ECE 307 DC Lecture 7 (1)

1ECE 307 – Basic Electrical Engineering

ECE 307 - Lecture 7Superposition, Maximum Power Transfer, and Dependent Sources

Department of Electrical and Computer Engineering

Clemson University


Electric Circuit• Performs a function:oProcess InformationoTransfer Power• Characterized by:oVoltagesoCurrentsoPoweroEnergy

Circuit Components• Resistor• Voltage Source• Current Source• Switch• Dependent Sources

Connections• Terminal• Node• Branch• Loop• Mesh

SimilarElectric Circuit

Reductions• Source Transformation• Parallel• Series• Thevenin• Norton

Analysis ToolsKirchhoff's Current Law

• Node Voltage MethodKirchhoff's Voltage Law

• Mesh Current MethodSuperposition

Overview of DC Electric Circuits

DC Lecture 7– Superposition, Maximum Power Transfer, Dependent Sources


Thevenin Equivalent Circuit to Understand Biological System

Journal of Physiology“ In order to demonstrate the relative importance of extracellular shunt pathways upon epithelial parameters of membrane potential and resistance, a simple Thevenin equivalent model was used … “

Thevenin equivalent circuit used to lump the electrical behavior of part of the tissue (a subcircuit of

the whole circuit)

Cornea Membrane


General Definition of Superposition

• A general property of any linear system where the response to a sum of inputs is equal to the sum of the individual responses.

• For the system T( ) and the inputs x1..xN

1 2 1 2( ) ( ) ( ) ( )N NT x x x T x T x T x

Response to Sum of Inputs Sum of Responses to Individual Inputs

Example Linear System:

?

?

3x

(1 2) (1) (2)

3(1 2) 3 1 3 2

9 9

T x

T T T

Example Non-linear System:

2

?

?2 22

x

(1 2) (1) (2)

(1 2) 1 2

9 5

T x

T T T

Ohm’s Law Equations

Power Equations


Superposition in Linear Circuits

• In a circuit, each branch current and voltage is the cumulative effect of the individual voltage and current sources.

Source N

Source 1

Source 2…

+

-

voio

vo-1io-1

vo-2io-2

vo-Nio-Nio = io-1 +io-2…+…io-N

vo = vo-1 +vo-2…+…vo-N

1 2 1 2( ) ( ) ( ) ( )N NT x x x T x T x T x

ElectricCircuit

Response to SumSum of Responses


Procedure to Analyze a Circuit Using Superposition

• Choose one current or voltage source and remove all others– Replace voltage source by short circuit– Replace current source by open circuit

• Solve for voltages and/or currents of interest.

Repeat for EVERY source in the circuit


Example 1 : Compare Superposition to Node Voltage Result

4Ω1A 3A2A

a

b

+

-io

vo

Find io and vo by Superposition

out node a

1 2 3 04

aN

vi A A A

Use Node Voltage Method:

1 4 2 4 3 4

1 2 3 4

= 24V

av A A A

A A A

Solve:

1 2 3

1 2 3

( ) ( ) ( )( )

T x T x T xT x x x

Matches the original definition of superpositionT(xi)=4Ω*xi

Superposition works for this circuit

Aside: Show solution using a previous method

0 24av v V


Example 1 (cont) : Compare Superposition to Node Voltage Result

4Ω1A 3A2A

+

-io

vo

Repeat using Superposition:

Consider the 1A Source

4Ω1A+

-io-1 vo-1

1 11 , 1 4 4o oi A v A V

4Ω2A+

-

io-2 vo-2


2 22 , 2 4 8o oi A v A V


4Ω3A+

-

io-3 vo-3

3 33 , 3 4 12o oi A v A V

vo=vo-1 +vo-2+vo-3 =24V io=io-1 +io-2+io-3 =6A

Solve using Superposition

Same as NVM


Example 2: Solve Circuit by Superposition

Use superposition to find the currentthrough the 8 Ω resistor in the circuit.

Find the current in the 8 resistor due to each source, we consider two circuits:

av bv cvShort

8i

8i 8i

av bv cv

Basically we have to solve the circuit twice.

