ece 410 – spring 2012 lecture #24 review for exam #2

ECE 410 – Spring 2012

Lecture #24Review for Exam #2

Circuit Terminology

Practice

How Many Nodes? How Many Essential Nodes?How Many Branches? How Many Essential Branches?

How Many Meshes?

Node-Voltage Method

• 1st Steps in Node-Voltage Method– Make a neat layout of your circuit with no branches crossing over– Mark the essential nodes– Pick one of the essential nodes to be your reference node and mark it.

(typically the node with the most branches)– Define the node voltages (voltages from reference node to other essential

nodes)

Node-Voltage Method Continued

• With the node voltages labeled in your circuit you are ready to generate your node-voltage equations– This is done by writing each of the essential

branch currents leaving the essential nodes in terms of the node voltages and then summing them to zero in accordance with Kirchoff’s current law.

Write Node Voltage Equations

Write branch currents exiting each essential node in terms of the node voltages and sum them to zero in accordance with KCL.

0251

10 2111

vvvVv

(1) (2)(3)

02102

212

vvv

Using N-V Method with dependent sources

• In order to solve a circuit with dependent sources, the node-voltage equations must also include the constraint equation that governs the dependent source written in terms of the node voltage.

Supernode Technique

• Sometimes a current flowing out of a node cannot be expressed in terms of the node voltages due to a voltage source between essential nodes

• These nodes can then be treated as a single ‘supernode’.

Mesh Current Method

• Only applicable to planar circuits.• Mesh current is the current that exists around the

loop of a mesh.. It is not necessarily a branch current.

• Apply KVL around each mesh loop, expressing all voltages in terms of the mesh current.

• This will create enough equations to solve the circuit• Once mesh currents are determined, the branch

currents (and voltages) can be determined.

Mesh-Current Setup Example

0)()(

0)()(

0)()()(

0)(

22433474

43343823

612832242

5162111

RiiRiiRi

RiRiiRii

RiiRiiRii

RiRiiRivs

4 equations, 4 Unknowns (i1, i2, i3, and i4)

Example Continued

• With mesh currents found, group current terms in each equation

0)()(

0)()(

0)()()(

0)(

22433474

43343823

612832242

5162111

RiiRiiRi

RiRiiRii

RiiRiiRii

RiRiiRivs

)(00

0)(00

)(0

00)(

237433221

44383821

2483682261

43625611

RRRiRiRii

iRRRiRii

RiRiRRRiRi

iiRiRRRivs

Solve equations for four mesh

currents

Mesh Current Technique with Dependent Sources

• Much like the Node Voltage technique, solving circuits with a dependent source with the mesh current technique requires the inclusion of a constraint equation for the dependent source.

• These equations should be expressed in terms of the mesh currents.

Special Cases with Mesh Current

• When using the mesh current technique, current sources in your circuit can have similar effects as voltages sources in the node voltage technique

• Current sources can at times directly indicate a mesh current, simplifying your solution

• Sometimes a current source can prevent you from writing an expression for the voltage drop across it in terms of the mesh currents or anything else

• This situation leads to the concept of the supermesh (analogous to the supernode)

Supermesh

• If a current source lies in a branch of your circuit, you can ‘remove’ the current source when creating your mesh currents.. Avoiding the branch that contained the source

Source Transforms

– Another method like Delta-Wye and Parallel/Series combinations to simplify circuits

– Can switch between equivalent sources: a voltage source with series resistor or current source with parallel resistor

How to Transform• Each of the two equivalent sources must

produce the same current through any load connected across the source.

RLRL

Ls

L

s

RR

Ri

RR

v

Lsload RR

RiI

(current divider)L

sload RR

vI

Riv ss R

vi ss OR

Example

What is Power of 6V source?

i = (19.2V-6V)/16 = 0.825A P = i*v = 0.825A*6V = 4.95W

i

Extraneous Resistors

• Resistors in parallel with the voltage source, or in series with the current source have no effect on the load.

Terminal Behavior

• We are often interested in looking at the behavior of an electrical circuit at output terminals.

• If a load is hooked to a circuit, how does the circuit behave relative to the load?

• If the circuit is made of linear elements, the entire circuit can be simplified into a Thevenin (or Norton) Equivalent circuit.

Justification

• We can consider the entire circuit a black box with two output terminals. The behavior at these terminals can be modeled by a voltage source and a series resistor.

