Topic 1 b_basic_concepts_and_theorem

1

Topic 1B

Basic Concept and Theorem

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Equivalent Circuits – The ConceptEquivalent circuits are ways of

looking at or solving circuits. The idea is that if we can make a circuit simpler, we can make it easier to solve, and easier to understand.

The key is to use equivalent circuits properly. After defining equivalent circuits, we will start with the simplest equivalent circuits, series and parallel combinations of resistors.

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Equivalent Circuits: A Definition

Imagine that we have a circuit, and a portion of the circuit can be identified, made up of one or more parts. That portion can be replaced with another set of components, if we do it properly. We call these portions equivalent circuits.

Two circuits are considered to be equivalent if they behave the same with respect to the things to which they are connected. One can replace one circuit with another circuit, and everything else cannot tell the difference.

We will use an analogy for equivalent circuits here. This analogy is that of jigsaw puzzle pieces. The idea is that two different jigsaw puzzle pieces with the same shape can be thought of as equivalent, even though they are different. The rest of the puzzle does not “notice” a difference. This is analogous to the case with equivalent circuits.

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Equivalent Circuits: Defined in Terms of Terminal Properties

Two circuits are considered to be equivalent if they behave the same with respect to the things to which they are connected. One can replace one circuit with another circuit, and everything else cannot tell the difference.

We often talk about equivalent circuits as being equivalent in terms of terminal properties. The properties (voltage, current, power) within the circuit may be different.

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Equivalent Circuits: A Caution

Two circuits are considered to be equivalent if they behave the same with respect to the things to which they are connected. The properties (voltage, current, power) within the circuit may be different.

It is important to keep this concept in mind. A common error for beginners is to assume that voltages or currents within a pair of equivalent circuits are equal. They may not be. These voltages and currents are only required to be equal if they can be identified outside the equivalent circuit. This will become clearer as we see the examples that follow in the other parts of this module.

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How do we use equivalent circuits?• We will use these equivalents to simplify circuits,

making them easier to solve. In some cases, one equivalent circuit is not simpler than another; rather one of them fits the needs of the particular circuit better. The delta-to-wye transformations that we cover next fit in this category. In yet other cases, we will have equivalent circuits for things that we would not otherwise be able to solve. For example, we will have equivalent circuits for devices such as diodes and transistors, that allow us to solve circuits that include these devices.

• Equivalent circuits are equivalent only with respect to the circuit outside them.

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Delta-to-Wye Transformations

• The transformations, or equivalent circuits, that we cover next are called delta-to-wye, or wye-to-delta transformations. They are also sometimes called pi-to-tee or tee-to-pi transformations.

• These are equivalent circuit pairs. They apply for parts of circuits that have three terminals. Each version of the equivalent circuit has three resistors.

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Delta-to-Wye TransformationsThree resistors in a part of a circuit with three

terminals can be replaced with another version, also with three resistors. Note that none of these resistors is in series with any other resistor, nor in parallel with any other resistor. The three terminals in this example are labeled A, B, and C.

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

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Delta-to-Wye Transformations

The version on the left hand side is called the delta connection, for the Greek letter . The version on the right hand side is called the wye connection, for the letter Y. The delta connection is also called the pi () connection, and the wye interconnection is also called the tee (T) connection. All these names come from the shapes of the drawings.

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

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Delta-to-Wye Transformation EquationsWhen we perform the delta-to-wye transformation

(going from left to right) we use the equations given below.

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

1

2

3

B C

A B C

A C

A B C

A B

A B C

R RR

R R R

R RR

R R R

R RR

R R R

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Wye-to-Delta Transformation Equations

RC

RARB

A

C

B

R2

R3

R1

A B

C

Rest of CircuitRest of Circuit

1 2 2 3 1 3

1

1 2 2 3 1 3

2

1 2 2 3 1 3

3

A

B

C

R R R R R RR

R

R R R R R RR

R

R R R R R RR

R

Perform the Wye-to-Delta transformation

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Equation 1We can calculate the equivalent resistance between terminals A and B, when

C is not connected anywhere. The two cases are shown below. This is the same as connecting an ohmmeter, which measures resistance, between terminals A and B, while terminal C is left disconnected.