Superposition may require more work


Example 2 (cont): Solve Circuit by Superposition

Apply NVM:

av bv cv

10V 0Va cv v

i out node a

10V 02 8 4

b b bv v vi

40 V7bv

85 A78

biv

0V 20Va cv v

Apply NVM:

i out node a

20V 02 8 4

b b bv v vi

40 V7bv

85 A78

bvi

Short

8i 8i

av bv cv

Solve: Solve:

8 8 810 A7i i i

b b


Example 2 (cont): Solve Circuit by Superposition

A Check of the Node Voltage Method Solution using an Alternate Approach

4 84 8

84 83

83 40

8 7833

10 V2

v V

These are in parallel

Use Voltage Divider Rule:

Parallel resistors have same voltage

83

8 4v v v 83

857 A

8

vi

2 82 8

82 85

Use Voltage Divider Rule:85 40

8 7855

20 V4

v V

Parallel resistors have same voltage

85

8 2v v v 85 5

8 7 A8

iv

The Superposition approach requires that we consider two circuits:

8i 8i

8 8 810 A7i i i


Example 3: Solve Circuit with Voltage and Current Sources by Superposition

Use superposition to find the currentthrough the 10 resistor in the circuit.

Open circuit

Short circuit

Turning the current source off results inan open circuit (zero current) in the branchcontaining the 8 resistor.

Turning off the voltage source resultsin a short circuit in series with the 2 resistor.

This is Circuit A This is Circuit B

Find the current in the 10 resistor due to each source, we consider two circuits:

10i

10i 10i


Example 3 (cont): Solve Circuit with Voltage and Current Sources by Superposition

9 4 13

Resistors in series

av bv

40V40V

02 10 13

solve to get:2600 V88

a

b b b

b

vv v v

v

10

260 A10 88

biv

Zero current through 8 resistormeans zero voltage drop.

Apply NVM

Circuit A

10i 10i

10i 10i



Resistors in parallel

Use current divider

9 53

4 121A A4 9 44

i

Next, expand circuitback out and usecurrent divider againto find current through10 resistor.

9102 2 12 1 A

10 2 12 44 22i i

Use current divider

Circuit B

10i 10i

By making this parallel connection we have lost the identity of i10

10i



10260 A88

i 101 4A= A22 88

i

10 10 10260 488 88

264 3A88

i i i A A

A

Circuit A Circuit B


Maximum Power Transfer

This is also a good model of a “real” battery, i.e., a battery has internal resistance

Thevenin Equivalent

Maximum Power Transfer:Given that the Vth and Rth are fixed, what is the value of RL such that PL (power to the load) is maximized ?



Lth

th L

RV VR R

Voltage Divider

2221 L

L thL L th L

RVP VI VR R R R

Power Dissipated in load, RL:

Differentiate PL and set to zero find extremum:

22 2

2 2

20th L L th LL L

th thL L th L th L

R R R R RdP Rd V VdR dR R R R R

Thevenin Equivalent



L thR R

22

2

2 2

2 2 2 2

2 2 2

20

2 0

2 2 2 0

0

th L L th Lth

th L

th L L th L th

th L th L L th L th

th L th

R R R R RV

R R

R R R R R V

R R R R R R R V

R R V

2 22

max 2

2

4 4

th

th

th thL th th

th th

ViR

V VP i R RR R

Solve for condition on RL for maximum power transfer:

Maximum power delivered to load:

Multiply by denominator

Expand and collect

Condition for Maximum Power Transfer:

+

-va

i+-

Rth

RL=RthVth


Example 4: Calculate Load Resistance for Maximum Power Transfer

Find the value of the variable resistor that draws maximum power from the circuit. Find Pmax.

Approach: 1) Find Thevenin Equivalent2) Use Formula for R0 and power

The arrow means that the value of the resistance can be changed – a variable resistor

Remove load and turn off all sources (open current sources, short voltage sources) to find Rth

10000 600010000 6000

1.25k 10k 2k 4k

12501250 375050005k

thR


Example 4 (cont): Calculate Load Resistance for Maximum Power Transfer

Mesh 1 1 9mAi

Mesh 2 2 1 2 22000 4000 10000 50 0i i i i

Solve 2 2mAi Use KVL to find Vth

2 12550 10000 0k thi v V

Remove load and find open circuit voltage, Voc=Vth.

Mesh Current Method

2

3

50 10000

50 10000 2 10

50 20 30V

thV i

Current in the 1.25k resistor is zero and therefore the voltage is v125k=0V

Solve

1i 2i


Example 4 (cont): Calculate Load Resistance for Maximum Power Transfer

22

max

304 4 5000

0.045W45mW

th

th

VP

R

0 5kthR R

Formula for maximum power transfer:

i+-

Rth=5k

R0=RthVth=30V

0

2

20 0 0

0

2

00

305000

th

th

th

th

Vi

R R

VP i R R

R R

V RR

Power Transfer:


Example 5: Calculate Load Resistance for Maximum Power Transfer (a little more difficult)

Find the value of load resistance that results in maximum power transfer to the load from the circuit.