Finding the Thevenin Voltage

• If we look at the terminal with no load attached (open circuit output), the voltage between the output terminals is VTh.

With an open circuit, no current flowsThis means no voltage drop across RTh

Hence Vab = VTh

Finding Thevinen Resistance

• Now lets look at what happens when the output is short circuited

• If we can find the short circuit current and have already found VTh, then we can easily calculate RTh

isc

sc

ThTh

Th

Thsc

i

VR

R

Vi

Example

• Find Vab (open circuit voltage) in the following circuit.

Use node voltage:

Vv

v

vv

vv

32

1605

601004

03205

25

1

1

11

11

v1

Vab = VTh = 32V

Example Continued

• Find closed circuit current and calculate RTh

Vv

v

vvv

vvv

16

16010

6051004

04

3205

25

1

1

111

111

Again… Use node voltage:

84

32

44

16

A

V

i

VR

AV

i

sc

ThTh

sc

v1

Final Equivalent Circuits• The resulting Thevinin Equivalent Circuit is shown below:

• Using a source transform the Thevenin circuit can be represented as a current source in parallel with the same RTh. This is the Norton Equivalent Circuit.

Alternate Methods

• Other methods exist for finding the Thevenin equivalent. The simplest is simply using source transforms and parallel/series simplifications.

Source Deletion

• Here is another method that works well for circuits containing only independent sources.

• Find RTh by ‘deactivating’ all the independent sources in the circuit and finding the resulting equivalent resistance from terminal a to terminal b.

• ‘Deactivating’ a votage source turns it into a short circuit, ‘deactivating’ a current source turns it into an open circuit.

Modification for Dependent Sources

• If dependent sources exist in the circuit, you can deactivate the independent sources and connect a test voltage source or test current source to the output terminal.

• The voltage delivered by the test source divided by the current supplied by the test source is equal to RTh.

ThT

T Ri

v

Basics of Power Transfer

• We have some network of sources and linear elements and will connect this network to a load RL. We want to maximize power in the load.

• This process is easiest to analyze by using the Thevenin equivalent circuit.

Analyzing Thevenin Circuit with Load

• The power dissipated in the load can be written as P=iL

2*RL • By using the Thevenin Equivalent circuit,

finding an expression for this power is straight-forward.

LLTh

Th

L

LTh

Th

RRR

VP

RiP

RR

Vi

2

2

Finding condition for Max Power Transfer

• To find the load value that give maximum power dissipated in the load, you must take the derivative of the Power expression with respect to the load resistor and set it equal to zero:

0)(

1

)(

2

0)()(2

0)(

23

232

22

LThLTh

L

LThLLThThL

LLThL

ThL

RRRR

R

RRRRRVdR

dP

RRRdR

dV

dR

dP

LTh

LTh

LTh

LLTh

LTh

LTh

LTh

L

RR

RR

RR

RRR

RR

RR

RR

R

0

0)(

2

0)(

)(

)(

2

3

33

Max Transfer when Load and Thevenin Resistance are the

same

How much power is transferred?

• Using our relation for max power transfer (RL=RTh) we can use our relation for power dissipated in the load to find the total power transferred.

• Power transferred increases as RTh decreases.

Th

Th

L

Th

L

LTh

LL

LTh

R

VP

R

V

R

RV

RR

RVP

4

4)2()(

2

max

2

2

2

2

2

max

Since RTh=RL

Superposition

• All linear systems obey the principle of superposition.

• This means that whenever a linear system is driven by multiple sources, the total response of the system is the sum of the responses to each individual source

• This principle is used in all areas of engineering and physics and is of course useful in Circuit Theory

What does this mean for solving circuits?

• The value of voltage or current relating to any element in a circuit can be thought of as the sum of the voltages and currents resulting on that element from each of the independent sources in the circuit individually.

=

+v2’ v2’’

v2

v1’ v1’’

v1

'''

'''

222

111

iii

iii

'''

'''

444

333

iii

iii

'''

'''

222

111

vvv

vvv

Uses for superposition

• Superposition is highly useful if you have a circuit that you have already solved that is altered by adding a new source of some type.