1 2 1 2

1 2

Ohmmeter #1 reads || ( ). Ohmmeter #2 reads .

These must read the same value, so || ( ) .

EQ C A B EQ

C A B

R R R R R R R

R R R R R

RC

RARB

A

C

B

R2

R3

R1

A B

C

Ohmmeter #1 Ohmmeter #2

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Equations 2 and 3So, the equation that results from the first situation is

1 2|| ( ) .C A BR R R R R

RC

RARB

A

C

B

R2

R3

R1

A B

C

Ohmmeter #1 Ohmmeter #2

We can make this measurement two other ways, and get two more equations. Specifically, we can measure the resistance between A and C, with B left open, and we can measure the resistance between B and C, with A left open.

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All Three EquationsThe three equations we can obtain are

1 2

1 3

2 3

|| ( ) ,

|| ( ) , and

|| ( ) .

C A B

B A C

A B C

R R R R R

R R R R R

R R R R R

This is all that we need. These three equations can be manipulated algebraically to obtain either the set of equations for the delta-to-wye transformation (by solving for R1, R2 , and R3), or the set of equations for the wye-to-delta transformation (by solving for RA, RB , and RC).

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Why Are Delta-to-Wye Transformations Needed?

• They are like many other aspects of circuit analysis in that they allow us to solve circuits more quickly and more easily. They are used in cases where the resistors are neither in series nor parallel, so to simplify the circuit requires something more.

• One key in applying these equivalents is to get the proper resistors in the proper place in the equivalents and equations. We recommend that you name the terminals each time, on the circuit diagrams, to help you get these things in the right places.

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ExampleConvert the Δ network to an equivalent Y network.

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Solution

5

151025

10251

CBA

CB

RRR

RRR

5.7

151025

15252

CBA

AC

RRR

RRR

3

151025

10153

CBA

BA

RRR

RRR

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Example

Transform the wye network into the delta network.

A B

C

R1=10 R2=20

R3=40

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Solution

70

20

401040202010

2

313221

R

RRRRRRRB

140

10

401040202010

1

313221

R

RRRRRRRA

35

40

401040202010

3

313221

R

RRRRRRRC

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1. Choose a reference node (“ground”)Look for the one with the most connections!

2. Define unknown node voltagesthose which are not fixed by voltage sources

3. Write KCL at each unknown node, expressing current in terms of the node voltages (using the I-V relationships of branch elements)

Special cases: floating voltage sources

4. Solve the set of independent equations N equations for N unknown node voltages

Node-Voltage Circuit Analysis Method

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Example Calculate the node voltages in the circuit shown.

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Solution

• Labeling the current.

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Solution• At node 1, applying KCL and Ohm’s law gives

• At node 2, applying KCL and Ohm’s law gives

• Solve the simultaneous equation gives v1= 13.33V and v2 = 20V.

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121321

320 2

0

45

vv

vvviii

21

2215142

5360 6

0510

4

vv

vvviiii

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A “floating” voltage source is one for which neither side is connected to the reference node, e.g. VLL in the circuit below:

Problem: We cannot write KCL at nodes a or b because there is no way to express the current through the voltage source in terms of Va-Vb.

Solution: Define a “supernode” – that chunk of the circuit containing nodes a and b. Express KCL for this supernode. Incorporate voltage source constraint into KCL equation.

R4R2 I2

Va Vb

+ -

VLL

I1

Nodal Analysis w/ “Floating Voltage Source”

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supernode

R4R2 I2

Va Vb

+ -

VLL

I1

Nodal Analysis

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Example

For the circuit shown, find the nodes voltages.

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Solution• The supernode contains the 2-V source, node 1 and

2, and the 10- resistor.

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Solution

• Applying KCL to the supernode, gives

• Applying KVL

• Solve , gives v1 = -7.333 V and v2 = -5.333V

202

74

0

2

02

72

21

21

21

vv

vv

ii

12 2 vv

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Node-Voltage Method and Dependent Sources

• If a circuit contains dependent sources, what to do?

Example:

–+

–+

80 V5i

20 10

200 2.4 A

i

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Node-Voltage Method and Dependent Sources

• Dependent current source: treat as independent current source in organizing and writing node eqns, but include (substitute) constraining dependency in terms of defined node voltages.