2

max 4if

th

th

L th

VP

R

R R

Recall what we know about maximum power transfer:

Approach: 1) Find Thevenin equivalent Circuit2) Formula for RL and power


Example 5 (cont): Calculate Load Resistance for Maximum Power Transfer (a little more difficult)

Remove load and find Vth Use Mesh Current Method

drops, CW

100 4 4 20 0k a b av i i i

drops, CW

30 4 4 4 0k b b b av i i i i

Solve27 A2

7A

a

b

i

i

Use KVL for Vth

20 4 4 0

20 4

102V

a b th

th a b

th

i i V

V i i

V

ai

bi

Possibly confusing point:The purple line does not define a Mesh Current and does not contribute to the voltages in the 4 resistors.


Example 5 (cont): Calculate Load Resistance for Maximum Power Transfer (a little more difficult)

Turn off sources and find Rth

4 4 2

4 4 2 2.4

2.4th

L

RR

2

max

2

4

1021084W

4 2.4

th

th

VPR

These are resistors in parallel

Maximum Power Delivered to the Load

Resistance for maximum power transfer:


New Components – Dependent (or Controlled) Sources

• The constant source values are dependant on a voltage or current elsewhere in the circuit– Dependent (Controlled) Voltage Source– Dependent (Controlled) Source

Xvα

Xiβor

Xiα

Xvβor

Controlled Voltage Source

Controlled Current Source


Dependent (Controlled) Voltage Source

Voltage Controlled Voltage Source

Xvα

numberyou must be given this location in the same circuit

Current Controlled Voltage Source

XIβ


+

-vx

ix

Behavior: produces voltage proportional to a voltage or current in the circuit


Dependent (Controlled) Current Source

Behavior: produces current proportional to a voltage or current in the circuit

Voltage Controlled Current Source

Xvα


Current Controlled Current Source

XIβ


+

-vx

ix


Circuit Analysis With Dependent Sources

• Each of these device will add another equation to the Mesh Current or Node Voltage analysis. – Extra equation describes the controlling

relationship, e.g. aix

– Extra equation for each dependent device in the circuit, e.g., 2 dependent sources will add 2 equations


Example 6: Example 3.12 from Textbook

vx=2v3

The value of v3 determines the value of vx.Example: If v3=2.5V then vx=5V

Find the Node Voltages


Example 6 (cont): Example 3.12 from Textbook

1. Label all Nodes and select Reference Node

2. Identify dependent nodesNo KCL at Node 1 since it is a

dependent node (Note that vx is a dependent source )

n=4

m=1

3. Write n-1-m=2 KCL eqns +1 eqn to describe dependent node + 1 eqn to describe dependent source

Reference Node

1 2 3

4A dependent node is not the same as a dependent source

2 32 1

out Node 2 1 2

3 2 3

Node 3 2 3

0

0

n

nout

v vv vi IR R

v v viR R

32Xv v


Equation for the dependent node:

1 1 0X Xv v v v

for use in simultaneous equations

New equation for dependent source: 3 32 2 0X Xv v v v

4 equations / 4 unknowns

1 1 2 2 1

2

32 1 2

1 1 1 1 0

01 1 10 0 001 0 0 1

0 0 2 1X

R R R R v IvvR R Rv

4. Solve

Reference Node

1 2 3

4


32Xv v


Example 7: Example 3.13 from Textbook

1. Label all meshes

2. Identify dependent meshes (current sources)

3. Write n-m=3 KVL eqns. + 1 eqn to describe dependent sources

n=3

m=0 The value of v determines the value of source (2v)

Find the Mesh Currents



M2:

M3:

1 1 1 1 2 2v drops, CW

2 1 2 2 3 2 3 4v drops, CW

3 2 4 3 5v drops, CW

( ) 0

( ) ( ) 2v 0

( ) 2v 0

k

k

k

v v i R i i R

v i i R i R i i R

v i i R i R

M1:

New: Describe dependent current source in terms of loop currents

1 2 2( )v i i R

4. Solve

4 equations with 4 unknowns


Summary

• Superposition– Total response = sum of response to each

individual source.• Maximum Power Transfer

– In a Thevenin equivalent circuit, maximum power is transferred to an adjustable load when RTH=RL

• Dependent Sources– Circuit elements– An external voltage or current controls the amount

of voltage or current produced by the source.


Molten Carbon Material

Santanu Kundu and Martin Beagley

Description:Photograph, this material is called mesophase pitch, a liquid-crystalline material that is used to manufacture carbon fibers and composites.

Science As Art at Clemson (http://geo.ces.clemson.edu/gallery/main.php)

http://geo.ces.clemson.edu/gallery/main.php?g2_itemId=247

ECE 307 DC Lecture 7 (1)

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