electonics final handouts

8/17/2019 Electonics Final Handouts

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RReevviieeww CCoouurrssee

ELECTRONICS ENGINEERING

N107 MIT, Muralla St., Intramuros, Manila ℡ (02) 247-5000 local 2100 or (02)524-5572

Electronics EngineeringReview Course

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EELLEECCTTRROONNIICCSS

• Science dealing with the development and application of devices and system involving flow of electrons in a vacuum, in gaseous media, in plasma and/or in semiconductors.

Application of Electronics:1. Communication Electronics2. Electric Power3. Digital Electronics

Basic Electrical Components:1. Active – devices that can be used for amplification, rectification or change energy

one another.

a. semiconductorb. electron tubesc. visual display devices

2. Passivea. Resistors – limits the flow of current or divide the voltages in the circuit.b. Capacitor – concentrates the electric field of voltage applied to a dielectc. Inductor – concentrates the magnetic field of electric current in the coil.

ATOMIC STRUCTURE

• Atom – the smallest particle of an element that retains the characteristics of the elemen

BOHR MODELFigure 1-1. - The composition of a simple helium atom.

Subatomic Particles:MASS CHARGE

Electrons 9.1 x 10-31kg 1.67 x 10-19 CProtons 1.67 x 10-27kg 1.67 x 10-19 C

Neutrons 1.67 x 10-27 kg none


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Atomic number – number of protons

Atomic Mass – number of protons and neutrons

SHELL & SUBSHELL AROUND THE NUCLEUS

SHELL

Numerical Literal Subshells Capacity

1 K s2 2 e-

2 L s2, p6 8 e-

3 M s2, p6, d10 18 e-4 N s2, p6, d10, f14 32 e-

• • • •

• • • •

• • • •

ELECTRICAL CLASSIFICATION OF MATERIALS

CONDUCTOR

• A material that easily conducts electric current• Number of valence electrons is 1-3

INSULATOR

• A material that DOES NOT conducts electric current under normal conditions• Number of valence electrons is 5-8• Valence electrons are tightly bound to the atoms.

SEMICONDUCTOR

• A material that is between conductors and insulators in its ability to conduct electricalcurrent

• Number of valence electrons is 4• Neither a good conductor nor a good insulator in its pure form

ENERGY BANDS

Figure 1-4. - The energy arrangement in atoms.


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Figure 1-5. - Energy level diagram.

TYPES OF SEMICONDUCTOR

1. Intrinsic Semiconductor – semiconductor in its pure (elemental) form.a. Siliconb. Germanium

Figure 1-7. - A two-dimensional view of a silicon cubic lattice.

Conduction Electrons – free electrons or electrons in the conduction band


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Hole – vacancy left by the electron in the valence band or the absence of electron

Electron-Hole Pair – for every electron raised to the conduction band by external energy, there isone hole left in the valence bandRecombination – occurs when a conduction-band electron loses energy and falls back into a hole inthe valence band.Electron Current –movement of free electrons in a semiconductive materials.Hole Current – movement of holes in a semiconductive material

2. Extrinsic Semiconductor – semiconductor with impuritiesDoping – The process of adding impurities to an intrinsic semiconductor.

a. Pentavalent atoms –atoms with 5 valence electrons.

1. Arsenic (As)2. Phosphorus (P)3. Bismuth (Bi)4. Antimony (Sb)

b. Trivalent atoms – atoms with 3 valence electrons1. Aluminum (Al)2. Boron (B)3. Indium (In)4. Galium (Ga)

a. N-Type Semiconductor – doped with pentavalent atomFigure 1-9. - Germanium crystal doped with arsenic.

b. P-Type Semiconductor – doped with trivalent atomFigure 1-11. - Germanium crystal doped with indium.


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PN JUNCTION

• The boundary between two different types of semiconductive material• Formed when a piece of intrinsic silicon is doped so that half is n-type and the other ha

p-typeFigure 1-13. - The PN junction barrier formation.

DEPLETION REGION

• The area near a pn junction on both sides that has no majority carriers• Region that is depleted of carriers.

BARRIER POTENTIAL

• The potential difference of the electric field across the depletion region• The amount of energy required to move electrons through the electric field

• Typically equal to0.7V for Silicon and 0.3V for Germanium at 25oC


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BIASING THE PN JUNCTION

• Application of external dc voltage across the pn junctionFORWARD BIAS – is the condition that allows current through a pn junction

Figure 1-14. - Forward-biased PN junction.

REVERSE BIAS - is the condition that prevents current through the pn junction

Figure 1-15. - Reverse-biased PN junction.

THE DIODE

• A semiconductor device that allows the flow of current in one direction only.• A single pn junction device with conductive contacts and wire leads connected to each region

anode cathode

Figure 1-16. - PN junction diode characteristic curve.


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Dynamic Resistance – the nonlinear internal resistance of a semiconductive material.

DIODE MODELS

The Ideal Diode Model

• A simple switch, FB is like a closed (on) switch, RB is like an open (off) switch• Barrier potential, forward dynamic resistance and the reverse current are all neglected

The Practical Diode Model

• Adds the barrier potential to the ideal switch model• When FB acts like a closed switch in series with a small voltage.• When diode is reverse bias it acts like an open switch

IF

VFVB

VR

IF

VFVB

VR


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The Complex Diode Model

• Consist of the barrier potential, the small dynamic resistance (r’d) and the large internalreverse resistance (r’R) • When FB acts like a closed switch in series with the barrier potential voltage and the small

forward dynamic resistance (r’d).

• When diode is reverse bias it acts like an open switch in parallel with the large internalreverse resistance (r’R).

Review Questions

IF

V0.7

V

Small reversecurrent due to hi h

IR

Slope due to low

forward


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1. Which of the following is currently being

used to describe the atomic structure?a. JJ Thomson’s modelb. Rutherford’s modelc. Bohr’s modeld. Armstrong’s model

2. Which of the following subatomicparticles, according to Bohr’s model, issmallest?

a. protonb. electron

c. neutrond. magnetron

3. ____ is 1800 times as heavy as _____.a. electron, protonb. proton, neutronc. neutron, protond. proton, electron

4. Which of the following statements istrue?

a. electron has a unit charge of +1b. proton has a unit charge of -1c. neutron has a unit charge of 0d. none of the above.

5. Which determines the atomic number ofan element?

a. number of electronsb. number of protonsc. number of neutronsd. number of protons and neutrons

6. Which determines the atomic mass ofan element?

a. number of electronsb. number of protonsc. number of neutronsd. number of protons and neutrons

7. 1H3 is an example of a Hydrogen atom.Which statement is true?

a. Hydrogen has an atomic number of

3b. Hydrogen has an atomic mass of 1

c. Hydrogen has number of proton

equals 3d. Hydrogen has number of neutroequals 2

8. Atoms with same atomic number budifferent atomic mass is called

a. isotopesb. isobarsc. cationd. anion

9. 18 Ar 40 and 20Ca40 atoms are consider

asa. isotopesb. isobarsc. cationd. anion

10. 3rd quantum number is also known aa. K-shellb. L-shellc. M-shell

d. N-shell11. How many subshells are there in N-

shell?a. 5 subshellsb. 4 subshellsc. 3 subshellsd. 2 subshells

12. f-subshell is the 4th subshell. What dof means?

a. falseb. factualc. fundamentald. finite

13. f-subshell can accommodate maximof ____ electrons.

a. 14b. 10c. 8d. 6

14. L-shell can accommodate maximum ____ electrons.


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a. 3117 μFb. 3207 μFc. 311.7 μFd. 320.7 μF

85. Find the voltage regulation giving a dcvoltage of 67 V without load and with fullload current drawn the output voltagedrops to 42 V.

a. 59.5%b. 62.7%c. 15.9%

d. 32.5%86. What is the voltage across a reverse

biased diode in series with a 10V DCsource and a 1kΩ resistor?

a. 0Vb. 0.7Vc. 0.3Vd. 10V


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TRANSISTOR

• A semiconductive device used for amplification and switching applications.• William Shockley, John Bardeen and Walter Brattain were the co-inventors of transisto

Bell Laboratories in 1947.

BIPOLAR JUNCTION TRANSISTOR (BJT)The term Bipolar is because two type of charges (electrons and holes) are involved in

flow of electricity• The term Junction is because there are two pn junctions

• There are two configurations for this device

NPN and PNP Transistors

NPN Transistor PNP Transistor


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Parts and Proper Bias1) Collector- moderately doped (collector carrier)

-the doping is between that of the emitter and the base- largest of the three regions

2) Base - Lightly doped (control)- controls the flow of carriers from emitter and collector

-smallest3) Emitter - Heavily doped

- 2nd largest- emits carrier

PROPER BIAS OF TRANSISTORFor proper operation of the circuit

1) Emitter base junction should be forward bias2) Collector-base junction should be reversed bias

Transistor Operation


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BJT CONFIGURATIONS

Relationship of β and α

IE = IB + ICMultiply with 1/ Ic[IE = IB + IC] 1/ IcIE IB IC

IC IC IC1 / α = 1 + 1/ β

Example No 1.Determine βDC, IE, and αDC for a transistor where IB = 50μ A and IC = 3.65mA.Current and Voltage Analysis

VCE = Voltage drop from collector to emitterVCB = Voltage drop from collector to baseVBE = Voltage drop from base to emitterIC = Current into the collectorIB = Current into the baseIE = Current out of the emitter.