• Dependent voltage source: treat as independent voltage source in organizing and writing node eqns, but include (substitute) constraining dependency in terms of defined node voltages.

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–+

–+

80 V5i

20 10

200 2.4 A

i

Example:

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Mesh analysis

• A loop is a closed path with no node passed more than once.

• A mesh is a loop that does not contain any other loop within it.

• Mesh analysis applies KVL to find unknown current.

• Mesh analysis apply to planar circuit.• A planar circuit is one that can be drawn in a plane

with no branches crossing one another.

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NODAL ANALYSIS

(“Node-Voltage Method”)

0) Choose a reference node

1) Define unknown node voltages

2) Apply KCL to each unknown node, expressing current in terms of the node voltages

=> N equations forN unknown node

voltages

3) Solve for node voltages

=> determine branch currents

MESH ANALYSIS

(“Mesh-Current Method”)

1) Select M independent mesh currents such that at least one mesh current passes through each branch*

M = #branches - #nodes + 1

2) Apply KVL to each mesh, expressing voltages in terms of mesh currents

=> M equations forM unknown mesh

currents

3) Solve for mesh currents

=> determine node voltages

Formal Circuit Analysis Methods

*Simple method for planar circuitsA mesh current is not necessarily identified with a branch current.

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1. Select M mesh currents.

2. Apply KVL to each mesh.

3. Solve for mesh currents.

Mesh Analysis:

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Example

• For the circuit shown, find the branch currents I1, I2 and I3 using mesh analysis.

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SolutionFor mesh 1, using KVL,

For mesh 2, using KVL,

Solve, i1 = 1A, i2 = 1A

I1= 1 A, I2=1 A, I3 = 0A.

123

010)(10515

21

211

ii

iii

12

046)(1010

21

2212

ii

iiii

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Problem: We cannot write KVL for meshes a and b because there is no way to express the voltage drop across the current source in terms of the mesh currents.

Solution: Define a “supermesh” – a mesh which avoids the branch containing the current source. Apply KVL for this supermesh.

Mesh Analysis with a Current Source

ia ib

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Eq’n 1: KVL for supermesh

Eq’n 2: Constraint due to current source:

ia ib

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ExampleFor the circuit, find the current i1 and i2.

40

Solution

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Solution

• Apply KVL to the supermesh,

• Apply KCL to a node in the branch where the two meshes intersect.

• Solving, i1= -3.2 A and i2 = 2.8 A.

20146

0410620

21

221

ii

iii

612 ii

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Mesh Analysis with Dependent Sources

• Exactly analogous to Node Analysis

• Dependent Voltage Source: (1) Formulate and write KVL mesh eqns. (2) Include and express dependency constraint in terms of mesh currents

• Dependent Current Source: (1) Use supermesh. (2) Include and express dependency constraint in terms of mesh currents

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What did you learn from you notes?

• That all circuits cannot be solved with simple approaches and more formal methods are sometimes needed

• Superposition is one such theorem

• You now can use node or mesh or superposition to solve the circuits

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Sample Problem

200 V 50 V

80 Ohms 40 Ohms

80 Ohms

Find the voltage across the resistor in black using Superposition

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Kill Source 1

50 V

80 Ohms 40 Ohms

80 OhmsShortcircuit

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Solve for V across Rblack

• You can do this by combining the resistors in parallel (40 Ohms) and then using the voltage divider equation

• Answer 25 Volts

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Kill Source 2

200 V

80 Ohms 40 Ohms

80 OhmsShortcircuit

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Solve for V across Rblack

• You can do this by combining the resistors in parallel (26.6) and then using the voltage divider equation

• Answer 50 Volts

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Combine 2 Answers

• Using Superposition

• Answer = 25 + 50 = 75 Volts

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Example

12 A 100 Ohms20 V

100 Ohms 50 Ohms

Find the voltage across the resistor in black .

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Partial Solution

• Kill each source individually

• Killing current => 10 V across it

• Killing voltage=> 3 amps flows through 100 Ohms => 300 V across it

• A total of 310 V across it

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Superposition Example 1 (1/4)

Find I1, I2 and Vab by superposition

10 V

20

5 5 AI1

b

I2

a

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Superposition Example 1 (2/4)Step 1: Omit current source.

By Ohm’s law and the

voltage divider rule:

10 V

20

5 5 AI11

b

I21

a

11 21 0.4I I A

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Superposition Example 1 (3/4)Step 2: Omit voltage source.

By the current divider rule and Ohm’s law :

20

5 5 AI12

b

I22

a

12

2

1.0

20ab

I A

V V

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Superposition Example 1 (4/4)

Combining steps 1 & 2, we get:

10 V

20

5 5 AI1

b

I2

a

1 11 12

1 2

0.6

22ab ab ab

I I I A

V V V V

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Superposition Example 2 (1/5)

16 A 6

16

16 A 10 64 V

Ix

Find Ix by superposition

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Superposition Example 2 (2/5)

16 A 6

16

10 Ixa

16 A

Activate only the 16 A Current source at the left. Thenuse Current Divider Rule:

10 1616 13

6 10 16xaI A A

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Superposition Example 2 (3/5)

6

16

16 A 10 Ixb

16 A

Activate only the 16 A Current source at the right. Thenuse Current Divider Rule:

1016 5

6 10 16xbI A A

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Superposition Example 2 (4/5)

6

16

10 64 V

Ixc16 A16 A

Activate only the 64 V voltage source at the bottom. Thenuse Ohm’s Law:

642

6 10 16xc

VI A

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Superposition Example 2 (5/5)

16 A 6

16

16 A 10 64 V

Ix

Sum the partial currents due to each of the sources:

13 5 2x xa xb xcI I I I A A A

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Superposition Example 3 (1/4)

• Solve for Vab by Superposition method

• Dependent source must remain in circuit for both steps

12 V 1.5ix 30 V

4 2

ix

a

b

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Superposition Example 3 (2/4)

1.5ix 30 V

4 2

ix

a

b

1 1 300 1.5

4 2ab ab

x

V Vi

1 1

1 1 1 915 1.5

4 4 2 8ab abV V

1

40

3abV V

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Superposition Example 3 (3/4)

12 V 1.5ix

4 2

ix

a

b

2 2 120 1.5

2 4ab ab

x

V Vi

2 2

1 1.5 1 97.5

2 4 4 8ab abV V

2

20

3abV

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Superposition Example 3 (4/4)

Combining the solutions:

12 V 1.5ix 30 V

4 2

ix

a

b1 2

40 20

3 3

ab ab ab

ab

ab

V V V

V V V

V

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Thevenin’s Theorem

• Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor.

• Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis.

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Implications

• We use Thevenin’s theorem to justify the concept of input and output resistance for amplifier circuits.

• We model transducers as equivalent sources and resistances.

• We model stereo speakers as an equivalent resistance.

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Independent Sources (Thevenin)

Circuit with independent sources

RTh

Voc

Thevenin equivalent circuit

+–

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No Independent Sources

Circuit without independent sources

RTh

Thevenin equivalent circuit

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Example: CE Amplifier

1kVin

2k

+10V

+

–Vo

+–

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Small Signal Equivalent

1kVin 100Ib

+

–

Vo

50

Ib

2k+–

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Thevenin Equivalent @ Output

1kVin 100Ib

+

-

Vo

50

Ib

2k

RTh

Voc

+

–

Vo

+–

+–

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Computing Thevenin Equivalent

• Basic steps to determining Thevenin equivalent are– Find voc

– Find RTh (= voc / isc)

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Example of Thevenin’s Theorem

Find the Thevenin equivalent of the circuit to the left of a-b. Then find the current through RL = 6 , 16 and 36.

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Solution

• Find RTh by turning off the 32 V voltage (replacing it with a short circuit) and the 2A current source (replacing it with an open circuit). The circuit becomes what is shown.

4116

124

112//4ThR

75

.30)25.0(12)(12

.5.0,ifor Solving

.2i .0)(12432

21

1i

2211

ViiV

Ai

Aiii

Th

• To find VTh, consider the circuit. Applying mesh analysis to the two loops.

Solution

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Solution

• The Thevenin equivalent circuit is shown.

When RL = 6, IL = 30/10 = 3A

When RL = 16, IL = 30/20 = 1.5A

When RL = 36, IL = 30/40 = 0.75A

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Thevenin/Norton Analysis

1. Pick a good breaking point in the circuit (cannot split a dependent source and its control variable).

2. Thevenin: Compute the open circuit voltage, VOC.

Norton: Compute the short circuit current, ISC.

For case 3(b) both VOC=0 and ISC=0 [so skip step 2]

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Thevenin/Norton Analysis

3. Compute the Thevenin equivalent resistance, RTh.

(a) If there are only independent sources, then short circuit all the voltage sources and open circuit the current sources (just like superposition).

(b) If there are only dependent sources, then must use a test voltage or current source in order to calculate

RTh = VTest/Itest

(c) If there are both independent and dependent sources, then compute RTh from VOC/ISC.

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Thevenin/Norton Analysis

4. Thevenin: Replace circuit with VOC in series with RTh.

Norton: Replace circuit with ISC in parallel with RTh.

Note: for 3(b) the equivalent network is merely RTh, that is, no voltage (or current) source.

Only steps 2 & 4 differ from Thevenin & Norton!

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Method 3: Thevenin and Norton Equivalent Circuits

vTH= open circuit voltage at terminal (a.k.a. port)

RTH= Resistance of the network as seen from port(Vm’s, In’s set to zero)

Any network of sources and resistors will appear to the circuit connected to it as a voltage source and a series resistance

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Norton Equivalent Circuit

Any network of sources and resistors will appear to the circuit connected to it as a current source and a parallel resistance

Ed Norton – Bell Labs, 1898-1983

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Calculation of RT and RN

• RT=RN ; same calculation (voltage and current sources set to zero)

• Remove the load.

• Set all sources to zero (‘kill’ the sources)

– Short voltage sources (replace with a wire)

– Open current sources (replace with a break)

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Calculation of RT and RN continued• Calculate equivalent resistance seen by the load

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Calculation of VT

• Remove the load and calculate the open circuit voltage

SROC VRR

RVV

21

22

(Voltage Divider)

85

Example• Use Thevenin’s theorem to calculate the current

through Resistor R6. (solution I=0.72A)

86

Exercise: Draw the Thevenin Equivalent

• To find RTH remove the load, kill the sources (short voltage sources, break current sources) and find the equivalent resistance.

• To find VTH Remove the load and calculate the open circuit voltage

87

Exercise: Draw the Thevenin Equivalent

• To find RTH kill the sources (short voltage sources, break current sources) and find the equivalent resistance.

• To find VTH Remove the load and calculate the open circuit voltage

VAB = 20 - (20Ω x 0.33amps) = 13.33V

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Exercise: Draw the Thevenin Equivalent

• To find RTH kill the sources (short voltage sources, break current sources) and find the equivalent resistance.

• To find VTH Remove the load and calculate the open circuit voltage

89

Calculation of IN• Short the load and calculate the short circuit current

(R1+R2)i1 - R2iSC = vs

-R2i1 + (R2+R3)iSC = 0

(KCL at v)

(mesh analysis)

RN=RTH

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Source Transformation

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Summary: Thevenin’s Theorem

• Any two-terminal linear circuit can be replaced with a voltage source and a series resistor which will produce the same effects at the terminals

• VTH is the open-circuit voltage VOC between the two terminals of the circuit that the Thevenin generator is replacing

• RTH is the ratio of VOC to the short-circuit current ISC; In linear circuits this is equivalent to “killing” the sources and evaluating the resistance between the terminals. Voltage sources are killed by shorting them, current sources are killed by opening them.

92

Summary: Norton’s Theorem• Any two-terminal linear circuit can be replaced

with a current source and a parallel resistor which will produce the same effects at the terminals

• IN is the short-circuit current ISC of the circuit that the Norton generator is replacing

• Again, RN is the ratio of VOC to the short-circuit current ISC; In linear circuits this is equivalent to “killing” the sources and evaluating the resistance between the terminals. Voltage sources are killed by shorting them, current sources are killed by opening them.

• For a given circuit, RN = RTH

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Maximum power transfer

• Maximum power is transferred to the load resistance when the load resistance equals the Thevenin resistance as seen from the load.

P

RLRTh0

Pmax