• All you need to do is deactivate all the original sources and find the values resulting from your new source

• The true altered values in your circuit are just the original values plus what results from the new source

Example

• Problem 4.9

With only 24V source:

With only 40mA source:

VVVvo 8.4125

2524

258020

2524'

VmAmAvo 8.010025

)100(2540100//2540''

Vvvv ooo 48.08.4'''

Sensitivity Analysis

• When creating a circuit you must use real components with inexact values

• Resistors vary from their nominal value by up to 20% depending on the type specified

• Resistors with more accuracy cost more money.• You need a way to be able to know what tolerance in

resistor values your circuit can accept and still manage to meet specifications

• Spending extra money only on resistors that create the biggest change in system performance is highly desired.

Example Sensitivity Analysis

• Express output value in terms of the system components:

+

-

vout

321

313

321

1

RRR

RViRv

RRR

Vi

out

i

R 1

5k

R 2

1 k

R 32 . 5 k

V 110V dc

0

Continued

• To find the variation caused by a specific component, take the derivative of the output variable with respect to that component.

2321

31232131

1

132131

11

321

31

RRR

RVRRRRV

dR

dv

RRRRVdR

d

dR

dv

RRR

RVv

out

out

out

Numerical solution

• Put variable parameters into the equation to find out how quickly vout changes with respect to the resistance of R1.

• This can be done for every component in the circuit

VE

dR

dv

kkk

k

RRR

RV

dR

dv

out

out

446.3

)5.215(

)5.2(10

1

22321

31

1

Output Data for our circuitDC SENSITIVITIES OF OUTPUT V(N00465)

ELEMENT ELEMENT ELEMENT NORMALIZED NAME VALUE SENSITIVITY SENSITIVITY (VOLTS/UNIT) (VOLTS/PERCENT)

R_R1 5.000E+03 -3.460E-04 -1.730E-02 R_R2 1.000E+03 -3.460E-04 -3.460E-03 R_R3 2.500E+03 8.305E-04 2.076E-02 V_V1 1.000E+01 2.941E-01 2.941E-02

This agrees with our calculated value

Nominal Output Voltage = 2.941V

Calculating Effects due to Component Change

• If we want to know how much the output will change due to the change in a component value we can just multiply the sensitivity by the amount of change.

• EXAMPLE– If we increase the value of resistor R3 by 100Ω

VVVvvv

VVEv

outoutout

out

024.308305.941.2

08305.0100*4305.8

'

Effects of multiple changes

• Using the superposition principle, we can see that the total change in output variable due to changes in multiple elements is just the sum of the changes due to each element change.

• EXAMPLE– R3 increases by 100Ω, R1 and R2 increase by 100Ω each.

VVV

vvvvv

VVEv

VVEv

VVEv

RoutRoutRoutoutout

Rout

Rout

Rout

955.20346.0346.08305.941.2

0346.0100*4460.3

0346.0100*4460.3

08305.0100*4305.8

)()()('

)2(

)1(

)3(

213

Comparison to Exact Value

• Sensitivity analysis only gives an estimate based on a linear assumed change and is less accurate the larger the variation you use. Here is the exact PSpice simulation for the changes we made:

Sensitivity analysis shows Vout=2.955V

Match is exact in our case

Resistor Tolerances

• Real resistors have tolerance values that define the amount of variation in their resistance value that can be expected.

• Resistors typically come with tolerances of 20% (rare), 10%, 5%, 2% or 1%.

• A 1k resistor with 10% tolerance can have a value anywhere between 900Ω and 1100Ω.

• A 1k resistor with 1% tolerance may have a value anywhere between 990Ω and 1010Ω

Example Circuit

• Here’s a circuit simulated in PSpice:

The output voltage is nominally 34.27VWhat range would the output voltage have if we used 10% resistors?What if we used 5%What about 1%

PSpice Sensitivity Output

The normalized sensitivity for each element can be multiplied by the % tolerance of the resistor to determine what he maximum shift in output voltage could be.

To find Vout max, choose the sign of your percentage tolerance to result in a positive voltage shift, to find Vout min, choose the sign of your percentage tolerance to result in a negative voltage shift.