E

C

I

I =α Forward

Current Gain

Typical values: 0.001 to 0.9999

Common Base Common Collector

B

E

I

I =+1 β Forward

Current Gain

α = β / (β + 1 )


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Example 2.Determine IB, IC, IE, VBE, VCE, and VCB in the circuit. The transistor has a βDC = 150

Collector Characteristic Curves


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Examples.Sketch an ideal family of collector curves for the circuit for IB = 5μ A to 25μ A in 5μ A increments. β= 100 and that VCE does not exceed breakdown.

Cutoff, Saturation, and The DC load line

Example 4.Determine whether or not the transistor is in saturation. Assume VCE(sat) = 0.2 V.

RB

VCC

VBB

RC

IB

IC

. βDC = 100


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Example 5) The transistor has the following maximum ratings: PD(max) =800mW, VCE(max) = 15V,and IC(max) = 100mA. Determine the maximum value to whichh VCC can be adjusted withoutexceeding a rating.

Derating PDmax)Example 6) A certain transistor has PD(max) of 1W at 25°C. The derating factor is 5mw/C° . What isthe PD(max) at a temperature of 70°C ?The Transistor as a switch

+VCC

RC

C

E

RB

+VCC

R

0

IB = 0

I = 0

+VCC

R

C

E

a) Cutoff

R

+V

R

+V

I

I

+VCC

RC

C

E

B) Saturation


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Example 7)a) For the transistor, what is VCE when VIN = 0V?b) What minimum value of IB is required to saturate this transistor if βDC is 200? Neglect VCE(sac) Calculate the maximum value of RB when VIN = 5V.

DC Operating point

DC Load Line


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Linear Operation

Waveform Distortion


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Example 1) Determine the Q-point, and find the maximum peak value of base current for linearoperation. Assume βDC = 200.

BASE BIAS


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Example 2) The Base bias circuit is subjected to increase in temperature from 25°C to 75°C. If βDC= 100 at 25°C and 150 at 75°C, determine the percent in Q-point values (IC and VCE) over thetemperature range. Neglect any change in VBE and the effects of leakage current.

VOLTAGE DIVIDER BIAS


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Analysis of Voltage-Divider Circuit

Example 3) Determine VCEand IC. Assume βDC = 100

+VCC

R

1

R2

RIN(base)

a)Unloaded

b) Loaded

Simplified Voltage-Divider Input Resistance at the Base

IE RE

+VCC

RC

VIN

+

-

IIN

+ -VBE

RIN(base) = βDCRE

+

R

+VCC

+VCC

10kΩ

5.6kΩ

1kΩ

560Ω


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77. Method used in producing thick filmcomponents.

a. Evaporationb. Epitaxialc. Screeningd. diffusion

78. Method(s) used in producing thin films.a. Vacuum evaporationb. Cathode sputteringc. Screeningd. A and b

79. A method used to deposit thin filmcomponents on a single substrate in ahighly evacuated chamber.

a. Diffusionb. Epitaxialc. Cathode sputteringd. Vacuum evaporation

80. A ____ is an IC that is used to processanalog signals.

a. ROMb. CMOSc. Linear ICd. VMOS

81. The most widely used digital logic familya. DTLb. TTLc. ECLd. RTL

82. ECL is high speed because

a. Operation is in the low noisenegative supply region

b. Construction in small geometriesc. The use of gallium arsenide

conductorsd. The operating transistors being

unsaturated83. What is a CMOS IC?

a. A chip with only bipolar transistors

b. A chip with n-channel transistors

c. A chip with p-channel(pmos) and n-channel(nmos) transistors

d. A chip with p-channel transistors84. Digital IC that contains 10 to 100 gates.

a. SSIb. MSIc. LSId. VLSI


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FIELD EFFECT TRANSISTOR

GENERAL VOLTAGE AMPLIFIER

MODEL

BJT AMPLIFIER CONFIGURATION


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BJT EQUIVALENT CIRCUIT

1. Hybrid model – uses h-parameters of a two-

port network system

2221212

2121111

V h I h I

V h I hV

+=

+=

HYBRID MODEL


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PERFORMANCE FACTORS

oio

oii

V h I h I

V h I hV

2221

1211

+=

+=

BJT EQUIVALENT CIRCUIT

2. re-model – considers the ac or dynamic resistance

of a forward bias pn junction

E

e

I

mV r

26=


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PERFORMANCE FACTORS

FIELD EFFECT TRANSISTORS

1. Junction Field Effect Transistor (JFET)

- a type of JFET that only operates at reverse bias tocontrol the flow of current

- acts as a normally close switch


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2. Metal Oxide Semiconductor Field EffectTransistor (MOSFET)

- a type of FET with insulated gate whichprovides a very high input impedance

a. Depletion type MOSFET (D-MOSFET)

- a MOSFET with channel when gate

is open

- acts a normally close switch


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b. Enhancement type MOSFET (E-MOSFET)

- a MOSFET without channel when gate

is open

- acts a normally open switch

JFET OPERATION

• When VGS = 0V:

- As VDD increases, carrier flow increases

from source to drain.

- Gate to Drain junction is reverse biased,resulting to a high channel resistance

- Drain current becomes saturated when

VDS equal pinch-off voltage (VP)


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JFET OPERATION

• When VGS ≠ 0V (reverse bias):

- As the reverse bias voltage at the gate

increases, channel resistance increases

- Drain current is easily saturated

- When VGS = |VP|, drain current is zero

DMOSFET OPERATION

• When VGS = 0V:

- As VDD increases, carrier flow increases

from source to drain.

- Gate to Drain junction is reverse biasedresulting to a recombination of carriers

between the channel and the substrate

- Drain current becomes saturated when

VDS equal pinch-off voltage (VP)


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DMOSFET OPERATION

When VGS ≠ 0V (reverse bias voltage):

- As the reverse bias voltage at the gateincreases, channel resistance increases

- Drain current is easily saturated

- When VGS = |VP|, drain current is zero

When VGS ≠ 0V (forward bias voltage):- As the forward bias voltage at the gateincreases, carriers are enhanced resultingto increase in drain current

EMOSFET OPERATION

When VGS = 0V:

- ID is zero

When VGS < 0V:

- ID is zero

When VGS > 0V:- Channel starts to build up

- When VGS equals VTH, carriers start toflow form source to drain


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REVIEW QUESTIONS:1. The term use to describe suddenreverse conduction of an electroniccomponent cause by excess reversevoltage across the device.

a. cut-offb. saturationc. avalanched. revertion

2. What phenomenon in electronicsdoes an avalanche breakdown primarily

dependent?a. dopingb. recombinationc. ionizationd. collision

3. The primary use of zener diode inelectronic circuits.

a. resistance regulatorb. rectifier

c. voltage regulatord. current regulator4. How do zener diodes widely used?

a. Current limitersb. Power collectorsc. Variable resistorsd. Voltage regulators

5. _______ is the type of biasrequired by an LED to produceluminescence.

a. Forward biasb. Zero biasc. Reverse biasd. Inductive bias

6. Which semiconductor material iscommonly used in the conduction of LED?

a. Silicon (Si)b. Gallium Arsenide (Ga As)c. Germanium (Ge)

d. Gallium (Ga)

7. If an arrow next to anoptoelectronic device schematic symbo

points away from the symbol, the devicconsidered to be

a. photoemissiveb. a p-type semiconductorc. photosensitived. an n-type semiconductor

8. _____ is a pn junctionsemiconductor device that emits non-coherent optical radiation when biased

the forward direction, as a result of arecombination effect.

a. LASERb. JFETc. LEDd. MOS

9. Find the normal operating voltaand current of LED

a. 60V and 20mAb. 5V and 50mAc. 0.7V and 60mAd. 1.7V and 20mA

10. A photodiode is normallya. forward-biasedb. reverse biasedc. neither forward nor reverse biasd. emitting light

11. The reverse current flowing throa photodiode with no light input

a. Saturation currentb. Dark currentc. breakdown currentd. Light current

12. The capacitance of the varactodiode increases when the reverse voltaacross it

a. decreasesb. increases

c. breakdown


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ADVANCED ELECTRONICS

INTEGRATED CIRCUIT FABRICATION

The silicon planar process of fabrication includes the following steps:a. Silicon Wafer (Substrate) Preparationb. Epitaxial Growthc. Oxidationd. Photolithographye. Diffusion

f. Ion Implantationg. Isolationh. Metallizationi. Packaging

The individual steps are explained in brief.a. Silicon Wafer Preparation. The basic material required for making the substrate, i.e.

silicon, is cut into thin sheets, or wafers. This step includes the substeps like crystal growth,doping, slicing into wafers, and polishing and cleaning the wafer.

Silicon wafer preparation.

b. Epitaxial Growth. ‘Epi’ in Greek means ‘upon’ and ‘teinon’ in Greek means ‘arranged.’

In reality, epitaxy is simply a process to grow a single-crystal layer on a single-crystalsubstrate.

Types: homoepitaxy- single-crystal layer on a single-crystal are of exactly the same material; heteroepitaxy- single-crystal layer on a single-crystal are different in any aspect material

Applied to all kinds of thin-film depositions as long as they arranged in order ; employed toprepare III-V and II-VI compound semiconductor materials and devices.


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Epitaxial growth.

Note : The thickness and other dimensions of the IC will be in mils.

where

c. Oxidation. Here an oxide layer is grown over the epitaxial layer. The SiO 2 layer formedoxidation prevents diffusion of almost all impurities. Oxidation is accomplished by placin

the silicon vertically into a quartz boat in a quartz tube, which is slowly passed through aresistance-heated furnace, in a presence of oxygen operating at about 1000oC.

After oxidations the structure has 3 layers, i.e. the substrate, the epitaxial layer, and theoxide layer.

Oxidation.

El t i E i i El t i E i i


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d. Photolithography. The purpose of this step is to remove portions of the SiO 2 layer sodiffusion can occur in selected areas.

Functional Components of Lithography• Energy—cause (photo)chemical reactions that modify resist dissolution rate• Mask—Pattern (or direct) energy to create an aerial image of mask in resist• Aligner—Align mask to previous patterns on wafer (to a tolerance level)• Resist—Transfer image from mask to wafer, after development Positive resist reproducesthe mask pattern, Negative resist reproduces inverse mask pattern• Substrate—Has previous mask patterns

The structure before applying the photolithography process is shown below

Input structure for photolithography.

The photolithography process involves many steps which are described in detail below.

The first step is to make a photoresist layer above the oxide layer.

Structure with photoresist layer.

The second step is to make masks, the locations of which are determined by the finalstructure. The masks will protect the photoresist layer from the ultraviolet light applied in stepthree.


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Structure with masks.

Step three involves passing an ultraviolet light over the masks. Wherever the mask is nopresent, the photoresist layer is ‘polymerized.’

Structure with mask layer during ultraviolet exposure.

Structure with mask layer after ultra-violet exposure.

Etching Techniques - The process of selective removal of regions of a semiconductor, metasilicon dioxide.

Types:Wet Etching: the wafers are immersed in a chemical solution at a predeterminedtemperature. In this process, the material to be etched is removed equally in all directio

Electronics Engineering Electronics Engineering


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Dry (Plasma) Etching: the wafers are immersed in gaseous plasma created by a radiofrequency electric field applied to a gas such as argon.

The structure is dipped into a solution of hydrofluoric acid in the fourth step. The masks and thenon-polymerized portions of the photoresist layer will dissolve in the hydrofluoric acid solution.

Structure after dissolution in hydrofluoric acid.

The fifth step involves dipping the structure into a photosensitive emulsion. The oxide layernot protected by the polymerized photoresist will dissolve in the solution.

Structure after dissolving in photoemulsive solution.

The sixth and final step is to remove the polymerized photoresist. At the end of this step, theoxide layer will be exposed.

Final structure after photolithography.

Chemical Mechanical Polishing (CMP) - is a process that planarizes the wafer. This is donebefore patterning of the wafer to provide flat surface to expose the mask image on.


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e. Diffusion. Diffusion is the process by which the N- type or P-type impurity silicon atoms

can be diffused into the epitaxial layer, through the holes in the oxide layer.

Diffusion.

Diffusion process occur in two steps:Predeposition Step: a high concentration of dopant atoms are introduced at the silsurface by a vapor that contains the dopant at a temperature of about 1000Predeposition tends to produce, near the silicon surface, a shallow but heavily dolayer.Drive-in: Used to drive the impurity atoms deeper into the surface, without addingmore impurities, thus reducing the surface concentration of the dopant.

f. Ion Implantation. In this process, an alternative to diffusion, the epitaxial layer can beimplanted with impurity ions.

g. Isolation. Since a number of different circuits are manufactured in a single planar proceit becomes essential to differentiate the circuits. This process checks whether any shortcircuit is present between different circuits and, if so, the corresponding part is identified unusable.

h. Metallization. This process provides electrical metal contacts to the different diffusedareas, where the terminals of the devices should be taken. Wherever the terminals shoube short-circuited always, the metal contacts will be short-circuited and a single lead termwill be taken ‘out.’



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Metallization.

i. Packaging. The circuits manufactured in a single process will be scribed and cut down intoseparate structures. Each structure will be packed as a separate IC. The packaging will beused to give output leads to users.

Different package configurations available:1. To-5 Glass Metal Package2. Ceramic Flat Package3. Dual-In-Line Package (DIP).4. Molded Matrix Array Package5. Flip-Chip Ball Grid Array

Example of IC packages.

Cleanrooms - are classified by the cleanliness of their air. The method most easily understood anduniversally applied is the one suggested in the earlier versions (A to D) of Federal Standard 209 inwhich the number of particles equal to and greater than 0.5 m m is measured in one cubic foot of airand this count is used to classify the room.


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Federal Standard 209 - This standard was first published in 1963 in the USA and titled "Cleanrand Work Station Requirements, Controlled Environments". It was revised in 1966 (209A), 1

(209B), 1987 (C), 1988 (D) and 1992 (E).

Federal Standard 209D Class Limits

DIGITAL INTEGRATED CIRCUITS

Properties and Definition

Ideal Logic Inverter

Operates from a single power supply

Voltage Transfer Characteristics

Voltage Transfer Characteristic (VTC)

The transfer characteristics of a logic gate are represented by a curve relating the output voltagethe input voltage. The curve is plotted on a graph where the input voltage is on the x-axis and theoutput voltage is on the y-axis.

CLASSMEASURED PARTICLE SIZE (MICROMETE

0.1 0.2 0.3 0.5 5

1 35 7.5 3 1

10 350 75 30 10

100 NA 750 300 100

1,000 NA NA NA 1,00010,000 NA NA NA 10,000

100,000 NA NA NA 100,000



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In order that the high and low voltage levels always be distinguishable, must always have

VOH > V IH and V OL > V IL

Where:VOH—The nominal, or minimum, logic-1 state output voltage.VOL—The nominal, or maximum, logic-0 state output voltage.ViL—The nominal, or maximum, input voltage required for logic-0 input.ViH—The nominal, or minimum, logic-1 input voltage.

Midpoint Voltage VM, sometimes referred as threshold voltage Vth , is defined as the point on VTCwhere VIN=VOUT and ideally appears at the center of the transition region.

Logic Swing – magnitude of voltage difference between the output high and output voltage levels.

Transition Width – amount of voltage change that is required of the output voltage from the high tothe low level (or vice versa)Noise Margins – terminology used to describe fluctuations/variations in the high and logic low levels.


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The effects of input variations are quantified in terms of the Noise Sensitivities. The high andnoise sensitivities are defined as the difference between the input and midpoint voltage for VIN at and VOL, respectively.

The quantity Noise Immunity is the ability of a gate to reject noise.

The term fan-in is used to describe the number of inputs to the gate. Similarly, the term fan-oused to describe the number of outputs of a gate. The maximum fan-out of a digital circurestricted by its input and output currents.

The maximum fan-out possible during the driving gate’s logical 1 output gate isNhigh = IOUT(high) / I’IN (high)

The maximum fan-out possible during the driving gate’s logical 0 output gate isNlow = IOUT(low) / I’IN (low)



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Transient Characteristics

Switching Speed Definitions

td = delay timetr = rise timets = storage timetf = fall time

tON = turn on timetOFF = turn off time

Propagation delay - symbolized tpd, is the average time required for a digital signal to travel from theinput(s) of a logic gate to the output.


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Power Dissipation

PCC (avg) = ( PCC(OH) + PCC(OL) ) / 2 PCC (avg) = VCC ( ICC(OH) + ICC(OL) ) / 2


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g gReview Course

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Simple TTL NAND gate.

The standard form of a TTL NAND gate is shown in below

Standard TTL NAND gate.

The standard form of TTL NAND gate is also called a modified TTL NAND gate. In this circuit, atotem pole or active pull-up stage is added to the simple TTL NAND gate to increase fan-out.

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Transfer characteristics of a TTL NAND gate. Region I is for logic-0 input. Region II is the transition stage. Region III is for logic-1 input.

EMITTER COUPLED LOGIC (ECL)

Also referred to as Emitter Coupled Transistor Logic (ECTL) this logic employs an emitter coupdifferential amplifier. The basic gates of this family are OR, NOT, and NOR.

This family has the minimum propagation delay because the output is not driven into saturation.

A standard ECL OR/NOR gate is shown.



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Review Course

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ECL OR/NOR gate.

COMPLEMENTARY METAL OXIDE SEMICONDUCTOR (CMOS) LOGIC

MOS Logic

This logic is similar to the other transistor logic families except it uses metal oxide semiconductorfield effect transistors instead of bipolar junction transistors. The advantages offered are easierfabrication process, increased operating speed, and low power comsumption.

Generally, the N-channel MOSFET is used in this family. Hence the logic is named NMOS logic. Thebasic gate used is a NOT gate.

CMOS Logic

Here, N-channel and P-channel MOSFETs are used. This increases the systems complexity andchip area compared to NMOS logic. The great advantage of CMOS logic is that the powerconsumption in a steady state is almost zero. Power consumption occurs only when there is aswitching action from one state to another. The basic gates used are NOT and NAND.

Review Course

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CMOS NOT gate.

Logic-1 = VCC and Logic-0 ~0 volts.

CMOS NAND gate.

CMOS Fabrication

CMOS (complementary metal oxide semiconductor field effect transistor) fabrication needs an N-channel MOSFET and P-channel MOSFET which are connected together as shown below



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Review Course

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N-channel and P-channel connection for CMOS fabrication.

Final structure of CMOS.

Review Course

70

RESONANCE

Resonance - condition wherein the inductive reactance (X L) of a coil equals the capacitive react(XC) of a capacitor in a circuit.

The resonant frequency f r can be computed from the condition above. Since,XL = 2 f r L

andXC = 1/2 f r C

then,

f r =

where: f r = resonant freq. in HzL = inductance in HenryC = capacitance in Farad

Series Resonance – L and C are in series.

I

Z

E

E

1

2π√LC

Electronics EngineeringR i C



36/84

Review Course

71

We can write the series impedance in rectangular form as:Z = R + j (XL – X C)

Taking the magnitude of the equation:lZl = R2 + (X L – XC)2

We can calculate the current I in the series:

I = =

But since XL = X C at resonance:then:

Z = RI = E/R

Characteristics of series resonant circuit:(1) Z=R minimum resistance(2) I = E/R maximum current(3) current is in phase with voltage(4) power factor is unity(5) power is maximum

Quality factor Q – figure of merit or factor of merit in sharpness of resonance.- ratio of the reactive power (energy stored) to the true power developed in the circuit

(energy dissipated per cycle).

Q = = in a

series resonant circuit. Therefore,

Q =

since XL = 2pif r Land

f r =

E E

lZl R 2 + (XL – XC)

2

Pa I2XLS

P I2 R

XLS

R

1

2π√LC

Review Course

72

thenQ =

Q rise in voltage across a series L or C:The Q of the resonant circuit can be considered a magnification factor that determines

much the voltage across L or C in increased by the resonant rise of current in a series circuit:VL = V C = IX C = IXL

butI = E/R

VC = EX C / R

VC = QE or V L = QE

Bandwidth:- any resonant frequency has an associated band of frequencies that provide resona

effects. The width of the resonant band of frequencies centered around f r is called the bandwidtthe tuned circuit.

I = 0.707 Imax

P = 0.5 Pmax

BW = f r /Q and BW = R/2piL

- From the equation, an increase in Q-factor will correspond to a decrease in BW.- As the L/C ratio increases, the response becomes sharper. Also, as R is decreased, the betterresponse.

1 L

R C

Imax

0.707 Imax

f 1 f f 2




37/84

Review Course

73

Parallel Resonant Circuit(Antiresonant Ckt.):

a b

Review Course

74

where Yac = 1/(RL + jX L)and Ybc = 1/(RC + jX C)simplify, then equate the susceptance: BL = BC

therefore,f ar =

Basic Variations:(a) RC is negligible

f ar = 1 –

(b) RL is negligible

Z = Yac + Y bc

c

1 R L2 – L/C

2π LC R C2 – L/C

1 R L2C

2π LC L




38/84

Review Course

75

f ar =

(c) RL and R C are both negligible

f ar =

Circuit conditions:1. total susceptance is zero2. inductive susceptance is equal to the capacitive susceptance of the circuit

3. impedance is maximum4. total current through the circuit is minimum5. power factor is unity

1 1

2π LC 1 – (R C2C/L)

1

2π LC

Review Course

76

Theoretical antiresonant circuit:

if BL = B C then Y = Gtherefore Z = Rand

f ar =

Q-factor:

Q = =

Q =

Resonant rise of the current:

IL = I T

since BL = B C

IL = I T

Z

1

2π LC

PQ E2/XL

P E2/R

R

XL

G + j BC

G + j (BC - BL)

G + j BC

G + j (BC - BL)




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Review Course

77

Bandwidth:

BW =f ar

Q

Review Course

78

- The above graph shows the variation in the current through the parallel RLC circuit. Noticethe current is a minimum at the resonant frequency. This is an example of a band-stop cresponse.- The shape of the band-stop response of a parallel RLC circuit depends on the value of R andas shown in the next graphs. Note that the band-stop characteristic becomes narrower as the vof R increases.

can be used to block frequencies near the resonant frequency, while allowing others to pass.




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Review Course

79

FILTERS

FILTERS - a network that possesses the ability to discriminate undesired frequency and allows thepassage of desired frequency.

• Generally, inductors and capacitors are used for filtering, because of their opposite frequencycharacteristics.

• A filter is usually a combination of capacitors, coils and resistors.

Passband – band of frequencies that the filter does not attenuate.Stopband – band of frequencies that the filter attenuates.

Classifications of Filters according to the frequency response:

1. Low-pass Filter (LPF) – filter that rejects or attenuates frequencies above the cut-off frequencyand passes frequency components below the cut-off frequency.

2. High-pass Filter (HPF) – filter that rejects or attenuates the frequencies below the cut-offfrequency and passes frequencies above the cut-off.

f c

PB SB

f c

SB PB

Review Course

80

3. Band-pass Filter - a filter that rejects or attenuates frequencies not within the two cufrequency.

4. Band-reject Filter - a filter that attenuates or rejects frequencies within the two cufrequencies.

Simple RC Low-pass Filter:

• XC at low frequency is very high, capacitor is open, V S = V O, AV = 1.

P S

f c1

S P

f c2

PB SB PB

f c1 f

R

CVS

VO




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81

• XC at high frequency is very low, V O is smaller than VS.

With RS and R L connected:

Simple RC High-pass Filter:

• At very low frequencies, C is open, therefore VO ≈0, AV ≈ 0.• At high frequencies, C is shorted, therefore, VO = V S, AV = 1.

A = 1

√1 + (ωRC) 2f c =

1

2πRC

A =R L

R + R L f c =1

2πRC

VS

C

R VO

82

with RS and RL connected:

Simple RC Bandpass Filter:

• not suitable for narrowband applications because too much interaction takes place betweesections if the cut-off frequencies are close together:

f C2 > f C1

A = f c =1

2πRC

A =R//R L

R S+R//R L f c =

1

2πRC

VS

R 1 C2

C1 R 2

LPF HPF

VO

VS R S C

R R L

VS

1

√1 + (1/ωRC) 2




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83

R2 > 10R1

• at frequencies below the passband, the circuit behaves like a HPF. For frequencies above the

passband, the circuit behaves like a LPF. Therefore,

At passband,

AV = R2 / (R1 + R2)

Band-stop Filter (or Notch Filter):

Wien-Bridge Circuit:

f c2 =1

2πR1Cf c1 =

1

2πR2C

R 1

R 2 VS

84

at high frequencies, C is shorted:

VO = voltage across R1/2 ; Av = 1/3

at low frequencies, C is open,

VO = voltage across R1/2 ; Av = 1/3

Somewhere between very low and very high frequency, we will find a frequency where bridge balances. In other words, VO will equal zero for a specific frequency.

Z1 R 1

Z2 R 1/2

R

VS

Z1 R 1

Z2 R 1/2 R

VO

VO

f f c2f o




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85

Notch filter completely rejects one particular frequency.

If A = 0, the circuit is balanced:

then f O =

and A =

BASIC LC FILTERS

Low pass:

f C =

Z1 Z3=

Z2 Z4

1

2πRC1 ωRC – (1/ωRC)

3 9 + [ωRC – (1/ωRC)]2

L

C

1 1 + √ 2

2π LC

86

High-pass:

f C =

Constant K-filter:

If we consider an L-type as a basic example, the values of inductance and capacitance cadesigned to make the product of XL and XC constant at all frequencies. The purpose is to havefilter present a constant impedance at the input and output terminals.

RK = characteristic image impedance of the filter networkZ1Z2 = RK2 = L/C (or nominal impedance)

For a T-network:

ZOT = RK + 1

C

L

1

2π √ LC (1 + √ 2 )

Z1/2 Z1/2

Z2

Z1

4Z2




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87

For a pi-network:

ZO

at cut-off, ZOT = 0 ; ZO = pi

for ZOT to be zero, let = -1 (1)

since RK = √ Z1 Z2

Z1 = (2) Z2 = (3)

Z1

2Z2 2Z2

R K

1 +Z1

4Z2

LPF HPF

ZOT ZOT

ZOΠ ZOΠ

Z1

4Z2

R K

Z2

R K 2

Z1

88

Substitute (2) and (3) in (1):

= = = -1

Z22 = ; Z2 = ± j

Z12 = - 4RK2 ; Z 1 = ± j 2RK

Use (+) inductance( - ) capacitance

therefore, for the T-network,

= j Z2 =

L1 = C2 =

f C =

Z1

4Z2

R K

4Z22Z1

4R K 2

- R K

4

R K

2

Z1/2 Z1/2

Z2

L1/2 L1/2

C

Z1 L1 1

4Z2 2 jω C2

R K 2 1

πf C πR K f C

1

π √ L1C2




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89

Formulas:

1. Low Pass

T-section

-section

L1 = C2 =

f C = R K = √ L1/C2

2. High Pass

T-section -section

L2 = C1 =

f C = RK = √ L2/C1

Z1/2 Z1/2 L1/2 L1/2

Z2 C2

Z1 L1

2Z2 2Z2 C2/2 C2/2

R K 1

πf C πR K f C

1π √ L1C2

2C1 2C1 C1

L2 2L2 2L2

R K 1

4πf C 4πR K f C

1

4π √ L1C2

90

3. Band Pass

T-section

-section

L1 = L2 =

C1 = C2 =

RK

= √ L1/C1 = √ L2/C2 f O

= √ f1f2

f2 = + +

f1 = + +

L1/2 2C1 2C1 L1/2

L2 C2

L1 C1

2L2 C2/2 2L2 C2/2

R K R K (f2 – f1) π(f2 – f1) 4πf1 f2

(f2 – f1) 14π R K f1 f2 πR K (f2 – f1)

1 1 1 12π L1C2 L1C2 L1C1

1 −1 1 12π L1C2 L1C2 L1C1




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4. Band Stop

T-section

-section

RK = √ L1/C1 = √ L2/C2

L1 = L2 =

C1 = C2 =

L1/2 L1/2

L1

C12L2 2L2

C2/2 C2/2

R K (f2 – f1) R K πf1 f2 4π(f2 – f1)

(f2 – f1) 1 π R K f1 f2 4πR K (f2 – f1)

L2

C2

92

The m-derived Filter:

This is the modified form of the constant k-filter. The design is based on the ratio of the filter cufrequency to the frequency of infinite attenuation. This ratio determines the m-factor. The m-defilter also can be high pass or low pass. The advantage is very sharp cut-off.

m=1m = 0.6±

m = 0.2±

m =almost zero

ZOT

ZOΠ

m-derived




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93

m-derived Equivalent Circuits:

Low-Pass Filter

m = √ 1 – (fc – f ) fc < f

High-Pass Filter

mL1/2 mL1/2

1 – m

4mL1

m C2

f c f α

2C1/ m 2C1/ m

L2/m

4m

1 – m2

C1

94

m = √ 1 – (f – f c) fc > f

f α f c




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95 96

DC CIRCUITS

Atom - the smallest particle of an elementthat still retains the properties of the element.Components of the Atom:

Proton - a positively charged particle of 1.6 x10 C and has a mass of 1.672x10 E-27.

Electron -a negatively charged particle of 1.6x 10 C and has a mass of 9.107x10E-32.

Neutron - A neutral particle with a mass of1.672 x 10E-27 C.

Valence electrons – electrons in theoutermost orbit of an atom.

Electric charge – exists if a body is deficientor has an excess number of electrons than its

normal values due to sharing.

Coulomb(C) is the unit of electric chargewhich was named after the French Physicist ,Charles A. Coulomb. 1 Coulomb of Charge isequivalent to 6.25 x 10 E18 electrons orprotons.

Ampere(A) - the unit of electric current. One

ampere is equivalent to once coulomb ofcharge passing a particular point in onesecond. The unit was named after the FrenchPhysicist Andre M. Ampere.

Resistance(R) – The property of a materialthat limits the current flow when subjected toa potential difference.

Electromotive Force – is the enesupplied to charge by some active de

such as a battery. A 2V of emf means tdevice supplies 2J of energy to eCoulomb of charge. EMF maintains potedifference.

Potential Difference – exists when energconverted to work as charges move frompoint to another point. A 2V potedifference between points A and B methat each Coulomb of charge will give u

energy of 2 J in moving from A to B.

Volt(V) – the unit of potential difference emf . One Volt of potential difference/eequal to one joule of work/endone/supplied per one coulomb of chThe unit was named after the ItPhysicist, Alessandro C. Volta.

Electric Current – electrons in motion. rate at which charges are moving ovperiod of time or the rate of change of chper unit time. It is caused by potedifference.

Ohm(Ω) – the unit of electrical resistanamed after the German Physicist GeoOhm.

Circular Mil –the area of a circle whdiameter is one mil.

Mil – a unit of length equivalent to 1/100an inch.


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99

RESISTANCE: The opposition to the flow of current.

where:

R – the resistance in ohms,Ω

ρρ – – tthhee r r eessiissttiivviittyy oor r ssppeecciif f iicc r r eessiissttaannccee oof f tthhee ccoonndduuccttoor r

L – the length of the conductorA – the area of the conductorV – the volume of the conductor

Resistivityρ

Length L Area

A

Ω-m m sq.m

Ω-cm cm sq.cmΩ-CM/ft ft CM

CONDUCTANCE:

where:

G – the conductance of the conductor,Siemensσ – the conductivity of the conductor

Rectangular Conductor:

Cylindrical Conductor:

if d is expressed in mils

where:

CM = circular-mils

R = ρA

L= ρ

V

L2 = ρ

2A

V

G =R

1 σ=

ρ

1

Area = d2 (CM)

Area =π

/4

[d2] sq. unit

1 inch = 1000 mils

a

b

Figure

Figure

d

1o

F-lb

BTU

1oC-kg

kcal

100

Resistivity of SomeCommon Materials

MaterialResistivity ρ

Ω-m Ω-CM/ft

Aluminum 2.83 10-8 17.02

Copper 1.724 10-8 10.371

Gold 2.44 10-8 14.676

Iron 98 10-8 589.4

Silver 1.629 10-8 9.805

EFFECT OF TEMPERATURE ON

RESISTANCE OF CONDUCTORS:

where:

R2

– resistance at temperature t2

R1 – resistance at temperature t1T – inferred absolute zero temperature

α1- temperature coefficient of resistance temperature t1

Inferred Absolute ZeroTemperature

MaterialT (oC)

Aluminum 228

Annealed Copper 234.5

Hard-Drawn Copper 241.5

Iron 180

Silver 243

INSULATION RESISTANCE OF

HIGH-VOLTAGE CABLES:

where:

ρ – resistivity of the insulating material

(Ω

-m)

l – length of the cable (m)

r 1 – radius of the conductorr 2 – radius of the insulating material

2tT

2R

1tT

1R

+=

+

R 2 = R 1 [1 + α1 (t2 - t1)]

11

tT

1α

+=

R 2

R 1

t1

t2

t

R =1

2r

rln 2π

ρ

l

r1

r2 Cable

Insulator




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101

COLOR CODING:

Color

NumberBand1,Band2

Multiplier

Band3

ToleranceBand4± %

Black 0 100 -

Brown 1 101 1

Red 2 102 2

Orange 3 103 -

Yellow 4 104 -

Green 5 105 0.5

Blue 6 106 0.25

Violet 7 107 0.1Gray 8 108 0.05

White 9 109 -

Gold - 10-1 5

Silver - 10-2 10

None - - 20

OHM’S LAW:

In an electrical circuit, the current isdirectly proportional to voltage and

is inversely proportional to

resistance.

where:

I – current in amperesV – voltage in volts

R – resistance in ohms, Ω G – conductance in Siemens

Kirchhoff’s Voltage Law (KVL)

The algebraic sum of all voltages in acircuit taken around a closed path is zero.

Convention:

First Digit

Second Digit MultiplierTolerance

igure

I = R

V I = V G

V

a

b

igure

Path b-a: Potential Rise = +V

Path a-b: Potential Drop = -V

Path b-a: Potential Rise = +Vr

Path a-b: Potential Drop = -Vr

R

I

Vr +

a

b

102

Kirchhoff’s Current Law (KCL)

The algebraic sum of all currents entering and leaving a node is zero.

Convention:

RESISTANCES IN SERIES:

The same current I exist on each resistor

The source voltage V is the summation ofvoltages across each resistor.

The total resistance RT is the sum of theindividual resistances.

RESISTANCES IN PARALLEL:

II == 1

1

R

V==

2

2

R

V==

nR

nV

V = V1 + V2 + … Vn

R T = R 1 + R 2 + .. R n

R T =TG

1=

1G

1+

2G

1+..

nG

1

I1 =1

R

V; I2 =

2R

V; In =

nR

V

I1 I4

I3 I2

node

Current entering a node: + sign

Current leaving a node: - sign

I1 – I2 – I3 + I4 = 0

I

V1 V2 Vn

R 1 R 2 R n

V

R n

I1

R 1 R 2 V

I2 In

IT




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103

The same voltage V exists across eachresistor.

The total current IT is the sum of individualcurrent passing through each resistor.

TWO RESISTANCES IN PARALLEL:

SERIES-PARALLEL CIRCUIT:

PARALLEL-SERIES CIRCUIT:

CURRENT DIVIDER:

VOLTAGE DIVIDER:

IT = I1 + I2 + .. In

GT = G1 + G2 + .. Gn

GT =TR

1=

1R

1+

2R

1+…

nR

1

R R TT ==21

21

R R

R R

+

R R

TT ==

R R

11

++

21

21

R R

R R

+

R R TT == 321

321

R R R

]R R [R

+++

I1 =21

2T

R R

]R [I

+

I2 =21

1T

R R

]R [I

+

VR 1

R 2 V2

V2 =21

2

R R

]R [V

+

R 2R 1

R 2

I1

R 1

I2

IT

R 2

R 1

R 3

R 2 R 1

R 3

104

ELECTRICAL POWER:

THEVENIN’S THEOREM:

where:

VTH – the open-circuit voltage measured

across terminals a and b with RL

removed.

RTH – the equivalent resistance with all

voltage sources shorted and all cur

sources opened, across terminals

and b with RL removed.

RL – load resistance.

IL – load current.

MAXIMUM POWER TRANSFER:

Maximum power transferred to the R L occurs when R L = R TH

P = I2

R (watts)

P = V I (watts)

P =R V

2

(watts)

VT

Circuit

Network R L

a

b

IL

R L

R T a

b

IL

igure

R L

R T

VTH

a

b

IL

Maximum Power:

Pmax =TH

2TH

2TH

L2

TH

4R

V

]2R [

R V

=

IILL == LTH

TH

R R

V

+




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105

NORTON’S THEOREM:

where:

ISC – the short-circuit current passingthrough terminals a and b with RL removed and terminals a and b short-circuited.

RTH – the equivalent resistance with allvoltage sources shorted and all currentsources opened, across terminals a and b with RL removed.

RL – load resistance.IL – load current.

SOURCE TRANSFORMATION:

NODAL ANALYSIS:

In this method, a solution is possible withn-1 equations, where n represents thenumber of nodes.

At node a :

CircuitNetwork

a

b

IL

R L R T

ISC

a

b

IL

IILL == LTH

THSC

R R

R I

+

⋅

VVTTHH == R R TTHH IISSCC

V2

zero potential

(Common Node)

R 1

V1

I1

R 2

R a

I3

Ia Ib

I2

R 3

R b

ba

Node Node

I1 = Ia + I2

21

1

R

VV

R

V

R

VV ba

a

aa −+=−

R T

VTH

a

b

ISCb

R T

a

106

At node b:

Once the node voltages are known ( Va, Vb ),all the branch currents can be calculated.

MILLMAN’S THEOREM:

Any combination of parallel connectedvoltage sources can be represented as asingle equivalent source usingThevenin’s and Norton’s theoremsappropriately.

where:V1, V2, ..Vn – Voltages of the individual

voltage sources.R1, R2,..Rn – internal resistances of the

individual voltage sources.VL – load voltageRL – load resistor

DELTA-WYE TRANSFORMATION:

WYE-DELTA TRANSFORMATION

R A =321

32

R R R

R R

++

R B =321

13

R R R

R R

++

R C =321

21

R R R

R R

++

Ib = I2 + I3

3

b2ba

b

b

R

VV

R

VV

R

V

2

−+

−=

VL=

Ln21

n

n

2

2

1

1

R

1

R

1

R

1

R

1

R

V

R

V

R

V

+++

++

....

....

R 1

V1

R 2

R LV2

VL

R 3

R 1 R 2

1

32

1

R A R B

R

2 3

R 3

R 1 R 2

1

32

1

R A R B

R C

2 3


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109

where:C – capacitance in FaradA – area of each plate sq. metersd – thickness of the dielectric material in

metersCAPACITANCE OF “n” PARALLELPLATES CAPACITOR with the sameDIELECTRIC MATERIAL

and thickness of INSULATION:

where:C – capacitance in Faradn – the number of plates

CAPACITANCE of several PARALLELPLATES CAPACITOR with DIFFERENTDIELECTRIC MATERIAL and THICKNESSof INSULATION:

PROPERTIES OF VARIOUSINSULATING (DIELECTRIC)

MATERIALS

Vacuum 1

Air 1.0006

Asbestos 2

Bakelite 5

Cellulose film 5.8

Marble 7

Mica 6

Paper (Dry) 2.2

Paper (Treated) 3.2

Glass 6

Porcelain 5.7Pressboard 6.2

Quartz, fused 3.5

Rubber 2.6

Silica, fused 3.6

Water 70

Wax, paraffin 2.2

Major Types of Capacitors: 1 Electrolytic

2 Dielectric

3 Plastic film types

4 Metallized plastic types

5 Glass and ceramics

6 Mica and mica/paper types

7 Air/vacuum types

C =d

Aεε ro ⋅⋅

d

εr

d d

εr εr

igure

C =d

Aεε1)(n ro

⋅⋅−

d1

εr1

d2 d3

εr2 εr3

igure

C =

321 r

3

r

2

r

1

o

εd

εd

εd

Aε

++

⋅

110

CAPACITANCE OF ANISOLATED SPHERE:

CAPACITANCE OF A SPHERICALCAPACITOR:

CAPACITANCE OF COAXIAL CABLE:

CAPACITANCE BETWEENTWO-PARALLEL WIRE:

CAPACITORS IN SERIES:

The same current I exist in each of capacitors.

The charge on each capacitor is the sam

The source voltage VT is the summatiothe voltages across each capacitor.

r

+

+

++

+

+

+

+

εr

igure

C = 4π · εo · εr · r

r 1

+

εr

r 2

+

+

+

+Q +-

-

-

-

--

-

igure

C = 4π·εo·εr· 12

21

rr

rr

−⋅

igure

rr

C =

1

2

ro

r

rln

εε2π ⋅⋅ (Farad/meter)

C =

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡−+

⋅⋅

1

2

2r

D

2r

D

rεoεπ

ln

(F/m)

IT = I1 = I2 = … In

QT = Q1 = Q2 = … Qn

VT = V1 + V2 + … Vn

igure

D

igure

V1

C1

V2 Vn

C2 Cn

IT

VT




56/84

111

For a number of capacitors in series,

For two capacitors in series:

The total elastance S:

CAPACITORS IN PARALLEL:

The total current IT is the sum of all currentsin each capacitor.

The total charge QT is the sum of all chargesin each capacitor.

The same voltage V exists across each

capacitors.

For a number of capacitors in parallel,

Computing for the elastance S:

SERIES-PARALLEL CIRCUIT:

PARALLEL-SERIES CIRCUIT:

TC

1=

1C

1+

2C

1+…

nC

1

CT =21

21

CC

CC

+⋅

ST = S1 + S2 + … Sn

IT = I1 + I2 + … In

QT = Q1 + Q2 + … Qn

VT = V1 = V2 = Vn

CT = C1 + C2 + … Cn

TS1 =

1S1 +

2S1 +…

nS1

CT =321

321

CCC

]CC[C

+++⋅

CT = C1 +32

32

CC

CC

+⋅

igure

Cn VT

IT

I1 I2 In

C1 C2 igure

C2

C1

C3

C2 C1

C3

igure

112

ENERGY STORED IN A CAPACITOR:

where:

W – stored energy in (Joules)C – capacitance in (Farads)V – voltage across a capacitor

in (Volts)

Q – charge in (Coulombs)

REVIEW QUESTIONS

1. If the number of valence electronsof an atom is greater than 4, thesubstance is usually

a. semiconductorb. an insulator

c. a conductord. none of the above

2. Electric current in a wire is the flowof

a. free electronsb. valence electronsc. bound of electronsd. atoms

3. EMF in a circuit is a forma. powerb. energyc. charged. none

4. The cgs unit of specific resistanceis

a. mho

b. ohm-m

c. ohm-sq.-md. ohm-cm

5. The resistance of a material is _____ its length.

a. directly proportionab. inversely proportio

toc. independent ofd. none of these

6. The value of α, i.e., temperature coefficient resistance depends upon _______ of the material.

a. lengthb. volumec. X-sectional aread. Nature

temperature

7. 7. The value of α0o

C oconductor is 1 / 236o C. The v

of α180C isa. 1 / 218 Cb. 1 / 272 Cc. 1 / 254 Cd. 1 / 265 C

8. Electrical appliances are connected in series because

a. series circuit complicated

b. power loss is greatc. appliances

different current rad. none of these

9. Electrical appliances are connein parallel because it

a. is a simple circuit

W = ½ C·V2 = ½ CQ2

= ½V·Q


57/84


58/84


59/84


60/84


Resonance the condition existing in a phase angle in (degrees)


where: A


61/84

121

Resonance – the condition existing in acircuit containing at least one resistor, aninductor and a capacitor wherein the current

behaves as if it is purely resistive.

Resonance characteristics:- total current is in phase with theimpressed voltage- power factor of the circuit is unity.- total reactive power is zero.- imaginary component of the totalimpedance ( or admittance ) in

complex form is zero.

SINUSOIDAL SOURCE:

A source (voltage or current) that variessinusoidally with time.

where:

Vm – maximum value (amplitude)w - angular velocity in (radians/sec)

t – time in (seconds)

θ, β - phase angle in (degrees)T- period in (seconds)f – frequency in (hertz)

Converting the sine function to cosine

function or vice-versa is done by using the

trigonometric identities as shown.

The PHASOR: A complex number that carries both the

amplitude and phase angle information ofany given sinusoidal function.

A sinusoidal wave can be represented by aPhasor.

TIME DOMAIN:

Polar Form:

Rectangular Form:

T =1 / f

V

π

2

π 2π

w t

igure

v( t ) = Vm cos ( wt + θ )

v( t ) = Vm sin ( wt +β

)

w = 2π

f

Vmcos(wt+θ) = Vmsin(wt+θ+90o)

Vmsin(wt+β) = Vmcos(wt+β-90o)

v( t ) = Vm cos ( wt ± θ )

V = Vm / ±θo

V = a ± jb

122

where:

Vm – maximum value (amplitude)θ - phase angle in (degrees)a – real part

jb – imaginary part

The Average Value of DifferentWaveforms: The Effective Value (RMS) of Different

Waveforms:

A ave = 0

Aave = π2

Am

Am

2ππ

One Cycle

0

Am

2ππ

Half- Cycle

0

Aave

Aave = π2

Am

Am

2ππFull-Wave Rectifier

0

Aave

Arms =2

mA

Am

2π π

Half- Cycle

0

Arms

Am

2π π

Full-Wave Rectifier

0

A rms

Am

2π π One Cycle

0

A rms

Arms =

2

mA

Arms = 2m

A

Am

2π π 0

Aave

Half –Wave Rectifier Full-Cycle

Arms =π

mA



III CAPACITOR: R-L SERIES CIRCUIT:


62/84

123

Voltage-Current Relationship of Different

Passive Circuit Elements:

I. RESISTOR:

In time-domain:

if

then

In phasor-domain:

II. INDUCTOR:

In time-domain:

if

then

In phasor-domain:

where:

XL = inductive reactance in Ω L = inductance in Henry

i = Im cos (wt + θi )

v = R

Im cos (wt +θ

i )

V = R I

i = Im cos( wt + θi )

v = wLIm

cos (wt +θ

i + 90

0

)

v = L dt

di

V = jwLI = XL I

XL = jwL = j2πfL

Am

2π π Half –Wave Rectifier Full-Cycle

0

Arms

Arms =2

mA

igure

i R

v

igure

I

R

V

i

v

L

I

V

L

124

III. CAPACITOR:

In time-domain:

if

then

In phasor-domain:

where:

XC = capacitive reactance in Ω C = capacitance in Farad

R-L SERIES CIRCUIT:

where:

Z – impedance in Ω

i = C dt dv

v = Vm cos (wt + θv )

i = wCVm cos (wt + θv + 900)

V = -jC1

wI = XC I

XC = -j C1

w= -j

fC2π1

Z =

2

L

2

XR +

Z = R + jXL

V =2

L

2

R VV +

V/

= V R + jV L

V = I Z

VL

VR

V ~

IR

L

VR

VL

I

V

θ 90

0

igure

The current I lags the voltage V by θ

V

X C

I

igure

i

v

C


R-C SERIES CIRCUIT: R-L-C SERIES CIRCUIT:


if VC < VL V I Z I Z I Z


63/84

125

R C SERIES CIRCUIT:

where:

Z – impedance inΩ

R L C SERIES CIRCUIT:

if VC > VL

Z =22

CXR +

Z = R – jXC

V = 22R C

VV +

V/-θ = VR – jVC

V = I Z

VR

V~

I R

L

C

VL

VC

igure

igure

V~

IR

C

VR

VC

VC V

I

θ 90

VR

The current I leads the voltage V by θ

VC - VL

VC

V

I

θ

VR

VL

igure

Current I leads volta e V b θ

Z = 2LC

2]XX[R −+

V =2

LC

2 ]VV[V

R −+

V = I Z

126

if VC < VL

IMPEDANCES IN SERIES:

IMPEDANCES IN PARALLEL:

where:

Y - Admittance in SiemensG - Conductance in SiemensB - Susceptance in Siemens

Two Impedances in Parallel:

Z =2

CL

2]XX[R −+

V =2

CL

2 ]VV[V

R

−+

V = I Z

igure

Current I lags

volta e V bθ

VR

VL

VL - VC

VC

V

I

θ

Vab

Z1 Zn Z2

I

a

b

igure

Vab = V1 + V2 +…Vn

Vab = I Z1 + I Z2 +…I Zn

Vab = I Zab

Zab = Z1 + Z2 +…Zn

I = I1 + I2 +… In

n21ab Z1...

Z1

Z1

Z1 ++=

Y =Z1 = G + jB

Yab=Y1 + Y2 +…Yn

Z

V

Z

V

Z

V

Z

V

n

ab

2

ab

1

ab

ab

ab ...++=

2Z

1Z

Z

1

Z

Z +

⋅= 2

igure

b

I1 I2

Vab Z1 Z2

In

Zn

a I


64/84


65/84


ELECTRICAL MACHINES l – the average length of the magnetic


MAGNETIC FLUX INTENSITY: where:


66/84

131

ELECTRICAL MACHINES

MAGNETIC CIRCUIT:

MAGNETOMOTIVE FORCE (MMF):

In SI:

where:ℑ – mmf in Ampere-Turns, At

N – the number of turns ( t )I – the current in the coil in Amps

In CGS:

where:ℑ - mmf in Gilberts

N – the number of turns ( t )I – the current in the coil in Amps

RELUCTANCE:

where:

ℜ – reluctance in At/Wb

circuit in meters

A – the cross-sectional area of the magneticcircuit in sq. meters

μ – permeability of the materialv – reluctivity

μ0 – permeability of free space

μ0 – 4π 10-7 (H/m)

μr – relative permeability

– flux in Webers

ℑ – mmf in Ampere-Turns, At

MAGNETIC FLUX:

where:In SI:

Φ – Flux in Weber

In CGS: Φ – Flux in Maxwells

MAGNETIC FLUX DENSITY:

where:

In SI: B – Wb/sq.meter = Tesla

In CGS: B – Maxwells/sq.cm = Gauss

l

Aμ I N

I N Φ

⋅⋅⋅=

ℜ⋅

=ℜℑ

=

A

Φ Β =

Iℜ

N

Φ

igure

ℑ = N·I

ℑ = 0.4π·N·I

AμμA v

Aμ

Φ

r o ⋅==

⋅=

ℑ=ℜ

lll

132

where:In SI: H – A-t/meter

In CGS: H – Oersted

PERMEABILITY:

ΦIΡ

μ = Henry/meter

RELATIVE PERMEABILITY:

μr = unitless

PERMEANCE:

P - Wb/A-t

FORCE ACTING on a CURRENT-CARRYING CONDUCTOR in aUNIFORM MAGNETIC FIELD:

F – force in NewtonB – flux density, normally perpendicular t

in TeslaI – current in the conductor in AmperesL – length of the conductor in meters

θ - the angle between B and I if not 90o

FORCE between PARALLELCURRENT-CARRYING CONDUCTORS

where:F – force in NewtonI1, I2 – conductor currents in Amperesl – length of the conductors in meters

r – the distance between the 2 paralleledconductors in meters

TRACTIVE FORCE of a MAGNET:

where:F – force per gap in NewtonB – flux density in Tesla

A – area in sq. metersμo – 4π 10-7 (H/m)

MAGNETIC FIELD ENERGY:(In the air-gap)

μ

ΒI N H =

⋅=

ℑ=

ll

μμH

Β μ ro ⋅==

(unitless) μμ μo

r =

AμΦ1

Pl

⋅=

ℑ

=

ℜ

=

F = B·I·L sinθ

F = 2 10-7

I1·I2

r

l

F =)(μ2

AB

o

2

⋅⋅

W =

o

2

μ2

AB

⋅⋅⋅l


where:W i J l

Series Magnetic Circuit:


Parallel Magnetic Circuit: d /dt – the time rate of change of flux


67/84

133

W – energy in JoulesB – flux density in Tesla

A – area in sq. metersμo = 4π 10-7 (H/m)l – length in meters

COMPARISON between MAGNETIC andELECTRIC CIRCUITS:

Magnetic Circuit Electric Circuit

MMF ℑ EMF V

Flux Current I

Reluctance ℜ Resistance R

Permeance P Conductance GReluctivity v Resistivity ρ

Permeabilityμ

Conductivity δ

agcT ℑ+ℑ=ℑ

agcT ΦΦ ℜ⋅+ℜ⋅=ℑ

agcT ℜ+ℜ=ℜ

agcT Φ=Φ=Φ

Magnetic Circuit

Electric Circuit

Iℜ

N

Φ

igure

V

I

R

I

Air-gapΦ

cℜ

agℜ

N

R ag IV

R C

Magnetic Circuit

Electric Circuitigure

134

Faraday’s Law of Electromagnetic

Induction:

An emf is induced in a conductor if suchconductor cuts the magnetic lines offorce.

The magnitude of the induced emf isproportional to the rate of change of flux-linkages.

AVERAGE INDUCED EMF:

In SI:

where:

N – the number of turns linking Φ - the flux in Weber

t – time in seconds

linkages

In CGS:

where:

- the flux in Maxwell or Line

The minus sign merely represents that

voltage is induced.

EMF Induced in a Short Length

Straight Conductor:

In SI:

where:

B – the flux density in Tesla

l – the length of the conductor in meter

v – the velocity in meters/second

In CGS:

where:

B – the flux density in Gauss

l – the length of the conductor in

centimeters

e = -Ndt

dΦ (volts)

e = B l v (volts)

e = -Ndt

dΦ

10-8 (volts)

e = B l v 10

-8 (volts)

ΦT

Φ1 Φ2

R 2 R 1 V

I2 I1

IT

Magnetic Circuit

Electric Circuitigure


v – the velocity in centimeters/second ℜ - reluctance4 10 7


ENERGY STORED in the MAGNETICM – the mutual-inductance in Henryk coefficient of coupling (k < 1)


68/84

135

EMF of SELF-INDUCTION:

where:L – the inductance of the coil in Henry

di/dt - the rate of change of current inAmps/sec

SELF-INDUCTANCE:

where:L – the self -inductance of the coil in HenryN – number of turnsA – cross-sectional area of the coil in sq. m

- flux in Webers

μo = 4π 10-7

μr – relative permeability of the core

l - length of the magnetic core in meters

Inductance of a Coaxial Cable:

where:L – the inductance in HenryD – outer diameter

d – inner diameter

Inductance of Two Long CylindricalConductors in Parallel

where:L –the inductance in HenryD – distance between the two linesr 1, r 2 – radius of the wires

e = Ldt

di (volts)

L =I

ΦN (Henry)

L =l

ANμμ

2

ro

⋅⋅ (Henry)

L =ℜ

2N (Henry)

L = 2

10-7

lnd

D

L = 4

10-7

ln

21rr

D

1Weber = 108 Maxwell

urns of wire

igure

d

D

igure

igure

r2r1

D

136

ENERGY STORED in the MAGNETICFIELD of an INDUCTOR:

where:W – energy stored in JoulesL –the inductance in HenryI – current in the inductor in Amperes

Mutual Inductance:

where:

k – coefficient of coupling (k < 1)I1 – the primary current in Amperes

N1 – primary turns N2 – secondary turns L1 – primary inductance in HenryL2 – secondary inductance in Henry

The DOT Convention:

Mutual Flux Aiding(Additive Polarity)

Mutual Flux Opposing(Subtractive Polarity)

W = ½ L I2 (Joules)

e2 = N2dt

2d Φ = kN2

dt

1d Φ

M = k 21 LL ⋅

Lta = L1 + L2 + 2 M

Lto = L1 + L2 – 2 M

M = 4

LL tota −

igure

Φ1

N1

e1ℜe2

N2

I1

e2 = Mdt

1dI

M = k N2

1dI

1d Φ = k

ℜ21NN

igure

igure



e = β l v 10-8


69/84

137

where:

Lta – the total-inductance; series-aiding inHenry

Lto – the total-inductance; series-opposingin Henry

M – mutual inductance between the coils

in Henry

Inductance in Series:

Inductances in Parallel:

Inductance in Series-Parallel:

Inductance in Parallel-Series:

DC GENERATOR:

A dc generator is an electrical machine thatconverts mechanical energy to electrical

energy.

Average EMF Induced in a Coil :In SI:

E = N t

Φ

LT = L1 + L2 + .. Ln

nL

1

2L

1

1L

1

TL

1...++=

LT = L1 +32

32

LL

LL

+

LT =321

321

LLL

]LL[L

+++

igure

L1 L2 Ln

igure

Ln L1 L2

igure

L3

L1

L2

igure

L3L1

L2

138

where:E - Average EMF induced in a coil in

volts

N - the number of turns in the coil

- flux in Webers


In CGS:

where:E - Average EMF induced in a coil in

volts

N - the number of turns in the coil

- flux in Maxwells or Lines


IInndduucceedd EEMMFF iinn aa CCoonndduuccttoor r ::

In SI:

where:β - Flux density in Webers/sq.m = Tesla

l - length of the conductor in meters

v - velocity in meters/sec

In CGS:

where:

β - Flux density in Linews/sq.cm =Gauss

l - length of the conductor in

centimeters

v - velocity in centimeters/sec

NOTE:β, l , v must be mutually perpendicul

not, the equation must be multiplied bysine of an angle between pairs of quantiti

DIRECTION OF THE INDUCED EMF:(Fleming’s Right Hand Rule)

Forefinger – represents the direction of tflux (pointing N to S)

Thumb – represents the direction of motio

the conductor movesMiddle Finger – represents the directionthe induced EMF.

TYPES of ARMATURE WINDING

I. LAP WINDING:

E = N

t

Φ

10-8

e = β

l

v

β

yb

yf

yc

yc = m



Wave Winding 2 mThe generated emf E is thus proportional


70/84

139

where:yb – back pitch (an odd number)

yf – front pitch (an odd number)

+ ⇒ progressive type of winding

- ⇒ retrogressive type of winding

Type of Winding

myc

Simplex Lap

(SL)

1 1

Duplex Lap(DL)

2 2

Triplex Lap

(TL)

3 3

Quadroplex Lap(QL)

4 4

II. WAVE WINDING:

where:y – average pitch

Z – the number of elements or coil sides

P – number of poles

C – number of commutator segments

+ ⇒ progressive type of winding

- ⇒ retrogressive type of winding

All other terms have been previously

defined.

Number of Brushes:

Type of Winding

Lap Winding P

Wave Winding 2

Number of Armature Parallel Paths:

Type of Winding a’

Lap Winding m P

yb = yf ± 2

m

yf yb

yc

yc =

2P

mC ±

y =P

2Z ±

y =2

yy f b +

140

DC-GENERATOR VOLTAGE

EQUATION BETWEEN BRUSHES:

In SI:

where:E – generated emf in the armature in

volts

P –the number of poles

–the flux per pole in Webers

Z – the number of armature conductors

SS – – tthhee ssppeeeedd iinn RRPPMM a’ – number of armature parallel paths

In CGS:

where:

–the flux per pole in Maxwells or

Lines

All other terms have been previously

defined.

After the machine has been assembled, a’,P, Z could be taken as a constant, thus

The generated emf E is thus proportional

- the flux

- the speed S

TYPES of DC GENERATORS :

Separately-Excited DC Generator:

SHUNT GENERATOR:

E =)(a'60

SZΦP

⋅⋅⋅⋅

E =)(a'60

SZΦP

⋅⋅⋅⋅

10-8

E = K

S

+

_

R f

If

DC E

R a

V

IL

R L

Po = V IL

Pa = E Ia

Pa = Po + Losses

E = V + Ia R a

Ia = IL + If

EV

R f

R a

If

+

_ IL Pa

Ia R L

Po




71/84

141

where:Po – power output in wattsPa – power developed in the armature in

watts

V – output (terminal/load) voltage in volts E – generated emf in the armature in voltsIL – load current in amperes If – field current in amperes Ia – armature current in amperes RL – load resistanceRf – shunt field resistanceRa – armature resistance

EXTERNAL CHARACTERISTIC OF ASHUNT GENERATOR:

A shunt generator is considered as having afairly constant output voltage.

SERIES GENERATOR:

where:Rs – series field resistance

All other terms have been previously defined.

EXTERNAL CHARACTERISTIC OF ASERIES GENERATOR:

A series generator could either be used as avoltage booster or a constant currentgenerator.

If = f R

V

EV

R s R a +

_ IL Pa

Ia

R L

Po

Po = V IL

Pa = E

Ia

Pa = Po + Losses

E = V + Ia (R a + R s)

Ia = IL

LLooaadd CCuurrrreenntt

L L o o a a d d V V o o l l t t a a g g

e e

142

LONG SHUNT

COMPOUND GENERATOR:

where:

All other terms have been previously defined.

SHORT-SHUNT

COMPOUND GENERATOR:

All other terms have been previously defi

EXTERNAL CHARACTERISTIC

COMPOUND GENERATORS:

Po = V IL

Pa = E Ia

Pa = Po + Losses

E = V + Ia (R s + R a)

Ia = IL + If

If = f R

V

Po = V IL

Pa = E Ia

Pa = Po + Losses

E = V+ IL

R s + Ia

R a

Ia = IL + If

If = f

SL

R

R IV ⋅+

LLooaadd ccuurrrreenntt

RRaatteedd

L�

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