26
2.9 RLC System Response .................................................................................................................... 2-28
2.9.1 RLC Equations .......................................................................................................................... 2-28
2.9.2 System Response .................................................................................................................... 2-29
2.9.3 Characteristic Transfer ............................................................................................................ 2-29
2.9.4 Resonance ............................................................................................................................... 2-29
2.9.5 Series Parallel Duality .............................................................................................................. 2-32
2.9.6 First Order ............................................................................................................................... 2-32
Chapter 2 Problems ............................................................................................................................. 2-33
Chapter 2 Waveforms ................................................................................................................................ 2-1
2.1 Introduction ..................................................................................................................................... 2-1
2.2 Waveforms .................................................................................................................................... 2-1
2.3 Transients ......................................................................................................................................... 2-2
2.3.1 First Order Transients ............................................................................................................... 2-2
2.3.2 RL Circuits .................................................................................................................................. 2-4
2.3.3 RC Circuits ..................................................................................................................................... 2-5
2.4 LaPlace ............................................................................................................................................. 2-6
2.5 LaPlace Operational Rules ................................................................................................................ 2-8
2.5.1 Partial Fraction Expansion ......................................................................................................... 2-9
2.5.2 Alternate Approach ................................................................................................................. 2-10
2.6 Fourier Series ................................................................................................................................. 2-11
2.7 Signals - Modulation ...................................................................................................................... 2-15
2.7.1 Modulation Types ................................................................................................................... 2-15
2.7.2 Amplitude Modulation (AM) ................................................................................................... 2-17
2.7.3 Angle Modulation ................................................................................................................... 2-19
2.7.4 Frequency Modulation (FM) ................................................................................................... 2-20
2.7.5 Phase Modulation ................................................................................................................... 2-21
2.7.6 Sampled Messages .................................................................................................................. 2-22
2.7.7 Digital - Pulse Modulation ....................................................................................................... 2-22
2.8 Signal transmission ........................................................................................................................ 2-23
2.8.1 dBm ......................................................................................................................................... 2-23
2.8.2 Noise ....................................................................................................................................... 2-23
2.8.3 Propagation - Transmit ........................................................................................................... 2-25
2.8.4 Reflections ............................................................................................................................... 2-26
2.9 RLC System Response .................................................................................................................... 2-28
2.9.1 RLC Equations .......................................................................................................................... 2-28
2.9.2 System Response .................................................................................................................... 2-29
2.9.3 Characteristic Transfer ............................................................................................................ 2-29
2.9.4 Resonance ............................................................................................................................... 2-29
2.9.5 Series Parallel Duality .............................................................................................................. 2-32
2.9.6 First Order ............................................................................................................................... 2-32
Chapter 2 Problems ............................................................................................................................. 2-33
Chapter 1 - Circuit Elements and Analysis ................................................................................................. 1-1
1.1 Introduction ............................................................................................................................... 1-1
1.2 Circuit Element Description & Construction .............................................................................. 1-3
1.2.1 Resistor ..................................................................................................................................... 1-3
1.2.2 Inductor ..................................................................................................................................... 1-3
1.3 Circuit Basics .............................................................................................................................. 1-5
1.3.1 Element Manipulation .............................................................................................................. 1-5
1.3.2 Element Combinations .............................................................................................................. 1-6
1.3.3 Sources ...................................................................................................................................... 1-7
1.3.4 Circuit Laws ............................................................................................................................... 1-8
1.4 Circuit Analysis ........................................................................................................................... 1-9
1.4.1 Loop Current Method ............................................................................................................. 1-10
1.4.2 Node Voltage Method ............................................................................................................. 1-12
1.4.3 Equivalent Circuit Method ...................................................................................................... 1-13
1.5 Single Source and Impedance .................................................................................................. 1-14
1.6 Steady State AC Analysis .......................................................................................................... 1-18
1.7 Electric Fields – Electrostatics .................................................................................................. 1-20
1.8 Magnetic Fields ........................................................................................................................ 1-21
1.9 Maxwell ................................................................................................................................... 1-24
Chapter 1 Problems ............................................................................................................................. 1-25
Chapter 3 - Power Analysis ........................................................................................................................ 3-1
3.1 Introduction ..................................................................................................................................... 3-1
3.2 Power Definitions ............................................................................................................................. 3-2
3.2.1 – Z,R,X,θ; S,P,Qpf Conversions .................................................................................................. 3-4
3.3 Three-Phase AC ................................................................................................................................ 3-5
3.3.1 Y-∆ Relationships ....................................................................................................................... 3-6
3.3.2 Phase Sequence ........................................................................................................................ 3-7
3.4 Power Transfer across a Reactance ................................................................................................. 3-8
3.5 Power One-Line ................................................................................................................................ 3-9
3.6 Power Problem Plan ....................................................................................................................... 3-10
3.7 Mechanical Power .......................................................................................................................... 3-11
3.8 Electric Machinery ......................................................................................................................... 3-13
3.8.1 Basics ....................................................................................................................................... 3-13
3.8.2 Machine Models ..................................................................................................................... 3-14
3.8.3 Machine Tests ......................................................................................................................... 3-18
3.8.4 Rotating Machines .................................................................................................................. 3-19
3.8.5 Power Conversion ................................................................................................................... 3-21
3.8.6 Losses ...................................................................................................................................... 3-22
3.8.7 Performance............................................................................................................................ 3-23
3.9 Transformer ................................................................................................................................... 3-24
3.9.1 Transformer Tests ................................................................................................................... 3-25
3.9.2 Transformer Turns .................................................................................................................. 3-28
3.10 DC Machines ................................................................................................................................ 3-31
3.11 AC Machines - Synchronous ......................................................................................................... 3-33
3.11.1 Synchronous Machine ........................................................................................................... 3-33
3.11.2 Induction Machine .................................................................................................................... 3-35
3.12 Transmission Lines ....................................................................................................................... 3-37
3.13 Per Unit Notation ......................................................................................................................... 3-39
3.14 Short Circuit Considerations ........................................................................................................ 3-41
3.14.1 Introduction .......................................................................................................................... 3-41
3.14.2 Fault Analysis ........................................................................................................................ 3-43
3.14.3 Symmetrical Components ..................................................................................................... 3-44
3.14.4 Ratings & Reactances ............................................................................................................ 3-45
3.14.5 Short Circuit Study ................................................................................................................ 3-46
3.14.6 Unbalanced Faults ................................................................................................................. 3-48
3.14.7 Faults with Rotating Machines .............................................................................................. 3-49
3.14.8 Fault Illustrations .................................................................................................................. 3-50
Chapter 3 Problems ............................................................................................................................. 3-55
Chapter 4 - Electronics ............................................................................................................................... 4-1
4.1 Introduction ..................................................................................................................................... 4-1
4.1.1 Solid State Device Characteristics ............................................................................................. 4-2
4.2 Boundary Conditions ........................................................................................................................ 4-4
4.3 Diodes & Rectifiers ........................................................................................................................... 4-6
4.3.1 Diodes ....................................................................................................................................... 4-6
4.3.2 Rectifiers & Clippers .................................................................................................................. 4-7
4.4 Operational Amplifiers ..................................................................................................................... 4-8
4.5 Transistors ...................................................................................................................................... 4-10
4.5.1 Bias vs. Small Signal ................................................................................................................. 4-10
4.5.2 Transistor Mathematical Relationships .................................................................................. 4-13
4.5.3 General Two-port Models ....................................................................................................... 4-15
4.5.4 BJT Transistor Models ............................................................................................................. 4-15
4.5.5 FET Transistor Models ............................................................................................................. 4-29
4.6 State Space ..................................................................................................................................... 4-35
4.6.1 The 6-Minute Approach .......................................................................................................... 4-35
4.6.2 Description .............................................................................................................................. 4-37
Chapter 4 Problems ............................................................................................................................. 4-42
Chapter 5 - Controls ................................................................................................................................... 5-1
5.1 Introduction ..................................................................................................................................... 5-1
5.2 Controls Basics ................................................................................................................................. 5-2
5.2.1 Introduction .............................................................................................................................. 5-2
5.2.2 Block Diagrams .......................................................................................................................... 5-2
5.2.3 Steady State Errors .................................................................................................................... 5-4
5.2.4 Time Response .......................................................................................................................... 5-5
5.3 Routh-Hurwitz Criteria ..................................................................................................................... 5-6
5.3.1 Introduction .............................................................................................................................. 5-6
5.3.2 Rules .......................................................................................................................................... 5-6
5.3.3 Routh-Hurwitz – Special Cases .................................................................................................. 5-6
5.4 Root Locus ..................................................................................................................................... 5-11
5.4.1 Introduction ............................................................................................................................ 5-11
5.4.2 Object of Root Locus ............................................................................................................... 5-12
5.4.3 Rules for Root Locus Construction .......................................................................................... 5-13
5.5 Frequency Response Plots ............................................................................................................. 5-18
5.5.1 Introduction ............................................................................................................................ 5-18
5.5.2 Bode Plots – Basic rules .......................................................................................................... 5-19
5.5.3 – Phase Margin and Gain Margin ........................................................................................... 5-22
5.5.3 Polar - Nyquist Plot. ................................................................................................................ 5-23
5.6 Analog Filters ................................................................................................................................. 5-25
Chapter 5 Problems ............................................................................................................................. 5-30
Problem 5-1 ...................................................................................................................................... 5-30
Chapter 6 Digital ........................................................................................................................................ 6-1
6.1 Introduction ..................................................................................................................................... 6-1
6.2 Binary (Digital) Systems ................................................................................................................... 6-1
6.2.1 Number Systems ....................................................................................................................... 6-1
6.2.2 Binary System Wiring ................................................................................................................ 6-4
6.2.3 The Huntington Postulates ....................................................................................................... 6-6
6.2.4 Basic Digital Gates ..................................................................................................................... 6-7
6.3 Karnaugh Maps ................................................................................................................................ 6-8
6.3.1 Construction / Simplification of Karnaugh Maps: ..................................................................... 6-9
6.4 Design w/ Multiplexer .................................................................................................................... 6-11
6.5 Decoder .......................................................................................................................................... 6-14
6.5.1 74156 DECODER ...................................................................................................................... 6-15
6.6 Flip Flops / Latch ............................................................................................................................ 6-16
6.6.1 General – Flip Flop Types ........................................................................................................ 6-17
6.6.2 Counter ................................................................................................................................... 6-18
6.6.3 Sequence Detector (Random) ................................................................................................. 6-23
Chapter 7 - Economics – Time Value of Money ......................................................................................... 7-1
7.1 Introduction ..................................................................................................................................... 7-1
7.2 Time and Interest ............................................................................................................................. 7-2
7.2.1 Uniform series ........................................................................................................................... 7-3
7.2.2 Gradient .................................................................................................................................... 7-5
7.2.3 Nominal Interest or APR ........................................................................................................... 7-6
7.2.4 Perpetual and Rule of 72........................................................................................................... 7-7
7.3 Rate of Return .............................................................................................................................. 7-8
7.3.1 Incremental Analysis ................................................................................................................. 7-9
7.3.2 Payback ................................................................................................................................... 7-10
7.3.3 Benefit Cost Ratio ................................................................................................................... 7-11
7.3.4 Tax Implications ...................................................................................................................... 7-12
7.4 Table of Terminology ..................................................................................................................... 7-13
7.5 Commentary .................................................................................................................................. 7-13
7.6 Review ............................................................................................................................................ 7-13
Chapter 7 Problems ............................................................................................................................. 7-15
Bibliography ......................................................................................................................................... 7-16
Chapter 8 Business Ethics .......................................................................................................................... 8-1
240.15 Rules of Professional Conduct ................................................................................................... 8-1
Chapter 9 Codes and Standards ................................................................................................................. 9-1
9.1 Introduction ..................................................................................................................................... 9-1
9.2 NEC Synopsis .................................................................................................................................... 9-1
General: Article 100 ........................................................................................................................... 9-1
Identification of grounded conductors: Article 200........................................................................... 9-1
Branch circuits: Article 210 ................................................................................................................ 9-1
Feeders: Article 215 ........................................................................................................................... 9-1
Branch-circuit, feeder, and service calculations: Article 220 ............................................................. 9-2
Overcurrent protection: Article 240 .................................................................................................. 9-2
Grounding: Article 250 ....................................................................................................................... 9-2
Conductors for general wiring: Table 310.16 ff ................................................................................. 9-2
Motors: Article 430 ............................................................................................................................ 9-2
Tables: Chapter 9 ............................................................................................................................... 9-2
Examples: Annex D ............................................................................................................................. 9-3
9.3 Motor Installation Tables ................................................................................................................. 9-4
NEC 240.6(A) Standard Ampere Ratings for Fuses & Circuit Breakers ........................................... 9-4
Table 310.16 Allowable Ampacities of Insulated Conductors Rated 0 Through 2000 Volts, 60°C
Through 90°C (140°F Through 194°F), Not More Than Three Current-Carrying Conductors in
Raceway, Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30°C (86°F) ........... 9-5
Table 310.15(B)(2)(a) Adjustment Factors for More Than Three Current-Carrying Conductors in a
Raceway or Cable ............................................................................................................................... 9-6
Table 310.15(B)(6) Conductor Types and Sizes for 120/240-Volt, 3-Wire, Single-Phase Dwelling
Services and Feeders. ......................................................................................................................... 9-6
Table 430.7(B) Locked-Rotor Indicating Code Letters .................................................................... 9-8
Table 430.52 Maximum Rating or Setting of Motor Branch-Circuit Short-Circuit and Ground-Fault
Protective Devices .............................................................................................................................. 9-8
Table 430.91 Motor Controller Enclosure Selection ...................................................................... 9-9
Table 430.248 Full-Load Currents in Amperes, Single-Phase Alternating-Current Motors ......... 9-11
Table 430.250 Full-Load Current, Three-Phase Alternating-Current Motors .............................. 9-12
Table 430.251(A) Conversion Table of Single-Phase Locked- Rotor Currents for Selection of
Disconnecting Means and Controllers as Determined from Horsepower and Voltage Rating ....... 9-13
Table 430.251(B) Conversion Table of Polyphase Design B, C, and D Maximum Locked-Rotor
Currents for Selection of Disconnecting Means and Controllers as Determined from Horsepower
and Voltage Rating and Design Letter.............................................................................................. 9-14
Table 8 Conductor Properties ...................................................................................................... 9-15
Table 9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C
(167°F) — Three Single Conductors in Conduit ............................................................................... 9-16
Table C.4 Maximum Number of Conductors or Fixture Wires in Intermediate Metal Conduit (IMC)
(Based on Table 1, Chapter 9) .......................................................................................................... 9-18
NEMA Controller Size for Motors, Transformers, & Capacitors ...................................................... 9-21
NEMA Table 11 Typical Characteristics and Applications of Fixed Frequency Small and Medium AC
Squirrel-Cage Induction Motors ....................................................................................................... 9-23
NEMA MOTOR DIMENSIONS ........................................................................................................... 9-24
NEMA MOTOR DIMENSIONS – 2 ...................................................................................................... 9-25
Chapter 9 Codes and Standards ................................................................................................................. 9-1
9.1 Introduction ..................................................................................................................................... 9-1
9.2 NEC Synopsis .................................................................................................................................... 9-1
General: Article 100 ........................................................................................................................... 9-1
Identification of grounded conductors: Article 200........................................................................... 9-1
Branch circuits: Article 210 ................................................................................................................ 9-1
Feeders: Article 215 ........................................................................................................................... 9-1
Branch-circuit, feeder, and service calculations: Article 220 ............................................................. 9-2
Overcurrent protection: Article 240 .................................................................................................. 9-2
Grounding: Article 250 ....................................................................................................................... 9-2
Conductors for general wiring: Table 310.16 ff ................................................................................. 9-2
Motors: Article 430 ............................................................................................................................ 9-2
Tables: Chapter 9 ............................................................................................................................... 9-2
Examples: Annex D ............................................................................................................................. 9-3
9.3 Motor Installation Tables ................................................................................................................. 9-4
NEC 240.6(A) Standard Ampere Ratings for Fuses & Circuit Breakers ........................................... 9-4
Table 310.16 Allowable Ampacities of Insulated Conductors Rated 0 Through 2000 Volts, 60°C
Through 90°C (140°F Through 194°F), Not More Than Three Current-Carrying Conductors in
Raceway, Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30°C (86°F) ........... 9-5
Table 310.15(B)(2)(a) Adjustment Factors for More Than Three Current-Carrying Conductors in a
Raceway or Cable ............................................................................................................................... 9-6
Table 310.15(B)(6) Conductor Types and Sizes for 120/240-Volt, 3-Wire, Single-Phase Dwelling
Services and Feeders. ......................................................................................................................... 9-6
Table 430.7(B) Locked-Rotor Indicating Code Letters .................................................................... 9-8
Table 430.52 Maximum Rating or Setting of Motor Branch-Circuit Short-Circuit and Ground-Fault
Protective Devices .............................................................................................................................. 9-8
Table 430.91 Motor Controller Enclosure Selection ...................................................................... 9-9
Table 430.248 Full-Load Currents in Amperes, Single-Phase Alternating-Current Motors ......... 9-11
Table 430.250 Full-Load Current, Three-Phase Alternating-Current Motors .............................. 9-12
Table 430.251(A) Conversion Table of Single-Phase Locked- Rotor Currents for Selection of
Disconnecting Means and Controllers as Determined from Horsepower and Voltage Rating ....... 9-13
Table 430.251(B) Conversion Table of Polyphase Design B, C, and D Maximum Locked-Rotor
Currents for Selection of Disconnecting Means and Controllers as Determined from Horsepower
and Voltage Rating and Design Letter.............................................................................................. 9-14
Table 8 Conductor Properties ...................................................................................................... 9-15
Table 9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C
(167°F) — Three Single Conductors in Conduit ............................................................................... 9-16
Table C.4 Maximum Number of Conductors or Fixture Wires in Intermediate Metal Conduit (IMC)
(Based on Table 1, Chapter 9) .......................................................................................................... 9-18
NEMA Controller Size for Motors, Transformers, & Capacitors ...................................................... 9-21
NEMA Table 11 Typical Characteristics and Applications of Fixed Frequency Small and Medium AC
Squirrel-Cage Induction Motors ....................................................................................................... 9-23
NEMA MOTOR DIMENSIONS ........................................................................................................... 9-24
NEMA MOTOR DIMENSIONS – 2 ...................................................................................................... 9-25
Page | 1-1
Chapter 1 - Circuit Elements and Analysis
1.1 Introduction The book is an outline and review of electrical engineering. The material is designed with four
purposes.
A Study Guide to provide assistance with preparation for professional examinations.
A Reference for investigating those occasional problems outside your regular specialty.
A Curriculum for professional classes for maintaining your license.
A Synopsis for university students in electrical engineering and other technical disciplines.
A generic electrical system covers equipment from a generator or power supply through controls to
a motor or load. In some problems, the system is analyzed as a whole. In small signal analysis, models
are employed.
Electrical systems always convert an available energy source to electrical energy. The electricity is
then conveniently transferred to a load which converts the electrical energy back to another energy
form.
Electrical systems, as all physical systems, operate based on the Trinity Principle which states: Any
item than can be uniquely identified can be further explained by three components. A corollary states:
Two of the three components that identify a system are similar and project into one plane, while the
third component is dissimilar and operates in an orthogonal plane.
The necessary terms for an electrical system can be identified using this grouping of three
quantities.
In a system, only 3 things are measured: voltage (V), current (I), and time (t)
i = through = flow rate = Amps = dq
dt
v = across = potential = Volts = 2 1v v
t = time
GENERATOR METER TRANSFORMER CONTROLLER MOTOR
Page | 1-2
From these three measurements, only three things can be calculated.
Z = impedance = ratio = Ohms (Ω) V
I
S = power = product = Watts (w) *VI =dE
dt
= angle = time diff = degrees (seconds) = 2 1t t
Page | 1-3
1.2 Circuit Element Description & Construction Impedance has only three linear (passive) elements available in a circuit: resistor, inductor, and
capacitor
1.2.1 Resistor
A resistor converts electrical energy into mechanical energy in the form of heat.
Resistance is measured in ohms (Ω). The voltage across a resistor, VR, is the product
of the resistance and the current through the resistor.
RV Ri
A resistor is simply a length of conductive material. The resistance of a piece of material is given by
T
lR
A
Where
T - resistivity of construction material
l is length of material, A is cross-sectional area
The resistivity of the construction material, T , is
temperature dependent.
20 201 20T T
1.2.2 Inductor
An inductor converts electrical energy into magnetic energy. The inductor stores
energy in a magnetic field. The inductance is measured in Henry (H). Voltage across an
inductor is the product of the inductance and the derivative of the current through the
inductor
L
diV L
dt
The energy stored in an inductor comes from
212LdW Vdq W Li
Because of the magnetic fields, the current in an inductor cannot change
instantly.
At its most basic form, an inductor is a coil of wire wrapped around a
closed magnetic path. The inductance created is calculated from
2 2N A NL
l
R
n-# of
turns
l – average path length
820 1.8 10cu 3
20 3.9 10cu
620 1.08 10nichrome 5
20 17 10nichrome
R
+ VR - i
+ VL -i
L
Page | 1-4
Where
μ – permeability of magnetic path
r o
74 10 /o henry m
1.2.3 Capacitor
A Capacitor stores electrical energy in an electric field. The capacitance is measured in Farads (F).
The voltage across a capacitor is
1 1cv dq i dt
C C
The current through a capacitor is found from
cdvi C
dt
The amount of energy stored in a capacitor is shown.
212L cdW Vdq CVdv W CV
Because of the electric field, the voltage across a capacitor (vc) cannot change instantaneously.
A capacitor is constructed of two conductors, separated by some dielectric material. The
capacitance generated is a product of the permittivity and the size of the conductors.
AC
l
Where
A – cross-sectional area of conductors
l – separation of conductors
ε – permittivity of dielectric material
The permittivity of the dielectric material is referenced to the permittivity of free space
o r
Where
o - permittivity of free space - 9110
36 ; r - dielectric constant (relative permittivity)
(copper) 2r
(amorphous steel) 2000r
(laminated steel) 6000r
( ) 5 10r glass
( ) 16r germanium
(titaniumdioxide) 86 173r
+ VC -i
C
Page | 1-5
1.3 Circuit Basics
1.3.1 Element Manipulation
Because of circuit combinations, reciprocal values of elements often arise. Where impedance (Z) is
opposition to current flow, the reciprocals of impedance indicate the ease of current flow. Impedance is
made up of two components, the resistance (real portion) and the reactance, which is the combination
of inductance and capacitance (imaginary portion). Impedance is the sum of the resistance, and the
reactance*j.
Z R jX
Admittance (Y) is the reciprocal of impedance, and is measured in mhos ( ) 1
YZ
Conductance (G) is the reciprocal of resistance: 1
GR
Susceptance (GB) is the reciprocal of reactance: 1
BGX
Elastance is the reciprocal of capacitance
Reluctance ( ) is the reciprocal of inductance: 1
L
1
1
1
1
B
Y mhoZ
G mhoR
G siemens SX
amp turns
L weber
Impedance is the sum of resistance and reactance: Z R jX
Admittance is the sum of conductance and susceptance: BY G jG
Page | 1-6
1.3.2 Element Combinations
The elements defined in the previous topics can be combined into combinations of impedances. The
configuration of the combinations determines the mathematical treatment of the impedances (Z). The
bases of all combinations are the series and parallel connections.
Series Impedances are added,
1 2 3totalZ Z Z Z
For resistors, the impedance is equal to the resistance, RZ R
1 2 3R totalZ R R R
The impedance of inductors is frequency dependant and is proportional to the inductance, LZ L .
Series inductors at the same frequency are treated the same as resistances.
1 2 3( )L totalZ L L L
Capacitors are a different animal. The impedance of capacitors is frequency dependant, but is
inversely proportional to the capacitance, 1
CZC
. For this reason, when capacitors are in series, the
series resistance is the sum of the elastances, or 1
C
1 2 3
1 1 1C totalZ
C C C
To calculate the impedance of elements in parallel, sum the reciprocals of the impedances, or the
admittance of each element. This gives the admittance of the entire circuit. This
is the reciprocal of the impedance of the circuit.
1 2 3
1 1 1 1total
total
YZ Z Z Z
The special case of two parallel impedances can be found by taking the product of the impedances
and dividing by the sum of the impedances.
1 2
1 2
total
Z ZZ
Z Z
For resistors, the admittance of the circuit is the sum of the conductances of the individual
elements.
1 2 3
1 2 3
1 1 1 1
R total
G G GZ R R R
For inductors, the admittance of the circuit is proportional to the sum of the reluctance of the
individual inductors.
Z1 Z2 Z3
Z1 Z2 Z3
Page | 1-7
1 2 3
1 1 1 1
L totalZ L L L
Again, capacitors are a special case. For capacitors at the same frequency, the admittance is
proportional to the sum of the capacitances.
1 2 3
1
C total
C C CZ
1.3.3 Sources
There are is a lower and upper limit on impedance. A short circuit is near zero impedance. An open
circuit is near infinite impedance. The source is the limit on voltage or current.
A short Circuit is defined as v=0 for any i
i
v=0
An open Circuit is defined as i=0 for any v
i=0
v+
A voltage source maintains v=Vo for any i
+v0
i
Current sources maintain i=Io for any v
I0
DC or constant sources are represented by
DCor
AC sources where cosov V t are
represented by
AC
Page | 1-8
1.3.4 Circuit Laws
Two circuit laws are used in all circuit analysis. These circuit laws adhere to the Conservation of
Energy – There is nothing new under the sun; or, more traditionally, the sum of the energy in a closed
system is zero
0E
pqE
t
pv
t
qi
t
Kirchhoff’s Current Law (KCL) relates to currents entering a node. KCL states that the sum of
currents entering a node is equal to zero (0).
i1
i2
i3
i4
1 2 3 40 0ni i i i i
By convention, if current enters a node, it is considered negative. Current leaving a node is
considered positive.
Kirchhoff’s Voltage Law (KVL) relates to voltages in a circuit loop. KVL states that the sum of voltages
in a loop is equal to zero (0)
1 20 0n R Lv v v v v
1 2
diV V iR L
dt
By convention, if current goes into “+” of voltage source, then the
voltage is considered positive. If current goes into “–“ of voltage
source, then the voltage is considered negative.
V1
RL
V2
i
VLVR
Page | 1-9
1.4 Circuit Analysis Five rules can aid in any circuit analysis.
Assume a consistent group of currents & voltages for each element (R, L, C)
Conservation: Write equations using Kirchhoff’s Law (KCL or KVL)
Use element definitions (Ohm’s Law)
Combine equations in terms of unknowns
Solve simultaneously using elimination or Cramer’s rule
Circuit analysis generally has a source (external forcing function) and elements (opposition). The
“answer” to a circuits analysis problem is voltage & current across an element, or a derivative such as
power or energy.
Three methods of solving circuit analysis
Loop current
Node voltage
Equivalent circuits
Page | 1-10
1.4.1 Loop Current Method
Assign loops – “window pane” method
Assume voltage polarity (+ into Z)
Write KVL around loop
Substitute Ohm’s law (element) for the voltage across the Z
Solve for unknowns
Example
6V
2Ω
10Ω
4Ω
8Vi1 i2
Find the current in the 10Ω resistor
KVL Left Loop
6 2 10 0VV V V KVL left loop
2 12V i Ohm’s law
10 1 210( )V i i
1 1 26 2 10( ) 0i i i Substitute
KVL Right Loop
8 10 4 0VV V V KVL right loop
4 24V i Ohm’s law
10 1 210( )V i i
2 1 28 10( ) 4 0i i i Substitute
1 2
1 2
12 10 6
10 14 8
i i
i i
Solve equations simultaneously
Page | 1-11
Cramer’s Rule
If solving for in – substitute RHS for in in numerator
Use matrix of coefficients in denominator
Manipulate using cross-multiplication
1
2
6 10
8 14 (6)(14) ( 10)( 8) 40.059
12 10 (12)(14) ( 10)( 10) 68
10 14
0.529
i A
i A
Now calculate the desired value.
10 1 2 0.588i i i A
Page | 1-12
1.4.2 Node Voltage Method
Assign node voltage @ each connection (Ground = ref = 0V)
Assume branch current direction (e.g. always leaves node)
Write KCL @ each node
Substitute elements (Ohm’s law) for current
Solve
Example
6V
2Ω
10Ω
4Ω
8V
a cb
Find the current in the 10Ω resistor
KCL @ node b
Assume all current leaves node
2 10 4 0i i i KCL node b
2
6
2 2
b a bV V Vi
Ohm’s Law
10
4
0
10
8
4 4
b
b c b
Vi
V V Vi
6 0 80
2 10 4
b b bV V V
Substitute
60 10 40 5 2 0b b bV V V Solve equation
17 100
5.88
b
b
V
V v
Now calculate desired value
10
0 5.880.588
10 10
bV vi A
Page | 1-13
1.4.3 Equivalent Circuit Method
Three types of equivalent circuit analysis can be used to reduce problem complexity. These are
combining series impedance, combining parallel impedance, and using ratios of impedance to divide
current or voltage.
For series impedances, use KVL - the current is the same, so sum
voltages in the path.
1 2
1
2
1 1
1 2
1 1 2
1 2
( )
R R
R
R
eq
eq
v v v
v i R
v i R
v i R R iR
R R R
For parallel impedances, use KCL – the voltage is the same,
so sum the current.
1 2
1
1
2
2
1 2
1 2
1 2 1 2
1 1( )
R
1 1 1
eq
eq
eq
i i i
vi
R
vi
R
vi v
R R
R RR
R R R R R
Voltage or current dividers are used to determine voltage across or current through a circuit
element.
A current divider is used in parallel circuits with the same voltage applied on the branches
21
1 2
( )Ropposite
i i iR R
A voltage divider is used in series circuits with same current in elements
1
adjacentv v
R1 R2
i1
R1
R2
i
i1
i2
Page | 1-14
1.5 Single Source and Impedance Thevenin & Norton equivalents are used to simplify any circuit to an equivalent impedance and a
single source (driver). These are calculated by taking the sources to the limit. L is manipulated the same
as R, C in series is manipulated like R in parallel. The following rules help in finding the Thevenin or
Norton equivalent circuits.
Find Zeq
Replace independent V sources by short
Replace independent I sources by open
Calculate resistance – use series / parallel rules
If the voltage supplied at the node and the current into the node are known, then V
ZI
Find Vth
Open circuit terminals
Leave sources active
Calculate the voltage across the open terminal. This will be the voltage across an impedance.
Find Is
Short circuit terminals
Leave sources active
Calculate the current through the shorted terminal. This will be the current that shorts (bypasses) an
impedance
Page | 1-15
Illustration
R1
R2
1
2
v
Replace voltages
R1
R2
1
2
1 2
1 2
eq
R RR
R R
Vth = voltage across R2 (Open terminals and calculate v)
R1
1
2
VTH
v1
Series circuit with voltage divider
2
1 2
th
RV v
R R
Alternatively
1 2
22
1 2
th
vi
R R
vRV iR
R R
Isc=I through terminals → R2=∞, it is bypassed, and Isc= I in series circuit
Page | 1-16
R1
1
2
ISC
v1
1
SC
vI I
R
Check
2
1 2 1 2
1 2
1
THeq
SC
Rv
R RV R RR
vI R R
R
Page | 1-17
Example
6v
2Ω
10Ω
4Ω
8v
Equivalent to
6v
2Ω
10Ω
4Ω
8v
Convert
2Ω 10Ω4Ω3A 2A
Simplify
4/3 Ω 10Ω3A+2A=5A
i
2 / /4
2* 4 4
2 4 3
THR
Current Divider for 10Ω
4
35 0.5884
103
i A A
Page | 1-18
1.6 Steady State AC Analysis Steady state AC analysis is used when the source is operating at a constant frequency. Because of
inductors and capacitors, an angle change or phase shift is created in the voltage and current drops
associated with the impedance.
Frequency ( f ) is related to the time it takes for a waveform to repeat. It is the number of
repetitions or cycles in a second (Hz).
1
2f
T
where
ω = angular frequency in radians per second
On many power systems the frequency is fixed or constant for normal, or steady
state operations. The phasor transform can be applied to provide a concise phasor
notation at the fixed frequency.
P *Vpeak cos (ωt + θ)+ = VRMSθ
2peak RMSV V
where
θ = phase shift between voltage and current cause by L and C.
Because of the phase shift, complex numbers are often used to manipulate the mathematics.
Complex numbers involve both a real and imaginary part. The imaginary part is defined as multiplied by
“j” which is referred to as the imaginary unit.
;
Euler’s formula allows conversion of rectangular coordinates to polar. It gives a representation of
the angle impact on the sinusoids.
cos sinje j
cos2
j je e
sin2
j je e
j
Time varying voltage (AC) is defined based on the magnitude and angles described above.
21j
j e
21 j
j ej
V
θ
Page | 1-19
( 2 )j j te RMSv R V e e
2 cos( )v V t
Impedance is the phasor representation of resistors, inductors, and capacitors.
Z R jX
V IZ
For a resistor
RV IR
RZ R
For an Inductor
LV j LI
LZ j L
2LZ j fL
For a Capacitor
C
IV
j C
1CZ
j C
2
C
jZ
fC
The combination of resistor, inductor, and capacitor can be shown in a phasor of voltages. In series
circuits, the current will be the same in all elements, so the impedance is proportional to the voltage. In
a parallel circuit the voltage is the same across all elements, so the impedance is proportional to the
current.
VC VR
VL V
I
R
+ VR - i
+ VL -i
L
+ VC -i
C
+ VL -
iLR C
+ VR - + VC -
+V
Page | 1-20
1.7 Electric Fields – Electrostatics
W(Energy) Fs Vq NI energy conversion
(Electricintensity) /F V
V meterq s
E (closed loop) 2s r
2coulomb(electric density)
qE
A m D
q = charge 1 electron = 1.6021x10-19 Coulomb
Permittivity is the dielectric or charge insulation material property.
r o
9136 10
o
Force on charge 2 due to charge 1
1 2
24
q qF
r
Electric field intensity at point 2 due to point charge at point 1.
1
24
qF
q r E
Radial electric field due to line charge on z-axis
( / )
2
L C m
r
E
coulombcharge density -
meterL
Electric field due to sheet charge in x-y plane
2( / )
2
s C m
E
2
coulombsheet charge density -
meters
Energy is dependent on the electric field, which can be converted to voltage.
W = q E l = ½ ε E2
W = qV = ½ CV2
Page | 1-21
1.8 Magnetic Fields
The fundamental relationships for magnetic devices are listed.
(magneto-motive force) mmf = amp-turns voltageNI H dl F R=
F(Magneticintensity) /
NIAmp meter
s H
(closed loop) 2 ( )s r S dL
2
NI
RH
2weber(field density) H
A m
B
2
(reluctance)l N
A L R
Permeability is the magnetic property of material.
r o
74 10 /o henry m
(copper) 2r
(amorphous steel) 2000r (laminated steel) 6000r
Force on pole 2 due to pole 1
1 2
24F
r
(fluxlinkage) N LI
The magnetic circuit for a machine includes the current and turns, the ferromagnetic metal, and air
gaps.
Cross sectional area of air is ~5% more than steel.
The magnetization curve shown below is a non-linear relationship for the magnetic circuit, and is
typical of all magnetic circuits. It is used to show the conversion between representations of magnetic
energy.
i
N
l
Air gap
lg
A – cross
sectional area
Page | 1-22
The first portion of the curve has a physical anomaly near zero. The portion less than 5000 is in the
unsaturated region. There is an approximate proportional change in the vertical axis as the horizontal
changes. About 5000 is called the knee. That is the transition region. The top portion of the curve above
about 5000 is the saturated region. There is very little change in the vertical parameter as the horizontal
is increased.
The values are strictly representative. Different material alloys will yield other range of values.
Nevertheless, the general shape and form can be used for a variety of problems.
The curve then can represent a number of different relationships. Some of the more common are
shown in the table below.
X name unit Y name Units Function name unit
F mmf A-turns flux Weber R reluctance
H intensity A-turns/m B density Wb/m2 µ permeability H/m
I field I Amps V terminal Volts synchronous
F field mmf A-turns a Internal gen Volts dc
The location and direction of a magnetic field are determined by the configuration of the
conductor. Note that the intensity and density are related by the permeability μ. Therefore,
either can be determined from these equations.
Magnetic flux lines are continuous about a source, and perpendicular at all points to source,
in a parallel plane
2
I
rH
A magnetic field is produced by straight conductor carrying
current I
0
50
100
150
200
250
300
350
0 2000 4000 6000 8000 10000 12000
I
r
θ1 θ2
r
B
I
Page | 1-23
Ffield direction follows the right hand rule.
( )F i l B
Thumb – B
First Finger – F
Middle Finger – I, l
Cross product vector directions are determined by the curl of
the right hand.
The motor relationship comes from a current flow through a
field.
The generator relationship comes from a wire moving
through a field.
Energy is dependent on the magnetic field, which can be
converted to current.
W = H l φ = ½ μ D2
W = Iφ = ½ LI2
1 2(sin sin )4
I
r
B
( )F i l B
(voltage)e v l B ,
B
I
F
Page | 1-24
1.9 Maxwell
Maxwell’s Equations are a summary of all that is electrical and magnetic in a Calculus form. Although
they are not easily used, they do provide a mathematically clever summary. The integral form is tedious,
but the point form supplies definitions.
0
BEt
DH Jt
D
E
The electric-magnetic energy equation contains all the information in one equation. [1] Note that all
the definitions are in the point equation. In addition, all the field information is in the distributed
equation. W is energy and V is volume in these equations.
z yqW
t
z y ys t y
r y
q b d sW
t V
*1+ “Applications Engineering Approach to Maxwell and Other Mathematically Intense Problems“,
Marcus O. Durham, Robert A. Durham, and Karen D. Durham, Institute of Electrical and Electronics
Engineers PCIC, September 2002.
Page | 1-25
Chapter 1 Problems
Practice Problem 1-1 (Old Style)
SITUATION:
The circuit shown in the figure below is the pi representation of a transmission line.
REQUIREMENTS:
Write a set of nodal equations for the circuit the “Ref” as the reference or common terminal.
Solve the equations of requirement (a) above for Vo if 100 0inV .
Vin
Ri=50Ω
-j50Ω
j100Ω
-j50Ω 50Ω
Ref
+
Vo
-
SOLUTION:
Simple circuits problem
a) Nodal equations use current – convert voltage sources to current sources.
50 -j50 -j50 50
a b
Vout
inV
50
j100
I1 I2 I4 I5
I3
1
0
50
avI
2
0
50
avI
j
3100
a bv vI
j
4
0
50
bvI
j
5
0
50
bvI
Page | 1-26
KCL @ a KCL @ b
1 2 350
50 50 50 100 100
(2 1) 1 2
in
in a a a b
a b in
VI I I
V v v v v
j j j
j v j v V
3 4 5 0
0100 100 50 50
1 (2 1) 0
a b b b
a b
I I I
v v v v
j j j
j v j v
b outv v
Solve using Cramer’s Rule – Substitute RHS for vb
2 1 2
1 0 2 2 90
2 1 1 4 4 1 1 4 2 45
1 2 1
in
in inout
j V
j j V Vv
j j j
j j
0.3535 135*out inv V
b) For Vin=1000
100 0*0.3535 135 35.35 135outv
Page | 1-27
Practice Problem 1-2 (Old Style)
SITUATION:
A starting circuit is needed that will limit the starting current in a dc motor to two and one half time
(2.5pu) the normal full load current.
The switches S1 and S2 in the starting circuit shown below are to close sequentially when the current
has dropped to normal full load current (1pu)
Both switches are open when the main breaker SM is closed.
Sm
EA
RA R1 R2
S1 S2
Vt
SOLUTION:
On Closing of Sm
1 2
1 2
2.5 0.4t tA A
A A
V VI pu R R R pu
R R R I
With Vt=1.0,
1 2
10.4
2.5
tT A
A
VR R R R pu
I pu
t At a A t t
A
V EV E I R R
I
When IA drops to 1.0 pu
1(0.4 ) 0.4 1 0.4 0.6a A tVt E pu E V pu
Close S1 and IA raises to 2.5pu, at that instance EA = 0.6pu
2
1 0.60.16
2.5
t aT A
A
V ER R R pu
I
When IA again drops to 1.0 pu
Page | 1-28
1 1*0.16 0.84A t A TE V I R pu
Close S2 and IA raises to 2.5pu, at that instance EA = 0.84pu
1 0.84
0.0642.5
t aT A
A
V ER R pu
I
When IA again drops to 1.0 pu
1 1*0.064 0.936A t A TE V I R pu
2 2
2 1 1
0.064
0.16 0.096
0.4 0.24
A
A
A
R pu
R R pu R pu
R R R pu R pu
Signals that are encountered can be a constant, direct current (DC); they can be repetitive,
alternating current (AC); or they can be short term, transients. The circuit elements respond differently
to each type signal. This chapter will address waveforms and tools to analyze their impact on systems
performance. The time domain signal response or solution contains all the components.
Page | 2-1
Chapter 2 Waveforms
2.1 Introduction Signals that are encountered can be a constant, direct current (DC); they can be repetitive,
alternating current (AC); or they can be short term, transients. The circuit elements respond differently
to each type signal. This chapter will address waveforms and tools to analyze their impact on systems
performance. The time domain signal response or solution contains all the components.
2.2 Waveforms By far, the sinusoid is the most common
repetitive waveform in electrical systems. It is the
physical result due to the rotational motion of
machines in a magnetic field.
The waveform definitions follow.
cosovoltage v V t
1( )frequency f hertz
T
2 ( / sec)f radians
0
10for sinusoid
T
DCaveragevalue V dtT
2
0
1
2
T
oRMS
Veffectivevalue V v dt
T
For multiple waveforms, use superposition. For effective or root mean square (RMS), this is square
root of the sum of the squares.
Generally, AC values of V & I are given in RMS. The frequency is assumed constant.
For a 100Volt, 60Hz voltage waveform
2 2
1 2 ...RMS RMS RMSV V V
100 2 cos(2 60 )v t
100 2oV
141OV volts
Vo = peak
Vrms = effective
v(V)
Time
T (sec)
Page | 2-2
2.3 Transients Transients are waveforms that exist for a short period of time. Waveforms are determined by the
circuit elements. Since there are only three elements, the most complex circuit is a second order. The
characteristic solution for a systems circuit is the time varying equation that describes the exponential
decay after a signal is applied. The variable, y, can represent either current or voltage.
( ) ( ) cost
y t F I F e t
where
F = final value (t=∞)
I = Initial Value (t=0)
τ = time constant
2.3.1 First Order Transients
First order systems are very common, since they are the model of a simple system. First order
systems have a resistor and either a capacitor or an inductor.
First Order Circuits
RC or RL
Form:
div L Ri
dt
dv vi C
dt R
Characteristic Solution
Response to a step input (DC)
( ) ( )t
y t F I F e
1LC
Page | 2-3
For Capacitor: dv
i Cdt
Voltage does not change instantaneously
Open circuit under DC conditions
Capacitor discharges to ( ) 0CV
Initial voltage = source voltage
For Inductor: di
v Ldt
Current does not change instantaneously
Short circuit under DC conditions
Inductor dissipates to ( )LI I
Initial current = source current
Process:
Find τ
-1
time constant= or /
1 time for exponent to be e
RC L R
e
use equivalent circuit w/o source to get RC or RL (Thevenin Impedance)
deactivate all the sources and replace with internal Z
reduce to single equivalent RC or RL
Find y(0)
use circuit (KVL) w/ element as source
Find y(final)
use circuit (KVL) w/ element as limit
Plot: Initial slope = F I
Transfer function = response
excitation=
output
input
Page | 2-4
2.3.2 RL Circuits
Standard calculus form
o
diV L Ri
dt
Inductor is short circuit in final state.
0( ) 0i L
( ) of
Vi L
R
L
R
General solution
( ) ( )t
y t F I F e
Current solution
(0 )
t
o oV Vi e
R R
(1 )
Rt
o LV
i eR
Stability reached after 3 – 5 time constants
i
Sec
oV
R
diL
dt
L
R
VOR
t=0
i
L
Page | 2-5
2.3.3 RC Circuits Standard calculus form
dv v
i Cdt R
(calculus form)
Capacitor is open circuit in final state.
i, vc cannot change instantaneously
0( ) ov C V
( ) 0fv C
RC
General solution
( ) ( )t
y t F I F e
Voltage solution
( )t
F I Fv V V V e
t
c ov V e
sec
Vc
Vo
RC
vo
R
Ct=0 i
Page | 2-6
2.4 LaPlace A standard waveform is defined in terms of time and frequency. A mathematical transform is often
used to provide a different mathematical tool. The phasor representation is one transform that applies
to steady state alternating circuits. LaPlace transforms are used for many manipulations of the inductor
and capacitor elements. The function can be transformed from time to the s domain, which represents a
stationary and rotational component.
s j
The most used transform pairs are illustrated.
( )f t ( )f t ( )F s ( )mI s
1 1
s
t 2
1
s
2
te 1
s
-α
( ) tf t e
10.37
e
1
( )f t t
( ) 1f t
Page | 2-7
tte 2
1
( )s
-α
2
sin t 2 2s
j
j
cos t 2 2
s
s
j
j
sinte t
2 2( )s
j
j
coste t
2 2( )
s
s
j
j
2
1
coste t
sinte t
2
1
-2
-1
0
1
2
2
sin t
-2
-1
0
1
2sin t
2
( ) tf t te 1
e
1
Page | 2-8
2.5 LaPlace Operational Rules The mathematical manipulation of the time function and the LaPlace transform follows defined
rules.
1 2 1 2( ) ( ) ( ) ( )f t f t F s F s
( ) ( )a f t a F s
( )( ) ( ) (0)
d f tf t sF s f
dt
2( ) ( ) (0) (0)f t s F s sf f
0
1( ) ( )
tf t dt F s
s
Page | 2-9
2.5.1 Partial Fraction Expansion
1
1 2 1 2
( ) ( ) 2( )
( ) ( )( ) ( ) ( )
KN s N s KF s
D s s p s p s p s p
1
11 1 2 3
1 2 1 3
( )( ) ( )
( )( )s p
N pK s p F s jK jK
p p p p
1 21 2( ) p t p tf t K e K e
Page | 2-10
2.5.2 Alternate Approach
1 2( ), ( )F s F s are from the table, known transforms
1 1 2 2( ) ( ) ( )F s K F s K F s
Right hand side to common denominator, equate numerators, solve for K1, K2…
1 1 2 2( ) ( ) ( )f t K f t K f t
Impedance in s-domain
Impedance V
ZI
RZ R LZ sL 1
CZsC
Admittance 1 I
YZ V
1
RYR
1
LYsL
CY sC
Page | 2-11
2.6 Fourier Series Any alternating waveform can be represented by the summation of a fundamental sine wave and its
multiples called harmonics. This summation is called a Fourier series.
0 1 1 2 2sin(1 ) sin(2 ) sin( )n ny Y Y t Y t Y n t
The term y is the instantaneous value at any time. It can be either current or voltage.
The Y0 term is the constant offset, average, or DC component. The Y terms are the maximum
amplitude for each of the harmonic frequencies. The angular frequency ω is 2πf. The phase shift angle
represents the time delay between the reference voltage waveform and the current. The n subscript
and coefficient of frequency indicates the harmonic number.
The time domain is a plot of the Y amplitude versus time for the curve. The frequency spectrum is a
plot of harmonic amplitude versus harmonic frequency number.
An odd function is created with the sum of the odd harmonics. A sine wave is the basic example. If
the waveform has the pattern of a fundamental sine wave, then it is odd.
( ) ( )y t y t
An even function is created with the sum of the even harmonics. A cosine is the basic example. If the
waveform has the pattern of a fundamental cosine wave, then it is even.
( ) ( )y t y t
A function that contains both even and odd harmonics will have spikes. A pulse and sawtooth are
examples.
Page | 2-12
Page | 2-13
The very definition of Fourier series indicates the series can take several forms. A cosine is an
orthogonal shift to a sine wave. As a result, a common representation is to use cosine terms for the even
harmonics and sine terms for the odd. Then the even harmonics become odd coefficients for the cosine
terms. Although this is a common representation, it is not as easy to visualize or to obtain a spectrum as
the simple sinusoidal form.
The Fourier series can be decomposed into the sum of even and odd parts.
( ) ( ) ( )e of t f t f t
The even part can be represented by the Fourier series
1
( ) cos( )2
oe n
n
af t a n t
Page | 2-14
The odd part can be represented by the Fourier series
1
( ) sin( )o n
n
f t b n t
The coefficients are similar.
0
1( )
T
oa f t dtT
0
0
2( )cos( )
T
na f t n t dtT
0
0
2( )sin( )
T
nb f t n t dtT
Page | 2-15
2.7 Signals - Modulation Modulation is the process of combining two signals into one combined waveform. The combination
can be through an adder or multiplier. The mathematical result looks very much like the Fourier series.
Modulation is similar to wrapping a paper note around a rock and tossing the combination. The
carrier wave, or rock, provides a vehicle for passing the information. The information is on the paper
note.
In its basic form, the carrier is a single waveform.
( ) sin(2 )c c cx t A f t
Ac = amplitude, f = frequency, and θ is the phase shift. Therefore only the amplitude, frequency, or
phase can be changed or modulated.
The message, information, or baseband has a similar form. Usually the message has a fixed or 0
phase shift.
( ) sin(2 )m m mm t A f t
2.7.1 Modulation Types
There are numerous variations to the types of modulation.
There are three analog modulation techniques based on the variables in the waveform.
Amplitude modulation (AM)
Frequency modulation (FM)
Phase modulation (PM)
Special variations of these techniques have unique characteristics that affect bandwidth and power.
Angle modulation includes both frequency and phase modulation, since they are operated
on by the sinusoid.
Double sideband modulation (DSB) is AM with the carrier removed.
Single-sideband modulation (SSB) is DSB with one of the sidebands removed.
There are three fundamental sampling or digital modulation techniques.
Pulse amplitude modulation (PAM)
Pulse frequency modulation (PFM)
Pulse phase modulation (PPM)
Variations of these techniques result in a variety of keying processes. The process of modulation and
demodulation is called a modem.
Pulse code modulation includes both frequency and phase.
Page | 2-16
Amplitude shift key modulation (ASK)
Frequency shift key modulation (FSK)
Binary-phase shift key modulation (BPSK)
Quadrature-phase shift key modulation (QPSK)
Quadrature amplitude modulation (QAM)
Page | 2-17
2.7.2 Amplitude Modulation (AM)
Amplitude modulation mixes the information or message with the carrier amplitude. The general
form of amplitude modulation is to add a function of the message to the carrier amplitude.
( ) [ ( )]sin(2 )c a cy t A k m t f t
For a single waveform, ka is unity.
1ak
The amplitude varies with the carrier and the signal. The expanded form illustrates the three
components, carrier + lower sideband - upper sideband.
( ) [ ( )]sin(2 )c cy t A m t f t
( ) [ sin(2 ]sin(2 )c m c cy t A A f t f t
1 1( ) sin(2 ) cos(2 2 ) cos(2 2 )
2 2c c m c m m c my t A f t A f f t A f f t
The modulation index is the depth of the variation around the original level of the carrier, Ac. When
multiplied by 100, it is the percent modulation.
mam
c c
AAm
A A
The power in an AM signal is the sum of the power in the carrier and the power in the signal.
22( )(( ( )) 1)
2
c m
cave
P P P
Akm t
AM signals can be demodulated with an envelope detector or a synchronous demodulator.
A double sideband (DSB) signal would contain the upper and lower sideband information but would
not have the carrier. DSB signals can be demodulated with a synchronous demodulator. A Costas loop is
a common technique.
Single sideband (SSB) can be either the lower or upper sideband information only without the
carrier or the other sideband. AM signals can be demodulated with a synchronous demodulator or by
carrier reinsertion and envelope detector.
Page | 2-18
The bandwidth has a lower frequency of fc-fm, center frequency fc, and an upper frequency of
fc+fm.
h lBW f f
Page | 2-19
2.7.3 Angle Modulation
Angle modulation mixes the signal as a component of the carrier sinusoid which includes the
frequency and phase terms. In essence the signal becomes the phase term.
( ) sin(2 )c c cx t A f t
( ) sin[2 ( )]c cy t A f t m t
( ) sin(2 )cos( ( )) cos(2 )sin( ( ))]c c c cy t A f t m t A f t m t
( ) sin(2 )cos( sin(2 )) cos(2 )sin( sin(2 ))]c c m m c c m my t A f t A f t A f t A f t
This is obviously a very complex function with numerous frequency components. There are infinite
sidebands to the signal. However, the amplitude of most deteriorates quickly. Frequency modulation
and phase modulation each use select components of this waveform.
The phase deviation or shift is a function of the message or information. As discussed earlier, it is
assumed that the message phase shift is zero. The function, kp, is the phase modulation index.
( ) ( )pt k m t
The instantaneous phase is the carrier angle added to the signal. This is the angle within the carrier
wave sin term.
( ) 2 ( )i ct f t t
The instantaneous frequency is the change of instantaneous phase with time. The instantaneous
frequency is the carrier frequency plus the frequency deviation.
2 ( )
i i
c
c
d
dt
df t
dt
( )
i i
c
c
df
dt
df t
dt
f f
The frequency deviation is the change in the phase, which is the change in the message with time.
( ) ( )d d
t km tdt dt
The message bandwidth is the frequency of modulation, fm.
m mBW f
The bandwidth of an FM & PM signal is approximated using Carson’s rule.
2( )
2( 1)
y m
fm m
BW f f
m f
Page | 2-20
2.7.4 Frequency Modulation (FM)
Frequency modulation mixes the information or message with the carrier frequency. The amplitude
is constant. The result is the carrier varies above and below its idle or normal frequency, fc. As the
voltage amplitude of the modulating signal increases in the positive direction from A to B, the frequency
of the carrier is increased in proportion to the modulating voltage.
Frequency modulation is adding the carrier frequency and a function of the message.
( ) ( )i c ff t f k m t
The modulation index or factor is the maximum deviation in frequency, Δf, divided by the
modulation frequency. When multiplied by 100, it is the percent modulation.
fm
m
fm
f
The frequency modulator constant is the frequency deviation divided by the amplitude of the
modulating or message signal.
f
m
fk
A
Page | 2-21
2.7.5 Phase Modulation
Phase modulation is another component of angular modulation that is sometimes referred to
indirect FM. Note that the phase is part of the sinusoid. Here, the amount of the carrier frequency shift
is proportional to both the amplitude and frequency of the modulating signal. The phase of the carrier is
changed by the change in amplitude of the modulating signal. The modulated carrier wave is lagging the
carrier wave when the modulating frequency is positive.
Phase modulation is manipulation of the angle of the carrier and a function of the signal.
( ) 2 ( )i c pt f t k m t
The modulation index is the peak phase variation.
( )pm
m
fm t
f
The phase modulation constant depends on both the frequency and amplitude. It is the ratio of the
phase deviation to the message.
( )
( )p
tk
m t
Page | 2-22
2.7.6 Sampled Messages
A message, m(t) can be recreated from uniformly spaced samples. The sampling frequency, called
the Nyquist frequency fN, must be at least twice as fast as the highest frequency being recreated.
1
2N
s
f fT
2.7.7 Digital - Pulse Modulation
Pulse or digital modulation is frequently used to transmit sampled messages. Analog to digital
conversion is a two step process. First, sampling changes the analog source to a series of discrete values,
called sample. Second, quantization, converts each sample to a number. The number of quantization
levels, q, is the two power of the number of bits.
2nq
The bandwidth required is inversely proportional to the inverse of twice the pulse length or
duration, T. This is called the Shannon bandwidth when the Dimensionality, D is included. For minimum
bandwidth, D=1.
2S
DBW
T
The message bandwidth, W, and the number of bits determine the minimum modulated bandwidth,
BW.
22 logBW nW W q
Page | 2-23
2.8 Signal transmission
2.8.1 dBm
Signal power can be measured in watts. However, comparison values and small signals are
measured in decibels.
1010log ( )signal
ref
Pdb
P
When the reference is on milliwatt, then the decibels are reference as dBm.
1010log ( )1
signalPdBm
mW
As a result a 1 milliwatt signal is 0dbM.
0 1 dBm mW
For amplitude measurements (Amps & Volts)
1020logsignal
noise
AdB
A
2.8.2 Noise
Noise is a random or background signal that may interfere with the message or information. Signal-
to-noise ratio is an indication of the power ratio between the desired information and the background
noise. The symbols are SNR or S/N.
/signal
noise
PS N
P
Often the expression is in terms of decibels (dB).
10 10/ ( ) 10log ( ) 20log ( )signal signal
noise noise
P AS N db
P A
In a digital signal, the number of bits in each value determines the SNR. Noise in a digital signal is
dependent on the conversion process. The dynamic range is an expression of the SNR.
10/ ( ) ( ) 20log (2 )nS N db DR db where n = # of bits
White noise creates a thermal noise power, P, in watts that is dependent on the bandwidth, Δf in
Hertz and temperature, T in degrees Kelvin. This is also the thermal noise that will be created by
electron activity in a resistor and is called Johnson noise.
Page | 2-24
-23
K
1.3806503 × 10 Boltzmann's constant
273.15
T B K
B
OC
P K T f
JK
K
T T
For current or voltage across the resistor the power has the standard relationships.
22V
P I RR
Thermal noise at room temperature is dependent on the bandwidth. The units are decibels.
( ) 174 10log( )P dB f
The total noise figure for a series of transfer functions or amplifiers is based on the ratio of the noise
figure for each stage, F, to the gain ratio of each stage, G. The noise figure and gain must be converted
to the power ratio from dB.
31 2
1 1 2
11...
1T
FF FF
G G G
Page | 2-25
2.8.3 Propagation - Transmit
The velocity of propagation of a wave is the distance the wave will travel in one time period. If the
distance is one wavelength, λ, then the velocity is the ratio of the wavelength to the frequency.
p
du
t f
In free space, the propagation velocity is the speed of light.
82.99 792 458 x 10c
The velocity of a wave on a transmission line is simply the ratio of the distance to the time it takes
for the wave to propagate. For a reflected wave, the distance is twice the length because of the trip
length and back.
p
du
t
Transmission of waves involves the power density in Watts per square meter. It is the ratio of the
power transmitted to the orthogonal area that the waveform strikes. A spherical shape is the normal
pattern of an omni-directional wave.
transmitteddensity 24
range from antenna, radius of sphere
xP PP
A R
R
Antennas can direct power in specific directions. The gain of the antenna is the radiation intensity in
a particular direction divided by the power that would be radiated from an omni-directional or isotropic
antenna.
Effective radiated power
Isotropic rated powerG
Power is dissipated as a waveform propagates. The attenuation or loss in free space depends on the
velocity of light. In other mediums, the velocity of propagation should be used. The loss is dB, distance is
m, and frequency is Hz.
420log
/fs
dP
c f
Characteristic impedance is the opposition in a circuit that connected to the output terminals of a
line will cause the line to appear infinitely long. It is the electric and magnetic property of the material
that impacts the velocity of propagation.
0
1p
p
Z uu
The electric property is permittivity in Farads per meter, Fd/m. It is a factor of the free air.
Page | 2-26
0
-120 8.854x10 Fd/m
r
The magnetic property is permeability in Henries per meter, Hy/m.
0
70 4 x10 Hy/m
r
From these three preceding concepts, the impedance of free space air is calculated.
0 377Z
Because of the definitions of inductance and capacitance in relation to permeability and
permittivity, characteristic impedance can be found in terms of circuit elements.
0
LZ
C
2.8.4 Reflections
Maximum power transfer occurs when the load is equal to the source or characteristic impedance.
When a discontinuity occurs on a line or a load is connected that does not match the characteristic
impedance, the waveform will be reflected and oppose the message signal. The reflection coefficient
describes both the magnitude and phase shift of the reflection. The coefficient is the ratio of the
complex forward voltage to the complex reverse wave voltage.
0
0
fL
L r
VZ Z
Z Z V
Standing wave ratio is the maximum power over the minimum power transferred. SWR is
dependent on the reflection coefficient.
1
1SWR
Voltage SWR is the maximum voltage over the minimum voltage nodes. VSWR only contains the
magnitude of reflection coefficient.
max
min
1
1
VVSWR
V
The reflection coefficient has the following range of values.
Γ = − 1: maximum negative reflection, line is short-circuited,
Γ = 0: no reflection, when the line is perfectly matched,
Γ = + 1: maximum positive reflection, line is open-circuited.
At the maximum nodes the waves interfere positively and add. At the minimum nodes, the waves
are colliding and subtract.
Page | 2-27
max
min
(1 )
(1 )
f r f
f r f
V V V V
V V V V
Transmission line properties are defined in terms of propagation constant. Propagation constant is
inversely proportional to the wavelength. The distance is measured from the load.
2
00
0
( )
( )
tan( )
tan( )
j d j d
j d j d
Lin
L
V d V e V e
I d I e I e
Z jZ dZ Z
Z jZ d
Page | 2-28
R L C
2.9 RLC System Response
2.9.1 RLC Equations
The three elements, RLC can be arranged in series or its dual parallel. This is a second order system.
The analysis of the circuit can be made in many domains. Typically the time domain is the starting point.
However, the Calculus required makes the mathematic interpretation difficult. For that reason
numerous transforms are use d. The math of the transforms will not be developed, but the
correspondence is apparent from the table. The duality of the circuits is intriguing.
Function Series Parallel
Reference Same current through all elements Same voltage across all elements
Diagram
Fundamental 2
2
1( )
d q dqv t L R q
dt Cdt
2
2
1 1( )
d di t C
R dt Ldt
Time 1( )
div t L Ri idt
dt C
1( )
dvi t C Rv vdt
dt C
LaPlace 1( ) ( ) ( )V s Ls R I s
Cs
1 1( ) ( ) ( )I s Cs V s
R Ls
Sinusoidal Steady State
1( ) ( ) ( )V j j L R I j
j C
1 1( ) ( ) ( )I j j C V j
R j L
Several observations can be made about the relationships.
ds j
dt
1 1dt
j s
'dq
q idt
'd
vdt
R L C
Page | 2-29
2.9.2 System Response
The system response is the solution to the second order equation.
Time constant is the time it takes for a signal to settle so that the exponential decay.
time constant L
RCR
2.9.3 Characteristic Transfer
Transfer functions are often used as a model for a system.
Function Series Parallel
Transfer function
( )( )
( )
I sY s
V s
( )( )
( )
V sX s
I s
Characteristic 1( )
1Y s
Ls RCs
1
( )1 1
Z s
CsR Ls
Standard form
2
/( )
1
s LY s
Rs s
L LC
2
/( )
1
s CZ s
ss
RC LC
Resonance 2 2
0
/( )
s LY s
s s
2 20
/( )
s CZ s
s s
2.9.4 Resonance
Frequency is inversely related to time. Angular frequency is one complete revolution of cycle of the
frequency.
2 f
Resonance is a very significant concept that may be a boon or ban to electrical systems. Resonance
is the frequency where the magnetic (or inductor) energy equals the electric (or capacitor) energy.
0
1
LC
Since the energies are balanced, it flows from one to the other resulting in a sinusoidal frequency.
The natural frequency is the oscillation determined by the physical properties. Resonant frequency is a
created oscillation that matches the natural frequency. Resonance is the frequency at which the input
impedance is purely real or resistive.
The frequency response has a roll-off on either side. The transition is called the cut-off frequency.
20 cL cH
Page | 2-30
Bandwidth, Δω, is the range between the upper and lower cut-off
frequencies. The bandwidth is also called the pass band or bandpass.
cH cL
02
cL
02
cH
Quality factor or selectivity is the sharpness of the peak at
resonance.
0Q
oLQ
R
Damping is the effect of resistance on the rate that a signal is stabilized to steady state. Undamped
implies that there is no resistance, R=0. The damping coefficient is dependent on the natural frequency
and is inversely proportional to twice the quality factor. Some authors use the symbol alpha, α, rather
than zeta, ς. Note this is also the real term of the LaPlace, σ.
0 actual damping
2 critical damping2 /
R
Q L C
The range of values for the camping coefficient reflects how quickly the waveform will settle and
whether it will overshoot. Under-damping results in oscillations or ringing, over-damping results in a
slow exponential approach to stability, critical-damping is the transition between oscillations and
exponential.
1 under-damped = oscillation
1 critical-damped = transition
1 overdamped = exponential
The relationship between the various factors can be described in terms of the quality factor.
0 0
2Q
Damped resonance, ωd, is a shift from the resonant frequency caused by the damping.
2 2 20d
The root of the characteristic equation has the real part as damping coefficient and the imaginary
part as the damped resonance. For the second order, there are two roots.
Page | 2-31
1,2 ds j
Page | 2-32
2.9.5 Series Parallel Duality
Comparison of the standard form and the resonance equation reveal the duality of impedance and
admittance. The symmetry of the duality resolves to a reciprocal form at resonance.
Function Series Parallel
Quality factor XQ
R
RQ
X
Quality factor 1 LQ
R C
CQ R
L
2.9.6 First Order
A first order system has a resistor and either a capacitor or inductor. Therefore, there is no
oscillation. However, there is still a cut-off frequency that is the inverse of the time constant.
1 1( ) 0 cR j
C RC Time Constant = RC
2( ) ( ) 0 c
RL j R j
L Time Constant =
L
R
End of chapter
Page | 2-33
Chapter 2 Problems Problem 2-1
Consider the circuit shown below. R1 and R2 are 5Ω
resistors. R3 is a 10Ω resistor and R4 is a 15Ω resistor.
Z1 is a 20μF capacitor, and V1 is a 120V source. The
time constant of the circuit is most nearly
(A) 85 μS
(B) 138 μS
(C) 550 μS
(D) 400 μS
SOLUTION:
Redraw the circuit to make it easier to see
The resistances can be combined to
determine the equivalent resistance of the
circuit.
4 3 1 2( / / )eqR R R R R
15 10 (5 / /5 )eqR
The time constant of a RC circuit is
27.5 20 550eq eqR C F S
The answer is (C)
25 2.5 27.5eqR
Z1R1
R3
R2
R4
Z1
R1 R3
R2
R4
AC
Page | 2-34
Problem 2-2
Consider the circuit shown in the problem above, and recreated below. R1 and R2 are 15Ω resistors.
R3 is a 20Ω resistor and R4 is a 15Ω resistor. Z1 is a 20mH capacitor, and V1 is a 120V, 60Hz source. The
switch has been closed for a significant period of time. The voltage across the inductor is most nearly.
R1=15Ω
R2=15Ω
R3+R4
20Ω + 15Ω=35Ω120V
60Hz
Z1=j7.54Ω
I
1
A
(A) 25.455°
(B) 10.580°
(C) 50.7-60°
(D) 61.890°
SOLUTION
Impedance of the Inductor Z1
1 2 60(20 ) 7.54Z j mH j
Redraw with all impedances
The answer is (B)
R1=15Ω
R2=15Ω
R3+R4
20Ω + 15Ω=35Ω120V
60Hz
Z1=j7.54Ω
I
1
A
3 4 1 35 7.54AR R R Z j
15 // 10.6 0.664B AR R j
1
120( ) 120 (10.6 0.664 )49.73 1.822
( ) (25.6 0.664 )
BA
B
R V jV j
R R j
1
1
1
( ) (49.73 1.822 )( 7.54 )1.83 10.31 10.5 79.94
( ) (35 7.54 )
AZ
A
V Z j V jV j V V
R Z j
Page | 2-35
Problem 2-3
What is the time constant of the figure shown?
0.2μF
3MΩ
12V
SOLUTION:
The time constant of an RC circuit is
6 63 10 0.2 10
0.6 seconds
RC
Page | 2-36
Problem 2-4
In the figure below, the switch has been open for a significant period of time and is closed at t=0.
What is the current in the capacitor at t=0+?
0.3μF
6MΩ
12V
SOLUTION:
The capacitor, at t=0+, acts as a short circuit. The current through the capacitor then is determined
by the voltage and the resistance
6
60
122 10
6 10c t
V Vi A
Z
Page | 2-37
Problem 2-5
In the figure below, the switch has been open for a significant period of time, and is then closed at
t=0. What is the current through the two capacitors at t=0+?
10V
75Ω
0.001F
0.001F 500Ω 200Ω
SOLUTION:
If the switch is opened for a significant period of time the capacitor on top of the circuit is charged
to 10V, and the capacitor in the middle of the circuit is discharged to 0V. At t=0+, the capacitors are
modeled as voltage sources with the charged voltages. The equivalent circuit is shown below
10V
75Ω
500Ω 200Ω0V
10V
is
iA iB
i2
The voltage across the 500Ω resistor is 0V, so iB=0A.
KVL on the left loop is
10 (75 ) 0 0
100.133
75
s
s
V i V
Vi A
KVL on the right loop is
2
2
10 0 (200 ) 0
100.05
200
V V i
Vi A
KCL
Page | 2-38
2 0
0.133 0 0.05 0
0.133 0.05 0.083
s A B
A
A
i i i i
A i A
i A
The current through the top capacitor is i2=0.05A
The current through the middle capacitor is iA = 0.083A
Page | 2-39
Problem 2-6
In the figure below, the switch has been open for a significant period of time. The switch is closed at
t=0. Find the current through the resistor at t=0+, and at t=1.25s. Find the energy in the inductor at
t=1.25s.
50V
20Ω
8H
SOLUTION:
The current in an inductor cannot change instantaneously, so
(0 ) 0Li A
The general solution for a first order RL circuit is
( ) 1
Rt
LV
i t eR
20 1.25
850
(2) 120
2.39
s
HV
i e
A
The energy in the inductor is found using
212LW Li
212(8 )(2.39) 22.85LW H J
Page | 2-40
Problem 2-7
A carrier wave of 12 MHz is amplitude modulated by an audio signal of 1.5 kHz. What are the upper
and lower limits of the resulting modulated signals bandwidth?
SOLUTION:
612 10cf Hz
31.5 10Mf Hz
Lower sideband frequency - 6 312 10 1.5 10 11,998,500C Mf f Hz Hz
Upper sideband frequency - 6 312 10 1.5 10 12,001,500C Mf f Hz Hz
Page | 2-41
Problem 2-8
A 110 MHz carrier is frequency modulated by a 65kHz information signal. The information signal has
1V amplitude, and a frequency modulator constant of 100Hz/V. What is the bandwidth?
SOLUTION:
Carson’s Rule
2( )f mBW ak f
32 1 100 65 10
130,200 130
HzBW V Hz
V
Hz kHz
Page | 3-1
Chapter 3 - Power Analysis
3.1 Introduction Power analysis looks at the energy conversion segment of electrical systems. Machines can be
modeled as a Thevenin equivalent voltage and impedance with a magnetizing circuit consisting of an
inductor with its resistance. Three types of problems are encountered.
Model parameters and losses require the complete model using circuit theory.
Transients and load flow use the Thevenin equivalent.
Steady state uses the terminal conditions with complex power.
The time domain signal representation contains all the components of the responses including DC,
transient, and sinusoidal.
( ) cost
y t F I F e t
Page | 3-2
3.2 Power Definitions Use per unit calculations if multiple voltages are used in the problem (i.e. transformers).
Use the following equations to obtain the desired quantity.
Complex apparent power – volt-amps, VA
* (cos sin )S VI P jQ S j
Real Power – watts, W
P VI pf S pf vi
resistance only, Q=0
Reactive Power – volt-amp reactive, VAR
1sin(cos ) sinQ VI pf VI
Q > 0 for inductive load, lagging pf
Q < 0 for capacitive load, leading pf
Power Factor
*
cosP vi
pfS VI
pf=1 when or L CZ R jX jX
X/R
tan LX
R
Convert Hp to real power (KW).
? 0.746 1
* *Hp kW
PHp eff
Alternately, convert Hp to apparent power.
? 0.746 1 1
* * *Hp kW
SHp eff pf
The angle has several relationships. It is also the time delay between the voltage and current
crossing the axis (going through zero value). The conversion is
2π radians = 360 degrees = 1 cycle
1cos 2V I Z S pf ft
Page | 3-3
2
tf
The relationship between the power terms is shown graphically below.
The relationship of the power terms is often illustrated with complex triangles.
Impedance is the ratio of voltage and current (Ohm’s Law)
2 2( )V
Z R jX R XI
Complex Numbers are easiest to manipulate in the following manner:
Add & Subtract P’s and Q’s
Multiply & Divide: Convert P & Q to S &, multiply (divide) magnitudes, add (subtract) angles
Phasor rotation is used to explain the relationship between the lines of a three phase power system
kW
kVAR kVA
θ
I
Vref θ
X
R
Z
Impedance Triangle
Q
P
S
Power Triangle
Page | 3-4
3.2.1 – Z,R,X,θ; S,P,Qpf Conversions
Often, when performing power calculations, all of the components of a problem are not known. Out
of the four components to any problem, only two must be known to calculate all four. The base
presumptions are the equations discussed above.
Before conversion, move power factor to angle.
1cos ; cos ( )pf pf
For Impedances
If know Z R X θ
R, X 2 2Z R X R R X X 1tanX
R
R, θ cosZ R R R tanX R
R,Z Z Z R R 2 2X Z R 1cosR
Z
Z, θ Z Z cosR Z sinX Z
Z, X Z Z 2 2R Z X X X 1sinX
Z
X, θ
sin
XZ
tan
XR
X X
For Power
If know S P Q θ
P, Q 2 2S P Q P P Q Q 1tan
Q
P
P, θ cosS P P P tanQ P
P, S S S P P 2 2Q S P 1cosP
S
S, θ S S cosP S sinQ S
S, Q S S 2 2P S Q Q Q 1sinQ
S
Q, θ
sin
QS
tan
QP
Q Q
Page | 3-5
3.3 Three-Phase AC Three-phase indicates 3 lines, 3 connections, 3 sources and 3 loads
When balanced, a single relationship for the phase values can be
related. Phase voltages are the voltage difference between each phase
and neutral.
The relationship between the lines and phases are simple vector
calculations. The reference voltage starts with an angle of zero. The line
voltage is the difference in the corresponding phase voltages.
3 32 2
0 120
(1 0) ( 0.5 ) (1.5 )
3 30
ab an bn p p
P p P
p
V V V V V
V j V j V j
V
The other two line voltages are simply shifted by 120°.
3 30 120
3 30 240
bc p
ca p
V V
V V
The relationships are visually shown in a phasor diagram.
The delta connection is shifted by 120° between phases.
From these considerations, the following relationships hold,
whether delta or wye.
303
303
N A B C
LLLN
LP
I I I I
VV
VV
Specific conditions impact the significance of the
equations.
ab bc
for , 0
Phase lags 30 behind line
is 30 behind V , and 90 ahead of V
N
an
I
V
A
C B
Line, IA
Neutral
0P anV V
120P bnV V 240P cnV V
Line, IB
Line, IC
Van
Vab
Vbn
Vcn
Vbc
Vca
30o
30o
30o
Line, IA
0L abV V
120L bc
V V
240L caV V
Line, IB
Line, IC
A
BC
Page | 3-6
3.3.1 Y-∆ Relationships
For a three-phase system, there are only two possible connections, wye and delta. The magnitude of
the voltage and current changes by a factor of 3 .
Wye Delta
Power can be determined using phase or line values.
1 Power cosP PV I
3 Power 3 cos easier for wyeP PV I
3 Power 3 cos delta or wyeL LV I
Apparent power contains both real and reactive components. Real (Watts) is dependent on
resistance and reactive (VARS) is dependent on inductors and capacitors.
Power factor = cos
Complex power (S) = VA = *3 L LV I
The complex value can be expanded to magnitude and angle.
cosP S cosQ S
2 2( )S P Q
Impedance is only a single phase relationship. There is no relationship to line values.
P
P
VZ
I
Neutral current is the sum of the current in the phases.
for wye w/ neutralN A B CI I I I
NI 0 for delta or wye w/o neutral
A key principle of three phase power affirms that regardless of whether the current is Y or ∆, the line
currents lag the line-to-neutral voltages by the phase (impedance) angle.
Z R jX 2 2( )Z R X 1tanX
R
3 P L
P L
V V
I I
3
P L
P L
V V
I I
Van
Vcn Vbn
a
bc
Ia
Ib
Ic
Page | 3-7
3.3.2 Phase Sequence
Phase sequence is used to determine the direction of rotation of machines and compatibility for
connection to other circuits.
“Look” in the zero axis
Rotate phase in counter-clockwise direction
Write down phasors as they come by
Sequence of voltage phases is every other letter
ab,bc,ca → abc
an,bn,cn → abc
Vab
Van
VcaV
cn
Vbc
Vbn
acb sequence
ab,ca,bc → acb
an,cn,bn → acb
Vab
Van
VbcV
bn
Vca
Vcn
abc sequence
Page | 3-8
3.4 Power Transfer across a Reactance Power is converted to heat in a resistor; however, power is transferred across a reactance and
causes a phase shift in the voltage.
EA
V2V1
The standard power computation is the starting relationships.
*S VI
P S pf
The real and reactive power transfer contains the phase shift.
1 2 1 21 2
sin( )VVP
X
2
1 2 1 1 2 1 2
1cos( )Q V V V
X
To create a consistent analysis, assume the angle relationships for voltage and apparent power.
1 0V V
S Z
From this, the remaining angles and the current are found.
I S
cospf
(cos sin )I I j
Page | 3-9
3.5 Power One-Line Most power analysis uses a one-line diagram. The sources or generators (G), loads or motors (M),
and transformers (T) are identified with a note about wye or delta connection. Transmission lines (Z) are
described by impedance. A connection is called a bus (B). Voltage, current, and power transfer is
calculated at each bus.
G1
T1Z1
Z2
B1 M2
M1
The description and specification of each component is displayed on the one-line. If there is
inadequate room, a table may be used. Impedance is often in terms of Ohms per distance.
Location/ condition
kVA Pf or /Z P Q VLL /V IL /I Z distance R X
G1
T1
B1
Z1
M1
Z2
M2
The known values are placed in the table. Other values in the table are calculated as required based
on the known parameters. The voltage, current, and power transfer is simply calculated at each node.
Page | 3-10
3.6 Power Problem Plan There is a sequence that assists in solving complex electrical problems. Although all these may not
be required, they provide a process to assure all conditions are considered. Simpler problems may
require only one of these steps.
Reduce to one port for simple circuits.
Thevenin or Norton equivalents are a boundary condition.
Alternatively use a two port network.
For the network, get a model: π(∆), T(Y), or transfer function.
It has input and output of voltage or current.
Solve input & output independently.
Use conservation (V, I)
The sum of voltage or currents is 0 (KVL, KCL).
Use Ohm’s law
Ratio V
ZI
Use fields.
This includes, electric, magnetic, and inverse-square.
Use power – thermal – conversion
Product *S VI
Power transfer across reactance gives phase shift.
Make a table of values known & missing at each point of interest in circuit.
Frequency response always depends on reactance X.
Remember the goal!
Page | 3-11
3.7 Mechanical Power Power is the product of potential across and flow rate (number/time) through the machine
*S P Q T FV vi
S = pressure F = Force T = torque v = volts
Q = vol/time V = distance/time ω = revolutions/time i = amps (coul/sec)
A conversion factor (K) may arise because of different units in the measured values. A performance
factor (PF) may be necessary to scale the machine rating.
*convert
PP
K PF
The conversion factor depends on the units
Pconvert = Hp S=psi Q=gpm k=1714 1 gpm=5.45 m3/day
Pconvert = Hp S=feet Q=bpd k=135663 pu k=56,000
Pconvert = Hp T=ft-lb w=rpm k=5250
Pconvert = Hp v=volt i=amp k=746
Pconvert = watt v=volt i=amp k=1
Efficiency is the ratio of output to input power terms
( )
out out
in out loss
P Peff
P P P
The efficiency (eff) depends on the pump (machine) design
triplex=0.9 duplex=0.85 centrifugal=0.7
rod=0.9 beam system=0.4 motor=0.98-0.94
The brake horsepower (BHP) is the required power into the mechanical machine
*
*
S QBHP
K eff
The speed ratio (SR) determines comparative speeds
motor speed pump sheavediameterSR
pump speed motor sheavediameter
Page | 3-12
Load Torque (TL) is the shaft effort required to be input to the load by the motor
* *L
BHP PF KT
The pump performance factor (PF) depends on the pump design
Duplex – 1.5 Triplex – 1.31 Quintuplex – 1.27
Starting Torque (TS) is the effort required by the motor during starting
* * *rating
S
HP PF SR KT
The motor performance factor depends on the motor torque design characteristic.
NEMA B–1.5 NEMA C–2.25 NEMA D–2.75 NEMA E–1.5
Equipment is sized by using the following sequence:
Determine the horsepower of the fluid,
Convert to brake horsepower of the pump
Determine the pump torque
Select a motor with horsepower rating greater than BHP and a NEMA torque design rating greater
than the load torque
Size motor and pump sheaves to maintain appropriate pump speed
Page | 3-13
3.8 Electric Machinery
3.8.1 Basics
Alternating current is created in a coil of wire by a magnet rotating very close to the wire. As the
magnetic pole distance varies, the magnitude of voltage induced on the coil changes.
The chart illustrates the magnet at four positions with the fifth position the same as the starting
point.
V
N
S
V
N S
V
S
N
V
S N
V
N S
+V
-V
Page | 3-14
3.8.2 Machine Models
Machines can be modeled as a Thevenin equivalent voltage and impedance with a magnetizing
circuit consisting of an inductor with its resistance. The model is a two-port network. Three types of
problems are encountered.
Model parameters and losses require the complete model using circuit theory.
Transients and load flow use the Thevenin equivalent without the magnetizing.
Steady state uses the terminal conditions with complex power.
In determining the model parameters, place the model of machine into a circuit then perform circuit
analysis.
What is the difference between the machines? There are four fundamental classes – DC,
synchronous, induction, and transformer.
The input energy and output energy determine the use. A motor has electrical in and mechanical
out. A generator has mechanical in and electric out. The same machine can be used in either form. It
simply depends on the driver input and the driven output.
Generator: Mechanical In – Electrical Out
Motor: Electrical In – Mechanical Out
Transformer: Electrical In – Electrical Out
Page | 3-15
Machine Two-Port Models
RfAdj
Rf
-
+A
Ra
Lf
mag flux
Tω
Ea
Internal
Generated
+
-
Vt (DC)
DC
Machine
+
-
Vf (DC)
If Ia
Rf
Adj
Rf
-
+A
Ra
Lf
mag flux
Tω
Ea
Internal
Generated
+
-
Vp (AC)
Synchronous
Machine
+
-
Vf (DC)
If Ia
jXs
Li R Uses curve of
f AI vs E
1A fE K K i
fK I
A actual actual
A reference reference
E n
E n
( ) DCf f f fadjV I R R
t A A AV E I R
( )p A A A sV E I R jX
Page | 3-16
R1 R2
R2 (1-s)/s
Mechanical
Rc
Core
jX1 jX2
TωjXm
magnetizing
Air
Gap
Vp
(AC)
+
-
I2I1
E1
Induction
Machine
R1 R2
Rc
Core
jX1 jX2
jXm
magnetizing
Air
Gap
Vp1
(AC)
+
-
I2I1
E1
Vp2
(AC)
+
-
Transformer
Page | 3-17
Machine Loss Diagrams
I2R
Losses
Core
Loss
Mechanical
Loss
Stray
Loss
in t LP V I
convP
A A ind indE I
out app mP
DC Motor
I2R
Losses
Core
Loss
Mechanical
Loss
Stray
Loss
3 cosin P AP V I
convP
ind ind out load mP
Synchronous Motor
I2R
LossesCore
LossMechanical
LossStray
Loss
in app mP
convP
ind ind
3 cosout p AP V I
Synchronous Generator
I2R
Losses
Stator
CU
Core
Loss
Mechanical
Friction &
Windage
Stray
Loss
3 cosin P PP V I
AGP
ind ind out load mP
Induction Motor convP
I2R
Losses
Rotor
CU
Page | 3-18
3.8.3 Machine Tests
Tests on machines are conducted with the following instrumentation connections. The source is a
variable voltage such as dc power supply or ac variac that can adjust the voltage into the machine. The
ammeter and the wattmeter may have a shunt that is used to bypass excessive current.
Read power, voltage, and current at each test.
For open circuit tests, connect as shown with no load. Run at rated voltage.
For short circuit tests, short the terminals of the transformer or block the rotor of the machine. Start
at low voltage and increase until near rated current.
PS
Ammeter
Voltmeter
WattMeter
Load
No load / open circuit test
Set rated voltage, frequency
Reduced I
Read I through core
100% Reactiveno loadI
Blocked Rotor / short circuit test
Set rated current, reduced V
Read reduced V
I through rotor
2 2full loadI R X
Page | 3-19
3.8.4 Rotating Machines
Rotating machines can involve numerous different analyses. These include the torque and voltage
generation, power output, power losses, and performance.
Torque & Voltage Generation
Rotate a coil inside a magnetic field to develop a voltage and torque. The flux = φ and K = a constant
of machine design.
induced K I
inducedarmature
E K
ind
de N
dt
2
ZpK
a (a=turns ratio)
Rather than angular speed, the revolutions are often preferred where n = speed in rpm.
2 n
AE K n
'
60
ZpK
a
AC machines cause flux to change in a sinusoidal fashion.
sine N t
Sinusoids are analyzed using effective or RMS values. Although the machines may be three phase,
the analysis operates on single-phase at a time.
max 2c cE N N f
max
2A
EE
AE E
3y AE E
The number of magnetic poles and the frequency determine the speed.
# poles – p
/ 2p repetitions in one rotation
Frequency, f, is measured in Hertz.
Page | 3-20
2elec mech
pf f
60
mm
nf
60 2 120
m me
n n ppf
Page | 3-21
3.8.5 Power Conversion
Machines are about energy conversion from one form to another. The conversion is generally
investigated using power which is energy in some time.
Mechanical Power Converted
conv shaft mP
Electrical Power Converted
conv in stray mechanical coreP P loss loss loss
conv A AP E I
Output Power
2out conv I RP P loss
AC Machine
3 cos 3 cosout p p L LP V I V I
DC Machine
out L LP V I
Page | 3-22
3.8.6 Losses
Losses are encountered anytime there is a resistance or opposition. These are electric or copper,
magnetic or ferrous core, and mechanical or friction. The inaccuracies of measurement are
compensated as stray losses.
Electric - Copper Losses (I2R)
DC Machine - (armature) (field)
AC Machine - (stator) (field)
Magnetic - Core Losses – Hysteresis & Eddy
m
c
E
R
Mechanical losses – friction & windage
No load rotational = (mechanical losses + core losses)
~Proportional to n3
Stray losses
miscellaneous - ~1% of output power
Brush losses (DC Only)
BD BD AP V I ( 2 )BDV V
2
A A AP I R2
f f fP I R
23s A AP I R2Pf f fI R
Page | 3-23
3.8.7 Performance
Performance is a comparison of maximums and minimums. They are normalized, so they are
dimensionless.
Efficiency
N = efficiency = 100 100out in loss
in in
P P P
P P
Voltage regulation
1 1
1
100%n f
R
f
V VV
V
Speed regulation
1 1
1
100%n f
R
f
n nS
n
Positive SR means speed drops with load
Page | 3-24
3.9 Transformer A transformer is a machine that does not rotate. Otherwise, it is
very similar to an induction AC machine. In application, the ideal
transformer is represented simply as two coils.
The model of a transformer fits the Thevenin equivalent output
with a magnetizing circuit all within a two port network
The relationship between input and
output sides are dependent on the turns
ratio, a.
in out
out in
V Ia
V I
The equivalent elements can be
referred to the primary or input.
1 pR R 1 pjX jX
22 sR a R 2
2 sjX ja X
Alternately, the circuit can be referred to the output or secondary.
1 2
pRR
a 1 2
pjXjX
a
2 sR R 2 sjX jX
R1 R2
Rc
Core
jX1 jX2
jXm
magnetizing
Air
Gap
Vp1
(AC)
+
-
I2I1
E1
Vp2
(AC)
+
-
Transformer
I1 I2
V1 V2
Page | 3-25
3.9.1 Transformer Tests
Transformer tests are conducted according to the standard test procedures. The model is first
determined as a one-port circuit.
Open Circuit test
Conduct with the secondary or output
disconnected. Apply rated full voltage.
Most voltage drop is across the
excitation coil, so the tests yield values of
the core impedance.
Measure ocV , ocI , ocP
The calculations can refer to primary.
1 1
E
E c c
jY
Z R X
ocE
oc
IY
V
The angles are determined from measured magnitudes of power, voltage, and current.
cos oc
oc oc
Ppf
V I
EZ
EY
Short Circuit Test
Conduct with secondary winding shorted. Apply reduced voltage from a variac, and increase the
voltage until rated current is measured.
Most current flowing is low resistance series path. This shows values of copper impedance. Very
little current flows through excitation branch.
Measure scV , scI , SCP
Then calculate the impedance values.
scSE
sc
VZ
I
cos sc
sc sc
Ppf
V I
EZ
Core
Loss
Mag
I loss
R1
n2R2'n2jX2'
jX1'
IL I2
n2Z2V2'Non-ideal
approximation
Page | 3-26
2 2( ) ( )
SE eq eq
p s p S
Z R jX
R a R j X a X
Because of the connections, we cannot separate
primary & secondary impedance, but usually this is not
necessary.
The impedances cause a shift in the angles
associated with the current and voltage drops.
Page | 3-27
Transformer Characteristic Example
Ratings Open Circuit Values Short Circuit Values
20 kVA V=8,000V V=489V
8000/240V I=0.214A I=2.5A
60 Hz P=400W P=240W
Tests on primary
Open Circuit
Find values associated with magnetizing current.
400cos 0.234 lagging=76.5
(8000)(0.214)
oc
oc oc
Ppf
V I
0.21476.5 0.0000062 0.0000261
8000
ocE
oc
IY j
V
1 1
159 , 38.4
E
c m
c m
Y jR X
R k X k
Short Circuit
Find values associated with series current.
240cos 0.196 lagging=78.7
(489)(7.5)
sc
sc sc
Ppf
V I
48978.7 38.4 1.92
2.5
scse
sc
VZ j
I
38.4 1.92eq eqR X
These are the values associated with both windings in series– primary and secondary are not
separated
Page | 3-28
3.9.2 Transformer Turns
Transformers consist of two inductors that are closely coupled. Usually an iron core provides an
improved magnetic path. Laminations are used in the iron to reduce the hysteresis and eddy current
losses. There are no moving parts to a transformer. It simply converts the voltage on one side to a
different voltage dependent on the number of turns on each side. The current is converted inversely to
the turns.
Transformer windings are identified either by location or by terminal markings. Primary windings are
labeled with "H". Secondary windings are identified with "X".
Subscripts identify the separate terminals
The coupling between the turns is determined by the polarity.
Normal polarity is subtractive. The same subscripts are aligned
between the primary and secondary. Additive polarity has the
opposite subscripts aligned.
The voltage (V) ratio between the primary and secondary is equal to the corresponding turns (N)
ratio.
p p
s s
V N
V N
The inverse of the current (I) ratio between the primary and secondary is equal to the turns (N)
ratio.
pS
P s
NI
I N
The same transformer can be used as a step-up or step-down unit. A step-up transformer has a
higher voltage and a lower current on the secondary. Conversely, a step-down transformer has a lower
voltage and higher current on the secondary.
An autotransformer has the secondary and the primary connected together. The voltage is placed
on the primary. One terminal becomes common with the output. The other
primary terminal is connected to one of the secondary terminals. The remaining
secondary terminal becomes the second output terminal. If the secondary is
connected with additive polarity, it is a boost connection. If the secondary is
connected with subtractive polarity, it is a buck connection.
The input is the common coil, cN , while winding 2 becomes the series coil,
SEN , which is added or subtracted from the input.
cL
H c se
NV
V N N
c seL
H c
N NI
I N
IL
I2V2
VHVL V1
I1 I2
V1 V2
Page | 3-29
The apparent power into and out of the transformer must be equal.
in out ios s s
The apparent power in the windings is the same in the common and the series winding.
w c c se ses V I V I
So the ratio of the apparent power gives a “gain” or apparent power advantage.
io se c
w c
s N N
s N
Page | 3-30
Example transformer turns:
Given: A transformer has a 120 volt primary and a 12 volt secondary. Primary current is 10
amps.
Find: Turns ratio
Secondary current
VA rating of each winding
120
12
p p p
s s s
V N N
V N N Turns ratio – 10:1
10
10 1
ps s
p s
NI I
I N
10*10
1001
sI
120*10 1200p p s sV I V I VA
Example autotransformer:
Given: Connect the transformer in the above example as a boost autotransformer with 120 volt
primary.
Find: output voltage
output current
output power
120 12 132Vout c sev v v
100out seI I A
132*100 13,200io H Hs V I VA
Page | 3-31
Vt
IL
IARA Field
weak
3.10 DC Machines DC Machines are the typical example of a Thevenin equivalent for the armature and a magnetizing
inductance for the field. The field is a stationary stator and the armature is a rotating rotor.
The field can be connected separately, parallel, or in series. Other than changing the value of the
voltage, the calculations are very similar. However, the performance is very different.
Generic Relationships
The magnetic flux is determined by the field current.
Li R
mmf F FN IF
FF
F
VI
R
The flux couples to the armature to produce voltage and
torque.
indind A AK I I
K
1A FE K K i
A actual actual
A reference reference
E n
E n
Separately Excited
A separately excited machine has different
dc voltage on the field and the armature.
t A A AV E I R
L AI I
)(
FF
F adj
VI
R R
( )F F F FadjV I R R V
Shunt Connected
A shunt excited machines has the field connected across the terminals.
A F LI I I
F TV V
RfAdj
Rf
-
+A
Ra
Lf
mag fluxEa
Internal
Generated
+
-
Vt (DC)
DC
Machine
+
-
Vf (DC)
If Ia
Page | 3-32
The terminal voltage controls flux, torque, and speed.
T A A AV E I R
T i F A AV K i I R
indT AV K R
K
2
T Aind
V R
K K
Series Connected
( )t A A A SV E I R R
A S LI I I
-
+A
RA +
-
IA Is&IL
Rs
Vt
Radj
RF
LF
compound
-
+A
RaRf adj
+
-
Vt
IA IL
Lf
Page | 3-33
3.11 AC Machines - Synchronous AC machines are the typical example of a Thevenin equivalent for the armature and a magnetizing
inductance for the field. The armature is stationary and the field is on the rotor. The armature is AC and
is connected directly to the line. The field may be dc which creates a synchronous machine or ac which
creates an induction machine.
Synchronous machine field is created by a separate DC source on the rotor.
Induction machine field is creates magnetic induction on the stator.
Stator has 3φ or 1φ currents.
120º equal magnetic current. in a 3φ machine
2 pole (1N-1S) is the minimum number and results in a synchronous speed of 3600 rpm. The
synchronous speed (nsync) of the machine can be determined from the number of poles. There is
always an even number of poles.
120
#sync
fn
poles
2 f speed
3.11.1 Synchronous Machine
A synchronous machine rotation locks onto the power line frequency which results in a constant
speed, dependent only on the frequency. The model is conducted per phase. The model is for a motor
or generator.
A p S A A AE V jX I R I
sin
cos AA
S
EI
X
3 sinp A
out
S
V EP
X
The variables have the following meanings.
Xs=Synchronous Reactance
δ = torque angle:
max τ @ 90º
15 20
Power and torque are available on the shaft.
P
Rf
Adj
Rf
-
+A
Ra
Lf
mag fluxEa
Internal
Generated
+
-
Vp (AC)
Synchronous
Machine
+
-
Vf (DC)
If Ia
jXs
Page | 3-34
3 sinp A
m s
V E
X
fK I
A generator has additional considerations.
1A fE K K i
( )f f f fadjV I R R
( )t A A A sV E I R jX
Li R
Droop
1n sys
droop
f f
f
1n droop sysf f f
Page | 3-35
3.11.2 Induction Machine
The induction machine follows the Thevenin equivalent for the armature with a magnetizing
inductance for the field all within a two-port network; however, the magnetizing circuit cannot be
separated, but is an integral part of the machine. The field is on the stator and the armature is on the
rotor.
The induction machine is simply a transformer that has the secondary shorted and allowed to spin.
As such it is a purely AC machine with no DC field. The rotor field has shorted bars with voltage induced
from the stator field.
Because of the rotation, additional calculations are required compared to the transformer. The rotor
speed varies with load. The difference in the synchronous speed and the mechanical shaft speed is
called the slip.
Slip
Slip, s, is normalized or per unit.
100sync m
sync
s
100 100slip sync mech
sync synce
n n ns
n n
0 rotor @ sync speeds
1 rotor stationary (locked)s
slip sync mn n n - Slip Speed
(1 )m syncn s n
Example slip:
Speed = 1720, s = 0.04 = 4%
Losses
3 cosin T LP V I
air gap in stator cu coreP P P P
conv air gap rotor cu ind mP P P
R1 R2
R2 (1-s)/s
Mechanical
Rc
Core
jX1 jX2
jXm
magnetizing
Air
Gap
Vp
(AC)
+
-
I2I1
E1
Induction
Machine
Page | 3-36
&out conv f w stray load mP P P P
Power & Torque
Only real elements consume power and convert it to mechanical energy or heat.
213stator cuP I R
21
23core c
c
EP R
R
air gap in s cu coreP P P P
2 223ag
RP I
s
22 23rotor cu agP I R sP
The developed mechanical power is dependent
on the slip.
conv ag r cuP P P
2 2 222 23 3
RI I R
s
22 2
(1 )3
sI R
s
& ( )out conv f w misc strayP P P P
This allows calculating locked rotor and no load examples of an induction machine.
The results give a combined X for the rotor and stator. These are separated based on the machine
design, whether A, B, C, D.
Core
Loss
f&w
R1 R2'
R’L
jX2'jX1'
2
(1 )L
sR R
s
22 2
(1 )sL
R R sR R R
s s
IL I2
Page | 3-37
3.12 Transmission Lines Power transmission is investigated by making a two-port network for the line. The components are
distributed and are typically measured in /mile, /foot, or
/km. Work the problem on per- phase.
Reactance of most cables is published by the manufacturer.
The reactance in ohms per 1000 feet of aerial cables with one
foot spacing can be found with the following formula.
1
0.02298 lnLXGMR
GMR is the geometric mean radius in feet. It can be calculated by multiplying the wire O.D. in inches
by .03245.
0.25 0.3894 0.03245GMR R e D d
R = wire radius in feet
D = wire diameter in feet
d = wire diameter in inches
The reactance of aerial cable depends on the spacing between wires. Reactance at spacings other
than one foot can be calculated with the following formula.
ln
11
lnnew old
spacingX X
GMR
Cable operating temperature has an effect on the resistance of a cable. Most cables have a rated
operating temperature of 90 °C. Aerial cable is rated 75 °C. Cables have higher resistances at their rated
operating temperature ratings than at ambient temperature. The resistance at rated operating
temperature can be calculated from the resistance at ambient temperature using the following formula.
C/2 C/2
R L
Page | 3-38
2 1 2 11R R T T
R2 = resistance at operating temperature
R1 = resistance at ambient temperature
T2 = rated operating temperature
T1 = ambient temperature
α = temperature coefficient of resistivity corresponding to temperature T1 (0.00393 for copper at 20
°C)
The current that causes the cable to reach its highest temperature may not be the maximum
available current. The current depends on the cable’s resistance and the resistance depends on the
current. The maximum available current should be the current that causes maximum temperature.
Page | 3-39
3.13 Per Unit Notation Per unit notation is used to reduce the complexity when working with circuits that have multiple
voltage levels. Per unit is nothing more than performing calculations based on percentages. Both Ohm’s
law and the power relationship permit a third term to be calculated from only two terms.
Two parameters are selected as the reference or base values. These are generally S and V. A
different base V is used on each side of a transformer. The base current and base impedance can be
determined from these two values
basebase
base
SI
V
2base
base
base
VZ
S
All the circuit equipment voltages and currents are then converted to per unit (percentage) values
before normal circuit calculations are made
*100equip
pu
base
SS
S
*100equip
pu
base
VV
V
*100equip
pu
base
II
I
*100equip
pu
base
ZZ
Z
As an example, transformer impedance is usually rated in per unit values. To find the actual
impedance, combine the above equations
100
pu
equip base
ZZ Z
2
100
pu baseequip
base
Z VZ
S
An example illustrates the relationship between per unit values and short circuit capability.
Transformer, 10bases kVA , 120basev , 2%puZ
22120
1000.0288
10000equipZ
Page | 3-40
10010000 5000
2
base
pu
SSCC kVA
Z
4167basesc
equip base equip
VV SCCI A
Z V Z (use pre-fault voltage)
Page | 3-41
3.14 Short Circuit Considerations
3.14.1 Introduction
A short circuit condition differs from normal current operations only by virtue of a sudden decrease
in the circuit impedance. The decrease in impedance is caused by a fault.
The power source is generally rated by a short circuit capacity (SCC) rating in volt-amps. This is the
product of the pre-fault voltage and the post-fault current. Short circuit current is restricted only by the
source impedance, since the load is greatly reduced.
1scc pre scVA V I
3 3scc pre scVA V I
With the short circuit capability and the voltage rating, the source impedance can be determined.
The impedance calculated is for each phase, if the system is three-phase.
2p
source
VZ
SCC
The SCC of a magnetic device, such as a transformer or machine, can be found using the percent
impedance (Zpu) and the device rating
100
pu pu
kVASCC kVA
Z Z
The available fault current is also restricted by the SCC of the transformer
3
3
3SC
line
SCCI
V
1
1SC
line
SCCI
V
The available fault current is restricted by the source fault current and the transformer turns ratio.
sec
primary
ondary
VN
V
secondary primarySC SCI I N
The available fault current is the smaller value that is calculated using the two methods above.
Other impedance in the wiring will further restrict the fault current.
pre
SC
fault
VI
Z
Page | 3-42
Short circuit contribution from induction machines continues after a fault. Inertia causes the
machine to continue turning with a collapsing magnetic field. This results in approx 25% of the
machine’s capability contributing to the fault current.
Page | 3-43
3.14.2 Fault Analysis
Fault analysis is commonly called a short-circuit study. Fault current
differs from normal current only by an accidental decrease in circuit Z.
Under fault conditions, the load (ZL) may be 1 or more Ohm, while
the fault is ~0.0001 ohm.
The resulting Z is a parallel combination.
41 1 11 10 1
0.0001 1TZ
The load is negligible compared to the fault.
A Afault
T f
V VI
Z Z
Realistically, the ZT will provide a significant restriction on fault current. The first part of the process
is to select the base values for the one-line diagram.
Need complete one-line diagram
Convert to per unit (percent)
Normally pick bS & bV , and then calculate bI & bZ
bb
b
SI
V
2
2 3
3
bbbb
b b b
VVVZ
I S S
Because to the three-phase characteristics, we can use either per phase values or line values for 3-
phase current and voltage calculations
Do all calculations on single phase basis for Z. Impedance is a phase measurement only.
EA Zf ZL
Page | 3-44
3.14.3 Symmetrical Components
Faults by definition indicate there is an imbalance in the system. This impacts the impedance and
the resulting current and voltage drops. Because of the asymmetry, the interaction in the calculation
would be very convoluted. A preferred method is to make a transform that converts the system to
balanced, and separates the components by the 120o normally expected in three-phase power systems.
The transform is called symmetrical components. The operator is α
0.5 0.866 1 120j
2 0.5 0.866 1 120j
Sequence Currents
Sequence currents are positive, negative, and zero sequence. These are found from the transform
operating on the phase currents.
213 1( )A A B C pI I I I I I
213 2( )A A B C nI I I I I I
130 ( )A A B C o GI I I I I I
Line Currents
Phase currents can be obtained by taking the inverse transform on the sequence currents.
0A A A AI I I I
0B B B BI I I I
0C C C CI I I I
Symmetrical components are used to take any unbalance combination of V & I and make them
operate as balanced 3φ model.
The only use is to aid the algebra. Symmetrical components are not “real”.
Page | 3-45
3.14.4 Ratings & Reactances
Under fault conditions, the power system waveform goes through a transient condition until it
stabilizes. Three types of transient are used to describe the changes in state.
1. Momentary is the first cycle. Use all induction motors and subtransient (Xd’’) reactances
2. Interrupting is for contact parting. Neglect branches w/ pure induction motors and use only
transient (Xd’) reactances, except below 600V
3. Asymmetrical – use multiplier from tables (if not known, use 1.6)
Reactance values are dependent of frequency. Therefore, the reactance will change as the slope of
the waveform varies. There are three conditions.
Sub-transient is during the first cycle to 0.1 seconds of the fault.
Transient is during 30 cycles to 2 seconds.
Synchronous is the steady state rating.
Rules of Thumb
(steady state) 1.0dX
(transient) 0.33dX
(subtransient) 0.25 600dX for V
(subtransient) 0.2 600dX for V
Page | 3-46
3.14.5 Short Circuit Study
A short circuit study is dependent on how the fault is connected within the one-line diagram.
1. Draw single line diagram w/ all sources of fault current, such as generators & motors, and
utility connections.
2. Replace all components, including reactance, with resistors (impedance) symbol, and label
with letters.
3. Show all transformer secondary feeding induction motors, whether motors are indicated or
not.
4. Join all components by “infinite bus” (neutral)
5. The source is not the infinite bus, but is simply another reactance.
6. Rearrange impedances into series & parallel.
7. Reduce to single Z.
8. Convert Δ blocks to star to further reduce (Thevenin Z)
Page | 3-47
Example
The circuit in Figure 1 has a fault at X.
Calculate the Thevenin Z at the fault.
Use pre-fault V at the fault to find fault current, I.
1)
G
G1
M
M1
Induction
MS
M2
Synchronous
M
M3
Fault
2) Fault
T1
G1
M1 M2
M3
3) Fault
T1
G1M1 M2
M3
4) Fault
T1
M1//G1//M2 M3
5) Fault
M1//G1//M2+T1 M3
6) Fault
Ztotal=(M1//G1//M2+T1)//M3
Page | 3-48
3.14.6 Unbalanced Faults
The previous fault analysis development was for a 3-phase fault.
0A B CI I I
Unbalanced conditions are redefined in terms of 3 components.
Positive sequence (+, p, 1)
System with sources rotating
“Normal” conditions
Negative sequence (-, n, 2)
The negative sequence is the same circuit as positive, but without sources.
Z may have different value
Zero Sequence (0, Z, 0)
The zero sequence circuit is simply the ground path.
Δ has no ground
has no Ground
has a ground path
Draw three circuits, one for each sequence.
Leave sources in positive
Make negative w/o sources
Zero indicates ground paths
Connect each sequence with fault impedances for each component.
3Zf represents the fault impedance in pos, neg & zero sequences.
Page | 3-49
3.14.7 Faults with Rotating Machines
EA vA
neutral
AV E Z I
0V Z I
0 0 00V Z I
For 3φ faults, use the positive sequence only.
0A B CI I I
0A A A AI I I I
For one phase, calculate the symmetrical current..
13A AI I , since 0A A AI I I
The current is drawn at the fault.
The zero sequence is the path through ground. It will
change depending on the xformer connection or Y .
Z0 will be different since transformer & machine ground
path may not be connected
Z0 motor or gen = 3 nZ
Z0 = 0 for solidly connected neutrals
In zero sequence, include machine impedance to ground.
3Zn=3 times impedance of any phase
Use in place of source voltage
An assumption can be made for the sub-transient reactance.
120 0d motor dZ Z
EA ZL
Xs+
ZL
Xs-
Xs0
IA+
IA-
IA0
3Zf
Page | 3-50
3.14.8 Fault Illustrations
A one-line diagram shows a generator, transformers, transmission line, and motor load. Convert the
one-line into positive, negative, and zero sequences.
G M1 2
Xng
Xg XL
XTu XTd
Xm
Up dn
Xnm
EgEm
1 2
Xg1 XL1 Xm1Xu1 Xd1
Positive
Sequence
1 2
Xg2 XL2 Xm2Xu2 Xd2
Negative
Sequence
Xg0 XL0 Xm2Xu0 Xd0
Zero Sequence3Xng0 3Xnm0
If the connection on a transformer to ground is , then insert the connection and
impedance. Else, leave the ground connection open.
Illustrations are developed for four types of
faults.
Three phase fault at 2, use positive sequence
only.
0 2 0V V
1
f
eqt
VI
Z
0 2 0I I
Single phase line to ground at 2, use positive, negative, and zero sequence in series.
0 1 2 13 fV V V Z I
1
1 2 0 3
fault
f
VI
Z Z Z Z
1 2 oI I I
Line to line, use positive in series with negative
and connect with Zf.
Eg
Zg1 ZL1
3Zf
Zu1 Zd1
short
Eg
Zg1 ZL1
3Zf
Zu1 Zd1
short
Vpu
Z1 Z2Zf
+
-V1
+
-
V2
Page | 3-51
1
1 2
f
f
VI
Z Z Z
1 2I I
Double line to ground, use positive with negative in parallel
with zero.
1
1 2 0( || 3 )f
VfI
Z Z Z Z
1 2 0I I I
Vpu
Z1 3Zf
Z2
+
-
V0
+
-
V2 Z0
Page | 3-52
Symmetrical RMS Fault Currents and Voltages in Terms of Sequence Impedances
Three-phase fault through three-phase fault impedance Zf
Line-to-Line fault. Phase b and c shorted through fault impedance Zf
Line-to-line fault. Phase a grounded through fault impedance Zf
Double line-to-ground fault. Phases b and c shorted, then grounded through fault impedance Zf
aI 1
f
f
V
Z Z 0
0 1 2
3
3
f
f
V
Z Z Z Z 0
bI 2
1
f
f
a V
Z Z
1 2
3f
f
Vj
Z Z Z
0
0 2
1 2 1 2 0
33
3
f
f
f
Z Z aZj V
Z Z Z Z Z Z
cI 1
f
f
aV
Z Z
1 2
3f
f
Vj
Z Z Z 0
20 2
1 2 1 2 0
33
3
f
f
f
Z Z a Zj V
Z Z Z Z Z Z
aV 1
f
f
f
ZV
Z Z
2
1 2
2 f
f
f
Z ZV
Z Z Z
0 1 2
3
3
f
f
f
ZV
Z Z Z Z
2 0
1 2 1 2 0
3 2
3
f
f
f
Z Z ZV
Z Z Z Z Z Z
bV 2
1
f
f
f
a ZV
Z Z
22
1 2
f
f
f
a Z ZV
Z Z Z
22 0
0 1 2
3 3
3
f
f
f
a Z j Z aZV
Z Z Z Z
2
1 2 1 2 0
3
3
f
f
f
Z ZV
Z Z Z Z Z Z
cV 1
f
f
f
aZV
Z Z
2
1 2
f
f
f
aZ ZV
Z Z Z
22 0
0 1 2
3 3
3
f
f
f
aZ j Z a ZV
Z Z Z Z
2
1 2 1 2 0
3
3
f
f
f
Z ZV
Z Z Z Z Z Z
bcV
1
3f
f
f
Zj V
Z Z
1 2
3f
f
f
Zj V
Z Z Z
0 2
0 1 2
3 23
3
f
f
f
Z Z Zj V
Z Z Z Z
0
caV
2
1
3f
f
f
a Zj V
Z Z
22
1 2
33
f
f
f
a Z j Zj V
Z Z Z
20 2
0 1 2
33
3
f
f
f
a Z Z Zj V
Z Z Z Z
2 0
1 2 1 2 0
3 33
3
f
f
f
Z Z ZV
Z Z Z Z Z Z
abV
1
3f
f
f
aZj V
Z Z
2
1 2
33
f
f
f
aZ j Zj V
Z Z Z
0 2
0 1 2
33
3
f
f
f
a Z Z Zj V
Z Z Z Z
2 0
1 2 1 2 0
3 33
3
f
f
f
Z Z ZV
Z Z Z Z Z Z
Page | 3-53
Page | 3-54
Page | 3-55
Chapter 3 Problems Practice Problem 3-1(Old Style)
SITUATION:
A paper mill is supplied by a 13.8kV, 3-phase, 60 Hz system with appropriate transformers.
The total load is as follows:
Induction Motors 600Hp Efficiency – 85% power factor – 0.8 lagging
Heating and Lighting 100kW unity power factor
Synchronous Motors 200Hp efficiency – 90% leading power factor
The synchronous motors are being operated at rated kVA and are over excited to correct the plant
power factor to 0.95 lagging.
It is desired to increase the mill output by 20%
A plant survey indicates that the installed induction motor capacity is adequate for this increase, but
the synchronous machines are at rated kVA.
It is suggested that it might be possible to increase the output of the synchronous motors by a
sufficient amount by reducing the excitation to unity power factor and providing power factor
correction with a static bank of capacitors.
REQUIREMENTS:
Determine if it is possible to increase the output of the synchronous motors by 20% by reducing
excitation without exceeding kVA ratings. Explain your answer
Determine the capacitance per phase, delta connected, to correct the power factor to 0.95 lagging if
the capacitors are connected across the 13.8kV line.
Page | 3-56
SOLUTION:
Given Existing System:
Induction motors:
600 0.746 1* * 526.6
1 85%( )
Hp kWP kW
Hp eff
526.6
658.20.8( )
kWS kVA
pf
2 2658.2 526.6 394.9Q kVAR
Heat & Light:
100P kW
100
1001.0( )
kWS kVA
pf
2 2100 100 0Q kVAR
Synchronous Motors:
200 0.746 1
* * 165.81 90%( )
Hp kWP kW
Hp eff
a) Plant:
792.4P kW
792.4
834.10.95( )
kWS kVA
pf
2 2834.1 792.4 260.5Q kVAR
Synchronous Motor:
260.5 394.9 134.4
plant indQ Q Q
kVAR
165.8 134.4 213.4 39.09S j
cos( 39.09) 0.776pf
By reducing exciting of synchronous motors to give unity power factor, P of synchronous motors
increases to 213.4kW.
S P Q Pf
Induction Motors 658.2 526.6 394.9 0.8 Heat & Light 100 100 0 1.0 Synchronous Motors 165.8
Plant 792.4
S P Q Pf
Induction Motors 658.2 526.6 394.9 0.8 Heat & Light 100 100 0 1.0 Synchronous Motors 165.8
Plant 834.1 792.4 260.5 0.95
S P Q Pf
Induction Motors 658.2 526.6 394.9 0.8 Heat & Light 100 100 0 1.0 Synchronous Motors 213.4 165.8 -134.4 -0.78
Plant 834.1 792.4 260.5 0.95
Page | 3-57
% increase = (213.4 165.8)
28.7%165.8
So, synchronous load increased by more than 20%, planned upgrades are possible.
b) With 20% plant increase (20% increase in
induction motor load and 20% increase in
synchronous motor load)
Q needed for 95% pf = 1930.9tan(cos 0.95) 930.9tan(18.2 ) 306.1kVAR
95% 396.9 306.1 90.8cap total pfQ Q Q kVAR
/
90.830.3
3 3
cap
cap ph
Q kVARQ kVAR
2V
SZ
2
c
VQ
X
/
2 2
30,3000.422 /
13,800 *(2 60)
cap ph
ph
QC F ph
V
S P Q Pf
Induction Motors 789.9 631.9 473.9 0.8 Heat & Light 100 100 0 1.0 Synchronous Motors 213.4 199 -77 0.93
Plant 1012.0 930.9 396.9 0.92
Page | 3-58
Practice Problem 3-2 (old style) 136
A generator is connected through a step-up transformer to a very large system (infinite bus).
The generator is rated: The transformer is rated:
840 MVA
90% power factor
24kV
Xd = 160%
Xd' = 25%
Xd'' = 18%
The transformer is delivering 600MW at 85% power factor to the system at 500kV when the
operator increases the output to 750MW without changing the setting of the generator voltage
regulator so that the generator terminal voltage is unchanged.
The operator then adjusts the voltage regulator so that the output of the transformer is 800VA.
REQUIREMENTS:
Determine the generator terminal voltage and the new transformer power factor and MVA output
after the power output is increased, but before the voltage regulator is adjusted.
Determine the generator terminal voltage and the transformer output power factor after the
voltage regulator is adjusted
Note: Answer to at least four (4) significant figures.
SOLUTION:
Generator: 840MVA, 90% pf, 24kV
Transformer: 800MVA, 22kV pri/500kV sec, 14%tX
Secondary connected to infinite bus – V fixed, f fixed
Select Per Unit Base – use transformer
800baseS MVA
2 500baseV kV
1 22baseV kV
2
8001.6
500baseI kA
S VI (cos sin )I I j *P S pf
1 2 1 21 2
sin( )VVP
X
21 1 2 1 2
1 2
cos( )V V VQ
X
800 MVA
22 – 500 kV
X = 14%
G
V1 V2
jXs jXT
Page | 3-59
1 1 2 2;V V
For infinite bus, fixed V&f , vary P&Q
For fixed load, fixed P&Q, vary V&f
Start Solution
Initial Conditions:
600
705.880.85
MWS MVA
pf
1705.880.8824 (cos 0.85) 0.8824 31.79
800pu
MVAS
MVA
2 1.0 0puV (infinite bus)
* 0.8824 31.790.8824 31.79
1.0 0
pu
pu
pu
SI
V
0.8824 31.79 (0.75 4649)puI j
21 (0.75 0.4649)( 0.14) (0.6508 0.105)tV IjX j j
1 2 21 (1 0) (0.06508 0.105)
1.0651 0.105
1.0703 5.6303
V V V j j
j
Check using power transfer equation
1 2 1 21 2
sin( )
1.0702*1*sin(5.6303 0)0.75 600
0.14
V VP
X
MW
Page | 3-60
Increase power to 750MW – no change to voltage
750
0.9375800
puP
2 1 0V (infinite bus)
1 1.0702V
The angle will change since P increases, but voltage magnitude has not changed
1 2 1 2 11 2 1 2
1 2
sin( )sin( )
VV P XP
X V V
1 1
0.9375*0.14sin( 0) 7.0445
1.0702*1
1 1.0702 7.0445 1.0621 0.1313V j
1 2( ) (1.0621 0.1313) (1 0)
0.14
0.9375 0.4436 1.0375 25.31
t
V V j jI
jX j
j
25.31S Z I
1cos ( 25.31) 0.904pf
* 1 0*1.0375 25.31 1.0375 25.31outS VI
* 800*1.0375 830rate base puS S S MVA
1 1 1* 22 *1.0702 23.54rate base puV V V kV kV
Page | 3-61
Change voltage so transformer is at rated MVA (800MVA)
800 1.0puS MVA S
1
11
pu
pu
pu
SI
V
2 2
750750 0.9375
800puP MW P
2 2 2
2
2 2
cos
0.9375cos 0.9375
1*1
20.364
P V I
P
V I
(cos sin )
1(0.9375 0.3480) 1 20.364
I I j
j
1 2
1 (0.9375 0.3480)( 0.14)
1 0.0487 0.13125 1.0487 0.13125
1.05689 7.13363
tV V IjX
j j
j j
Check using power transfer equation
1 2 1 21 2
sin( )
1.05689*1*sin(7.133 0)0.9375 750
0.14
V VP
X
MW
1 22 *1.05689 23.252rateV kV kV
Page | 3-62
Practice Problem 3-3 (old style)
SITUATION:
A refinery is planning to install a compressor-expander connected to an induction machine.
Under certain conditions the induction machine, taking 8 MVA at 85% power factor, will act as a
motor to drive the compressor-expander to compress a certain gas.
At other times, the compressor-expander will act as a turbine to drive the electric machine as a
generator supplying 6 MW at 82% power factor.
Also connected to the plant bus:
Induction motors taking 4 MVA at 86% pf
Incandescent lighting taking 2MW
Synchronous motor taking 2 MW at 90% pf
The plant will receive the necessary power from the utility at 13.8kV
REQUIREMENTS:
In order to formulate the conceptual design of the new substation, determine the plant load in MW,
amperes and power factor that the utility must supply under the following conditions:
When the induction machine is driving the compressor-expander.
When the compressor-expander is driving the induction machine.
SOLUTION:
Keep track of P’s and Q’s – Solve for S, I and pf.
System – Induction motor driving compressor expander
Induction motors:
4S MVA
0.86pf
4 *0.86 3.44P MVA MW
2 24 3.44 2.041Q MVAR
Lighting
2P MW
1.0pf
2
2.01.0
MWS MVA
S P Q Pf
Induction Motors 4 0.86 Lighting 2 1.0 Synchronous Motors 2 0.9 leading Compressor-Expander 8 0.85
Plant
Page | 3-63
2 21 1 0Q MVAR
Synchronous Motors
2P MW
0.9( )pf leading
2
2.2220.9
MWS MVA
2 22.222 2 0.9686Q MVAR
Compressor – Expander
0.85pf
8S MVA
8
6.80.85
MVAP MW
2 28 6.8 4.215Q MVAR
a) 14.24plantP MW
14.24 5.286 15.189 20.37plantS j
cos20.37 0.9375pf
*
*
3
3
15.189635.5
3 *13.8
S VI
SI
V
MVAAmp
kV
b) Same as above, but compressor-expander driving induction generator
Compressor – Expander
S P Q Pf
Induction Motors 4 3.44 2.041 0.86 Lighting 2 2 0 1.0 Synchronous Motors 2.222 2 -0.9686 0.9 leading Compressor-Expander 8 6.8 4.214 0.85
Plant 14.24 5.286
Page | 3-64
6.0P MW
0.82pf
6.0
7.3170.82
S MVA
2 2( 7.317) ( 6.0) 4.188Q MVAR
1.44plantP MW
1.44 5.260 5.454 74.69plantS j
cos74.69 0.264pf
*
3
5.454228.2
3 *13.8
SI
V
MVAAmp
kV
S P Q Pf
Induction Motors 4 3.44 2.041 0.86 Lighting 2 2 0 1.0 Synchronous Motors 2.222 2 -0.9686 0.9 leading Compressor-Expander -7.317 -6.0 4.188 0.85
Plant 1.44 5.26
Page | 3-65
Practice Problem 3-4 (old style – similar to new style)
SITUATION:
In order to accommodate the needs of larger plant power systems, a switchgear manufacturer is
offering a new line of vacuum circuit breakers rated in accordance with ANSI C 37.04 “Rating Structure
for A-C High Voltage Circuit Breakers.
The following data apply:
Nominal voltage: 7.2kV
Nominal 3-phase MVA class: 700 MVA
Rated maximum voltage: 8.25kV
Rated voltage range factor (k): 1.3
Rated insulation level – low frequency: 36kV
Rated continuous current: 3000 A
Rated short circuit current: 46kA
Rated interrupting time: 5 cycles
REQUIREMENTS:
Determine the following related required capabilities (to three significant figures)
The symmetrical interrupting capability with the prefault voltage @ 7.2kV
The symmetrical interrupting capability with the prefault voltage @ 6.1kV
The symmetrical interrupting capability with the prefault voltage @ 8.5kV
Closing and latching capability operating voltage at 7.2kV
Closing and latching capability operating voltage at 6.1kV
Closing and latching capability operating voltage at 8.5kV
Three-second short time current carrying capability operating @ 7.2kV
If the circuit breaker is applied at a point where the system impedance is 0.08Ω/phase, what is the
change in margin of symmetrical interrupting capability over short circuit current available when the
operating voltage is changed from 7.2kV to 7.4kV?
Page | 3-66
SOLUTION:
Symmetrical interrupting capability at 7.2kV
Max kVA remains the same, so, as voltage goes down, current goes up, as long as voltage is within
voltage range (1.3)
8.25
1.157.2
, so keep constant kVA
3 3max 8.2546 10 52.71 10
7.2SC rated
nom
VI I Amp
V
Symmetrical interrupting capability at 6.1kV
Max kVA remains the same, so, as voltage goes down, current goes up, as long as voltage is within
voltage range (1.3)
8.25
1.356.1
, so limit voltage ratio to 1.3
3 3*1.3 46 10 *1.3 59.8 10SC ratedI I Amp
Equipment is rated at 8.25kV, do not use at 8.5kV
Max interrupting current: 3 31.3*46 10 59.8 10mI Amp
Refer to ANSY C 37.04 standards
Closing & latching is 1.6 times the max current (at any voltage below rated)
3& *1.6 95.68 10c l mI I Amp
Closing & latching is 1.6 times the max current (at any voltage below rated)
3& *1.6 95.68 10c l mI I Amp
Equipment is rated at 8.25kV, do not use at 8.5k
Three second current capability = max interrupting current (3s < 5s rating)
3 31.3*46 10 59.8 10mI Amp
7.2kV is L-L voltage, need L-N (phase) voltage
Page | 3-67
7200
4156.93 3
LLLN
VV V
System Z is 0.08,
3720051.96 10
3 *0.08scI Amp
From a) above, Max interrupting current at 7.2kV is 52.71x103 Amp, so margin is 748 Amp
at 7.4kV, short circuit current is
3740053.406 10
3 *0.08scI Amp
Max short circuit current is
3 3maxmax
8.2546 10 51.283 10
7.4SC rated
nom
VI I Amp
V
So, equipment is overrated by 2,122 Amps. Change in margin is 2,870 Amps.
Page | 3-68
Practice Problem 3-5(old style)
SITUATION:
An industrial plant has four generators connected to a 13.8kV bus as shown in Figure 433 below. The
neutral of each generator is connected through a neutral circuit breaker to a common neutral bus. This
bus is connected through a neutral resistor to ground.
G
G1
G
G2
G
G3
G
G3
Individual generator reactances in pu:
2
0
0.16
0.16
0.08
dX j
X j
X j
REQUIREMENTS:
a) Determine the value of the neutral resistor required to limit a line-to-ground fault to that of
a three-phase fault when only one of the units is operating.
b) If a 0.1 pu neutral resistor is used, determine the ground fault current in pu when all units
are operating, but with only one neutral circuit breaker closed.
c) Same as b., except all neutral circuit breakers closed.
SOLUTION:
Estimate positive, negative and zero sequence impedances from data given.
1
2 2
0 0
0.16
0.16
0.08
dZ X j
Z X j
Z X j
a) f resistorZ Z
Page | 3-69
3
1
f
ph
f
vI
z z
1
0 1 2
3
3
f
ph
f
vI
Z Z Z Z
Operating at 100% voltage (1.0 pu)
3
1.06.25
0.16 0phI j
j
1
3*1.0 3
0.16 0.08 0.16 3 0.4 3ph
resistor resistor
Ij j j Z j Z
Set 1phI = 3 phI and solve for resistorZ
36.25
0.4 3
30.4 3
6.25
3 0.48 0.4
0.026667
resistor
r
R
R
jj Z
j Zj
Z j j
Z j
b)
1
0 1 2
12 2 2 2
3 3*1
3 0.08 0.04 0.04 3* 0.1
3 3
( 0.16) (0.3)
8.8235
f
ph
f
ph
vI
Z Z Z Z j j j j
Ij X R
pu
c) With all breakers closed, Zo=0.02j
1
0 1 2
12 2 2 2
3 3*1
3 0.08 0.04 0.02 3* 0.1
3 3
( 0.1) (0.3)
9.487
f
ph
f
ph
vI
Z Z Z Z j j j j
Ij X R
pu
3 x 0.1
j0.16
j0.16
j0.08
j0.16
j0.16
j0.08
j0.16
j0.16
j0.08
j0.16
j0.16
j0.08
Positive sequence
impedance
Negative sequence
impedance
Zero sequence
impedance
One breaker
closed
Page | 3-70
Problem 3-6 (old)
SITUATION:
A 1000/1250 kVA, OA/FA, 13.2kV:4160V single phase transformer is part of a 3000/3750 kVA Y-Δ
bank.
Factory tests are made on this transformer at 25°C and the following data recorded.
DC Resistance: r1 = 0.40 Ω r2 = 0.035 Ω
With secondary open and 13.2kV applied to the primary: I1 = 10A, Pin = 5500W
With secondary shorted and 800V applied to the primary: I1 = 75.76A , Pin = 5800W
Assume the three single phase transformers are equal.
REQUIREMENTS:
For the operating temperature of 75°C, determine:
The percent effective resistance on the self-cooled rating base
The percent reactance on the self-cooled rating base.
The percent impedance on the self-cooled rating base
The no-load loss of the three-phase bank (kW)
The total loss of the three-phase bank (kW) with the transformer operating at its force cooled rating.
The efficiency of the bank carrying 3750 kVA at 85% pf
Background
1000/1250 kVA OA/FA
13.2kV/4.16 kV
DC Resistance: r1=0.40Ω r2=0.035Ω
Open Circuit Test: V1=13.2kV I1=10A Pin=5500W
Short Circuit Test: I1=75.76A Pin=5800W
Fan Load = 750W
Sbase=1,000 kVA Vbase=13.2kV Zbase=2 2(13,200)
174.241,0000,000
base
base
V
S
Turns Ratio: 13.2
3.1734.16
p
s
Va
V
Page | 3-71
Solution:
(a)Percent effective (ac) resistance on the self-cooled rating base
ac dc core mechr =r +r + r (rmech is 0 for transformer)
Equivalent dc resistance referred to primary:
2 21 2(25 ) 1.4 3.173 *0.035 0.7524dcr C r a r
Effective resistance from short circuit test
2 21
5800(25 ) 1.0105
(75.76)
inac e
Pr r C
I
The components of the ac resistance at test temperature
ac dc core
core
core
r (25 C)=r (25 C)+r (25 C) 1.0105
1.0105 0.7524 r (25 C)
r (25 C) 1.0105 0.7524
0.2581
Resistance changes with temperature.
rdc increases with temp (positive temp coeff)
rcore resistance decreases with temp (negative temp coeff)
∆R = α T0
∆T
or
R = R0 [ 1 – α (T – T0)]
For copper, the inferred absolute zero coefficient is -234.4.
So the equation reverts to
R = 234.4 + T
R0 234.4 + T0
Apply to both the copper and the core resistance.
ac
234.5 75 234.5 25r (75 C)=0.7524 0.2581
234.5 25 234.5 75
1.1138
Convert to per unit.
1.1138
( 75 ) 0.006392 0.6392%174.24
acr pu C
Page | 3-72
(b) Percent reactance on the self-cooled rating base
Impedance
Z = V = √R2 + X2
2 2
2 2
80010.56
75.76
10.56 1.1138
10.501
10.501( ) 0.060267 6.0267%
174.24
scac
sc
ac ac ac
ac
ac
VZ
I
X Z r
X
X pu
(c) Percent impedance on the self-cooled rating base
10.56( ) 0.0606 6.06%
174.24acZ pu
(d) No-load loss of 3 phase bank (from open circuit test)
Pno-load=3*Pin = 3*5500 = 16.5kW
(e) Total loss of 3 phase bank operating at FA rating
S = VI* → I = S / V
2
2
125094.697
13.2
3*( )
3*(94.697 *1.1138 5500)
46.46
FA
lossFA FA ac no load
kVAI A
kV
P I r P
kW
(f) Efficiency
3750 *0.85
3,187.5
3,187.598.54%
(3,187.5 46.46 0.75 )
out
out
in
P kVA pf
kW
P kWeff
P kW kW kW
Page | 3-73
Problem 3-7 (old) 436
SITUATION:
A generating station is connected as shown in Figure Problem 2-7 below. Transformer T2 was destroyed
and must be replaced; however, no records exist of the nameplate, and the proper phase relations must
be determined so that a new transformer can be specified.
REQUIREMENTS:
Neatly sketch and label phasors A'B'C', and state sequence A'B'C' or C'B'A'.
Neatly sketch and label phasors A''B''C'' and state sequence A''B''C'' or C''B''A''
Complete the nameplate for T2 – ratings not required.
H0 H1 H2 H3
X1 X2 X3
GENT2
H1 H2 H3
X0 X1 X2 X3
T1
A’B’
C’
ABC
N
B’’C’’N
A’’
T3 H0 H1 H2 H3
X0 X1 X2 X3
A
C
B
N
H3
H1H2
H0
X3
X1
X2
H3
H1
H2
H3
H1H2
H0
X3
X1X2
X0
T1
T2
T3
Page | 3-74
SOLUTION:
This is a problem about phase sequences. It illustrates the phase shifting between (1) wye and delta,
(2) between line-line and line-ground, and (3) between line and phase. Although these are obviously
related, the actual connections can be quite different.
Delta Delta Wye Wye
Phase Line Phase Line
L-L L-L L-N L-L
Phase sequence is drawn from the perspective of looking down the x-axis to the left. The phasors
rotate CCW. Record the phase sequence AN, BN, CN or CN, BN, AN or record the line sequence AB, BC,
CA or CA, BC, AB. Select every other letter. The sequence is positive ABC or negative CBA.
For a transformer the terminals are labeled on the primary and secondary.
Neutral
Primary H1 H2 H3 H0
Secondary Additive X1 X2 X3 X0
Transformers in a wye-delta configuration are shown. Note the
corresponding orientation that does not result in a phase shift. AN-
XY, BN-YZ, CN-ZX
Steps for determining transformer connection. Make a table of
the line connections and the transformer connections. Fill in the
rows of unknowns. Note the order that data is filled.
Order Action Options
1 Reference phase AN BN CN or AB BC CA
AN BN CN
2 Transformer primary connection H1 H2 H3 H0
3 Primary actual phase/line connection AN BN CN or AB BC CA
4 Transformer secondary connection X1 X2 X3 X0
5 Secondary actual phase/line connection AN BN CN or AB BC CA
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
7 Sequence ABC or CBA
A B
N
C
X
Y
Z
Page | 3-75
Transformer T3 is a wye-wye. The primary and secondary are aligned in phase.
Order Action Connection
1 Reference phase AN BN CN or AB BC CA
AN BN CN
2 Transformer primary connection H1 H2 H3 H0
H1H0 H2H0 H3H0
3 Primary actual phase/line connection AN BN CN or AB BC CA
AN BN CN
4 Transformer secondary connection X1 X2 X3 X0
X1X0 X2X0 X3X0
5 Secondary actual phase/line connection AN BN CN or AB BC CA
C”N B”N A”N
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
0 o 120 o 240o
7 Sequence ABC or CBA CBA
Transformer T1 is a wye-delta. The primary and secondary are shifted in phase.
Order Action Connection
1 Reference phase AN BN CN or AB BC CA
AN BN CN
2 Transformer primary connection H1 H2 H3 H0
H3H0 H2H0 H1H0
3 Primary actual phase/line connection AN BN CN or AB BC CA
AN BN CN
4 Transformer secondary connection X1 X2 X3 X0
X3X1 X2X3 X1X2
5 Secondary actual phase/line connection AN BN CN or AB BC CA
A’C’ B’A’ C’B’
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
0 o 120 o 240o
7 Sequence ABC or CBA ABC
Page | 3-76
Transformer T2 is a delta-wye. The primary & secondary are shifted in phase. The secondary
orientation is unknown
Order Action Options
1 Reference phase AN BN CN or AB BC CA
AN BN CN
4 Transformer primary connection H1 H2 H3 H0
H1H3 H2H1 H3H2
2 Primary actual phase/line connection AN BN CN or AB BC CA
A’C’ B’A’ C’B’
5 Transformer secondary connection X1 X2 X3 X0
X1X0 X2X0 X3X0
3 Secondary actual phase/line connection AN BN CN or AB BC CA
C”N B”N A”N
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
90o 210o 330o
7 Sequence ABC or CBA CBA
Page | 4-1
Chapter 4 - Electronics
4.1 Introduction Electronics involves a small signal operating on active devices to produce a different signal. Active
devices are non-linear energy amplifiers. Passive devices are simply modeled as RLC components.
Electronics is predominantly small signal variations that exist around a larger operating point. This fits
the standard response or solution, without the exponent, or essentially the cos( )t portion.
( ) cost
y t F I F e t
The constant part of the signal is the power supply, called the bias. The dc voltage and current
define an operating point , quiescent point (q-point), or bias. The bias dictates what portion of the curve
the device is operating on.
The small signal is the AC component. Generally AC analysis involves investigating a transfer
function operating on the input signal. The function is a two port model of the circuit.
The signal voltage and current cause slight variations around the operating point defined on the dc
bias line.
0 1 cosi I I t
ln1
1ln
new old
spacingX X
GMR
oI and oV are the DC operating point. The cos ωt term is the signal varying about the operating
point.
Electronic functions are by definition non-linear systems. In order to simplify the solution, it is
necessary to use assumptions.
Linearize with time varying about operating point is appropriate if
1 0cos( )I t I
For AC small signal, the DC supply voltage appears as a short
Page | 4-2
4.1.1 Solid State Device Characteristics
Electronics often use semiconductors. The conductivity of a semiconductor material is a physics
property.
n pq n p
electron mobilityn
hole mobilityp
electron concentrationn
hole concentrationp
-19charge on electron (1.6 10 )q C
The silicon material is doped to change the number of electrons (negative) or holes (positive).
p-type material; p ap N
n-type material; n dn N
Carrier concentrations at equilibrium are the product of the p and n concentration.
2( )( ) ip n n
in = intrinsic concentration
Built in potential or contact potential of a p-n junction depends on Boltzmann’s constant,
temperature and the electrical charge.
2
ln a do
i
N NkTV
q n
Thermal voltage is the constant, temperature, and charge components.
0.026T
kTV V
q at 300°K
Na = acceptor concentration
Nd = donor concentration
T = temperature (K)
k = Boltzmann’s Constant = 1.38x10-23 J/K
Page | 4-3
Capacitance of abrupt p-n junction diode is a function of the voltage.
( )1
o
bi
CC V
VV
Co = junction capacitance at V=0
V = potential of anode with respect to cathode
Vbi = junction contact potential
Resistance of a diffused layer is based on a sheet of material.
s
LR R
W
Rs = sheet resistance = d
in ohms per square
= resistivity
d = thickness
L = length of diffusion
W = width of diffusion
Page | 4-4
4.2 Boundary Conditions The response to a signal is investigated at three different conditions: the midpoint, the upper limit
or boundary, and the lower limit or boundary. Frequency changes the impedance and the time constant.
As a result, the voltage and gain of the circuit is changed.
The upper and lower boundary analysis determines the frequency of the transition.
High Frequency Cutoff – Shunt Capacitance dominates
f
Series capacitor is conducting 1
02
se
se
XfC
Frequency is determined by shunt capacitance& parallel resistors.
1 2
1 1 1 1c shC C
RC R R R
Low Frequency Cutoff – Series resistance and capacitance dominates
0f
Shunt capacitor is open 1
2sh
sh
XfC
Frequency is determined by series (coupling) capacitor and resistance in series.
1 2
1c serR R R C C
RC
3 dB Frequency
The response is the voltage that changes with frequency. It has an exponential rise to the low cutoff,
transitions to a bandwidth that is relatively flat, then transitions to an exponential decay past the upper
cutoff. The cut-off frequencies are called 3dB because the voltage has decayed to that value.
Vmax
.707 Vmax
ω ωoω3db
Page | 4-5
3 2 between +3db and -3dbndb
20log o
in
Vdb
V
30.15 log
20
o
in
Vdb
V
0.1510 2 o
in
V
V
3db point @ 2o inV V
Attenuation
Attenuation is the reduction in signal. It is influence by the cut-off of each stage in an amplifier.
ATF (Attenuation Factor) = 1
2 1n
n= # of stages
*high onef f ATF
fone =one stage frequency
fhigh = all stage high frequency
flow = all stage low frequency
onelow
ff
ATF
Page | 4-6
4.3 Diodes & Rectifiers Diodes are the simplest semi-conductor device that consists of a single p-n junction. Functionally,
these are rectifiers.
4.3.1 Diodes
Diodes are non-linear as noted in the waveforms. They are used for power supplies, wave shaping,
and logic. For most circuit analysis, the ideal model of the diode can be used. The assumptions for an
ideal diode are zero current in reverse direction and zero voltage drop in forward direction. Real diodes
are somewhat less idyllic.
Ideal Diode
I=0+ -
V=0- +
Real Diode
(Piecewise Linear)
+ - - +
The Shockley equation defines the V/I relationship in a real diode.
1D
T
vV
D si I e
sI = saturation current
η = emission coefficient, typically 1 for Si
TV = thermal voltage = kT
q
Page | 4-7
4.3.2 Rectifiers & Clippers
The application or performance of diodes is a rectifier. Representative circuits and their output are
shown. The input voltage Vin is a sinusoid.
Input Waveform
Half Wave Rectifier
RL VoAC
Full Wave Rectifier
RLVin
Vo
Voltage Clipper
V1
RLVo
AC
Page | 4-8
4.4 Operational Amplifiers
Ideal
An ideal op-am has infinite input impedance, zero output impedance, and almost infinite gain. The
gain, A is described by the relationship.
1 2( )ov A v v
A is large (>104) and v1 and v2 are small enough so the amplifier is not
saturated. For the ideal operational amplifier, assume that the input
currents are zero and that the gain A is infinite. Then, when operating
linearly, the voltage difference is 2 1 0v v .
Differential Amplifier
Two ungrounded input terminals create a
differential amplifier. The device amplifies the
voltage difference between the inputs.
The general form is shown. The significant
relationships are calculated.
out
in
VA
V
The op amp has a high Z input. That implies the
voltage between the terminals is essentially 0. So,
there is no current flow.
V- = V+
Compare the current on the negative and positive input branches. Use those functions to calculate
the output voltage.
01
1 2
V VV VI
Z Z
2
3 4
0V V VI
Z Z
The circuit is a voltage divider between the feedback side and balancing side.
2 4 1 20 1 2
1 1 3 4
Z Z Z ZV V V
Z Z Z Z
(Voltage divider)
-
+
v2
v1
vo
V1
Z1
V2
Z3
Z2
Z4
-
+
Vo
Page | 4-9
The connection of the op-amp changes the gain. Complex reactance creates mathematics functions
of integrator and differentiator.
Inverting Amplifier
Voltage on – terminal
2 2
1 1
Z RA
Z R
-
+
R1 R2
ViA
Non-Inverting Amplifier
Voltage on + terminal
1 2
1
R RA
R
-
+
R1 R2A
Vi
Integrator
Capacitor in feedback
2
1 1 1
1 1 1ZA
Z R sC sCR
-
+
R1 AVi
C
Differentiator
Capacitor on input
22 2
1
( )( )Z
A R sC sCRZ
-
+
AVi
R2
C
Page | 4-10
4.5 Transistors Transistors are the next level of complexity above diodes for semi-conductor devices. A transistor
consists of two junctions. Functionally, transistors are amplifiers. When operating at a boundary
condition, such as in digital circuits, a transistor is a switch.
I
vbe
Q-point
time
time
Vbe
cI
Input Signal
Output Signal
4.5.1 Bias vs. Small Signal
Bias is the DC voltage that determines the operating point. Small signal is the AC message that is
superpositioned on the DC.
Transistor Bias – External
Transistor biasing is a process of external connections similar to those in the figure. Because of the
connection, we know certain items.
Know:
Voltage @ ground = 0
Voltage @ Rc = Vcc
Want to find:
Voltage @ base. It is a divider of Vcc
The Thevenin equivalent input is a voltage divider of the power supply. If Rb
>>10R2, then the base current can be ignored and VBG can be easily determined.
1
1 2
CCBG Thevenin
V RV V
R R
R1
R2
Rc
RE
VL
VccB
C
E
Page | 4-11
The Thevenin resistance is obtained by shorting the power supply (VCC). The equivalent
circuit is shown in the figure.
where
Next, find the current.
Current @ base (input)
Current @ emitter (output)
The analysis must include loop elements. In essence do per unit conversion. The current through the
transistor has gain. IB is the base current. If Rbase (=hfeRE) is not >> than 210* R , then VBG must be
determined from this circuit.
The operating point of the transistor is referred to as the Quiescent point (Q point). This is the point
that the circuit will operate at when only the DC signal is in place. The Q-point is defined by the Ic
current at the Q-point (ICQ) and VCE at the Q-point (VCEQ). For an existing circuit, the Q-point can be
determined by using the Thevenin equivalents.
2
1 2
1 2
( )
TH CC
TH
TH BECQ
THE
FE
CEQ CC CQ C E
RV V
R R
R R R
V VI
RR
h
V V I R R
Often when designing an amplifier
circuit, a graphical approach is easier. The
Q-point is the intersection of the DC load
line and the AC load line. The DC load line is
the line drawn between the
1 2/ /THR R R
1CC
THCC TH
VR R
V V
2CC
THTH
VR R
V
( 1)
( 1)
E B C B
TH BE TH B E E
TH BEB
TH E
I I I I
V V R I R I
V VI
R R
R1
R2
VBGVcc
RTH
RE
IE
VTH
VBE
IB
DC
CCV
R
CQI
CEQVCCV
Page | 4-12
Collector/Emitter current when the transistor is conducting (shorted) and the Collector voltage (VCC).
Collector/Emitter current is VCC divided by RDC where
The load line is shown superimposed on the transistor characteristic curves in the figure to the right.
The next step is to select a Q-Point. The goal for most linear (Class A) amplifiers is to select a Q-point
near the middle of the load line. This allows for an equal amount of swing above and below the Q-point.
Once the Q-point is selected, VCEQ and ICQ can be determined.
The AC load line is the line that represents the small signal that is superimposed on the DC bias
circuit. AC signals bypass (short) VCC; it also includes the load impedance (RL). The endpoints of the ac
load line are found using the following
formulae
( )
( )
( )
( )
CEQ CE off
c sat CQ
C ac
CE off CEQ CQ C
ac C L E
V vi I
R R
v V I R
R R R R
The ideal situation for a Class A
amplifier is for the AC load line is for the Q-
point to be in the middle of the AC load
line. This can be adjusted by modifying RC.
(no AC coupling)dc C ER R R
DC
CCV
R
V
CQI
CEQV
( )CE off
ac
v
R
( )CE offvCCV
Page | 4-13
4.5.2 Transistor Mathematical Relationships
Name and Schematic Symbol Mathematical Relationships
NPN Bipolar Junction Transistor (BJT)
Relationships valid in active mode of operation.
PNP Bipolar Junction Transistor (BJT)
Relationships valid in active mode of operation.
N-Channel Junction Field Effect Transistor (JFET)
P-Channel Junction Field Effect Transistor (JFET)
Cutoff Region:
Triode Region:
Saturation Region:
C
E
B
iC
iE
iB
E B Ci i i
C Bi i
C Ei i
1
BE
T
VV
C si I e
emitter saturation currentSI
thermal voltageTV
C
E
B
iC
iE
iB
E B Ci i i
C Bi i
C Ei i
1
BE
T
VV
C si I e
emitter saturation currentSI
thermal voltageTV
D
S
G
iD
iS
D
S
G
iD
iS
GS pv V
0Di
GS Pv V
GD Pv V
2
22DSS
D DS GS P DS
p
Ii v v V v
V
GS Pv V
GD Pv V2
1 GSD DSS
P
vi I
V
Page | 4-14
N-Channel Depletion MosFET (NMOS)
IDSS = drain current with (in the saturation region
K = conductivity factor VP = pinch off voltage
P-Channel Depletion MosFET(NMOS)
Same as N-Channel, with current directions and voltage polarities reversed
D
S
G
iD
iS
B
N-Channel Enhancement MosFET (NMOS)
Cutoff Region:
GSv Vt
0Di
Triode Region:
GSv Vt
GD tv V
22D DS GS t DSi K v v V v
Saturation Region:
GS tv V
GD tv V
2
D GS ti K v V
IDSS = drain current with 0GSv (in the
saturation region K = conductivity factor Vt = pinch off voltage
D
S
G
iD
iS
B
P-Channel Enhancement MosFET(NMOS)
Same as N-Channel, with current directions and voltage polarities reversed
D
S
G
iD
iS
B
0GSv
2
DSS PI KV
D
S
G
iD
iS
B
Page | 4-15
4.5.3 General Two-port Models
Two-port models have external measurements which take three forms; impedance, admittance, or
hybrid. Impedance models have voltage in terms of current. Admittance models have current in terms of
voltage. Hybrid models have both voltage and current as outputs. The internal connections have three
arrangements: T or wye, Π or delta, and L.
4.5.4 BJT Transistor Models
Bipolar Junction Transistor – Use Common emitter as general example
The complete model is shown on the left, while the simplified version is on the right.
The hybrid two-port network is the generic form for the BJT transfer function and includes the
current, voltage, and hybrid parameters, h.
1 1 2
2 1 2
i r
f o
v h i h v
i h i h v
The definition of the hybrid parameters are in terms in input (1) , and output (2).
General Form Specific to Common Emitter
2
1
1 0
i in
v
vh R
i
(1 )ie ib b ich h i h
2
2
1 0
f
v
ih
i
1
cfe
b
ih
i
1
1
2 0
r
i
vh
v
= gain of circuit
1
2
2 0
1o
i
ih
v out
= gain of transistor
hi
hrV2 hfi11/ho
i1 i2B
E
C
V1 V2
hfe
ib ib
hfeib
B
E
C
Page | 4-16
The second subscript is determined by which terminal connects to the common lead for the two-
port model.
For common base – b input – e, output – c
For common emitter – e input – b, output – c
For common collector – c input – b, output – e
Common collector is also called an emitter follower.
Common collector circuits permit assumptions which greatly simplify the circuit.
2 0reh v
1
0 ignoreoe
o
hh
H- parameters are real at low frequencies and complex a high frequencies. At high frequencies, the
capacitance has more effect.
Since this is a small signal investigation on internal parameters, the DC supply appears as a short.
Page | 4-17
Common Emitter – Hybrid Parameters
The common emitter is the most common configuration for
transistor amplifiers.
The h-parameters for a common emitter model are listed.
4
Q point
10coe
ce
ih
v
Q point
cfe
b
ih
i
3
Q point
(25 10 ) febe Tie fe
b EQ EQ
hv Vh h
i I I
i feA h
The small signal circuit removes the DC components.
The equivalent circuit input and output impedance are found.
b iei ie
b ie
R hZ h
R h
If the base bias resistor Rb>>hie, then the output is greatly
simplified.
1
o
oe
Zh
Conditions Small Signal Equivalent Model
Complete
hi
hrV2 hfi11/ho
i1 i2B
E
C
V1 V2
As hre0
hie hfeib1/hoe
ib icB
E
C
vbe vce
iB
ic
+
V
be
-
+
Vce
-
iL
Rb
VBB Re
RL
Ce
VCC
i1
RbRL
ib
ic
ie
+
Vbe-
+
Vce-
iL
Page | 4-18
As hre0 and hoe e0
hie hfeib
ib icB
E
C
vbe vce
Complete common-
emitter amplifier
hie hfeib
ib icB
E
C
vbe vceRbRLii iL
ZiZo
Page | 4-19
Hybrid Model – Common Emitter Model at High Frequency
The hybrid is a two port network defined in terms of both h-parameters and circuit parameters.
As a result, it incorporates both the non-linear transistor and a two-port model as shown below.
B
C
E
B C
E
rbb’rb’e Cb’e
rb’crce
Cb’c
Eb’
e
Ebe Ece
b’
Because of the capacitance, the frequency has a major impact on impedance. The Miller effect
resolves the situation where the voltages at both ends of a capacitor change at the same time. Rather
than consider the actual effect, the model uses a large capacitor, CMiller . The Miller replaces Cb’c. At low
freq assume Cm is included in Cb’c. The circuit assumptions are noted.
Cmiller '(1 )Lb c
d
RC
r
' ' '
0.025@ 300
fe
ie bb b e bb
EQ
hh r r r T K
I
β = common emitter low frequency current gain = 1
feh
Ebe = diode drop forward biased b-e voltage 0.6V
The equivalent circuit has the input side coupled to the output by the input voltage operating on the
transconductance, gm. By using superposition, the input is calculated first. Then the results are used on
the output side.
Page | 4-20
Rbb’
ReVbe Zc
gmVbe
B
E
C
Input Side
Calculated 1st
Output Side
Use input to
calculate
rπ
gmVbe
B
E
C
Rce
The frequency effects are calculated. At the cutoff frequency, the gain is down 3dB.
Cutoff frequency = ' ' ' ' '
1 1
2 ( ) 2b e b e b c b e b e
fr C C r C
3 db bandwidth = fef h f 1
fb
i
ve
hA
jh
The internal circuit parameters have the following relationships.
'
40 300fe
m EQ
b e
hg I T K
r
' 10 50
2 5 (hi freq caps)
bbr
'
'
fe mb e
T b e T
h gC
r
'
0.26
40
fe
b e
EQ E
hr
I I
1 1' 2 3
5 pb c ob cbC C v pf p
Page | 4-21
Common Base – Hybrid Parameters
A common base is used for high frequency applications because the base separates the input and
output, minimizing oscillations at high frequency. It has a high voltage gain, relatively low input
impedance and high output impedance compared to the common collector.
1 0
11
e
cob
o cb i i
ih
Z v
0
1
cb
cfb
e v
ih
i
0
1cb
eb ieTib i
e EQ fev
v hVh Z
i I h
i fbA h
A simplified circuit model is developed.
Complete Circuit
ri
vi
VCC
iE iLRL
R2R1Cb
iC
Small Signal Circuit
ri
vi
iE
iL
RLiC
Page | 4-22
Small Signal Equivalent Model
ri
E
vi hrbvcb
hib
hebi1 1/hob
RL
B
C
iE
iL
iC
i1
Simplified Small Signal Equivalent
Model
0rbh
1fbh
0obh
ri
E
vi
hib
i1 RL
B
C
iE
iL
iC
i1
Page | 4-23
T Model – Common Base Model at High Frequency
The T-model is simply a configuration that uses conventional circuit analysis without special
parameters.
re
Ce
rb
Zc
IcIee
b
c
At low frequency, capacitors are high impedance and appear as open circuits.
e eZ R
'' is close to unity(1)b e
e m b e
e
VI i g V
r
c cZ R
At high frequency, capacitors are low impedance and appear as a short circuit which passes signals.
1 1
2e e e
e t e
Z sC CR f r
1 tc b
c
fZ I I
sC f
c E cBOI I I
The variables have the following meanings.
α = fraction of emitter current collected
Icbo = reverse current characteristics of collector base junction
IE = dc value of emitter current
ft = high frequency performance (where hfe decreases to unity)
The emitter resistance is the base to emitter diode resistance. It is also called rb’e & rd.
26 T
E
E e
Vr
I r at room temperature
The typical ranges of values are noted.
Rb range 10’s to 100’s ohms – base spreading resistor
α range 0.9 to 0.999
rc range hundreds megohms
Vb’e 0.6v (0.5 – 0.7v) which is diode drop forward bias voltage.
Page | 4-24
Page | 4-25
Common Collector (Emitter Follower)– Hybrid Parameters
The common collector amplifier is also called an emitter follower because the output is taken from
the emitter resistor. The connection has a gain of near unit. It can be an impedance matching device
since its input impedance is much higher than its output impedance. It is used as a buffer in digital
circuits.
'
1
io ib
fe
rZ h
h
( 1)i ie fe eZ h h R
1vA
Complete Circuit
ri
vi
iE
ib
+
vE
-
BC
E
RE
RB
VCC
Small Signal Circuit ri
viiE
Rb
ib
+
vE
-
BC
E
RE
+
vb
-
Small Signal Equivalent Model
Looking into base
ri
viiE
hib
Rb +
vE
-
B
C
E
RE
+ vbe -
Page | 4-26
Small Signal Equivalent Model
Looking into emitter
v’i
iE
hib
+
vE
-
B
C
E
RE
+ vbe -
'
1
i
fe
r
h
Page | 4-27
Emitter Follower (common collector) Model at High Frequency
High frequency performance is very different because of the capacitors created in the device and in
the circuit. Therefore, alternative analysis assumptions are needed.
Circuit
Ri
vi Cb’c
Cb’e
Re
i1 ib
ie
vb
iCb’e
iRe
Equivalent Circuit
Ri
vi Cb’c
Cb’e
Ze
i1
vb
rb’e
ve
The frequency impact on the impedance is readily observable.
1
( 1)1
Te fe
j
Z hj
Assumptions reduce the complexity of the problem.
Cb’c is relatively small, and '1fe e b e ih R r R
e
1' R
1( )
1 1
T
i
T
j
ev R
i
vA
v j
Upper cutoff frequency 11 i
e
Th R
R
ff
If Cb’c is large enough for its impedance to be equal to Ri at fh2<<fh1, then the gain is in terms of h-
parameters.
Page | 4-28
''
1 1( )
11
fe e
v
i b cb e fe e
h RA
j R Cr h R
Upper cutoff frequency 2
'
1
2h
i b c
f fR C
If 1 2h hf f the gain changes.
2 2
1
111
T
i
e
j
v RR
h T T h
Aj j
j
Page | 4-29
4.5.5 FET Transistor Models
Field Effect Transistors (FET) are devices that are less affected by frequency. They can also handle
more power, in the range of amps. They are also more expensive.
Circuit
Vdd
RdId
Drain
SourceGate
RsRs CsVin
Equivalent circuit
Rg
Rd
rd
gmVqVqin
CsRs
The following constraints apply.
There may be capacitor coupling which includes Cqd Cqs & Cds in the small signal.
Vp is the pinch off voltage, which is negative. Pinch-off is the condition that stops conducting ID but
still has leakage current IDSS.
VDS is the breakdown voltage where the transistor fails.
Keep the supply voltage, VDD, below the breakdown voltage, VDS.
Igs is the gate leakage current in nanoamperes. DI is the drain current.
For a JFET, the drain current is dependent on the leakage current and drain voltage.
2
D gs
D DSS
D
V VI I
V
Page | 4-30
Example
A FET circuit with self bias is desired. The FET limits
and the supply consumption are provided.
Vp = -3 volts with current of 6ma
VBD = 30 volts
Required: Determine the bias for 10V Drain to
Source and channel current of 4ma.
Given:
ID = 4 mA
IDSS = 6 mA
|VD| = 3V
Solution:
2 2
34 6
3
0.55
D gs gs
D DSS
D
gs
V V VI I
V
V V
Assumptions:
Select Rg = 1MΩ
This makes voltage drop across Rg0 (Input resistance >10X source)
Calculations:
Write Kirchhoff current law @ FET
0 0D g S g
D S
I I I I
I I
For 30BDV , use 30DDV = 24 volts (80%)
Write KVL around loop for Bias
0D SDD R DS RV V V V
Substitute using Ohm’s law & values
24 10 0D D D SI R I R
Vgs is determined by ID & RS
RdId
Drain
SourceGate
RgRsVin
VDD
Page | 4-31
Vgs = IDRS 3
0.55140
4 10
gs
S
D
VR
I
Substitute in KVL
3
14 0
14 4 10 ( ) 0
3500
D D D S
D S
D S
I R I R
R R
R R
Use RS = 140
3500 3.3kD SR R
Page | 4-32
Basic FET Models
The FET has several different models that are used for design purposes.
Current Source Model
rds
G
S
D
gmvgs
Voltage Source Model
rds
G
S
D
μvgs
Complete Circuit
Common source voltage amplifier
ri
vi
cc1
R1
R2
R3
Rd
Rs Cs
Cc2
RL
DC
Small Signal Equivalent ri
vi R3+(R1||R2) Rdrds RL+
vL
-
gmvgs
Page | 4-33
The equations for the FET are the normal big three: input impedance, output impedance, and gain.
3 1 2( || )iZ R R R
||o d dsZ R r
1 3 1 2
1( || )
1 / ( || )v m L oA g R Z
r R R R
Voltage Gain
usually 1 3 1 2( || )r R R R
If RL<<Zo then
v m LA g R
Page | 4-34
Source Follower (Common Drain) Amplifier
The FET source follower is similar to the BJT emitter follower configuration.
Complete Circuit
vi
cc R1
Rs1
Rs2
VDD
+
vs
-
ii
Small Signal analysis
vi
Rs=Rs1+Rs2
VDD
+
vs
-
ii
Small Signal Equivalent
rds
vi
μvgs
+
vs
- Rs=Rs1+Rs2
ii
G
S
D
Thevenin Equivalent Circuit
Zo
vi
Zi μvgs+
vs
-
Rsii
GS
D
Assumptions assist in solving for the coupling voltage.
1R
s gs o ds sv v i r v
The big three are then calculated.
1
1
dso
m
rZ
g
2
1 2
11
1 ( )
( 1)1 s
s s
g
i Ri R R
v RZ R
i
'1
vA
Page | 4-35
4.6 State Space State space models can be used for a wide range of problems from controls to communications and
computer modeling. State space is a technique to represent differential equations when applied to the
LaPlace s-domain.
4.6.1 The 6-Minute Approach
Definitions
The state equation has the form ( )
( ) ( )dx t
Ax t Bu tdt
The output equation has the form ( ) ( )y t Cx t
The state variables can be defined as 1 2( ) ( ), ( ) / ( )x t x t dx t dt x t
A, B, and C are matrices while u is input, y is output, and x is the transfer variable.
The transfer function, Y(s)/U(s) can be broken into parts by ( ) ( )
( ) ( )
X s Y s
U s X s. Then Y(s)/X(s) is the
numerator of the transfer function and X(s)/U(s) is 1 over the denominator.
The s-operator is defined as the first derivative. ' /s x dx dt
S-domain to Differential
Convert the numerator Y(s)/X(s) to ( ) ( )* ( )Y s numerator X s .
Substitute to obtain y(t)= f(x(t)), using #6 definition.
( ) / ( ) 4
( ) ' 4
Y s X s s
y t x x
Develop an equation for the highest derivative of x to be the same power as the highest power of s
in the denominator.
2
1
3 6s s
''x
Replace the highest power with the input.
'' ...x u
Replace the remaining s-values using #6 definitions.
'' 3 ' 6x u x x
The result is two differential equations.
Page | 4-36
( ) ' 4y t x x
'' 3 ' 6x u x x
Substitute 1 2( ) ( ) and ( ) / ( ) ( )x t x t d x d t x t into the matrix form.
'1 1
'22
1
2
0 1 0
6 3 1
4 1
x xu
xx
xy
x
Page | 4-37
4.6.2 Description
State space is used to solve ordinary differential equations. The process uses op-amps to create
summers and integrators. Note, differentiation produces noise. The output is limited to maximum and
input is limited to minimum because of noise. The values are selected so the constant coefficient has a
range of 0.1 to 10.0. Because of the op-amp circuit, each operation is always negative or inverting
Multiplier Summer Integrator
KVi Vo
V1
Vn
Vo
V1V2
V3
Vn
vo
o iV KV and 0 1K 1 1( )o n nV a V a V 1 1
0( )
t
o n nV a v a v
Third order differential equation
2 1 0 ( ) 0x a x a x a x f t
1 11x
x x x
a2
a1
a0
1
f(t)
Page | 4-38
State Space method
òS S ò òS
a2
a1
a0
- - -
f(t)x
Scaling and Sampling Time
Time – T=ht
h = scaling factor
T = simulation time
t = time between samples
3 2 2 1 0 ( ) 0Th x h a x ha x a x fh
Simply replace 3 by x h x etc.
Time scaling factor of 1
hreplaces 1 on the input of each integrator.
( ) ( )
( ) ( )( )
t kT
y t Fx t
y kT Fx kt
k = which sample
T = interval between samples
Magnitude scaling
Used to get full linear range from an op-amp.
X=kx
X = simulation variable
x = dependent variable
k = scaling factor
max
max
Xk
x
k is chosen and incorporated in each term so that it does not change the equation.
Page | 4-39
3 0 02 2 1 1
3 2 1 0
( )0
m
m
k x a k x f f ta k x a k x
k x k k k f
Simplify equations but keep terms in parentheses together. This is scaled variable X.
Example
0.82 18 250 400 0tx x x e
Initial conditions (0) 35 (0) 10x x
Maximum values max max max35 250 2500x x x
Maximum simulation V = 10 volts
Solution:
Find scaling factors
max2
max
100.004
2500
vk
x
max
max1
max
100.04
250
vk
x
max0
max
100.2
35
vk
x
maxmax
100.025
400
vf
f
Make differential w/ unity coefficient on USB
Divide by coefficient (example = 2)
0.89 125 200 0tx x x e
Substitute scaling factors
0.8200 0.0250.004 9 0.04 125 0.2
00.004 0.04 0.2 0.025
tex x x
Make differential w/ unity coefficient on USB
0.80.9 0.04 2.5 0.2 0.8 0tx x x e
Multiply scaling factor time initial value to obtain scaled initial values
1
0
(0) (0.04)(10) 0.4
(0) (0.2)(35) 7
k x
k x
Page | 4-40
Adders (summers) use round values (0.1, 1, 10)
coefficients are obtained by using multiplier k and
summer value
Example: 2.5(0.2x)
Adjust scale coefficients by integrator coefficient
Example:
Draw differential equation (from above)
100.004x 0.04x
5
0.9
0.25
0.8e-0.8t = force
0.2x
-0.4 7
0.25 10(0.2x)
summer
100.004x 0.04x
Page | 4-41
Integrator Realization – Draw it Out
Locate input x(t) and output y(t) along a line
Place as many integrators as order (n) of system
Place a summer before the output and one before each integrator. Total number of summers = n+1,
or the maximum number of y terms
Use summer to bring signals together. Use connectors for departing signals
Draw forward, x-input signals. Use scalars for gain on each signal. Terminate the signal after as many
integrators as the order. Feedback will prove the differentiation.
Draw feedback, y-output signals. Use scalars for gain on the signal. Terminate the signal after as
many of the integrators as the order. This represents the differentiation.
Draw the main line. Remove all operators that are null. Use as many feed-forward and feedback taps
as necessary to complete the system. If the previous two steps are consistent, this will be one line.
Page | 4-42
Chapter 4 Problems
Problem 4-1 (Old Style)
SITUATION:
The operational amplifier circuit shown below is connected to provide a variable output resistance
at terminals a-b.
-
+
VinRo
Ra
LOAD
a
b
REQUIREMENTS:
Find a Thevenin equivalent circuit as a function of the potentiometer setting α at terminals a-b.
Assume the op-amp to be ideal and that R is much larger than RA.
SOLUTION:
Circuit is a non-inverting amplifier (voltage on + terminal).
Redraw into standard form
Find Thevenin Equivalent voltage (VTH).
Open circuit terminals, leave sources active, calculate voltage across open terminals
( (1 ) )
TH ab
ab A
A
V v
v I R R
I R
abA
vI
R
-
+(1-a)R
RA Vin
aR
b Vo a
Page | 4-43
( (1 ) )
( (1 ) )
( (1 ) )
in A A
abA
inab A
V I R R
vR R
R
Vv R R
R
Find Thevenin Equivalent Impedance (ZTH)
Short voltage sources, open current sources, then calculate series/parallel resistances.
Alternately, short circuit the terminals, leave sources active, then find ISC, TH
TH
SC
VZ
I
inSC
A
VI
R
( (1 ) )
( (1 ) )
in
ATHTH
SC in
A
A
A
RVR RV
RI V
R
RR
R R
Page | 4-44
Problem 4-2 (Old Style)
A FET amplifier composed of three identical stages is shown in the Figure below. It may be assumed
that Cs adequately bypasses Rs. It may also be assumed that there is 40 pF of stray shunt capacitance per
stage. Determine a value of RD such that the mid-frequency amplification is 60dB if
32.5 10mg 8dr k 100gR k
Cb
Rg
Rd
CsRs
Cb
Rg
Rd
CsRs
Cb
Rg
Rd
CsRs
Cb
Rg
REQUIREMENTS:
Determine a value of RD such that the mid-frequency amplification is 60dB.
SOLUTION:
FET Model
ReId
Drain
SourceGate
RgRg CSVin
Cb
Rgout Cstray/stage
Page | 4-45
Rg
VDD
Rd
rd
gmVgsVgin
CsRs
Rgout
Assumptions:
Gain of FET is across rd. If CS shunts RS, then we can simply connect rd to ground.
At the signal frequency, VDD is a short.
Once input and output stages are isolated using model, each can be solved individually.
Rg
Rd
rd
gmVq
Vqin
Cb
Rgout
a
Given:
32.5 10mg 8dr k 100gR k 40strayC pf shunt
Gain relates the output and input voltage.
20logo o
in in
V VGain A db
V V
For gain of 60db overall:
3
60 20log
60log 3
20
10
o
in
o
in
o
in
V
V
V
V
V
V
For 3 stage amplifier, gain for each stage is 10 (10*10*10=103)
Page | 4-46
Circuit analysis:
For an AC signal, the Capacitor bC is a short, which means it conducts.
Calculate the nodal equation at “a”.
31 1 1( 0)( ) 2.5 10N out in
d D g
I V Vr R R
Use the gain relationship to find the resistance.
3
3
1 1 1
1 1 110 ( 2.5 10 )
8 100
18.7
1.15 10
outm
in d D g
D
D
vA g
v r R R
k R k
R k
Page | 4-47
Problem 4-3
In the figure of Problem 4-2 above determine the high frequency cutoff for the entire amplifier if
32.5 10mg 8dr k 100gR k 8.7DR k
SOLUTION
Cutoff for each stage is reciprocal of the time constant. The capacitance is given as the stray:
1
2hi cutoff
sh sh
fC R
40shC stray pf
The resistors are in parallel.
1 1 1 1
4
SH d D g
SH
R r R R
R k
Calculate the frequency.
12 3
10.995
2 (40 10 )(4 10 )hi cutofff MHz
Cutoff for entire amplifier is a root containing the number of stages.
1
1
36 6
2 1
0.995 10 2 1 0.995 10 0.51 0.501
nhi all hi one one
hi all
f f ATF f
f MHz
Page | 4-48
Problem 4-4
In the figure of Problem 4-2 above determine a value of Cb such that the low frequency cutoff for
the entire amplifier is not greater than 200Hz (indicate any further assumptions made) if
32.5 10mg 8dr k 100gR k 8.7DR k
SOLUTION:
Low frequency cutoff for each stage:
1
lo cutoff
series seriesC R ?seriesC
Calculate the equivalent resistance.
511.042 10
1 1series g
d D
R R
r R
The frequency range is reduced by multiple stages.
1
1
13
2 1
2 1
200 2 1 102
n
n
lo onelo all lo one lo all
ff f f
Hz Hz
Cut-off frequency is time constant dependent.
5
1 10.015
2 (102)(1.042 10 )series
lo series
C fR
Page | 4-49
Problem 4-5
A voltage amplifier is shown in the figure below. Determine the high-frequency cutoff. Use
appropriate approximations.
' 20b br ' 500b er ' 0.05b ec equiv F 0.004mg
100kΩ
50kΩ
5kΩ
5kΩ Xc=0
Vo
a a’
SOLUTION:
The circuit is a common emitter BJT amp, draw equivalent hybrid small signal model
Rbb’
Recbe
gmVbe
B
E
C
Input Side
Calculated 1st
Output Side
Use input to
calculate
+
Vbe
-
Ignore R’s which are used for Bias.
Large C will pass anything.
Resistors rb’e & rb’b are in parallel on input.
20*500
2020 500
inr
The input circuit to the transistor is shown at the right.
20
cbe
B +
Vbe
-
Page | 4-50
'
11
1 1b e in
sCV VsCRR
sC
The cutoff freq is inversely related to the time constant.
8
8
1 110 / sec
20 5 10c rad
RC
810
1592 2
f kHz
Page | 4-51
Problem 4-6
In the voltage amplifier shown in Problem 4-6, a 10μH coil is put in place between the a and a'.
Determine the upper frequency cutoff with this coil in place.
' 20b br ' 500b er ' 0.05b ec equiv F 0.004mg
SOLUTION:
The circuit is a common emitter BJT amp, draw equivalent hybrid small signal model.
Rbb’
Recbe
gmVbe
B
E
C
Input Side
Calculated 1st
Output Side
Use input to
calculate
+
Vbe
-
Ignore R’s, they are used for Bias.
Large C – pass anything
Resistors rb’e & rb’b are in parallel on input.
20*50020
20 500inr
The input circuit of the transistor is shown at right.
Use the standard bandpass or resonance equation for a second order
system. The equation can be stated in terms of circuit elements or in terms of
frequency as noted in chapter 2.
2
/( )
1
s LY s
Rs s
L LC
2 2
0
/( )
s LY s
s s
From the admittance, calculate the voltage drop across the capacitor.
12
' 2 6 122
1 2 10
1 2 10 2 10
in in
b e
V VLCV
R s ss sL LC
20
cbe
B
E
+
Vbe
-
10-5H
Page | 4-52
The denominator is the characteristic equation when the input is set to zero. Equate the frequency
form and the circuit element forms to find the frequency.
2 22 n ns s
2 12 61 2 10 1.414 10n nLC
6
0.7072 10
o nR QL Q
1
2n
R
Q L
The 3db cutoff frequency is 225.2n kHz
Page | 4-53
Problem 4-7
The circuit shown in the figure below is a cascade connection of a FET and a junction transistor.
100ieh 10DR k 10dr k 100feh 5s cR R k 310 (FET)mg
You can neglect hoe, hre and all capacitance.
Determine the overall small signal amplification out
in
VV
for this cascode connection. A cascode is a
high frequency connection with an input as a common source ant the output is a common gate.
+
Vin
-
RD
RS
Rc
+
Vout
-
SOLUTION
Draw the small signal models and solve.
Rg
Rd
rd
gmVqVqin
Rs
hie
Rc
hfeib
Page | 4-54
Simplify:
Since d ier h and D ieR h , in a parallel arrangement, you can drop both and leave only hie.
Determine the output voltage.
( )
(1 ( ))
o s s C C
m g fe m g s fe m g c
m g fe s c
v i R i R
g V h g V R h g V R
g V h R R
Assume the forward parameter is not a contribution.
For hfe>>1
( ( ))o m g fe s cv g v h R R
Next determine the input voltage so the gain relationship can be determined.
(1 )
(1 )
1
( )
1
g i m g fe s
in fe m g s
in fe m s g
ing
m fe s
fe m c sout
in m fe s
v v g V h R
V h g V R
V h g R V
VV
g h R
h g R RV
V g h R
Substitute values to obtain the gain.
3
3
( )
1
100 10 (5 5 )
1 10 100 5
2
fe m c soutv
in m fe s
h g R RVA
V g h R
k k
k
RSgmVg
hie
hfeib=-hfegmVg
vo
Rc
Page | 5-1
Chapter 5 - Controls
5.1 Introduction
Controls are predominantly about using transfer functions to describe the response of a system to signal
inputs.
The transfer function is usually a two-port network. The most complex circuit is a second order, which
has been discussed in everything from circuits and power to electronics. By having multiple transfer
functions which result in multiplication, the system transfer function can become very long. Controls
analysis is the process of deciphering these complex functions into a physically realizable performance in
terms of real and angular components.
The standard waveforms for signals are the big three: dc or step, exponential, and sinusoidal.
Special cases of the exponential are also considered. These are impulse and ramp.
Page | 5-2
5.2 Controls Basics
5.2.1 Introduction
Control basics set the profile. The profile includes manipulation of the one-line or block diagram, the
errors, and the overshoot. A characteristic equation is used for stability analysis. The roots of the
numerator are called zeroes while the roots of the denominator are called poles. The gain of the
characteristic equation can be manipulated to improve stability.
5.2.2 Block Diagrams
A block diagram is used as a one-line representation of the process. It consists of three components:
input, R; output, Y; and the system transfer function. For analog systems, the diagram is in the LaPlace
transformed or s-domain. Obviously, this can be transformed directly into the frequency domain.
s j
Transfer Function – Output / Input = F=Y/R
Forward Gain – 1 2G G G
Feedback Gain – H
Unity Feedback – H = 1
Error signal – E R HY
Output – Y GE
Open Loop Transfer – ( )
( )
K s zGH
s p
Closed Loop Transfer – 1
Y GF
R GH
Characteristic Equation – 1 0GH ; 1GH
Denominator – 0
Open Loop 1 1 180
Roots of Characteristic Equation are the eigenvalues
eigenvalues – poles of YR
- closed loop poles
If a system is stable, all roots are in the Left Half Plane
The system is marginally stable if 1 root at origin, or 1 complex conjugate on jω axis
For frequency questions - s j
S G2Gain
G1Plant
HFeedbac
k
EError
RInput
YOutput
+
-
Page | 5-3
Low frequency gain - K (DC, ω=0)
00
( )
( )
Y K j zF
R j p
Bandwidth ( BW ) is at amplitude equal to 12
* low frequency gain
0
1
2BW
BW
Y YF
R R
- Solve for BW
Page | 5-4
5.2.3 Steady State Errors
Steady state errors are the response of the system to inputs. A second order system will have three
error responses; position (t=0), velocity (t), and acceleration (t2). The inputs that cause these responses
are step (DC), ramp (constant slope), and parabolic (exponential).
Use unity feedback & transfer function
Use final value theorem
0 0
( )lim ( ) lim
1 ( )s s
sR sess sE s
G s
The signal must be stable – not oscillating.
Error constants are the results of stability.
Position Velocity Acceleration
constant 0
lim ( )ps
k G s
0
lim ( )vs
k S s
2
0lim ( )s
ka S s
Laplace ds
dt
22
2
ds
dt
Steady state calculation depends on the order of the input.
step ramp parabolic
Time ( ) 1r t t 212
t
s 1( )R ss
21
s 3
1s
ess 0
1
lim1 ( )s
ss
G s
2
0
1
lim1 ( )s
ss
G s
3
0
1
lim1 ( )s
ss
G s
0
1
1 lim ( )s
G s
0
1lim
( )s sG s
20
1lim
( )s s G s
1
1 pk
1
vk
1
ak
0 no erroress
no control, explodesess
0ess as add pure integrators (increase system type)
System type = power of s in denominator
Page | 5-5
5.2.4 Time Response
The time response of a system is the familiar three component DC, exponential, sinusoid.
From this, the time domain positions can be calculated. The waveform chapter in the section RLC
System Response has additional relationships.
Percent overshoot exists in under-damped systems only.
For second order system
211 ( )
% 100( )
e yos
y
( ) 1 for normalized systemy
21(max) 1 for normalized systemy e
Rise time is the amount of time to rise from 10% to 90% of final value
Settling time is the amount of time necessary to reach and stay within a band around the final value
The damping frequency is a shift from the natural or resonant where the wave is dropping off.
2 2d n
( ) ( )% 100 max
( )
y t yos
y
max ( )y t
p
d
nt
( )y
Page | 5-6
5.3 Routh-Hurwitz Criteria
5.3.1 Introduction
Controls problems are the result of multiplications which often yield functions that are higher than
second order. Consequently, traditional solutions become unworkable and alternative approaches are
necessary. Factoring is a possibility, but can be very tedious. The Routh-Hurwitz criteria were developed
to eliminate factoring. The coefficients of a function are used to find stability.
5.3.2 Rules
Determine the function to be analyzed for stability.
Put coefficients of the function in 1st two rows of array – as even/odd powers
Next row – use determinant of 1st column & column to right of location being evaluated
Complete the row (use -1 on determinants)
If have a zero as last element in row, simply move last element of previous row to present row
Interpretation
# of roots in right half plane = # sign changes in first column
Example
4 3 2( ) 6 13 12 4f s s s s s
Create Routh table.
4
3
22,1 2,2
11,1
00,1
1 13 4
6 12 0
11 4
9.8 0
4
s
s
s P P
s P
s P
2,1
1 13
6 121 11
6P 2,2
1 4
6 01 4
6P
6 12
11 41,1 1 9.8
11P
2,1
(1*12 6*13)1 11
6P
2,2
(1*0 6*4)1 4
6P
1,1
(6*4 11*12)1 9.81
11P
5.3.3 Routh-Hurwitz – Special Cases
Page | 5-7
A zero in the first column of the array implies division by zero – not kosher.
Two cases
First Case – zero in first column of row & some non-zero elements in the row.
Replace the zero by a small number (ε) and proceed.
Second Case – all elements in row are zero – This occurs if roots are on jω axis or roots are symmetrical
on axis about origin – ( p root ).
Form auxiliary polynomial fa using coefficients of row before zero row.
Take derivative of auxiliary polynomial and use as substitute for zero row.
Derivative gives max/min when zero.
Example 1
Determine the stability of the function.
3 2( ) 2 2f s s s s
Develop the Routh array from the coefficients.
3
2
11,1
0
1 1
2 2
4 0
2 0
s
s
s p
s
1,1
1 1
2 21 0
2p - would give zero row for s1
Instead, make fa using row above
2
1,1
2 2
4
4
a
a
f s
dfs
ds
P
Interpretation – first column positive – no sign change
no poles in RH plane, - stable
Example 2
Page | 5-8
Determine the value of k for stability.
ΣRInput
Y(s)Output
+
-
2
2
1
( 1)( 2 2)
k s
s s s
Closed Loop Characteristic Equation - 1 0GH
In this problem, H=1.
2
2
( 1)1 1 [1] 0
( 1)( 2 2)
k sGH
s s s
Expand to obtain the stability function.
RInput
Y(s)Output
1
G
GH
2 2
3 2
( 1)( 2 2) ( 1)
( 3) (4 2 ) ( 2)
f s s s s k s
s k s k s k
Create Routh table.
3
2
11,1
00,1
1 4 2
3 2
s k
s k k
s P
s P
To be stable, the first column must have no sign changes. In this case the signs remain positive.
2,1 0 3 0P k 3k
0,1 0 2 0P k 2k
2
1,1
2 3 100 0
3
k kP
k
Denominator will be positive if ( 3k ). For the coefficient to be positive, the numerator must be >0
22 3 10 0k k
Solve using quadratic equation.
0,1 2P k
2
1,1
1 4 2
3 2 2 3 10( 1)
3 3
k
k K k kP
k k
Page | 5-9
( 3) 9 (4)( 2)(10)
2( 2)
3 89
4
3.11&1.61
3.11 1.61
k
k
Combine all the constraints.
2 1.61k
If k = limits, have poles on jω @ the limits
If k outside limits, have instability
Page | 5-10
Example 3
For the problem above, what are the poles on the jω axis?
ΣRInput
Y(s)Output
+
-
2
2
1
( 1)( 2 2)
k s
s s s
Use Routh table with the k used to calculate the limits.
When k=-2 when k=1.61
3
2
1
0
1 8
1 0
8
0
s
s
s
s
3
2
1
0
1 1.68
4.61 3.61
0
3.61
s
s
s
s
If any rows are zero, develop an alternative function.
s0 row = zero row s1 row – zero row
8
@ 0
8( ) 0
0
a
a
f s
s j f
j
2
*
* 2
4.61 3.61
@ 0
4.61( ) 3.61 0
a
a
f s
s j f
j
The frequencies of the roots are available.
@ origin, 0s j 2 3.61
4.61j
3.61
4.61
0.7831
0.885s j j
Page | 5-11
5.4 Root Locus
5.4.1 Introduction
Root Locus is the process of determining the location of all the roots to a feedback function. The
location of the roots will change as the value of the gain, k changes. By adjusting k, the stability can be
improved or degraded. The function that is investigated is the characteristic equation, which is the
denominator of the transfer function.
Closed Loop Transfer Function 1
Y G
R GH
Open Loop Transfer Function – GH
When H=1. the open loop is the plant, G
Characteristic Equation
1 0GH
Solving characteristic equation implies 1GH
1GH
,( 1,3,5,7)
(2 1),( 0, 1, 2)
GH m m
or
GH k k
Adjust the gain, k, so the open loop transfer is one. Substitute the actual function values for GH and set
the equation to one.
1 2
1 2
( )( )1
( )( )
k s z s zGH
s p s p
Calculate the system gain, K.
product of vector length of poles
product of vector length of zeroesK
2( 1)GH s zi s pi k
Page | 5-12
5.4.2 Object of Root Locus
The object of the root locus process is to determine the movement of the roots as k changes. The
analysis goes through this development.
Make plot of open loop function for 0k to k .
See where the function is stable & unstable. Avoid or fix unstable points.
Make sketch of significant points to draw.
Use the six rules to aid in the construction.
Page | 5-13
5.4.3 Rules for Root Locus Construction
Each rule provides a significant factor to assist in the sketch of the roots or eigenvalues. The sketch then
shows the stability as the roots change due to the gain, k being adjusted.
1) Starting and Ending Points Root Locus plots start ( 0k ) on the open-loop poles and
end ( k or k ) on the open-loop zeros
2) Root Locus Segments on the Real Axis Root loci occur on a particular segment of the real axis if and only if there are an odd number of total poles and zeros of the open-loop transfer function lying to right of that segment.
3) Imaginary Axis Intersections Use the Routh-Hurwitz criterion to determine j axis crossings of the root locus plots. Both the
gain k and the value of ω* may be found from the Routh table.
4) Asymptotes (for p z )
Root locus plots are asymptotic to straight lines with angles given by
(2 1)
( )A
k
p z
as s approaches infinity. These straight lines intersect at a point 1 on the real axis specified by:
1
poles of ( ) zeros of ( )GH s GH s
p z
Where p is the number of finite poles of GH(s) and z is the number of finites zeros of GH(s).
5) Angles of Departure and Arrival Assume a point s arbitrarily near the pole (for departure) or the zero (for arrival) and then apply
the fundamental angel relationship
angles of zeros of ( ) angles of poles of ( ) (2 1)GH s GH s k
6) Breakaway Points Breakaway points may be determined by expressing the characteristic equation for the gain k as a function of s and then solving for the breakaway points sB from
( )
0
Bs s
dk s
ds
Page | 5-14
Example 1
Sketch the root locus plot for the system in the diagram.
ΣRInput
Y(s)Output
+
-
( 1)( 2)
( 4)
k s s
s s
Set up the open loop function.
( 1)( 2)
( 4)
k s sGH G
s s
Plot the locus from poles to zeroes.
X – poles – denominator
O – zeros – numerator
Use rule 1, 2
Stay on axis only when have odd # of p&z to right
Example 2
Plot the root locus from the plant and feedback functions.
( 2) ( 3)
( 10) ( 1)
k s sG H
s s
The location of the poles and zeroes are obvious. However, the path is somewhat more tedious.
Obviously the path needs to cross from the right to
the left plane. Use Routh with the characteristic
equation which contains the open loop transfer
function.
1 0GH
Substitute the values.
Set up the Routh table.
2 2
2
( 1)( 10) ( 2)( 3) 0
9 10 5 6 0
( 1) (5 9) (6 10) 0
s s k s s
s s ks ks k
k s k s k
k=0k=k=k=0
-1-2-4
k=0
-1-2-3 10
2
1
0
1 6 10
5 9
6 10
s k k
s k
s k
Page | 5-15
Make s1=0 with 95
k .
Develop the auxiliary function fa from the row s2
2( 1) (6 10)af k s k
Determine the frequency for the crossing the axis, jω*
*( 1)( ) (6 10) 0k j k
A complex conjugate results in complementary frequencies. When 95
k , the root locus crosses the
vertical axis.
2
95
6 10
1 k
k
k
0.54
Make s0 =0 for 106
k
When 106
k , the root locus crosses jω @ 0 on its path to the right half plane, RHP.
The root locus moves from one pole in the LHP, crosses into the RHP at 0 . It collides with a pole in
the RHP. The path then comes back @ 95
k in complex conjugate pairs. It stays close to the origin.
The track should be circle, since make polynomial is s2. If the roots are s3 or higher, the path becomes an
ellipse.
k=0
-1-2-3
Example 3
Develop the root locus for an open loop transfer function.
Page | 5-16
2( 2)( 2 2)
kGH
s s s
The second order poles can be factored.
( 2)( 1 )( 1 )
kGH
s s j s j
Observations can be made about the expected results.
Not all poles are real. jω
p>z asymptotes
complex angle of departure
Follow the rules to develop all the plotting parameters.
Rule 2 – on real axis
Rule 3 – not all real, use Routh to find k & jω
21 ( 2)( 2 2) 0GH s s s k
3 21 4 6 (4 ) 0GH s s s k
Make the Routh table.
3
2
1
0
1 6
4 4
20
4
4
s
s k
ks
s k
1,0
1 6
4 4 20
4 4
k kP
Determine the frequency to make a row zero.
1 0 @ 20s k
24 (4 )af s k
* 24( ) 24 0j
6 @ 20k
Rule 4 - p z 3 0 3 3zeros @ p z
Page | 5-17
, ,3 33 0
A
Intersection
( 2) ( 1 ) ( 1 ) 0 4
3 0 3i
poles zeros j j
p z
Rule 5 - The complex conjugates need to 180 to all other poles & zeros
Pole @ 1 902
s j
zeros of GH - poles of GH=(2 )k
from -1-j = 902
from -2= 454
All these are s from poles, from local pole = x
(0) ( ) (2 )2 4x k
4x
, the θ for conjugate is complex
Plot the calculated points and draw the root locus from poles to zeroes.
-1- j
-4/3-2
-1+ j
Page | 5-18
5.5 Frequency Response Plots
5.5.1 Introduction
Frequency plots indicate the performance of a system over a range of frequencies. The plots can be
linear, semi-log (Bode), log-log, or polar plots (Nyquist). All these tools have the same information, and
are used to determine relative stability.
In contrast, the transition from the RHP using Routh- Hurwitz gives absolute stability in a limited system
without time shift.
A comparison of the costs and benefits of each response plot assists in selecting the one most
appropriate for a particular problem.
Linear – direct, but cumbersome math
Bode – easiest. Use log10 of open loop frequency response
Magnitude values in decibels (db) are plotted on the vertical
1020log ( )G j
Perform for each pole & zero @ K, then add the sketches.
Phase angle is plotted under the magnitude plot
Start w/ ( )G j in standard form
2
2
1( ) ( )
( ) (1 2 ( )
n
n n
j
ms j j
kG j G s
j
Plot the asymptotes.
These differ from actual values by 3db @ limits (ωn is corner frequency)
Slope of asymptotes = 20*m on the db/decade
Slope is negative for poles and positive for zeros.
m is the exponent power of s associated with the root.
Determine the phase angle @ the values.
1tann
90 *m for all ω
At the root, cross the 45o angle on the way from 0o two decades before the root in route to 90o two
decades after the root.
Page | 5-19
5.5.2 Bode Plots – Basic rules
The basic rules for Bode are quite short. The examples are better illustrations than a wordy explanation.
1) Magnitude – make break @ ωn, plot asymptote 20*m db/decade, where m = power of s.
2) Phase angle – make 45 break @ ωn plot from 2 decades before to 2 after the root.
3) Break down for poles and up for zeros
4) Plot 2 decades before to two decades after
5) Σ plots to get effective
Example 1
Make a frequency plot of a plant function.
1000( 1)
( )( 2)( 10)( 20)
s j
sG s
s s s
Convert the form to a magnitude and phase angle.
2
2 2 2 2 2 2
1000 1( )
2 10 20G j
1 1 1 1( ) tan tan tan tan1 2 10 20
G j
The above equations are used for the linear plots.
Example 2
Make a frequency plot of a plant function above.
Convert the function to standard form for Bode & log-log.
1000(1)1
(2)(10)(20) 1( )
1 1 12 10 20
j
G jj j j
1000(1)
2.5(2)(10)(20)
K
Page | 5-20
20 log(2.5) = 8, DC portion – normalized
Magnitude is plotted for each root.
Break up + 20db for zero @ 1
Break down -20db for poles @ 2, 10, 20
@ DC, response is8 0 , start phase diagram at 0°
Phase angle is plotted for each root.
+45˚ break points for zero @ ω=1
-45˚ break points for poles @ ω = 2, 10, 20
Combine the magnitude with 3 db smoothing @ corners (1,2,10,20)
1
rad/sec0 dB
-40 dB
20 dB
rad/sec
0 dB
-180°
1 100 10k
+45°
10 10000.1
10 1000
1000100
Combine the angles with a 5.7o smoothing at each break.
Page | 5-21
Page | 5-22
5.5.3 – Phase Margin and Gain Margin
The phase margin and gain margin of a system are indicators of the relative stability of the system. The
phase margin basis of the system is the absolute value of the difference between the actual phase angle
and -180° when the gain is 0dB.
phasemargin ( 180 ) @0dB
The gain margin basis of the system is the absolute value of the difference between the actual gain and
0dB when the phase angle is -180°.
gain margin ( ) 0G j dB @-180°
Example:
For the system response shown below, determine the phase margin and the gain margin.
0.01
rad/sec0 dB
-80 dB
40 dB
0.1 100.001 1001
-160 dB
0.01
rad/sec0 dB
-90°
0.1 100.001 1001
-180°
At approx 0.01 rad/sec, the gain crosses 0dB. At that frequency, the phase angle is -90°
0.01
phasemargin 90 ( 180 ) 90
At approx 10 rad/sec. the phase angle crosses -180°. At that frequency, the gain is approx -70dB.
10
gain margin 70 (0 ) 70dB dB dB
Page | 5-23
5.5.3 Polar - Nyquist Plot.
Re ( )G j vs Im ( )G j
Use the Bode Plot from the previous example to determine points on the Plot. Only a couple of points
are needed to determine the overall curve
The following steps outline the process.
@ ω=0
(8/20)10 2.5
0
G
G
Plot the point (2.5,0)
At the point that the phase = 0 real axis, called ωpc (phase crossing)
@= 0
(12/20)10 3.98
0
G
G
Plot the point (3.98,0)
Page | 5-24
At the point when 1G , this is called ωgc , arc from origin
@ 1G
01 10
125
G
Plot the point (-0.574,-0.819)
If zeros of |G| in left HP, called minimum phase
Plot the mirror image for complex conjugate pairs
Page | 5-25
5.6 Analog Filters
Filters are classified according to the order of the transfer function, and according to function. The most
common functions are low-pass, high-pass, band-pass and band-reject filters. Other types are phase-
lead and phase-lag. A chart of filter information, including transfer functions, system response and
circuits are shown below for some of the most common type filters.
First Order Low-Pass Filters
Frequency Response
ωc0 ω
H
1( ) (0)
2cH j H
RC Voltage Source (Thevenin)
v1
R1
C R2
+
v2
-
2
1 1
1( )
1
eq
eq
RvH s
v R sR C
1 2
1 2
eq
R RR
R R
1c
eqR C
RL
v1
R1
R2
+
v2
-
L
2 2
1
1( )
1eqeq
v RH s
Lv R sR
1 2eqR R R eq
c
R
L
RC Current Source (Norton)
R1 R2C
i2
i1
2
1 1
1( )
1
eq
eq
RiH s
i R sR C
1 2
1 2
eq
R RR
R R
1c
eqR C
Page | 5-26
First Order High-Pass Filters
Frequency Response
1( ) ( )
2cH j H j
ωc0 ω
H
RC
v1
R1
R2
+
v2
-
C
2 2
1
( )1
eq
eq eq
sR Cv RH s
v R sR C
1 2eqR R R 1
c
eqR C
RL Voltage Source (Thevenin)
v1
R1
R2
+
v2
-L
2
1 1
( )1
eq eq
eq
LsR Rv
H sLv R s
R
1 2
1 2
eq
R RR
R R
eq
c
R
L
RL Current Source (Thevenin)
R1 R2L
i2
i1
2
1 1
1( )
1
eq
p
RvH s
v R sR C
1 2
1 2
eq
R RR
R R
1c
pR C
Page | 5-27
Band-Pass Filters
Frequency Response
ωo0 ω
H
ωL ωU
0
1( ) ( ) ( )
2L UH j H j H j
3-dB Bandwidth = u LBW
Series
v1
R1
R2
+
v2
-
CL
2
21 1
1( )
1eq
v sH s
Rv R Cs s
L LC
1 2eqR R R 1
0LC
2 2
1 2
( )o
eq
R RH j
R R R
eqRBW
L
Parallel
v1
R1
R2
+
v2
-
L C
2
21 1
1( )
1eq
v sH s
v R C s sR CLC
1 2
1 2
eq
R RR
R R
10
LC
2
1 2 1
( )eq
o
RRH j
R R R
1
eq
BWR C
Page | 5-28
Band-Reject Filters
Frequency Response
ωo0 ω
H
ωL ωU
1( ) ( ) 1 (0)
2L UH j H j H
3-dB Bandwidth = u LBW
Series
v1
R1
R2
+
v2
-
C
L
2
2 2
21
1
( )1
eqeq
sv R LCH ssv R s
R C LC
1 2eqR R R 1
0LC
2 2
1 2
(0)eq
R RH
R R R
eqRBW
L
Parallel
v1
R1
R2
+
v2
-
L
C
2
2
21 1
1
( )1
eq
eq
sRv LCH sRv R
s sL LC
1 2
1 2
eq
R RR
R R
10
LC
2
1 2 1
(0)eqRR
HR R R
1
eq
BWR C
Page | 5-29
Phase-Lead Filters Phase-Lag Filters
ω1 ωm ω2
ω
log|H(jω)|
ωm0 ω
φm
ω1 ω2
ω1 ωm ω2
ω
log|H(jω
)|
ωm0 ω
φm
ω1 ω2
v1
R2
+
v2
-C
R1
2 1 1 1
1 1 22
11
( )1 1
eq
eq
sRv sR C
H ssv R sR C
1 2
1 2
eq
R RR
R R
1
1
1
R C 2
1
eqR C
1 2m max ( )m mH j
2 1 2 1
1 2
arctan arctan arctan2
m
m
1
1 2
(0)eqR
HR
1
2
mH j
( ) 1H j
v1
R1
R2
+
v2
-
C
2 2 2
1 1
1 1( )
1 1
eq
eq eq
Rv sR C sR CH s
v R sR C sR C
1 2eqR R R 1
1
eqR C 2
2
1
R C
1 2m min ( )m mH j
1 2 1 2
2 1
arctan arctan arctan2
m
m
(0) 1H 1
2
mH j
2 1
2
( )eq
RH j
R
Page | 5-30
Chapter 5 Problems
Problem 5-1
The circuit shown below is constructed
from an ideal op-amp (infinite gain
bandwidth product)
RC1 -
+
R
C2
Vout
Vin
The phase shift between input and output signals, as used
below, does not include the inversion (180°) from the
operational amplifier circuit.
20R k
1 100C pF
2 0.01C F
5.1.1 Which of the following is most
correct?
A) This is not a high-pass active filter.
B) The input circuit time constant is 2 μS.
C) The feedback circuit time constant is
0.2mS.
D) The phase of the output lags the phase
of
the input signal
5.1.2 The poles and the zeros of the transfer function
are approximately
A) One pole at 5 kHz and one zero at 500 kHz
B) Two zeros, one at 800 Hz and one at 80 kHz
C) Two poles, one at 5 krad/sec and one at
500krad/sec
D) A paired pole and zero at 55 krad/sec
E) One pole at 800 Hz and one zero at
500krad/sec
5.1.3 The dc response (dB) is
A) 20
B) 0
C) -20
D) -30
E) -40
5.1.4 The response (dB) at 10 MHz is most nearly
A) 10
B) 0
C) -10
D) -20
E) -40
Page | 5-31
5.1.5 The frequency at which the response
is -20dB is most nearly:
A) 0 Hz
B) 800 HZ
C) 500 kHz
D) 50 krad/sec
E) 80 krad/sec
5.1.6 The maximum phase shift between input and
output signals occurs under approximately which of the
following conditions?
A) When the response is minimum
B) When the response is equal to the square root of the
ratio of the capacitances
C) When the response is equal to the ratio of the time
constants
D) When the response is maximum
E) At a frequency that is twice that of a pole
5.1.7 The phase shift is approximately half
of its maximum value under which of the
following conditions?
A) When the response is -20 dB
B) When the response is equal to the
square root of the ratio of the capacitances
C) When the response is equal to the ratio
of the time constants
D) At the frequencies of a zero and a pole
E) At a frequency that is half that of a zero
5.1.8 The phase shift is 90° under which of the following
conditions?
A) When the response is -10dB
B) When the response is equal to the square root of the
ratio of the capacitances
C) When the response is equal to the ratio of the time
constants
D) At the frequencies of 800Hz and 80kHz
E) Never
5.1.9 The phase shift is approximately 5%
of its maximum value under which of the
following conditions?
A) When the response is -10dB
B) When the response is equal to the ratio
of the capacitances
C) When the response is a maximum
D) At the frequencies of 500 rad/sec and 5
Mrad/sec
E) At a frequency that is half that of a zero
5.1.10 Which of the following is true of this circuit?
A) Its sinusoidal response is maximum at the pole and
minimum at its zero
B) Its response is maximum when it is equal to the ratio
of the square of the capacitances
C) Its response is maximum at dc and minimum at very
high frequency
D) Its response increases with increasing frequency from
its value at dc
E) Its output phase leads the input signal phase
Page | 5-32
SOLUTION:
Calculate the gain function.
2 1
2
1
11
2
2
5
3
11
11
1
1
5 100.01
5 10
out
in
RRsCV RsC
V RsCRRsC
SRCC
CS
RC
s
s
Convert to standard form in terms of frequency.
5) 5 5
3
3 3
1 10.01(5 10 5 10 5 10
15 10
1 15 10 5 10
out
in
j j
V
j jV
Observe poles and zeros.
Pole at 5x103 rad/sec = 796Hz – Break Down
Zero at 5x105 rad/sec = 7.96kHz – Break Up
Determine the DC valued from K.
1; 20log 0K K , DC response is 0dB
@ DC, the gain is 1 1 180 . The starting point for the Phase diagram is 180°.
Plot gain and phase angle asymptotes.
Page | 5-33
500 5x103 5x104 5x105 5x106
500 5x103 5x104 5x105 5x106
79.6 796 7.96k 79.6k 796k
rad/sec
Hz
0 dB
-40 dB
180°
90˚
-
3dB
-37dB
-5.7°
101.4°
5.1.1
This is not a high-pass active filter. The correct answer is A)
5.1.2
Zero at 5x105 rad/sec (79.6kHz)
Pole at 5x103 rad/sec (796 Hz)
The correct answer is E)
5.1.3
By inspection @ DC, the gain is 0 dB. The correct answer is B)
5.1.4
At 10 MHz, by inspection, the gain is -40dB. The correct answer is E)
5.1.5
By inspection, -20dB occurs at 50krad/sec (7.96kHz). The correct answer is D)
5.1.6
By inspection, max phase shift occurs at -20dB (0.1), which is 1 2C C . The correct answer is B)
Page | 5-34
5.1.7
By inspection, the phase shift is approx. half its max value at the zero and the pole. The correct answer is
D).
5.1.8
The phase shift never reaches 90°. The correct answer is E)
5.1.9
The phase shift is approx 5% of its maximum value at one decade above the zero frequency (5x106
krad/sec) and one decade below the pole frequency (500 rad/sec). The correct answer is D)
5.1.10
By inspection, the response is maximum at DC and minimum at very high frequencies. The correct
answer is C).
Page | 5-35
Problem 5-2
The response to a unit step test is shown in the
figure below.
0.58152
0.5
0.9069
Data List
Peak value of unit step response – 0.58152
Steady-state value of unit step response – 0.5
Time to peak value of unit step response – 0.9069
sec
5.2.1 Data from the data list that are necessary
and sufficient to determine the percent
overshoot are:
A) 1 only
B) 2 only
C) 1 and 2 only
D) 1 and 3 only
E) 2 and 3 only
5.2.3 The expression for the time to peak in terms
of damping ratio and natural frequency of
oscillation is
A) 21
p
n
t
B) 1
p
n
t
C) 1
p
n
t
D) 2
1
1p
n
t
E) p
n
t
5.2.2 The percent overshoot is most nearly
A) 0%
B) 0.163%
C) 16.3%
D) 58.2%
5.2.4 The dc gain is most nearly
A) 0
B) 0.5
C) 1
D) 2
Page | 5-36
E) 100% E) 8
5.2.5 Data from the data list that are necessary
and sufficient to determine the damping ratio are
A) 1 only
B) 2 only
C) 1 and 2 only
D) 1 and 3 only
E) 2 and 3 only
5.2.7 The natural frequency of oscillation n is
most nearly
A) 1
B) 1.414
C) 2
D) 4
E) 16
5.2.6 The damping ratio is most nearly
A) 0
B) 0.5
C) 1
D) 2
E) 8
5.2.8 The transfer function of the open loop system
is
A) 2
8
8 16s s
B) 2
0.5
8 16s s
C) 2
1
4 16s s
D) 2
8
4 16s s
E) 2
0.5
4 16s s
Page | 5-37
SOLUTION
5.2.1 max ( ) ( )
% overshoot = 100( )
y t y
y
Need 1 and 2 only. The correct answer is C)
5.2.2 0.58152 0.5
% overshoot = 100 16.304%0.5
. The correct answer is C)
5.2.3 21
p
n
t
. The correct answer is A)
5.2.4 The dc gain is the steady state value divided by the input (unit step), or 0.5. The correct answer is
B
5.2.5 The damping ratio may be found from the equation 21
% overshoot = 100e
. The data needed
are 1 and 2 only. The correct answer is C)
5.2.6 210.58152 0.5
% overshoot = 100 16.304% 1000.5
e
.
21
2
2 2 2
2 2
2
1006.135
16.304
ln 6.135 1.8141
3.29(1 )
( 3.29) 3.29
0.25; 0.5
e
The correct answer is B)
5.2.7
2
2
1
1
4.0 / sec0.9069 0.75
p
n
n
p
n
t
t
rad
The correct answer is D)
5.2.8 Form of the equation is 2
2 22
n
n n
Ks
2
160.5
4 16s s
The correct answer is D)
Page | 5-38
Problem 5-3
A control system has an open loop response of 2
8
4 16s s . A tachometer feedback system is installed
in the control system in order to make the system critically damped.
5.3.1 The rate gain needed for the tachometer
feedback to achieve a critically damped system is
most nearly
A) 0.125
B) 0.5
C) 1
D) 2
E) 4
5.3.2 The location of the critically damped poles is
most nearly
A) -16
B) -4
C) 0
D) 4
E) 16
SOLUTION
The block diagram, with the tachometer (rate) feedback is shown
5.3.1
The transfer function is developed in the circuit
element form.
.
The frequency form of the second order bandpass or resonance equation is 2
2 22
n
n n
Ks
.
Set the denominator characteristic equation in circuit form to the equivalent frequency form. Then
calculate the frequency variables.
.
For critical damping, ζ=1. Therefore, the gain b can be calculated.
The correct answer is B)
2 2
( ) 8 8
( ) 1 4 16 8 (4 8 ) 16T T
Y s G
X s GH s s K s s K s
2 16 4n n
4 8 2T nK
4 8 2 2(1)(4) 0.5T n TK K
X(s)Input
Y(s)Output
+
-2
8
4 16s s
TK s
S
Page | 5-39
5.3.2
The critical damped transfer function is
2 2
8 8
8 16 4s s s
. Therefore the poles are -4, -4.
The correct answer is B)
Page | 5-40
Problem 5.4
A control system is described by the block diagram
X(s)Input
Y(s)Output
+
-2
4
6 13
s
s s
Σ ( )cG s
controller
Data List:
Gs(s)=2
4
6 13
s
s s
Gc(s)=( )
( 1)
K s z
s s
Possible values of z = -3 or -5
5.4.1 The open loop transfer function is
A) K
B) ( )
( 1)
K s z
s s
C) 2
( )( 4)
( 1)( 6 13)
K s z s
s s s s
D) 2( 1)( 6 13)s s s s
E) 2
4
6 13
s
s s
5.4.4 The angles the asymptotes of the root loci
makes with the real axis of the s plane are
A) ±0°
B) ±90°
C) ±180°
D) ±270°
E) ±360°
5.4.2 The poles of the open loop transfer function
are:
A) z, -4 only
B) 0, -1 only
C) -3±j2 only
D) 0,-1, -3±j2 only
E) z, 0, -1, -3±j2, -4
5.4.5 Data from the data list that are necessary
and sufficient to determine the intersection of
the root loci with the real axis of the s plane are
A)1 only
B) 1 and 2 only
C) 1 and 3 only
D) 2 and 3 only
E) 1, 2, and 3
Page | 5-41
5.4.3 The zeros of the open loop transfer function
are:
A) z, -4 only
B) 0, -1 only
C) -3±j2 only
D) 0,-1, -3±j2 only
E) z, 0, -1, -3±j2, -4
5.4.6 The intersections of the root loci with the
real axis of the s plane, if 3 5z and are
A) 0 and 0
B) 0 and 1
C) 0 and +1 and -1
D) 1 and 1
E) -3 and -5
5.4.7 The characteristic function if z = -3 is
A) 4 3 27 19 13s s s s
B) 4 3 27 20 17s s s s
C) 4 3 27 20 20 12s s s s
D) 4 3 27 34 118 180s s s s
E) 4 3 27 (19 ) (13 7 ) 12s s K s K s K
5.4.9 What is the first term in the third row
Routhian array if z=-3?
A) (156+12K2)
7
B) 120
7
C) 19 K
D) 13 K
E) 12K
5.4.8 Which of the following values of z provide for
a stable system for any value of K?
A) -3 only
B) any z>-5
C) -3 and -5 only
D) any z>-3
E) Neither -3 nor -5
5.4.10 For with values of K is the system stable is
z = -3
A) All K>0
B) All K<0
C) All large K
D) All small K
E) All K<15
Page | 5-42
SOLUTION:
5.4.1
The open-loop transfer function ( )
( ) ( )( )
c s
Y sG s G s
X s
2
( )( 4)
( 1)( 6 13)
K s z s
s s s s
.
The correct answer is C)
5.4.2
The poles of the open lop transfer function are the roots of 2 6 13 1s s s s
Roots are 0, -1, 3 9 13 =0, -1, 3 2j
The correct answer is D)
5.4.3
The zeros of the open loop transfer function are the roots of ( )( 4)K s z s . Roots are z= -4
The correct answer is A)
5.4.4
Asymptotes – angle = (180 360 )
p z
k
n n
0, , p zk n n ; 4; 2p zn n . Therefore the angle = 90
The correct answer is B)
5.45
- c
p z
poles zeros
n n
H(s), Gc(s) and the value of z are needed.
The correct answer is E)
5.4.6
Page | 5-43
0 ( 1) ( 3 2) ( 4) - 1.5 0.5
4 2c
p z
j zpoles zerosz
n n
For 3, 0cz . For 5, 1cz
The correct answer is B)
5.4.7
2
2
( 1)( 6 13) ( )( 4)1 ( )
( 1)( 6 13)c
s s s s K s z sG s H
s s s s
The Characteristic equation = 4 3 27 (19 ) 13 (4 ) 4s s K s K z s Kz
For 4 3 23, 7 19 13 (4 ) 12z s s K s K z s K
The correct answer is E)
5.4.8
0 ( 1) ( 3 2) ( 4) - 1.5 0.5
4 2c
p z
j zpoles zerosz
n n
For 0c , there are poles in the right half of the s-plane – unstable.
The correct answer is D)
5.4.9
For z=-3, the Routh array is
2,1
19 (13 7 ) 120
7 7
K KP
2,2 12P K
2,2
1,1
2,1
13 7 713 7 4.9 13 2.1
K PP K K K
P
0,1 2,2 12P P K
The correct answer is B)
5.4.10
4
3
22,1 2,2
11,1
00,1
1 19 12
7 13 7 0
s K K
s K
s P P
s P
s P
Page | 5-44
From the Routh array above, 2,1 0P , 1,1 0 6.19P for K , 0,1 0 0P for K .
Therefore, the function is stable for all 0K .
The correct answer is A)
Page | 6-1
Chapter 6 Digital
6.1 Introduction Digital systems are the language of computers and microprocessor hardware. A basic understanding of
the principals associated with digital systems are presented.
6.2 Binary (Digital) Systems
6.2.1 Number Systems
Decimal numbers
When using numbers, we usually write only the coefficient and let the position indication the power of
10. The coefficient range = 0-9 for base 10
Binary numbers
The coefficient range = 0-1 for base 2
3 2 1 02 101010 1 2 0 2 1 2 0 2 10
Octal numbers
Octals are simply binary numbers combined into groups of 3 ( base 8=23)
388 2 1| 001 12
The coefficient range for octal numbers is 0-7
3 2 1 0
107392 7 10 3 10 9 10 2 10
Page | 6-2
Hexadecimal numbers
Hexadecimal numbers are binary numbers that are combined into groups of 4 (base 16 = 24)
41616 2 0 |1010 0A
The range for hexadecimal numbers is 0-F.
Hex # Decimal Equiv Hex # Decimal Equiv
0 1 8 9
1 2 9 10
2 3 A 11
3 4 B 12
4 5 C 13
5 6 D 14
6 7 E 15
7 8 F 16
Page | 6-3
Arithmetic
Arithmetic rules for all bases have the same basic set of rules.
Addition
111
101
1100
Subtraction
111
101
010
Multiplication
111
101
111
000
111
100011
Decimal to Binary Equivalence
0 0000 5 0101
1 0001 6 01101
2 0010 7 0111
3 0011 8 1000
4 0100 9 1001
Page | 6-4
6.2.2 Binary System Wiring
Equivalent Names
5V = 1 = high = True = on = closed
0V = 0 = low = False = off = open
Resistors
Pull-up – from power source –
2.2KΩ
Current limiting – in series with load
330Ω
Switch
When switch = open, output = 1 (True)
When switch = closed, output = 0 (False)
Called an inverting switch
2.2kΩ
output
5 Volts
switch
Transistor
Simply a digital switch with an electrical
input (inverter)
When base switch = off, output = 1
(True)
When base switch = on, output = 0
(False)
2.2kΩ
output
330Ω
5V
Page | 6-5
Output – LED
Generally connected to ground
Use current limiting resistor
LED Long leg is the ground
This is a common cathode arrangement.
A common anode would have inverted
diodes
330Ω
5 Volts
330Ω
TTL
Standard Arrangement
5V Power, VCC, on upper right pin
Ground on lower left pin
Notch at top
a11
a22
3a3
4a4
b1
b2
b3
b4
14
13
12
11
a15
a26
7a3
b1
b2
b3
10
9
8
5 volts
Problem:
Wire a switch circuit with an LED connected to the output pin.
Page | 6-6
6.2.3 The Huntington Postulates
Required entities to define an algebra: A defined group of coefficients (R, N, Q, Z [0,1]), Real Natural, Complex Binary
A defined group of operators and a table which defines how each operator works.
A defined group of axioms or postulates (unproven theorems) from which new theorems,
lemmas, corollaries, and propositions may be constructed.
[0,1] = B (coefficients)
OR AND NOT
For all x.y elements of group B: for each operator
P1 (a) x y B (b) *x y B (closure) only 1 or 0
P2 (a) 0x x (add Identity, I=0) (b) *1x x (multiply Identity, I=1) P3 (a) x y y x (b) * *x y y x (commutative)
P4 (a) *( ) ( * ) ( * )x y z x y x z (b) ( * ) ( )*( )x y z x y x z
(distributive)
P5 (a) 1x x (b) * 0x x (complement, unique) P6 (a) B contains at least 2 distinct elements
Useful theorems: listed in pairs that have correspondence – duality – every expression is valid if operator and identity elements are changed. To find the dual, exchange + to * and 1 to 0.
T1 (a) x x x (b) *x x x T2 (a) 1 1x (b) *0 0x
T3 (a) ( )x x (involution)
T4 (a) ( ) ( )x y z x y z (b) ( ) ( )x yz xy z (associative)
DeMorgan’s Theorem – complement function by interchanging AND & OR operators and complementing each literal
T5 (a) ( )x y x y (b) ( * )x y x y
T6 (a) x xy x (b) ( )x x y x (absorption)
T7 (a) x x y x y (b) ( )x x y xy
Operator Precedence
Parentheses () Not AND OR
Complement – take dual and complement each literal
+ 0 1
0 0 1 1 1 1
* 0 1
0 0 0 1 0 1
~
0 1 1 0
Page | 6-7
6.2.4 Basic Digital Gates
NAND
X
Y
Z
F=(xyz)’
And Invert
XOR
F=x’+y’+z’ = (xyz)’xyz
Invert Or
NOR
X
Y
Z
F=(x+y+z)’
Or Invert
XAND
F=x’y’z’=(x+y+z)’x
yz
Invert And
A one input gate acts like an inverter
To use NAND requires Sum of Products (SOP) form
To use NOR requires Product of Sums (POS) form
Page | 6-8
6.3 Karnaugh Maps
Definitions: Literal a variable of the problem Don’t Care Output values for which no value is necessary (situations that never occur, or if
they do, can be ignored SOP Abbreviation for Sum of Product POS Abbreviation for Product of Sum Reflected Code A binary code which has the property that the codes following every 2n-1 codes
are the reflection of the first 2n-1 codes in all bits except the most significant. The MSB of the first 2n-1
is a zero, and the MSB of the next 2n-1 codes is a 1.
Example Binary Code Reflected Code
00 00
01 01
10 11
11 10
PURPOSE:
A method of realizing a function in either of the two standard forms such that the number and
complexity of terms in the function is minimal.
Construction:
Draw a square or rectangular figure allowing 2n squares (for a problem with n variables).
The table should be drawn so that there are 2x rows and 2y columns where x y n .
Label the rows and columns with reflected code from left to right and top to bottom.
Fill in each square of the map with its corresponding truth table function value.
Simplification:
Circle the largest contiguous binary (2w, w n ) group of 1’s for SOP (0’s for POS) which is rectangular or square
Consider all the edges of the map to be physically adjacent.
Each circled block of 1’s (0’s) corresponds to one SOP (POS) term. The term is extracted by observing which literals do not change for the block. These can be complemented to create the POS terms using 0’s. The literals are then AND’ed (OR’ed) together to realize the function.
Finally, each term, which corresponds to each rectangular block, is OR’ed (AND’ed) together to realize the function
Each 1 (0) in the map must be circled at least once to realize the function. (Sometimes there is more than one way to do it).
Additional Rule for Don’t Cares: It is not required that don’t cares be circled, but treat them as 1 (0) if it will help with minimization.
Page | 6-9
6.3.1 Construction / Simplification of Karnaugh Maps:
Example 1:
( )( ')
AB A B
A B A B
Example 2:
A B
A B
Example 3:
( )( ' )
A B AB
A B A B
Example 4:
( )( )
A B C BC AC
A C A BC
Eliminate Races - Overlap
(SOP)F A C BC AB AC
A B F
0 0 0 0 1 1 1 0 1 1 1 0
B A
0 1
0 0 1
1 1 0
A B F
0 0 1 0 1 1 1 0 1 1 1 0
B A
0 1
0 1 1
1 1 0
A B F
0 0 1 0 1 0 1 0 0 1 1 1
B A
0 1
0 1 0
1 0 1
A B C F
0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 1
C AB
0 1
00 0 1
01 0 1
11 1 1
10 1 0
C AB
0 1
00 0 1
01 0 1
11 1 1
10 1 0
Page | 6-10
Example 5:
(0,1,3,4,6,7) (2,5)F (0,1,3,4,6,7)F (2,5)F
( )( ) ( )
(SOP)
F A B C A B C POS
F A B BC AC A C
(SOP)F B C A C AB
Eliminate Races
(SOP)F A B A C AB BC AC
C AB
0 1
00 1 1
01 0 1
11 1 1
10 1 0
A B C F
0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 1
C AB
0 1
00 1 1
01 0 1
11 1 1
10 1 0
C AB
0 1
00 1 1
01 0 1
11 1 1
10 1 0
C AB
0 1
00 1 1
01 0 1
11 1 1
10 1 0
C AB
0 1
00 1 1
01 0 1
11 1 1
10 1 0
Page | 6-11
6.4 Design w/ Multiplexer
Multiplexers are a mixed bag
Designers try to minimize gates good
Do clever things good
Results are not logical bad
Hard to troubleshoot bad
Use multiplexer to show sequence of states
3 level pattern
#1 – multiplexers that determine the next state of registers
#2 – Register that holds present binary state
#3 – Decoder that provides a separate output for each control state
Put “Select” line across top of Karnaugh map
With “C” Select With “A” Select
'
F A B AC BC
F A B AC B C
F A B B C AC
F A B B C AC
0
1
Select C
I AB AB AB
I A B
0
1
Select A
I B
I C
C AB
0 1
00 0 0
01 1 1
11 1 0
10 1 0
A BC
0 1
00 0 1
01 0 0
11 1 0
10 1 1
Page | 6-12
Three different implementations of the same function.
A valve header requires the following logic to control its actions:
(2,3,4,6)
(0,1,5,7)
F
For 8:1 Multiplexer
I0
I1
I2
I3
I4
I5
I6
I7
EN
S2 S1 S0
Z
+5V
+5V+5V
+5V
A B C
F
constants
For 4:1 Multiplexer
Only need
“A” inputs
I0
I1
I2
I3
EN
S2 S1
Z
A
+5V
A’
B C
B & C are select
Do not cross areas of constant BC
A B C F
0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0
BC A
00 01 11 10
0 0 0 0 1 1 3 1 2
1 1 4 0 5 0 7 1 6
BC A
00 01 11 10
0 0 0 0 1 1 3 1 2
1 1 4 0 5 0 7 1 6
E 0 1 3 2
Page | 6-13
For 2:1 Multiplexer
I0
I1
EN
S0
Z
ABA’B
C
F
Need A&B
Input
C=0 C=1 C is select
Do not cross areas of constant C Use I0 for C = 0 Area Use I1 for C = 1 Area
For 2:1 Multiplexer – different select
I0
I1
EN
S0
ZB
C’
A
F
Need B&C
Input
A=0 A=1 A is select
Do not cross areas of constant A Use I0 for A = 0 Area Use I1 for A = 1 Area
Choice of inputs to select lines determines how to partition K-Map
Select line – ABC, then each value is fixed, nothing changes
Select line = BC, then A value changes
Select line = C, then AB values change
BC A
00 01 11 10
0 0 0 0 1 1 3 1 2
1 1 4 0 5 0 7 1 6
B A
0 1
0 0 1
1 1 1
B A
0 1
0 0 1
1 0 0
BC A
00 01 11 10
0 0 0 0 1 1 3 1 2
1 1 4 0 5 0 7 1 6
C B
0 1
0 0 0
1 1 1
C B
0 1
0 1 0
1 1 0
Page | 6-14
6.5 Decoder
Function
SOP = (minterms) (OR of minterms)
POS = Π(maxterms) (AND of maxterms)
Each output of decoder is a minterm
For function in SOP
Use Active Hi output for OR gate
If use Active Lo output – must use invert-OR = NAND
For function in POS
Active Lo output – use AND gate
If use Active Hi output – must use invert-AND = NOR
Decoder Example
SOP Form
SOP uses 1’s
F= (1,2,4,6,7)
S0
A
S1
B
CS2
O0
O1
O2
O3
O4
O5
O6
O7
EN
0
Active Hi
Sum (OR)
F
S0
A
S1
B
CS2
O0
O1
O2
O3
O4
O5
O6
O7
EN
0
Active Lo
NAND
F
POS Form
POS Uses 0’s
F= Π(0,3,5)
S0
A
S1
B
CS2
O0
O1
O2
O3
O4
O5
O6
O7
EN
0
Active Lo
AND
F
S0
A
S1
B
CS2
O0
O1
O2
O3
O4
O5
O6
O7
EN
0
Active Hi
NOR
F
Page | 6-15
Note that the Functions are equivalent - F= (1,2,4,6,7) = Π(0,3,5)
6.5.1 74156 DECODER
The 74156 is a dual 2:4 decoder
The 2 input lines can be decoded into 4 output lines
An Active low device – when the input is selected low, the decoded output line will be low
Two separate 2:4 decoders may be individual, or connected as a 3:8
a11
O0
O1
O2
O3
7
6
5
4S1S0
0 0
Y4Y5Y6Y7
a11
O0
O1
O2
O3
9
10
11
12
S0
0 0
S1 Y0Y1Y2Y3
B(3)A(13)
Strobe W’ (2)
Data C (1)
Strobe W’ (14)
Data C’ (15)
The select lines are common
WCBA is sequence for complete address
W’ have write to output, this is enable line
C-C’ tie together – data
BA Address
RAM
Write – write line=0, then data (1 or 0) on data line is clocked into decoder
A1A0 – Same address connected to both decoders
DATA (C-C’)– Inverted on decoder b, direct on Decoder a, therefore output of
both each side of a FF will force the FF to change state
Page | 6-16
6.6 Flip Flops / Latch When S=0, R=1, Qt+1 = 0 When S=1, R=0, Tt+1 = 1
QR Flip Flop Function – 1tQ S R Q
S R Q Qt+1
0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 indeterminate
1 1 1 indeterminate
Q SR
0 1
00 0 1
01 0 0
11 IND IND
10 1 1
Page | 6-17
6.6.1 General – Flip Flop Types
Characteristic Table Excitation Table
Characteristic Equation
Symbol
Circuit
Asynchronous (RS Flip Flop)
S R t dtQ
't dtQ
tQ t dtQ
S R
0 0 tQ tQ 0 0 0 -
0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1 1 0 0 1 1 - 0
0
t dt tQ S R Q
SR
Q
QSET
CLR
S
R
R
S
Q
Q`
S R
S R
0 0 1 1 0 0 1 - 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 1 1
1 1 - 1
1
t dt tQ S RQ
S R
Q
QSET
CLR
S
R
R
S
Q
Q`
Synchronous
S R
S R
0 0 0 0 0 - 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1 1 0 0 1 1 - 0
1
0
t tQ S R Q
SR
Q
QSET
CLR
S
R
cp
S
R
cp
Q
QSET
CLR
S
R
J K
J K
0 0
0 0 0 -
0 1 0 1 0 1 1 - 1 0 1 0 1 0 - 1 1 1
1 1 - 0
1t t tQ JQ K Q
J
Q
Q
K
SET
CLR
cp
K
J
cp
Q
QSET
CLR
S
R
D
D
0 0 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 1
1tQ D
Q
QSET
CLR
D
cp
D
cp
Q
QSET
CLR
S
R
T
T
0
0 0 0
1
0 1 1
1 0 1 1 1 0
1t t tQ TQ T Q
T Q
Q`
Q
QSET
CLR
S
RT
cp
t dtQ 't dtQ tQ t dtQ
tQ tQ
1tQ 1'tQ tQ 1tQ
1tQ 1'tQ tQ 1tQ
tQ tQ
tQ tQ
1tQ 1'tQ tQ 1tQ
1tQ 1'tQ tQ 1tQ
tQ tQ
tQ tQ
Page | 6-18
6.6.2 Counter
Sequential circuit that goes through a prescribed sequence of states on application of input pulse
Binary counter – counter that follows binary sequence\
n-bit counter has n flip flops
only input is count pulse, clock is implied
Output = present state of FF
Counter completely specified by a list of the count sequence – is the sequence of binary states it
undergoes
Page | 6-19
Counter Design
Description – Specifications
Draw state diagram a. Used to show
progressive states of sequence events
b. Same as state table c. Present state in circle d. Vector shows next state e. Vector information
o First character indicates input that causes change of state. Use 1 for clock
o 2nd number(s) show output for present state. Only happens when arriving at next state, not during transition
Make State Table a. This contains the same information as the state diagram, and is
interchangeable.
Assign state variables a. Assign each desired output to a FF if have no conditional outputs
(no logic) b. Assign each state adjacent so that you do not change more than
one state variable for a single change in input – mirror code
c. n=# state variables → 2 #n states
i. Example 4 states = 2n → n = 2 state variables, which is the number of FFs
d. Identify 2 FFs as Q1Q2, QAQB, AB, PQ, depending on author e. Sometimes do not use abc, but numbers
Make transition table a. This is simply present state – next state with
state assignments b. Output = present state is use FF for output
Determine FF type & assign letters to each a. Make column for input to each FF b. From excitation table for the FF, what is the
input to the FF required to make transition from PS to NS. i.e. if PS Q1=1, NS Q1=0
c. FF input T1 must be 1 to cause toggle D FF = Next State
PS NS w=0
NS w=1
Out w=0
Out w=1
a a b 0 0 b c b 0 0 c d c 0 0 d d a 0 0
PS Assignment
a 00 AB b 01 AB c 11 AB d 10 AB
Inputs
w
Present State Q1Q2
Next State Q1Q2
Outputs
Z1
FF Input T1T2
0 00(a) 00 0 00 1 00(a) 01 0 01 0 01(b) 11 0 10 1 01(b) 01 0 00 0 11(c) 10 0 01 1 11(c) 11 1 00 0 10(d) 10 0 00 1 10(d) 00 0 10
a
/ 0w
b
/ 0w
c
d
/ 0w
/ 0w
/ 0w
/ 0w/1w
Use a 1
for
clock
Page | 6-20
Draw K-map for FF input @ this present state. Where is the T FF input required from excitation table to get next state. d. Use PS & Input to yield FF input
1 1 2 1 2T wQ Q wQ Q 2 1 2 1 2T wQ Q wQ Q 1 2z wQ Q
Start w/ specification – end w/ Boolean expression. Can draw diagram and wire circuit from this.
Q1Q2
W 00 01 11 10
0 0 1 0 0
1 0 0 0 1
Q1Q2
W 00 01 11 10
0 0 0 1 0
1 1 0 0 0
Page | 6-21
Example: 3 bit binary counter
State diagram
state Table (T Flip Flops)
Output = Present State = Q2Q1Q0
000
001
010
011
100
101
110
111
Karnaugh Maps
T2 T1 T0
2 1 0T Q Q 1 0T Q 0 1T
Draw Circuit
PS NS T2 T1 T0
000 001 0 0 1 001 010 0 1 1 010 011 0 0 1 011 100 1 1 1 100 101 0 0 1 101 110 0 1 1 110 111 0 0 1 111 000 1 1 1 000
Q1Q0
Q2 00 01 11 10
0 0 0 1 0
1 0 0 1 0
Q1Q0
Q2 00 01 11 10
0 0 1 1 0
1 0 1 1 0
Q1Q0
Q2 00 01 11 10
0 1 1 1 1
1 1 1 1 1
Page | 6-22
T0 Q0
T1 Q1
T2 Q2
5V
Page | 6-23
6.6.3 Sequence Detector (Random)
Sequence consists of n bits
number of states: Mealy = n, Moore = n+1
Make transition table with added columns of current sequence and desired sequence. Table columns are current sequence, desired sequence, input, present state, next state, output, FF input
Label current state & desired state
Write desired sequence under column for all rows
Place sequence up to this point under correct sequence
Draw line through as many as useable of current sequence to next state in desired sequence.
Same number of states must be crossed in both
Output = 1 when all states are crossed for Mealy. For Moore, add one more state when all states are crossed
Else, go to 7.
Note: it is easier to make input
Page | 7-1
Chapter 7 - Economics – Time Value of Money
7.1 Introduction Does money have the same value today that it did 10 years ago? Would you rather have your money
now or later? If you can get it sooner, would you be willing to give a discount on the amount. If you are
willing to get it later, how much interest do you need?
In the process of creating projects, it is necessary to balance time, money, and quality. Other chapters
look at the time and quality issues. This section discusses the relative value of money. The study of the
value of money is called economics.
Economics is often divided into two segments: macro and micro. Macroeconomics deals with large-scale
money manipulation that is often controlled by government. It attempts to look at an economy as a
whole. Microeconomics is local scale money that is controlled by a project, corporation, or other entity.
This chapter will focus on project economics.
Economics has been called the science of allocating scare resources. That definition is limiting. It
assumes there is a fixed sized pie. If one group gets a larger slice, then someone else must necessarily
receive a smaller share. That is a negative feedback analysis, and conflicts with the fundamental premise
of positive feedback.
Positive feedback philosophy is the basis for developing new technology. It assumes something that has
little value can have value added through technology, thereby creating more value or wealth. This
comes without reducing the wealth of others.
Consider a common example. Sand has little value. The most common component of sand, however is
silicon oxide; it can be modified to create the silicon wafers used in microelectronics. A previously
limited value material now has great value.
Page | 7-2
7.2 Time and Interest At the present time money has a known value, called the present value or purchase price, PV. At some
time in the future, the same money will have a different value called the future value, FV.
Two things relate the values. Interest rate, i, is the percentage of a sum of money charged for its use.
This is also called the discount rate. Interest is based on a period of time. In most cases it is one year, but
it can be any interval. The number of interest periods is n.
It is assumed that all money transfer is at the end of a period. If it is at the beginning, the present value
must be calculated from one period in the future.
The simplest calculation is for a single payment exemplified in the cashflow diagram. A value below the
line is negative or paid out. A value above the line is positive or received in. Hash marks without arrows
simply identify a time period.
FV
PV
Equations for calculating the values are shown below. In addition, many texts give an abbreviation for
the factor that relates the value to be calculated to the known value. In addition, we will show the
equivalent spreadsheet command, (Microsoft Excel 2003), since that is the most common way of
calculating value of money. [Excel] The Excel formulae start with an “=” sign
Find the amount of money that it would take in the future, FV, which has an equivalent value to the
present worth, PV.
*(1 )nFV PV i
( , ,0, ,)FV i n PV
Conversely, when the value in the future is known, determine the present value.
*(1 ) nPV FV i
( , ,0, ,)PV i n FV =PV(i,n,0,FV,)
Page | 7-3
7.2.1 Uniform series
Rather than single payments, a much more common practice is to have progress payment amounts, A. If
these are at regular time intervals, the process is called a uniform series. When uniform amounts are
applied to the future value and present value, four new relationships are established. Some authors
treat these as equivalent uniform annual cost (EUAC).
Future value and uniform amounts are illustrated in a cashflow diagram.
FV
A A A A A A A
Series compound amount is the future value found from the uniform amount.
(1 ) 1ni
FV Ai
( , , ,,)FV i n A
Sinking fund is the uniform payment found from the future value.
(1 ) 1n
iA FV
i
( , ,0, ,)PMT i n FV
The cashflow diagram demonstrates the present worth and the uniform payment amounts.
PV
A A A A A A A
Series present worth is the present value found from the uniform payment.
(1 ) 1
(1 )
n
n
iPV A
i i
( , , ,,)PV i n A
Page | 7-4
Capital recovery is the uniform payment found from the present value.
(1 )
(1 ) 1
n
n
i iA PV
i
( , , ,,)PMT i n PV
Capital recovery on salvage is a modification of the capital recovery relationship. The future value of the
salvage, S, is converted to a uniform payment and subtracted from the present value convert to a
uniform series. Alternately the salvage value is transferred through the present worth as developed
above.
(1 )
( )(1 ) 1
n
n
i iA PV S iS
i
( , , , )PMT i n PV S
Page | 7-5
7.2.2 Gradient
A gradient is a value that increases or decreases at a rate for each time interval. An arithmetic gradient
is an integer multiple during each interval. It can have either a uniform amount or a present value, PV.
G
A A A A
PV
AA
2G
3G
4G
(n-1)G
Arithmetic gradient uniform series is calculated from the gradient value, G, after the first interval.
1
(1 )n
nA G
i i i
In a similar manner the present value is derived from the first gradient, G.
(1 ) 1
(1 )
n
n
i inPV G
i i i
A geometric gradient grows by a rate, g, during each interval. Therefore the change in amount varies
according to an exponential curve.
A1
PV
A2
A3
A4
An
The present value depends on two rates, the interest and the gradient, as well as the first amount in the
series.
1 (1 ) (1 )n ng i
PV Ai g
Frequently, the gradient rate is equal to the interest rate. Then the equation reduces significantly.
11 (1 )PV A n i
Page | 7-6
7.2.3 Nominal Interest or APR
Nominal interest rate, r, is also called the annual percentage rate (APR). It is the reported rate that is
compounded multiple times M, per year. It may be stated as 18% APR or 1.5% per month. These are
equivalent. However, they are not the effective interest rate.
Effective annual interest rate, ia is also called annual effective yield (AYR). It is substantially greater than
the nominal rate.
1 1
m
a
Ri
M
The most insidious effect is continuous compounding at a nominal rate. This creates an exponential
growth.
For a single payment, the relationships are simple, but expensive.
( )rnFV PV e
( )rnPV FV e
Page | 7-7
7.2.4 Perpetual and Rule of 72
Present value of perpetuities is a common way to determine the worth of an ongoing, perpetual uniform
payment. this is valuable in evaluating annuities, insurance, and lotteries.
A
PVi
Rule of 72 is a quick technique to determine how long it takes for the present value of money to double.
The number Of periods is determined from 72 divided by the interest rate.
72
ni
Page | 7-8
7.3 Rate of Return
Project economics are based on recovery of money. A variation of present worth has been used in all
the calculations to this point. Projects typically can be implemented with a number of different
alternatives. Therefore, the economic analysis is a comparison of alternatives.
Internal rate of return (IRR) is the break-even interest rate i* at which the present worth of a project is
zero. With a present worth of zero, we are ambivalent and do not particularly care about the project.
( *) PV cas in - PV cash out =0PV i
The calculation of IRR is an iterative process. Begin with the number of time intervals, n, and the
present, future, or uniform amounts. Guess at an interest rate and calculate the present worth. If it is
positive increase the interest rate. If the present worth is negative decrease the interest rate. Continue
estimating interests until the present value is adequately close to zero.
It is not necessary to laboriously perform this calculation. Many calculators and spreadsheets have the i*
function built into the spreadsheet package (Microsoft Excel 2003).
( , )IRR values guess
Investors have a minimum acceptable rate of return, MARR. This is the interest that must be exceeded if
the project is to be funded. Compare the calculated rate of return with the MARR. If the IRR is less than
the MARR, then the project should not be done.
Typically the analyst will go back and tweak the costs associated with the project until an acceptable IRR
is obtained. This is a very important process since it forces the analyst to have a better handle on the
project, and, as a result, the project is more likely to perform as expected.
Page | 7-9
7.3.1 Incremental Analysis
There is an anomaly with simple IRR analysis. For projects with a short term and low investment, the IRR
may be very high. However, that project may not have the greatest present worth. This is because
present worth is an absolute dollar measurement while IRR is a relative measurement.
Incremental investment analysis is used to compare the difference in investments for projects.
Begin with a table for the number of years. In year zero, place the present worth or purchase price
under each project. In year one, place the future value or uniform amount under each project. Continue
for all remaining years.
Next create a column that is the difference in the cashflow between the two projects at each time
interval. This gives a sequence of incremental values.
n Project1 Project2 2 –1
0 P10 P20 P20 – P10
1 F11 F21 F21 – F11
2 F12 F22 F22 – F21
IRR1 IRR2 IRR2-1
Next calculate the rate of return on the differences. If the rate of return on the difference exceeds the
MARR, then the incremental investment should be made to do the project.
Page | 7-10
7.3.2 Payback
Payback is simply how long it takes for net receipts to equal investment outlays. Conventional payback is
calculated ignoring the time value of money.
Discounted payback includes the interest on the money. For a project with the same cash inflows each
year,
cost
Payback=uniform annual benefit
For projects with varying cash inflows, these inflows are added until the cost is equaled.
For payback analysis, all costs and all revenue prior to the payback are included without considering
differences in their timing. All consequences after the payback time are completely ignored.
This is obviously a simple technique. It also has some serious constraints. Nevertheless, it is still a
popular indicator since it gives the time to recover the investment. It is important to note again that the
simple payback ignores the time value of money, which may be very high for an expensive, or lengthy,
project.
Page | 7-11
7.3.3 Benefit Cost Ratio
Benefit is simply another way a describing the positive cash flow or value. Cost describes the negative
cash flow or value. When comparing projects, these are calculated using incremental analysis. The ratio
is the benefit to cost ratio. If the ratio is greater than 1, the incremental project should be done.
(benefits)
(costs)
PVBC
PV
Page | 7-12
7.3.4 Tax Implications
The taxes on income and investments will influence the value of the project. Calculate the value of the
project before income tax (BFIT). Financial types refer to this as EBITDA (Earnings Before Interest, Tax,
Depreciation and Amortization). Calculate the value of the project after income tax (AFIT). The process
can be used with all the techniques including present worth, incremental analysis IRR, and payback.
The value of the analysis is it reflects the real net worth, since different investments have alternative tax
implications. BFIT may be a better indicator of projects, but AFIT is the better indicator of value to the
organization. After all, the purpose of projects is not for the projects sake, but to increase the value to
the organization.
Page | 7-13
7.4 Table of Terminology
Numerous symbols are used for economic comparisons. These are shown in a table.
Symbol Definition I Interest rate per interest period, often one year N Number of interest periods PV Present value of money FV Future value of money A Amount of money at end of period in a uniform series for n periods G Arithmetic gradient, uniform period change in amount of money G Geometric gradient, uniform rate of cash flow change. R Nominal interest rate per interest period M Number of compounding subperiods S Salvage value is cash recoverable at the end of a project. PV*, FV* Amount of money flowing continuously during one period
7.5 Commentary
Determining which project is the best choice is less than straightforward. Many issues enter the
evaluation, besides the pure economic calculations.
There are biases and preferences from the management and the project team. These are difficult to
eliminate, and the biases may even be desirable. With practice, there can come that ‘gut feel’ that this is
right. However, gut feel can also be wrong.
Most organizations place some criteria on selecting projects. Traditionally over many years, an IRR of
15% AFIT has been acceptable, assuming the project was well defined. For higher risk ventures, the
MARR often moves toward 30%.
Other companies place great emphasis on payback. Traditionally, a 3-year term has been acceptable. As
risk increases, the term is frequently shortened toward 1 year.
With these more stringent criteria, one must wonder if any projects will ever be built. That is not the
point. The objective is to compensate for higher than normal risk. There is no future in investing money
in ventures that do not increase the wealth of the stakeholders.
7.6 Review
Page | 7-14
Money has different values, depending on the time it is obtained. The worth of the money is dependent
on present, future, and uniform amounts. There are multiple projects on which to invest money. The
preferred project is determined by incremental analysis that compares the projects. The methods are
investor rate of return, payback, and benefit to cost ratio. The return after taxes is a better indicator of
worth.
Page | 7-15
Chapter 7 Problems
1. Retirement savings. Given: Uninflated growth of capital is 3% per year. You begin paying into a retirement program when you turn 25. After25 years, you begin collecting $50,000 annually. You collect for 40 years. a. What is the present value of the annuity? b. What are your annual contributions to have that amount in the account? c. How much will be in the account when you begin collecting? d. Is the annual contribution a reasonable amount to expect you can contribute after you pay taxes
on a $50,000 salary? e. Wait 35 years to begin collection and receive for 30 years, do you contribute more or less each
month? f. To retire at 50, what is the best approach?
2. Auto Purchase. Given: A new university graduate purchases a new Whizmobile for $25,000. She borrows the money at an interest of 1%/mo. Loan period is 60 months. a. Find, future value, if a single balloon payment is made at 60 months. b. Monthly payments if made at the end of each month. c. Total amount paid (cash-flow) d. Total Interest paid e. Amount unpaid after 12 months a. Assume the same monthly payments were deposited in savings program. The account pays 1%
per month. How many months would it before you had $25,000?
3. Rule of 72: Given: $1000 to invest and the interest is compounded 10% per year. a. How long will it take to double your money? b. What is the effective monthly interest rate?
4. House Purchase. Given: Engineer with a salary of $40,000 per year. Federal and state income tax of 25%. Mortgage company will lend 2.5 times the annual salary for a house. Interest rate is 7% per year. Loan period is 30 years. Use uniform series of annual payments procedure.
Create a spreadsheet with the following columns. Calculate the values for the end of each year.
a. Principal remaining at start of year b. Interest owed for year c. Annual payment d. Amount of payment that is principal e. Total payments to date
Page | 7-16
f. Total interest to date g. Double the payment. h. Apply the extra amount to principal only. i. Tabulate the lower principal remaining. j. How long does it take to pay off the loan, with double payments?
Calculate the following items for a 15-year loan.
a. Total cash flow (amount paid). b. Total interest paid.
The next sections should be added to the spreadsheet after taxes.
c. Effect on income tax for interest paid d. Cash flow after taxes
Bibliography
Anthony Tarquin, Anthony Engineering Economy, McGraw-Hill Education, 1989
Excel 2003, Microsoft, Redmond, WA, 2003.
Page | 8-1
Chapter 8 Business Ethics Ethics and professional behavior are critical components for the practice of any profession. The
expected professional conduct is outlined in the NCEES Model Rules [1]. These are excerpted for review
and consideration and credit is acknowledged.
240.15 Rules of Professional Conduct A. Licensee’s Obligation to Society
1. Licensees, in the performance of their services for clients, employers, and customers, shall be
cognizant that their first and foremost responsibility is to the public welfare.
2. Licensees shall approve and seal only those design documents and surveys that conform to accepted
engineering and surveying standards and safeguard the life, health, property, and welfare of the public.
3. Licensees shall notify their employer or client and such other authority as may be appropriate when
their professional judgment is overruled under circumstances where the life, health, property, or
welfare of the public is endangered.
4. Licensees shall be objective and truthful in professional reports, statements, or testimony. They shall
include all relevant and pertinent information in such reports, statements, or testimony.
5. Licensees shall express a professional opinion publicly only when it is founded upon an adequate
knowledge of the facts and a competent evaluation of the subject matter.
6. Licensees shall issue no statements, criticisms, or arguments on technical matters which are inspired
or paid for by interested parties, unless they explicitly identify the interested parties on whose behalf
they are speaking and reveal any interest they have in the matters.
7. Licensees shall not permit the use of their name or firm name by, nor associate in the business
ventures with, any person or firm which is engaging in fraudulent or dishonest business or professional
practices.
8. Licensees having knowledge of possible violations of any of these Rules of Professional Conduct shall
provide the board with the information and assistance necessary to make the final determination of
such violation. (Section 150, Disciplinary Action, NCEES Model Law)
B. Licensee’s Obligation to Employer and Clients
1. Licensees shall undertake assignments only when qualified by education or experience in the specific
technical fields of engineering or surveying involved.
2. Licensees shall not affix their signatures or seals to any plans or documents dealing
with subject matter in which they lack competence, nor to any such plan or document not prepared
under their direct control and personal supervision.
3. Licensees may accept assignments for coordination of an entire project, provided that each design
segment is signed and sealed by the licensee responsible for preparation of that design segment.
Page | 8-2
4. Licensees shall not reveal facts, data, or information obtained in a professional capacity without the
prior consent of the client or employer except as authorized or required by law. Licensees shall not
solicit or accept gratuities, directly or indirectly, from contractors, their agents, or other parties in
connection with work for employers or clients.
5. Licensees shall make full prior disclosures to their employers or clients of potential conflicts of
interest or other circumstances which could influence or appear to influence their judgment or the
quality of their service.
6. Licensees shall not accept compensation, financial or otherwise, from more than one party for
services pertaining to the same project, unless the circumstances are fully disclosed and agreed to by all
interested parties.
7. Licensees shall not solicit or accept a professional contract from a governmental body on which a
principal or officer of their organization serves as a member. Conversely, licensees serving as members,
advisors, or employees of a government body or department, who are the principals or employees of a
private concern, shall not participate in decisions with respect to professional services offered or
provided by said concern to the governmental body which they serve. (Section 150, Disciplinary Action,
NCEES Model Law)
C. Licensee’s Obligation to Other Licensees
1. Licensees shall not falsify or permit misrepresentation of their, or their associates’, academic or
professional qualifications. They shall not misrepresent or exaggerate their degree of responsibility in
prior assignments nor the complexity of said assignments. Presentations incident to the solicitation of
employment or business shall not misrepresent pertinent facts concerning employers, employees,
associates, joint ventures, or past accomplishments.
2. Licensees shall not offer, give, solicit, or receive, either directly or indirectly, any commission, or gift,
or other valuable consideration in order to secure work, and shall not make any political contribution
with the intent to influence the award of a contract by public authority.
3. Licensees shall not attempt to injure, maliciously or falsely, directly or indirectly, the professional
reputation, prospects, practice, or employment of other licensees, nor indiscriminately criticize other
licensees’ work. (Section 150, Disciplinary Action, NCEES Model Law)
Model Rules, National Council of Examiners for Engineering and Surveying, Clemson, SC,
http://www.ncees.org/introduction/about_ncees/ncees_model_rules.pdf, 2005
Page | 9-1
Chapter 9 Codes and Standards
9.1 Introduction
There are minimum performance requirements that should be incorporated in any design. The actual
installation should meet or exceed these.
Standards are consensus documents typically developed by professional and technical societies. These
typically become an American National Standards Institute (ANSI) standard.
Codes are standards that have been incorporated as law.
Industry practices are common techniques used by knowledgeable individuals.
9.2 NEC Synopsis The National Electrical Code (NEC) is the standard for practical safeguarding of persons and equipment.
The areas that involve calculations are the components that are most frequently on the professional
exams. Articles are used as broad categories from NEC 2005. The specific section & table numbers may
change between versions of the NEC.
General: Article 100
Definitions, workmanship, and responsibility are discussed in this section.
Identification of grounded conductors: Article 200
Grounding is bare, green, or green with yellow stripe.
Grounded, or neutral, is white or gray.
High-leg or wild-leg of a three-phase delta system with a center-tapped winding that is grounded must be orange. (Article 110)
Branch circuits: Article 210
Branch circuits are everything but motors.
General lighting can be on 15 A breakers.
The minimum number of 20 A branch circuits in a residence is 2 for small appliances, 1 for bath, and 1 for laundry.
Ground-fault protection is required in bathrooms, kitchens, garages, outside, and any other place that contact with a ground is possible.
Arc-fault breakers are required for bedrooms.
Feeders: Article 215
Ampacity = (1.25 * largest Full Load Amps) + sum of all other current loads.
Page | 9-2
Branch-circuit, feeder, and service calculations: Article 220
Demand factors and calculation requirements are covered.
Start in Part III and step through the applicable sections.
Overcurrent protection: Article 240
Overcurrent protection for specific equipment: Table 240.3
Protection of conductors shall be in accordance with ampacity from Table 310.16 ff
Grounding: Article 250
Sizing of grounding electrode conductor: Table 250.66
Sizing of equipment grounding conductor: Table 250.122
The maximum resistance of a grounding electrode according to Art 250.56: 25 Ω,
Conductor to be grounded - neutral or common: Art 250.26
Conductors for general wiring: Table 310.16 ff
Maximum capacity is shown in table at 30C
* AWG 14, 12, & 10 have maximum rating based on Art 240.4D, which is 5 A less than Table 310.16
Ampacity is rerated for different temperatures by the correction factors.
NM cable (Romex) is sized as 60C, regardless of insulation type
Alternate ampacity can be used under engineering supervision using Art 310.60D
Motors: Article 430
Motor full load current rating: Table 430.247 ff
Short circuit current = 6* full load current
Overload current = 1.5 * full load current
Maximum setting of motor branch circuit protection & ground fault protection: Table 430.52
Maximum locked-rotor current for selection of disconnecting means: Table 430.251
Locked rotor code letters for KVA per horsepower: Table 430.7
Environmental selection of motor controller enclosures: Table 430.91
Capacitor added to motor terminal: reduces total current from overload, so reduce size of overload only
M
Tables: Chapter 9
Conduit dimensions and fill
Insulated conductors nominal dimensions for AWG, diameter, area
Page | 9-3
Conductor properties including AWG, area, Ohms/kft
Cable resistance and reactance in conduit
Power source limitations
Examples: Annex D
Single family dwelling with appliance, heating, air conditioning
Store building
Industrial feeders
Multifamily dwelling with demand factors on single-phase and three-phase
Generator field control
Mobile home
Page | 9-4
9.3 Motor Installation Tables This section lists excerpts from industry, national, and international standards. The excerpts are for
illustration and educational purposes. There are often several related tables and information for other
configurations and applications. In addition, the standards have added detail and information that
applies to all these excerpts. Therefore, any application should refer to the standard, rather than the
excerpts.
Standard Organization Application
National Electrical Code (NEC) NFPA 70-2005
National Fire Protection Association, Batterymarch Park, Quincy, MA
Electrical installations in occupancies
National Electrical Safety Code (NESC) IEEE C2-2002
Institute of Electrical & Electronics Engineers, New York, NewYork
Electrical supply stations, overhead, and underground lines.
Motors and Generators (MG1) NEMA MG1-2003
National Electrical Manufacturers Association
Performance of motors and generators
NEC 240.6(A) Standard Ampere Ratings for Fuses & Circuit Breakers
Amperes
15 20 25 30 35 40 45 50 60 70 80 90 100 110 125 150 175 200 225 250 300 350 400 450 500 - - - 600 700 800 1000 1200 1600 2000 2500 3000 4000 5000 6000 - -
Page | 9-5
Table 310.16 Allowable Ampacities of Insulated Conductors Rated 0 Through 2000 Volts, 60°C
Through 90°C (140°F Through 194°F), Not More Than Three Current-Carrying Conductors in Raceway,
Cable, or Earth (Directly Buried), Based on Ambient Temperature of 30°C (86°F)
Temperature Rating of Conductor (See Table 310.13.)
60°C (140°F)
75°C (167°F)
90°C (194°F)
60°C (140°F)
75°C (167°F)
90°C (194°F)
Size AWG or kcmil
Types TW, UF
Types RHW, THHW, THW,
THWN, XHHW, USE, ZW
Types TBS, SA, SIS, FEP, FEPB, MI, RHH, RHW-2, THHN, THHW, THW-2, THWN-2, USE-2, XHH,
XHHW, XHHW-2, ZW-2
Types TW, UF
Types RHW, THHW, THW,
THWN, XHHW, USE
Types TBS, SA, SIS, THHN, THHW, THW-2, THWN-2, RHH, RHW-2,
USE-2, XHH, XHHW, XHHW-2, ZW-2
Size AWG or kcmil
COPPER ALUMINUM OR COPPER-CLAD ALUM
18 — — 14 — — — — 16 — — 18 — — — —
14* 20 20 25 — — — — 12* 25 25 30 20 20 25 12* 10* 30 35 40 25 30 35 10*
8 40 50 55 30 40 45 8
6 55 65 75 40 50 60 6 4 70 85 95 55 65 75 4 3 85 100 110 65 75 85 3 2 95 115 130 75 90 100 2 1 110 130 150 85 100 115 1
1/0 125 150 170 100 120 135 1/0 2/0 145 175 195 115 135 150 2/0 3/0 165 200 225 130 155 175 3/0 4/0 195 230 260 150 180 205 4/0
250 215 255 290 170 205 230 250 300 240 285 320 190 230 255 300 350 260 310 350 210 250 280 350 400 280 335 380 225 270 305 400 500 320 380 430 260 310 350 500
600 355 420 475 285 340 385 600 700 385 460 520 310 375 420 700 750 400 475 535 320 385 435 750 800 410 490 555 330 395 450 800 900 435 520 585 355 425 480 900
1000 455 545 615 375 445 500 1000 1250 495 590 665 405 485 545 1250 1500 520 625 705 435 520 585 1500 1750 545 650 735 455 545 615 1750 2000 560 665 750 470 560 630 2000
CORRECTION FACTORS
Ambient Temp. (°C)
For ambient temperatures other than 30°C (86°F), multiply the allowable ampacities shown above by the appropriate factor shown below.
Ambient Temp. (°F)
21–25 1.08 1.05 1.04 1.08 1.05 1.04 70–77
26–30 1.00 1.00 1.00 1.00 1.00 1.00 78–86
31–35 0.91 0.94 0.96 0.91 0.94 0.96 87–95
36–40 0.82 0.88 0.91 0.82 0.88 0.91 96–104
41–45 0.71 0.82 0.87 0.71 0.82 0.87 105–113
46–50 0.58 0.75 0.82 0.58 0.75 0.82 114–122
51–55 0.41 0.67 0.76 0.41 0.67 0.76 123–131
56–60 — 0.58 0.71 — 0.58 0.71 132–140
61–70 — 0.33 0.58 — 0.33 0.58 141–158
71–80 — — 0.41 — — 0.41 159–176
* See 240.4(D).
Page | 9-6
Table 310.15(B)(2)(a) Adjustment
Factors for More Than Three Current-
Carrying Conductors in a Raceway or
Cable
Number of
Current-
Carrying
Conductors
Percent of Values in Tables
310.16 through 310.19 as
Adjusted for Ambient
Temperature if Necessary
4–6 80
7–9 70
10–20 50
21–30 45
31–40 40
41 and above 35
Table 310.15(B)(6) Conductor Types and Sizes
for 120/240-Volt, 3-Wire, Single-Phase Dwelling
Services and Feeders.
Conductor Types RHH, RHW, RHW-2, THHN, THHW, THW,
THW-2, THWN, THWN-2, XHHW, XHHW-2, SE, USE, USE-2
Conductor (AWG or kcmil)
Copper Aluminum or Copper-Clad Aluminum
Service or Feeder Rating (Amperes)
4 2 100 3 1 110 2 1/0 125 1 2/0 150
1/0 3/0 175 2/0 4/0 200 3/0 250 225 4/0 300 250 250 350 300 350 500 350 400 600 400
Page | 9-7
Figure 430.1 Article 430 Contents
Page | 9-8
Table 430.7(B) Locked-Rotor Indicating Code Letters
Code Letter
Kilovolt-Amperes per Horsepower
with Locked Rotor
A 0–3.14 B 3.15–3.54 C 3.55–3.99 D 4.0–4.49 E 4.5–4.99 F 5.0–5.59 G 5.6–6.29 H 6.3–7.09 J 7.1–7.99 K 8.0–8.99
Code Letter
Kilovolt-Amperes per Horsepower
with Locked Rotor
L 9.0–9.99 M 10.0–11.19 N 11.2–12.49 P 12.5–13.99 R 14.0–15.99 S 16.0–17.99 T 18.0–19.99 U 20.0–22.39 V 22.4 and up
Table 430.52 Maximum Rating or Setting of Motor Branch-Circuit Short-
Circuit and Ground-Fault Protective Devices
Percentage of Full-Load Current Type of Motor Non-time
Delay Fuse Dual Element (Time-Delay) Fuse
Instantaneous Trip Breaker
Inverse Time Breaker
Single-phase motors
300 175 800 250
AC polyphase motors other than wound-rotor Squirrel cage other than Design B energy-efficient
300 175 800 250
Design B energy-efficient
300 175 1100 250
Synchronous 300 175 800 250 Wound rotor 150 150 800 150 Direct current (constant voltage)
150 150 250 150
Page | 9-9
Table 430.91 Motor Controller Enclosure Selection
For Outdoor Use
Enclosure Type Number1
Provides a Degree of Protection Against the Following Environmental Conditions
3 3R 3S 3X 3RX 3SX 4 4X 6 6P
Incidental contact with the enclosed equipment
X X X X X X X X X X
Rain, snow, and sleet X X X X X X X X X X
Sleet2 — — X — — X — — — —
Windblown dust X — X X — X X X X X
Hosedown — — — — — — X X X X
Corrosive agents — — — X X X — X — X
Temporary submersion — — — — — — — — X X
Prolonged submersion — — — — — — — — — X
For Indoor Use
Enclosure Type Number1
Provides a Degree of Protection Against the Following Environmental Conditions
1 2 4 4X 5 6 6P 12 12K 13
Incidental contact with the enclosed equipment
X X X X X X X X X X
Falling dirt X X X X X X X X X X
Falling liquids and light splashing
— X X X X X X X X X
Circulating dust, lint, fibers, and flyings
— — X X — X X X X X
Settling airborne dust, lint, fibers, and flyings
— — X X X X X X X X
Hosedown and splashing water
— — X X — X X — — —
Oil and coolant seepage — — — — — — — X X X
Oil or coolant spraying and splashing
— — — — — — — — — X
Corrosive agents — — — X — — X — — —
Temporary submersion — — — — — X X — — —
Prolonged submersion — — — — — — X — — —
Page | 9-10
1Enclosure type number shall be marked on the motor controller enclosure. 2Mechanism shall be operable when ice covered. FPN: The term raintight is typically used in conjunction with Enclosure Types 3, 3S, 3SX, 3X, 4, 4X, 6, 6P. The term rainproof is typically used in conjunction with Enclosure Type 3R, 3RX. The term watertight is typically used in conjunction with Enclosure Types 4, 4X, 6, 6P. The term driptight is typically used in conjunction with Enclosure Types 2, 5, 12, 12K, 13. The term dusttight is typically used in conjunction with Enclosure Types 3, 3S, 3SX, 3X, 5, 12, 12K, 13.
Page | 9-11
Table 430.248 Full-Load
Currents in Amperes, Single-Phase
Alternating-Current Motors
Horse-power
115 Volts
200 Volts
208 Volts
230 Volts
4.4 2.5 2.4 2.2
¼ 5.8 3.3 3.2 2.9
7.2 4.1 4.0 3.6
½ 9.8 5.6 5.4 4.9 ¾ 13.8 7.9 7.6 6.9 1 16 9.2 8.8 8.0 1½ 20 11.5 11.0 10 2 24 13.8 13.2 12 3 34 19.6 18.7 17 5 56 32.2 30.8 28 7½ 80 46.0 44.0 40 10 100 57.5 55.0 50
Page | 9-12
Table 430.250 Full-Load Current, Three-Phase Alternating-Current Motors
Induction-Type Squirrel Cage and Wound Rotor (Amperes)
Synchronous-Type Unity Power Factor* (Amperes)
Horse-power
115 Volts
200 Volts
208 Volts
230 Volts
460 Volts
575 Volts 2300 Volts
230 Volts
460 Volts
575 Volts
2300 Volts
½ 4.4 2.5 2.4 2.2 1.1 0.9 — — — — — ¾ 6.4 3.7 3.5 3.2 1.6 1.3 — — — — — 1 8.4 4.8 4.6 4.2 2.1 1.7 — — — — — 1½ 12.0 6.9 6.6 6.0 3.0 2.4 — — — — — 2 13.6 7.8 7.5 6.8 3.4 2.7 — — — — —
3 — 11.0 10.6 9.6 4.8 3.9 — — — — — 5 — 17.5 16.7 15.2 7.6 6.1 — — — — — 7½ — 25.3 24.2 22 11 9 — — — — — 10 — 32.2 30.8 28 14 11 — — — — — 15 — 48.3 46.2 42 21 17 — — — — —
20 — 62.1 59.4 54 27 22 — — — — — 25 — 78.2 74.8 68 34 27 — 53 26 21 — 30 — 92 88 80 40 32 — 63 32 26 — 40 — 120 114 104 52 41 — 83 41 33 — 50 — 150 143 130 65 52 — 104 52 42 —
60 — 177 169 154 77 62 16 123 61 49 12 75 — 221 211 192 96 77 20 155 78 62 15 100 — 285 273 248 124 99 26 202 101 81 20 125 — 359 343 312 156 125 31 253 126 101 25
150 — 414 396 360 180 144 37 302 151 121 30 200 552 528 480 240 192 49 400 201 161 40 250 — — — — 302 242 60 — — — — 300 — — — — 361 289 72 — — — — 350 — — — — 414 336 83 — — — —
400 — — — — 477 382 95 — — — — 450 — — — — 515 412 103 — — — — 500 — — — — 590 472 118 — — — —
*For 90 and 80 percent power factor, the figures shall be multiplied by 1.1 and 1.25, respectively.
Page | 9-13
Table 430.251(A) Conversion Table of Single-Phase
Locked- Rotor Currents for Selection of Disconnecting
Means and Controllers as Determined from Horsepower
and Voltage Rating
Rated Horsepower
Maximum Locked-Rotor Current in Amperes, Single Phase
115 Volts 208 Volts 230 Volts
½ 58.8 32.5 29.4 ¾ 82.8 45.8 41.4 1 96 53 48 1 ½ 120 66 60 2 144 80 72 3 204 113 102 5 336 186 168 7 ½ 480 265 240 10 600 332 300
Page | 9-14
Table 430.251(B) Conversion Table of Polyphase Design B, C, and D Maximum Locked-Rotor
Currents for Selection of Disconnecting Means and Controllers as Determined from Horsepower
and Voltage Rating and Design Letter
Rated Maximum Motor Locked-Rotor Current in Amperes, Two- and Three-Phase, Design B, C, and D*
Horsepower 115 Volts 200 Volts 208 Volts 230 Volts 460 Volts 575 Volts
B, C, D B, C, D B, C, D B, C, D B, C, D B, C, D
½ 40 23 22.1 20 10 8 ¾ 50 28.8 27.6 25 12.5 10 1 60 34.5 33 30 15 12 1½ 80 46 44 40 20 16 2 100 57.5 55 50 25 20
3 — 73.6 71 64 32 25.6 5 — 105.8 102 92 46 36.8 7½ — 146 140 127 63.5 50.8 10 — 186.3 179 162 81 64.8 15 — 267 257 232 116 93
20 — 334 321 290 145 116 25 — 420 404 365 183 146 30 — 500 481 435 218 174 40 — 667 641 580 290 232 50 — 834 802 725 363 290
60 — 1001 962 870 435 348 75 — 1248 1200 1085 543 434 100 — 1668 1603 1450 725 580 125 — 2087 2007 1815 908 726 150 — 2496 2400 2170 1085 868
200 — 3335 3207 2900 1450 1160 250 — — — — 1825 1460 300 — — — — 2200 1760 350 — — — — 2550 2040 400 — — — — 2900 2320
450 — — — — 3250 2600 500 — — — — 3625 2900
*Design A motors are not limited to a maximum starting current or locked rotor current.
These tables for use only with 430.110, 440.12, 440.41 and 455.8(C).
Page | 9-15
Table 8 Conductor Properties Conductors Direct-Current Resistance at 75°C (167°F)
Size Stranding Overall Copper Aluminum
(AWG Area QTY Diameter Diameter Area Uncoated Coated
or kcmil)
mm2 Circular mils
mm in. mm in. mm2 in.2 ohm/ km
ohm/ kFT
ohm/ km
ohm/ kFT
ohm/ km
ohm/ kFT
18 0.823 1620 1 — — 1.02 0.040 0.823 0.001 25.5 7.77 26.5 8.08 42.0 12.8 18 0.823 1620 7 0.39 0.015 1.16 0.046 1.06 0.002 26.1 7.95 27.7 8.45 42.8 13.1
16 1.31 2580 1 — — 1.29 0.051 1.31 0.002 16.0 4.89 16.7 5.08 26.4 8.05 16 1.31 2580 7 0.49 0.019 1.46 0.058 1.68 0.003 16.4 4.99 17.3 5.29 26.9 8.21
14 2.08 4110 1 — — 1.63 0.064 2.08 0.003 10.1 3.07 10.4 3.19 16.6 5.06 14 2.08 4110 7 0.62 0.024 1.85 0.073 2.68 0.004 10.3 3.14 10.7 3.26 16.9 5.17
12 3.31 6530 1 — — 2.05 0.081 3.31 0.005 6.34 1.93 6.57 2.01 10.45 3.18 12 3.31 6530 7 0.78 0.030 2.32 0.092 4.25 0.006 6.50 1.98 6.73 2.05 10.69 3.25
10 5.261 10380 1 — — 2.588 0.102 5.26 0.008 3.984 1.21 4.148 1.26 6.561 2.00 10 5.261 10380 7 0.98 0.038 2.95 0.116 6.76 0.011 4.070 1.24 4.226 1.29 6.679 2.04
8 8.367 16510 1 — — 3.264 0.128 8.37 0.013 2.506 0.764 2.579 0.786 4.125 1.26 8 8.367 16510 7 1.23 0.049 3.71 0.146 10.76 0.017 2.551 0.778 2.653 0.809 4.204 1.28
6 13.30 26240 7 1.56 0.061 4.67 0.184 17.09 0.027 1.608 0.491 1.671 0.510 2.652 0.808 4 21.15 41740 7 1.96 0.077 5.89 0.232 27.19 0.042 1.010 0.308 1.053 0.321 1.666 0.508 3 26.67 52620 7 2.20 0.087 6.60 0.260 34.28 0.053 0.802 0.245 0.833 0.254 1.320 0.403 2 33.62 66360 7 2.47 0.097 7.42 0.292 43.23 0.067 0.634 0.194 0.661 0.201 1.045 0.319
1 42.41 83690 19 1.69 0.066 8.43 0.332 55.80 0.087 0.505 0.154 0.524 0.160 0.829 0.253 1/0 53.49 105600 19 1.89 0.074 9.45 0.372 70.41 0.109 0.399 0.122 0.415 0.127 0.660 0.201 2/0 67.43 133100 19 2.13 0.084 10.62 0.418 88.74 0.137 0.3170 0.0967 0.329 0.101 0.523 0.159 3/0 85.01 167800 19 2.39 0.094 11.94 0.470 111.9 0.173 0.2512 0.0766 0.2610 0.0797 0.413 0.126 4/0 107.2 211600 19 2.68 0.106 13.41 0.528 141.1 0.219 0.1996 0.0608 0.2050 0.0626 0.328 0.100
250 127 — 37 2.09 0.082 14.61 0.575 168 0.260 0.1687 0.0515 0.1753 0.0535 0.2778 0.0847 300 152 — 37 2.29 0.090 16.00 0.630 201 0.312 0.1409 0.0429 0.1463 0.0446 0.2318 0.0707 350 177 — 37 2.47 0.097 17.30 0.681 235 0.364 0.1205 0.0367 0.1252 0.0382 0.1984 0.0605 400 203 — 37 2.64 0.104 18.49 0.728 268 0.416 0.1053 0.0321 0.1084 0.0331 0.1737 0.0529 500 253 — 37 2.95 0.116 20.65 0.813 336 0.519 0.0845 0.0258 0.0869 0.0265 0.1391 0.0424
600 304 — 61 2.52 0.099 22.68 0.893 404 0.626 0.0704 0.0214 0.0732 0.0223 0.1159 0.0353 700 355 — 61 2.72 0.107 24.49 0.964 471 0.730 0.0603 0.0184 0.0622 0.0189 0.0994 0.0303 750 380 — 61 2.82 0.111 25.35 0.998 505 0.782 0.0563 0.0171 0.0579 0.0176 0.0927 0.0282 800 405 — 61 2.91 0.114 26.16 1.030 538 0.834 0.0528 0.0161 0.0544 0.0166 0.0868 0.0265 900 456 — 61 3.09 0.122 27.79 1.094 606 0.940 0.0470 0.0143 0.0481 0.0147 0.0770 0.0235 1000 507 — 61 3.25 0.128 29.26 1.152 673 1.042 0.0423 0.0129 0.0434 0.0132 0.0695 0.0212
1250 633 — 91 2.98 0.117 32.74 1.289 842 1.305 0.0338 0.0103 0.0347 0.0106 0.0554 0.0169 1500 760 — 91 3.26 0.128 35.86 1.412 1011 1.566 0.02814 0.00858 0.02814 0.00883 0.0464 0.0141 1750 887 — 127 2.98 0.117 38.76 1.526 1180 1.829 0.02410 0.00735 0.02410 0.00756 0.0397 0.0121 2000 1013 — 127 3.19 0.126 41.45 1.632 1349 2.092 0.02109 0.00643 0.02109 0.00662 0.0348 0.0106
Notes: 1. These resistance values are valid only for the parameters as given. Using conductors having coated strands, different stranding type, and, especially, other temperatures changes the resistance. 2. Formula for temperature change: R2 = R1 [1 + α (T 2 - 75)] where α cu = 0.00323, α AL = 0.00330 at 75°C. 3. Conductors with compact and compressed stranding have about 9 percent and 3 percent, respectively, smaller bare conductor diameters than those shown. See Table 5A for actual compact cable dimensions. 4. The IACS conductivities used: bare copper = 100%, aluminum = 61%. 5. Class B stranding is listed as well as solid for some sizes. Its overall diameter and area is that of its circumscribing circle.
Page | 9-16
Table 9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F)
— Three Single Conductors in Conduit
Ohms to Neutral per Kilometer Ohms to Neutral per 1000 Feet6
Size
XL (Reactance) for All Wires
Alternating-Current Resistance for Uncoated Copper Wires
Alternating-Current Resistance for Aluminum Wires
Effective Z at 0.85 PF for Uncoated Copper Wires
Effective Z at 0.85 PF for Aluminum Wires
Size
(AWG or kcmil)
PVC, Aluminum Conduits
Steel Conduit
PVC Conduit
Aluminum Conduit
Steel Conduit
PVC Conduit
Aluminum Conduit
Steel Conduit
PVC Conduit
Aluminum Conduit
Steel Conduit
PVC Conduit
Aluminum Conduit
Steel Conduit
(AWG or kcmil)
14 0.190 0.240 10.2 10.2 10.2 — — — 8.9 8.9 8.9 — — — 14
0.058 0.073 3.1 3.1 3.1 — — — 2.7 2.7 2.7 — — —
12 0.177 0.223 6.6 6.6 6.6 10.5 10.5 10.5 5.6 5.6 5.6 9.2 9.2 9.2 12
0.054 0.068 2.0 2.0 2.0 3.2 3.2 3.2 1.7 1.7 1.7 2.8 2.8 2.8
10 0.164 0.207 3.9 3.9 3.9 6.6 6.6 6.6 3.6 3.6 3.6 5.9 5.9 5.9 10
0.050 0.063 1.2 1.2 1.2 2.0 2.0 2.0 1.1 1.1 1.1 1.8 1.8 1.8
8 0.171 0.213 2.56 2.56 2.56 4.3 4.3 4.3 2.26 2.26 2.30 3.6 3.6 3.6 8
0.052 0.065 0.78 0.78 0.78 1.3 1.3 1.3 0.69 0.69 0.70 1.1 1.1 1.1
6 0.167 0.210 1.61 1.61 1.61 2.66 2.66 2.66 1.44 1.48 1.48 2.33 2.36 2.36 6
0.051 0.064 0.49 0.49 0.49 0.81 0.81 0.81 0.44 0.45 0.45 0.71 0.72 0.72
4 0.157 0.197 1.02 1.02 1.02 1.67 1.67 1.67 0.95 0.95 0.98 1.51 1.51 1.51 4
0.048 0.060 0.31 0.31 0.31 0.51 0.51 0.51 0.29 0.29 0.30 0.46 0.46 0.46
3 0.154 0.194 0.82 0.82 0.82 1.31 1.35 1.31 0.75 0.79 0.79 1.21 1.21 1.21 3
0.047 0.059 0.25 0.25 0.25 0.40 0.41 0.40 0.23 0.24 0.24 0.37 0.37 0.37
2 0.148 0.187 0.62 0.66 0.66 1.05 1.05 1.05 0.62 0.62 0.66 0.98 0.98 0.98 2
0.045 0.057 0.19 0.20 0.20 0.32 0.32 0.32 0.19 0.19 0.20 0.30 0.30 0.30
1 0.151 0.187 0.49 0.52 0.52 0.82 0.85 0.82 0.52 0.52 0.52 0.79 0.79 0.82 1
0.046 0.057 0.15 0.16 0.16 0.25 0.26 0.25 0.16 0.16 0.16 0.24 0.24 0.25
1/0 0.144 0.180 0.39 0.43 0.39 0.66 0.69 0.66 0.43 0.43 0.43 0.62 0.66 0.66 1/0
0.044 0.055 0.12 0.13 0.12 0.20 0.21 0.20 0.13 0.13 0.13 0.19 0.20 0.20
2/0 0.141 0.177 0.33 0.33 0.33 0.52 0.52 0.52 0.36 0.36 0.36 0.52 0.52 0.52 2/0
0.043 0.054 0.10 0.10 0.10 0.16 0.16 0.16 0.11 0.11 0.11 0.16 0.16 0.16
3/0 0.138 0.171 0.253 0.269 0.259 0.43 0.43 0.43 0.289 0.302 0.308 0.43 0.43 0.46 3/0
0.042 0.052 0.077 0.082 0.079 0.13 0.13 0.13 0.088 0.092 0.094 0.13 0.13 0.14
4/0 0.135 0.167 0.203 0.220 0.207 0.33 0.36 0.33 0.243 0.256 0.262 0.36 0.36 0.36 4/0
0.041 0.051 0.062 0.067 0.063 0.10 0.11 0.10 0.074 0.078 0.080 0.11 0.11 0.11
250 0.135 0.171 0.171 0.187 0.177 0.279 0.295 0.282 0.217 0.230 0.240 0.308 0.322 0.33 250
0.041 0.052 0.052 0.057 0.054 0.085 0.090 0.086 0.066 0.070 0.073 0.094 0.098 0.10
300 0.135 0.167 0.144 0.161 0.148 0.233 0.249 0.236 0.194 0.207 0.213 0.269 0.282 0.289 300
0.041 0.051 0.044 0.049 0.045 0.071 0.076 0.072 0.059 0.063 0.065 0.082 0.086 0.088
350 0.131 0.164 0.125 0.141 0.128 0.200 0.217 0.207 0.174 0.190 0.197 0.240 0.253 0.262 350
0.040 0.050 0.038 0.043 0.039 0.061 0.066 0.063 0.053 0.058 0.060 0.073 0.077 0.080
400 0.131 0.161 0.108 0.125 0.115 0.177 0.194 0.180 0.161 0.174 0.184 0.217 0.233 0.240 400
0.040 0.049 0.033 0.038 0.035 0.054 0.059 0.055 0.049 0.053 0.056 0.066 0.071 0.073
500 0.128 0.157 0.089 0.105 0.095 0.141 0.157 0.148 0.141 0.157 0.164 0.187 0.200 0.210 500
0.039 0.048 0.027 0.032 0.029 0.043 0.048 0.045 0.043 0.048 0.050 0.057 0.061 0.064
600 0.128 0.157 0.075 0.092 0.082 0.118 0.135 0.125 0.131 0.144 0.154 0.167 0.180 0.190 600
0.039 0.048 0.023 0.028 0.025 0.036 0.041 0.038 0.040 0.044 0.047 0.051 0.055 0.058
750 0.125 0.157 0.062 0.079 0.069 0.095 0.112 0.102 0.118 0.131 0.141 0.148 0.161 0.171 750
0.038 0.048 0.019 0.024 0.021 0.029 0.034 0.031 0.036 0.040 0.043 0.045 0.049 0.052
1000 0.121 0.151 0.049 0.062 0.059 0.075 0.089 0.082 0.105 0.118 0.131 0.128 0.138 0.151 1000
0.037 0.046 0.015 0.019 0.018 0.023 0.027 0.025 0.032 0.036 0.040 0.039 0.042 0.046
Page | 9-17
Notes: 1. These values are based on the following constants: UL-Type RHH wires with Class B stranding, in cradled configuration. Wire conductivities are 100 percent IACS copper and 61 percent IACS aluminum, and aluminum conduit is 45 percent IACS. Capacitive reactance is ignored, since it is negligible at these voltages. These resistance values are valid only at 75°C (167°F) and for the parameters as given, but are representative for 600-volt wire types operating at 60 Hz. 2. Effective Z is defined as R cos(θ) + X sin(θ), where θ is the power factor angle of the circuit. Multiplying current by effective impedance gives a good approximation for line-to-neutral voltage drop. Effective impedance values shown in this table are valid only at 0.85 power factor. For another circuit power factor (PF), effective impedance (Ze) can be calculated from R and XL values given in this table as follows: Ze = R × PF + XL sin[arccos(PF)].
Page | 9-18
Table C.4 Maximum Number of Conductors or Fixture Wires in Intermediate Metal Conduit
(IMC) (Based on Table 1, Chapter 9)
CONDUCTORS
Size Metric Designator (Trade Size)
Type (AWG/ kcmil) 16 (½)
21 (¾)
27 (1)
35 (1¼)
41 (1½)
53 (2)
63 (2½)
78 (3)
91 (3½)
103 (4)
RHH, RHW, RHW-2
14 4 8 13 22 30 49 70 108 144 186 12 4 6 11 18 25 41 58 89 120 154
10 3 5 8 15 20 33 47 72 97 124 8 1 3 4 8 10 17 24 38 50 65 6 1 1 3 6 8 14 19 30 40 52
4 1 1 3 5 6 11 15 23 31 41 3 1 1 2 4 6 9 13 21 28 36 2 1 1 1 3 5 8 11 18 24 31 1 0 1 1 2 3 5 7 12 16 20
1/0 0 1 1 1 3 4 6 10 14 18 2/0 0 1 1 1 2 4 6 9 12 15 3/0 0 0 1 1 1 3 5 7 10 13 4/0 0 0 1 1 1 3 4 6 9 11
250 0 0 1 1 1 1 3 5 6 8 300 0 0 0 1 1 1 3 4 6 7 350 0 0 0 1 1 1 2 4 5 7 400 0 0 0 1 1 1 2 3 5 6 500 0 0 0 1 1 1 1 3 4 5
600 0 0 0 0 1 1 1 2 3 4 700 0 0 0 0 1 1 1 2 3 4 750 0 0 0 0 1 1 1 1 3 4 800 0 0 0 0 0 1 1 1 3 3 900 0 0 0 0 0 1 1 1 2 3
1000 0 0 0 0 0 1 1 1 2 3 1250 0 0 0 0 0 1 1 1 1 2 1500 0 0 0 0 0 0 1 1 1 1 1750 0 0 0 0 0 0 1 1 1 1 2000 0 0 0 0 0 0 1 1 1 1
TW 14 10 17 27 47 64 104 147 228 304 392 12 7 13 21 36 49 80 113 175 234 301 10 5 9 15 27 36 59 84 130 174 224 8 3 5 8 15 20 33 47 72 97 124
RHH*, RHW*, RHW-2*, THHW, THW, THW-2
14 6 11 18 31 42 69 98 151 202 261 12 5 9 14 25 34 56 79 122 163 209 10 4 7 11 19 26 43 61 95 127 163 8 2 4 7 12 16 26 37 57 76 98
6 1 3 5 9 12 20 28 43 58 75
4 1 2 4 6 9 15 21 32 43 56 3 1 1 3 6 8 13 18 28 37 48 2 1 1 3 5 6 11 15 23 31 41 1 1 1 1 3 4 7 11 16 22 28
1/0 1 1 1 3 4 6 9 14 19 24 2/0 0 1 1 2 3 5 8 12 16 20 3/0 0 1 1 1 3 4 6 10 13 17 4/0 0 1 1 1 2 4 5 8 11 14
250 0 0 1 1 1 3 4 7 9 12 300 0 0 1 1 1 2 4 6 8 10 350 0 0 1 1 1 2 3 5 7 9 400 0 0 0 1 1 1 3 4 6 8 500 0 0 0 1 1 1 2 4 5 7
600 0 0 0 1 1 1 1 3 4 5 700 0 0 0 0 1 1 1 3 4 5 750 0 0 0 0 1 1 1 2 3 4
Page | 9-19
800 0 0 0 0 1 1 1 2 3 4 900 0 0 0 0 1 1 1 2 3 4
1000 0 0 0 0 0 1 1 1 3 3 1250 0 0 0 0 0 1 1 1 1 3 1500 0 0 0 0 0 1 1 1 1 2 1750 0 0 0 0 0 0 1 1 1 1 2000 0 0 0 0 0 0 1 1 1 1
THHN, THWN, THWN-2
14 14 24 39 68 91 149 211 326 436 562 12 10 17 29 49 67 109 154 238 318 410
10 6 11 18 31 42 68 97 150 200 258 8 3 6 10 18 24 39 56 86 115 149 6 2 4 7 13 17 28 40 62 83 107
4 1 3 4 8 10 17 25 38 51 66 3 1 2 4 6 9 15 21 32 43 56 2 1 1 3 5 7 12 17 27 36 47 1 1 1 2 4 5 9 13 20 27 35
1/0 1 1 1 3 4 8 11 17 23 29 2/0 1 1 1 3 4 6 9 14 19 24 3/0 0 1 1 2 3 5 7 12 16 20 4/0 0 1 1 1 2 4 6 9 13 17
250 0 0 1 1 1 3 5 8 10 13 300 0 0 1 1 1 3 4 7 9 12 350 0 0 1 1 1 2 4 6 8 10 400 0 0 1 1 1 2 3 5 7 9 500 0 0 0 1 1 1 3 4 6 7
600 0 0 0 1 1 1 2 3 5 6 700 0 0 0 1 1 1 1 3 4 5 750 0 0 0 1 1 1 1 3 4 5 800 0 0 0 0 1 1 1 3 4 5 900 0 0 0 0 1 1 1 2 3 4 1000 0 0 0 0 1 1 1 2 3 4
FEP, FEPB, PFA, PFAH, TFE
14 13 23 38 66 89 145 205 317 423 545 12 10 17 28 48 65 106 150 231 309 398 10 7 12 20 34 46 76 107 166 221 285
8 4 7 11 19 26 43 61 95 127 163 6 3 5 8 14 19 31 44 67 90 116
4 1 3 5 10 13 21 30 47 63 81 3 1 3 4 8 11 18 25 39 52 68 2 1 2 4 6 9 15 21 32 43 56
PFA, PFAH, TFE 1 1 1 2 4 6 10 14 22 30 39 1/0 1 1 1 4 5 8 12 19 25 32
2/0 1 1 1 3 4 7 10 15 21 27 3/0 0 1 1 2 3 6 8 13 17 22 4/0 0 1 1 1 3 5 7 10 14 18
Z 14 16 28 46 79 107 175 247 381 510 657 12 11 20 32 56 76 124 175 271 362 466 10 7 12 20 34 46 76 107 166 221 285 8 4 7 12 21 29 48 68 105 140 180 6 3 5 9 15 20 33 47 73 98 127
4 1 3 6 10 14 23 33 50 67 87 3 1 2 4 7 10 17 24 37 49 63 2 1 1 3 6 8 14 20 30 41 53 1 1 1 3 5 7 11 16 25 33 43
XHH, XHHW, XHHW-2, ZW
14 10 17 27 47 64 104 147 228 304 392 12 7 13 21 36 49 80 113 175 234 301
10 5 9 15 27 36 59 84 130 174 224 8 3 5 8 15 20 33 47 72 97 124 6 1 4 6 11 15 24 35 53 71 92
4 1 3 4 8 11 18 25 39 52 67 3 1 2 4 7 9 15 21 33 44 56 2 1 1 3 5 7 12 18 27 37 47
Page | 9-20
1 1 1 2 4 5 9 13 20 27 35
1/0 1 1 1 3 5 8 11 17 23 30 2/0 1 1 1 3 4 6 9 14 19 25 3/0 0 1 1 2 3 5 7 12 16 20 4/0 0 1 1 1 2 4 6 10 13 17
250 0 0 1 1 1 3 5 8 11 14 300 0 0 1 1 1 3 4 7 9 12 350 0 0 1 1 1 3 4 6 8 10 400 0 0 1 1 1 2 3 5 7 9 500 0 0 0 1 1 1 3 4 6 8
600 0 0 0 1 1 1 2 3 5 6 700 0 0 0 1 1 1 1 3 4 5 750 0 0 0 1 1 1 1 3 4 5 800 0 0 0 0 1 1 1 3 4 5 900 0 0 0 0 1 1 1 2 3 4
1000 0 0 0 0 1 1 1 2 3 4 1250 0 0 0 0 0 1 1 1 2 3 1500 0 0 0 0 0 1 1 1 1 2 1750 0 0 0 0 0 1 1 1 1 2 2000 0 0 0 0 0 0 1 1 1 1
Page | 9-21
NEMA Controller Size for Motors, Transformers, & Capacitors
Load Voltage
Continuous Current
Service Limit Current
Motor Maximum
Motor Maximum
Transformer Primary Switching
Transformer Primary Switching
Capacitor Switching
Circuit Closing Maximum
Non-plugging and Non-jogging Duty
Plugging and Jogging Duty
Inrush Current < = 20 times Continuous Amp
Inrush Current = 20 to 40 times Continuous Amp
Inrush Current Peak Including Offset
NEMA
V Amp Amp HP HP HP HP kVA kVA kVA kVA kVAR Amp
Size 1φ 3 φ 1 φ 3 φ 1 φ 3 φ 1 φ 3 φ 3 φ 3 φ
00 115 200 230 380 460 575
9 11 1/3 — 1 — — —
— 1-1/2 1-1/2 1-1/2 2 2
1/4 — 1/2 — — —
— 1 1 1 1-1/2 1-1/2
— — — — — —
— — — — — —
— — — — — —
— — — — — —
— — — — — —
87
0 115 200 230 380 460 575
18 21 1 — 2 — — —
— 3 3 5 5 5
1/2 — 1 — — —
— 1-1/2 1-1/2 1-1/2 2 2
0.6 — 1.2 — 2.4 3
— 1.8 2.1 — 4.2 5.2
0.3 — 0.6 — 1.2 1.5
— 0.9 1 — 2.1 2.6
— — — — — —
140
1 115 200 230 380 460 575
27 32 2 — 3 — — —
— 7-1/2 7-1/2 10 10 10
1 — 2 — — —
— 3 3 5 5 5
1.2 — 2.4 — 4.9 6.2
— 3.6 4.3 — 8.5 11
0.6 — 1.2 — 2.5 3.1
— 1.8 2.1 — 4.3 5.3
— — 6 — 13.5 17
288
1P 115 230
36 42 3 5
— —
1-1/2 3
— —
— —
— —
— —
— —
— —
— —
2 115 200 230 380 460 575
45 52 3 — 7-1/2 — — —
— 10 15 25 25 25
2 — 5 — — —
— 7-1/2 10 15 15 15
2.1 — 4.1 — 8.3 10
— 6.3 7.2 — 14 18
1 — 2.1 — 4.2 5.2
— 3.1 3.6 — 7.2 8.9
— — 12 — 25 31
483
3 115 200 230 380 460 575
90 104 7-1/2 — 15 — — —
— 25 30 50 50 50
7-1/2 — 15 — — —
— 15 20 30 30 30
4.1 — 8.1 — 16 20
— 12 14 — 28 35
2 — 4.1 — 8.1 10
— 6.1 7.0 — 14 18
— — 27 — 53 67
947
4 115 200 230 380 460 575
135 156 — — — — — —
— 40 50 75 100 100
— — — — — —
— 25 30 50 60 60
6.8 — 14 — 27 34
— 20 23 — 47 59
3.4 — 6.8 — 14 17
— 10 12 — 23 29
— — 40 — 80 100
1581
5 115 200 230 380 460 575
270 311 — — — — — —
— 75 100 150 200 200
— — — — — —
— 60 75 125 150 150
14 — 27 — 54 68
— 41 47 — 94 117
6.8 — 14 — 27 34
— 20 24 — 47 59
— — 80 — 160 200
3163
6 115 200
540 621 — —
— 150
— —
— 125
27 —
— 81
14 —
— 41
— —
6326
Page | 9-22
230 380 460 575
— — — —
200 300 400 400
— — — —
150 250 300 300
54 — 108 135
94 — 188 234
27 — 54 68
47 — 94 117
160 — 320 400
7 230 460 575
810 932 — — —
300 600 600
— — —
— — —
— — —
— — —
— — —
— — —
240 480 600
9470
8 230 460 575
1215 1400 — — —
450 900 900
— — —
— — —
— — —
— — —
— — —
— — —
360 720 900
14205
9 230 460 575
2250 2590 — — —
800 1600 1600
— — —
— — —
— — —
— — —
— — —
— — —
665 1325 1670
25380
Service-Limit Current Ratings - The service-limit current ratings shown represent the maximum rms current, in amperes, which the controller
shall be permitted to carry for protracted periods in normal service. At service-limit current ratings, temperature rises shall be permitted to exceed
those obtained by testing the controller at its continuous current rating. The current rating of overload relays or the trip current of other motor
protective devices used shall not exceed the service-limit current rating of the controller.
Plugging or Jogging Service - The listed horsepower ratings are recommended for those applications requiring repeated interruption of stalled
motor current encountered in rapid motor reversal in excess of five openings or closings per minute and shall not be more than ten in a ten minute
period.
Capacitor terminals - If maximum available current is greater than 3,000 amperes, consult NEMA ICS-2 Standard.
Page | 9-23
NEMA Table 11 Typical Characteristics and Applications of Fixed Frequency Small and Medium AC
Squirrel-Cage Induction Motors
Design Letter Locked Rotor Torque
Pull- up Torque
Break-down Torque
Locked Rotor Current
Slip Typical Applications Relative Efficiency
Polyphase Characteristics
Percent Rated Load Torque*
Percent Rated Load Torque
Percent Rated Load Torque
Percent Rated Load Current
Percent Sync Speed
Design A
High locked rotor torque High locked rotor current
70-275 65-190 175-300 Not defined
0.5-5% Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or high
Design B
Normal locked rotor torque Normal locked rotor current
70-275 65-190 175-300 600-700 0.5-5% Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or high
Design C
High locked rotor torque Normal locked rotor current
200-285 140-195 190-225 600-700 1-5% Conveyors, crushers, stirring motors, agitators, reciprocating pumps and compressors, etc., where starting under load is required
Medium
Design D
High locked rotor torque High slip
275 NA 275 600-700 5-8% High peak loads with or without flywheels such as punch presses, shears, elevators, extractors, winches, hoists, oil-well pumping and wire-drawing motors
Low
Design N
Small motor
- NA - - - Centrifugal loads where starting torque requirements are relatively low.
Low
Design 0
Small motor - NA - - NA
Design L
Medium motor
- 100% - - NA Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or low
Design M
Medium motor
- 100% - - NA Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or low
*Higher values are for motors having lower horsepower ratings.
Page | 9-24
Frame D E 2F H N* O* P* U V AA AB* AH AJ AK BA BB BD* TAP 42 2-5/8 1-3/4 1-11/16 9/32 slot 1-1/2 5 4-11/16 3/8 1-1/8 3/8 4-1/32 1-5/16 3-3/4 3 2-1/16 1/8 4-5/8 1/4-20 48 3 2-1/8 2-3/4 11/32 slot 1-7/8 5-7/8 5-11/16 1/2 1-1/2 1/2 4-3/8 1-11/16 3-3/4 3 2-1/2 1/8 5-5/8 1/4-20 56 3-1/2 2-7/16 3 11/32 slot 2-7/16 6-7/8 6-5/8 5/8 1-7/8 1/2 5 2-1/16 5-7/8 4-1/2 2-3/4 1/8 6-1/2 3/8-16 56H 5 2-1/8 143T 3-1/2 2-3/4 4 11/32 2-1/2 6-7/8 6-5/8 7/8 2-1/4 3/4 5-1/4 2-1/8 5-7/8 4-1/2 2-1/4 1/8 6-1/2 3/8-16
145T 5 182 4-1/2 2-11/16 7/8 2-1/4 2-1/8 5-7/8 4-1/2 1/8 6-1/2 3/8-16 184 4-1/2 3-3/4 5-1/2 13/32 2-11/16 8-11/16 7-7/8 7/8 2-1/4 3/4 5-7/8 2-1/8 5-7/8 4-1/2 2-3/4 1/8 6-1/2 3/8-16 182T 4-1/2 3-9/16 1-1/8 2-3/4 2-5/8 7-1/4 8-1/2 1/4 9 1/2-13 184T 5-1/2 3-9/16 1-1/8 2-3/4 2-5/8 7-1/4 8-1/2 1/4 9 1/2-13
213 5-1/2 3-1/2 1-1/8 3 2-3/4 215 5-1/4 4-1/4 7 13/32 3-1/2 10-1/4 9-9/16 1-1/8 3 3/4 7-3/8 2-3/4 7-1/4 8-1/2 3-1/2 1/4 9 1/2-13 213T 5-1/2 3-7/8 1-3/8 3-3/8 3-1/8 215T 7 3-7/8 1-3/8 3-3/8 3-1/8 254U 8-1/4 4-1/16 1-3/8 3-3/4 3-1/2
256U 6-1/4 5 10 17/32 4-1/16 12-7/8 12-15/16 1-3/8 3-3/4 1 9-5/8 3-1/2 7-1/4 8-1/2 4-1/4 1/4 10 1/2-13 254T 8-1/4 4-5/16 1-5/8 4 3-3/4 256T 10 4-5/16 1-5/8 4 3-3/4 284U 9-1/2 5-1/8 1-5/8 4-7/8 4-5/8 286U 11 5-1/8 1-5/8 4-7/8 4-5/8
284T 7 5-1/2 9-1/2 17/32 4-7/8 14-5/8 14-5/8 1-7/8 4-5/8 1-1/2 13-1/8 4-3/8 9 10-1/2 4-3/4 1/4 11-1/4 1/2-13 286T 11 4-7/8 1-7/8 4-5/8 4-3/8 284TS 9-1/2 3-3/8 1-5/8 3-1/4 3 286TS 11 3-3/8 1-5/8 3-1/4 3 324U 10-1/2 5-7/8 1-7/8 5-5/8 5-3/8
326U 12 5-7/8 1-7/8 5-5/8 5-3/8 324T 8 6-1/4 10-1/2 21/32 5-1/2 2-1/8 5-1/4 2 14-1/8 5 11 12-1/2 5-1/4 1/4 13-3/8 5/8-11 326T 12 5-1/2 16-1/2 16-1/2 2-1/8 5-1/4 5 324TS 10-1/2 3-15/16 1-7/8 3-3/4 3-1/2 326TS 12 3-15/16 1-7/8 3-3/4 3-1/2
364U 11-1/4 6-3/4 2-1/8 6-3/8 6-1/8 365U 12-1/4 6-3/4 2-1/8 6-3/8 6-1/8 364T 9 7 11-1/4 21/32 6-1/4 18-1/2 18-1/4 2-3/8 5-7/8 2-1/2 15-1/16 5-5/8 11 12-1/2 5-7/8 1/4 13-3/8 5/8-11 365T 12-1/4 6-1/4 2-3/8 5-7/8 5-5/8 364TS 11-1/4 4 1-7/8 3-3/4 3-1/2
365TS 12-1/4 4 1-7/8 3-3/4 3-1/2 404U 12-1/4 7-3/16 2-3/8 7-1/8 6-7/8 405U 13-3/4 7-3/16 2-3/8 7-1/8 6-7/8 404T 10 8 12-1/4 13/16 7-5/16 20-5/16 20-1/8 2-7/8 7-1/4 3 18 7 11 12-1/2 6-5/8 1/4 13-7/8 5/8-11 405T 13-3/4 7-5/16 2-7/8 7-1/4 7
404TS 12-1/4 4-1/2 2-1/8 4-1/4 4 405TS 13-3/4 4-1/2 2-1/8 4-1/4 4
NEMA MOTOR DIMENSIONS
Shaft – Key Dimensions Frame Dimensions
NEMA Shaft (U)
Key (R) Dimen (S)
3/8 21/64 flat
1/2 29/64 flat 5/8 33/64 3/16
7/8 49/64 3/16 1-1/8 63/64 1/4 1-3/8 1-13/64 5/16
1-5/8 1-13/32 3/8 1-7/8 1-19/32 1/2
2-1/8 1-27/32 1/2 2-3/8 2-1/64 5/8
2-1/2 2-3/16 5/8 2-7/8 2-29/64 3/4
3-3/8 2-7/8 7/8 3-7/8 3-5/16 1
Page | 9-25
444U 14-1/2 8-5/8 22-7/8 22-3/8 2-7/8 8-5/8 19-9/16 8-3/8 445U 16-1/2 8-5/8 22-7/8 22-3/8 2-7/8 8-5/8 19-9/16 8-3/8 444T 14-1/2 8-1/2 22-7/8 22-3/8 3-3/8 8-1/2 19-9/16 8-1/4
445T 11 9 16-1/2 13/16 8-1/2 22-7/8 22-3/8 3-3/8 8-1/2 19-9/16 8-1/4 447T 20 8-15/16 22-15/16 22-3/4 3-3/8 8-1/2 3 21-11/16 8-1/4 14 16 7-1/2 1/4 16-3/4 5/8-11 449T 25 8-15/16 22-15/16 22-3/4 3-3/8 8-1/2 21-11/16 8-1/4 444TS 14-1/2 5-3/16 22-7/8 22-3/8 2-3/8 4-3/4 19-9/16 4-1/2 445TS 16-1/2 5-3/16 22-7/8 22-3/8 2-3/8 4-3/4 19-9/16 4-1/2
447TS 20 4-15/16 22-15/16 22-3/4 2-3/8 4-3/4 4NPT 21-11/16 4-1/2 449TS 25 4-15/16 22-15/16 22-3/4 2-3/8 4-3/4 4NPT 21-11/16 4-1/2
NEMA MOTOR DIMENSIONS – 2
Frame Size Information Suffix letters after the NEMA frame size indicates that the frame differs in some way from the standard frame. Below is a list of suffixes that may be found after the frame size and their definition.
A DC Motor or Generator
C ―C‖ flange mounting on drive end **
D ―D‖ flange mounting on drive end **
E Shaft dimensions for elevator motors in frames larger than the 326U frame
H Frame with an ―F‖ dimension larger than a frame without (small framed motors)
J Jet pump motors
JM ―C‖-face mounted close coupling pump with mechanical seal
JP ―C‖-face mounted close coupling pump -packed pump
K Sump pump motor LP & LPH ―P‖ flange mounting vertical solid shaft pump
P & PH ―P‖ flange mounting vertical hollow shaft pump
S Standard short shaft T Included as part of a frame number-standard dimension
U Included as part of a frame number-standard dimension V Vertical mounting
Y Special mounting dimensions -manufactured specified Z Special shaft dimensions -manufactured specified
** If the face mounting is on the end opposite the drive, the suffix will be as follows: ―FC‖ or ―FD‖
Assembly Position F-1 VS. F-2
F-1 F-2
Notes From Front: -When a ―C‖ flange has been added to a NEMA motor the ―BA‖ dimensions are:
Frame: ―BA‖:
143-5TC 2-3/4 182-4TC 3-1/2 213-5TC 4-1/4 254-6TC 4-3 4
-Dimensions: C, N, O, P, & AB are specific to each manufacturer. Please call for exact dimensions.
Page | 9-26
Pre-NEMA Dimensions
FRAME U V D 2F E
203 3/4 2 5 5-1/2 4 204 3/4 2 5 6-1/2 4 224 1 2-3/4 5-1/2 6-3/4 4-1/2 225 1 2-3/4 5-1/2 7-1/2 4-1/2 254 1-1/8 3-1/8 6-1/4 8-
1/14 5
284 1-1/4 3-3/4 7 9-1/2 5-1/2 324 1-5/8 4-5/8 8 10-
1/2 6-1/4
324S 1-5/8 3 8 10-1/2
6-1/4
326 1-5/8 4-5/8 8 12 6-1/4 326S 1-5/8 3 8 12 6-1/4
364 1-7/8 5-3/8 9 11-1/4
7
364S 1-7/8 3-1/2 9 11-1/4
7
356 1-7/8 5-3/8 9 12-1/4
7
365S 1-5/8 3-1/2 9 12-1/4
7
FRAME U V D 2F E
404 2-1/8
6-1/8 10 12-1/4 8
404S 1-7/8
3-1/2 10 12-1/4 8
405 2-1/8
6-1/8 10 13-3/4 8
405S 1-7/8
3-/2 10 13-3/4 8
444 2-3/8
6-7/8 11 14-1/2 9
444S 2-1/8
4 11 14-1/2 9
445 2-3/8
6-7/8 11 16-1/2 9
445S 2-1/8
4 11 16-1/2 9
504 2-5/8
7-5/8 12-1/2
16 8
504S 2-1/8
4 12-1/2
16 8
505 2-7/8
8-3/8 12-1/2
18 9
505S 2-1/8
4 12-1/2
18 9
584 2-7/8
11-5/8
14-1/2
18 9
584S 2-3/8
4-3/4 14-1/2
18 9
Page | 9-27
Standard NEMA Frames vs Horsepower 900 RPM 1200 RPM 1800 RPM 3600 RPM
HP T-Frame U-Frame Original T-Frame U-Frame Original T-Frame U-Frame Original T-Frame U-Frame Original 1964-
Present 1952-Present
NEMA 1964-Present
1952-Present
NEMA 1964- Present
1952- Present
NEMA 1964- Present
1952- Present
NEMA
1 182T 213 225 145T 184 204 143T 182 203 . . . 1.5 184T 213 254 182T 184 224 145T 184 204 143T 182 203 2 213T 215 254 184T 213 225 145T 184 224 145T 184 204 3 5 7.5 10 15 20 25 30 40 50 60 75 100 125 150 200
215T 254T 256T 284T 286T 324T 326T 364T 365T 404T 405T 444T 445T . . .
254U 256U 284U 286U 326U 364U 365U 404U 405U 444U 445U . . . . .
284 324 326 364 365 404 405 444 445 504 505 . . . . .
213T 215T 254T 256T 284T 286T 324T 324T 364T 365T 404T 405T 444T 445T . .
215 254U 256U 284U 324U 326U 364U 365U 404U 405U 444U 445U . . . .
254 284 324 326 364 365 404 405 444 445 504 505 . . . .
182T 184T 213T 215T 254T 256T 284T 286T 324T 326T 364TS 365TS 404TS/405TS 405TS/444TS 444TS/445TS 455TS
213 215 254U 256U 284U 286U 324U 326U 364U 365US 404US/405US 405US/444US 444US/445US 445US . .
225 254 284 324 326 364 364/365 365/404 404/405 405S/444 444S/445S 445S/504S 504S/505S 505S . .
145T/182T 182T/184T 184T/213T 213T/215T 215T/254T 254T/256T 256T/284TS 284TS/286TS 286TS/324TS 324TS/326TS 326TS/364TS 364TS/365TS 365TS/405TS 404TS/444TS 405TS/445TS 444TS
184 213 215 254U 256U 284U/286U 286U/324U 324S/326S 326S/364US 364US 365US/405US 404US/444US 405US/445US 444US 445US .
224 225 254 284 324 326 364S/365S 364S/404S 365S/405S 404S/444S 405S/445S 444S/504S 445S/505S 504S 505S .
250 . . . . . . . . . 445TS . .
These charts are provided for reference use only. We are not responsible for any printing errors.
Page | 9-28
NEMA Motor Enclosure Type
Type Abbreviation
Description Designed for use in
Open Drip Proof ODG Open Drip-Proof, Guarded non-hazardous, relatively clean areas, most common type, ODG-FV Open Drip-Proof, Force Ventilated
ODG-SV Open Drip-Proof, Separately Ventilated ODP Open Drip-Proof
Totally Enclosed TEAO Totally-Enclosed, Air-Over extremely wet, dirty, or dusty areas TEBC Totally-Enclosed, Blower-Cooled
TECACA Totally-Enclosed, Closed Circuit,, Air to Air
TEDC-A/A Totally-Enclosed, Dual Cooled, Air to Air TEDC-A/W Totally-Enclosed, Dual Cooled, Air to
Water
TEFC Totally-Enclosed, Fan-Cooled TENV Totally-Enclosed Non-Ventilated TETC Totally-Enclosed, Tube Cooled TEWAC Totally-Enclosed, Water/Air Cooled TEXP Totally-Enclosed, Explosion-Proof
Weather Protected
WPI Weather Protected, Type I adverse outdoor conditions
WPII Weather Protected Type II
Special XE Premium Efficient improved efficiency XL Extra Life XP Explosion-Proof withstanding an explosion of a
specified dust, gas, or vapor XT Extra Tough Dust ignition proof preventing the ignition of a dust,
gas, or vapor surrounding the motor
IEC IP-22 Open Drip-Proof representative IEC designations IP-44 Totally-Enclosed IP-54 Splash Proof IP-55 Washdown
Page | 9-29
Electrical Power System Design Example
Pump: 20 Hp, 300 RPM, 18‖ sheave Motor: 3Φ, 460 V Power: 3Φ, 7200 LN Environment: ambient 98F, outdoor
Pump: Motor: Power: Environment:
# Measure Parameter Standard Table or Reference
Example Factor
Example Result
Problem Factor
Problem Result
1 Motor horsepower NEC 430.250 - 20 2 Full load Amps - FLA NEC 430.250 - 27 3 Lock letter code & kVA/hp NEC 430.7(B) F 5.59 112 4 Lock rotor amp calculate kVA*1000/1.732 V 112,000/1.732*460 141 5 Lock rotor amp for disconnect NEC 430.251(B) - 145 6 Wire rating:1.25*largest+other NEC 430.24 1.25*27 + 0 34 7 Insulation type NEC 310-16 - THHN 8 Insulation temperature NEC 310-16 - 90C 9 AWG / kcmil NEC 310-16 - 10 AWG 10 Temperature correction amp NEC 310-16 0.91 36 11 Max breaker rating & type NEC 430.52 800 instant 216 12 Actual breaker size NEC 240.6(A) - 200 13 Controller enclosure NEC 430.91 - 3R 14 Controller size NEMA Controller - 2 15 Controller max closing amp NEMA Controller - 483 16 Overload setting % - Amp - 105 28.4 17 Motor enclosure NEMA Enclosure TEFC 18 Motor NEMA design NEMA 11 - B 19 Motor sync speed 120 * freq / poles 120*60/4 1800 20 Motor slip - shaft speed NEMA 11 2% 1764 21 Motor frame NEMA Dimension 2 - 256T 22 Shaft diameter NEMA Dimensions U 1-5/8 23 Sheave diameter P(n*dia) = M(n*dia) 300*18/1800 3‖ 24 # wires - 3Φ + N 4 25 Conduit type & size NEC C4 et al 1/2 3/4 26 Total kVA 1.732 V * I / 1000 1.732*480*27/1000 22.5 26 Transformer kVA size NEMA - 3 – 7.5 28 Secondary volt & Y-Δ - 277 / 480 Y 29 Primary volt & Y-Δ - 12470 Δ 30 Transformers taps two 2-1/2 + 31 Transformer impedance PU