Analysis in ExcelELEMENT ELEMENT ELEMENT NORMALIZED % tolerance % tolerance % toleranceNAME VALUE SENSITIVITY SENSITIVITY 10 5 1

(VOLTS/UNIT) (VOLTS/PERCENT) voltage shift voltage shift voltage shift

R_R1 1.00E+02 7.19E-02 7.19E-02 0.72 0.36 0.07R_R2 2.00E+01 -2.64E-01 -5.28E-02 0.53 0.26 0.05R_R4 1.50E+01 1.38E+00 2.06E-01 2.06 1.03 0.21R_R5 2.50E+01 5.42E-01 1.35E-01 1.35 0.68 0.14R_R3 3.00E+00 -8.29E-01 -2.49E-02

sum 4.67 2.33 0.47Vout max 38.94 36.60 34.74Vout min 29.60 31.94 33.80

What if we want to keep the max output voltage under 37 volts,Which resistor(s) should we attack first in terms of reducing tolerances?

Analysis ContinuedELEMENT NORMALIZED ToleranceNAME SENSITIVITY %

(VOLTS/PERCENT)

R_R1 7.19E-02 10 0.72R_R2 -5.28E-02 -10 0.53R_R4 2.06E-01 5 1.03R_R5 1.35E-01 10 1.35R_R3 -2.49E-02 10 -0.25

sum 3.38Vout max 37.65

ELEMENT NORMALIZED ToleranceNAME SENSITIVITY %

(VOLTS/PERCENT)

R_R1 7.19E-02 10 0.72R_R2 -5.28E-02 -10 0.53R_R4 2.06E-01 5 1.03R_R5 1.35E-01 5 0.68R_R3 -2.49E-02 10 -0.25

sum 2.71Vout max 36.98

We can reduce the maximum output variation from +/- 4.67V, to only +/-2.71V, by only improving the tolerance on two most sensitive resistors. A complete change of all the resistors would have only reduced the output variation to +/- 2.33V

Op Amps

• Operational Amplifiers (Op Amps) are a basic building block of many electronic circuit designs.

OpAmp Terminals

• Op amps have at minimum five basic terminals that are uses– An inverting input (marked with -)– An non-inverting input (marked with +)– A positive power supply– A negative power supply– An output terminal

Output behavior

• In an ideal op amp, the open loop gain is infinite, so if the amplifier is to remain in the linear region Vn=Vp this is called the virtual short.

• In addition, an ideal op amp has no current flowing into the inverting and non inverting terminals. ip=in=0

• How does the amp remain in this condition?– Negative feedback is used to keep the input voltages equal– This means that there is a circuit path from the output of

the op amp to it’s inverting terminal

Example Negative Feedback Control System #1 - HVAC

• Lets look at other negative feedback systems to help with understanding

• One of the most ubiquitous is the thermostat for your AC/furnace• It compares two inputs (the desired temperature Tset and the

actual ambient temperature Troom)

+

-

Tset

Troom

AC/Furnace control

Feedback loop

roomset TTAHVACdrive

If Troom > Tset output is negative and AC cools room

If Troom < Tset output is high and furnace heats the room

Troom is forced to Tset What happens on a very cold or hot day when HVAC can’t keep up?

Method for solving circuits with negative feedback op amp

• Assume that it is operating in linear regime (i.e. that the output isn’t saturated)

• Thus assume vn=vp (remember in = ip = 0)

• Find vp it’s value will not depend on the output of the op amp

• Find an expression for vn this will include the output voltage of the op amp

• Set vn = vp… you now have a governing equation for the op amp circuit

• Use this equation to solve for whatever parameter is of interest

• If you find the output is saturated at the supply limit, use this limiting value for the output voltage in your final equation.

The Inverting Amplifier

In the linear region:

ss

fo v

R

Rv

The inverting gain is:

s

finv R

Rg

The type of circuit shown in the previous assessment problem is the general layout of an inverting amplifier

Summing Amplifier

• An extension of the inverting amplifier is the summing amplifier

• Superposition principle can be used to show that the output of the summing amp is the sum of outputs of each of the individual inputs.

cc

fb

b

fa

a

fo v

R

Rv

R

Rv

R

Rv

Non-Inverting Amplifier

• The non-inverting amplifier amplifies the input voltage buy a positive gain factor.

In the linear region:

ss

fo v

R

Rv

1

The gain is:

s

f

R

Rg 1

Differential Amplifier

• The differential amplifier is used to amplify the difference between two input voltages

In the linear region:

aba

bo vv

R

Rv

The gain is:

a

b

R

Rg

c

d

a

b

R

R

R

RIf: