5: Thévenin and Norton Equivalents

5: Thévenin and NortonEquivalents


• Equivalent Networks

• Thévenin Equivalent

• Thévenin Properties

• Determining Thévenin

• Complicated Circuits

• Norton Equivalent

• Power Transfer

• Source Transformation

• Source Rearrangement

• Series Rearrangement

• Summary

E1.1 Analysis of Circuits (2017-10110) Thevenin and Norton: 5 – 1 / 12

Equivalent Networks








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• Summary


From linearity theorem: V = aI + b.

Equivalent Networks








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• Summary



Use nodal analysis:KCL@X: X

1 − 6 + X−V2 = 0

KCL@V: V−X2 − I = 0

Equivalent Networks








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• Summary




1 − 6 + X−V2 = 0

KCL@V: V−X2 − I = 0

Eliminating X gives: V = 3I + 6.

Equivalent Networks








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1 − 6 + X−V2 = 0

KCL@V: V−X2 − I = 0


There are infinitely many networks with the same values of a and b:

Equivalent Networks








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• Summary




1 − 6 + X−V2 = 0

KCL@V: V−X2 − I = 0



Equivalent Networks








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1 − 6 + X−V2 = 0

KCL@V: V−X2 − I = 0



These four shaded networks are equivalent because the relationshipbetween V and I is exactly the same in each case.

Equivalent Networks








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• Summary




1 − 6 + X−V2 = 0

KCL@V: V−X2 − I = 0



These four shaded networks are equivalent because the relationshipbetween V and I is exactly the same in each case.The last two are particularly simple and are respectively called the Nortonand Thévenin equivalent networks.

Thévenin Equivalent








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• Summary


Thévenin Theorem: Any two-terminal network consisting of resistors, fixedvoltage/current sources and linear dependent sources is externallyequivalent to a circuit consisting of a resistor in series with a fixed voltagesource.









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• Summary



We can replace the shaded part of thecircuit with its Thévenin equivalentnetwork.









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• Summary












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• Summary




The voltages and currents in the unshadedpart of the circuit will be identical in bothcircuits.









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The new components are called theThévenin equivalent resistance, RTh, andthe Thévenin equivalent voltage, VTh, ofthe original network.









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• Summary





The new components are called theThévenin equivalent resistance, RTh, andthe Thévenin equivalent voltage, VTh, ofthe original network.

This is often a useful way to simplify a complicated circuit (provided thatyou do not want to know the voltages and currents in the shaded part).

Thévenin Circuit Properties








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• Summary


A Thévenin equivalent circuit hasa straight line characteristic withthe equation:

V = RThI + VTh









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• Summary



V = RThI + VTh

⇔ I = 1RTh

V −VTh

RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

I (m

A)









• Power Transfer




• Summary



V = RThI + VTh

⇔ I = 1RTh

V −VTh

RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

I (m

A)

Three important quantities are:

Open Circuit Voltage: If I = 0 then VOC = VTh. (X-intercept: o)









• Power Transfer




• Summary



V = RThI + VTh

⇔ I = 1RTh

V −VTh

RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

I (m

A)



Short Circuit Current: If V = 0 then ISC = −VTh

RTh

(Y-intercept: x)









• Power Transfer




• Summary



V = RThI + VTh

⇔ I = 1RTh

V −VTh

RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

I (m

A)




RTh

(Y-intercept: x)

Thévenin Resistance: The slope of the characteristic is dIdV = 1

RTh

.









• Power Transfer




• Summary



V = RThI + VTh

⇔ I = 1RTh

V −VTh

RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

I (m

A)




RTh

(Y-intercept: x)


RTh

.

If we know the value of any two of these three quantities, we can work outVTh and RTh.









• Power Transfer




• Summary



V = RThI + VTh

⇔ I = 1RTh

V −VTh

RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

I (m

A)




RTh

(Y-intercept: x)


RTh

.

If we know the value of any two of these three quantities, we can work outVTh and RTh.

In any two-terminal circuit with the same characteristic, the three quantitieswill have the same values. So if we can determine two of them, we canwork out the Thévenin equivalent.

Determining Thévenin Values








• Power Transfer




• Summary


We need any two of the following:

Open Circuit Voltage:

Short Circuit Current:

Thévenin Resistance:









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• Summary



Open Circuit Voltage:



Open Circuit Voltage:We known that I1k = 6 because there is nowhere else for the current to go.So VOC = 6× 1 = 6V.









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• Summary



Open Circuit Voltage: VOC = VTh = 6V



Open Circuit Voltage:We known that I1k = 6 because there is nowhere else for the current to go.So VOC = 6× 1 = 6V.









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• Summary






Short Circuit Current:The 2 k and 1 k resistors are in parallel and so form a current divider inwhich currents are proportional to conductances.So ISC = −

1/23/2 × 6 = −2mA









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• Summary




Short Circuit Current: ISC = −VTh

RTh

= −2mA


Short Circuit Current:The 2 k and 1 k resistors are in parallel and so form a current divider inwhich currents are proportional to conductances.So ISC = −

1/23/2 × 6 = −2mA









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• Summary





RTh

= −2mA


Thévenin Resistance:We set all the independent sources to zero (voltage sources → shortcircuit, current sources→ open circuit). Then we find the equivalentresistance between the two terminals.The 3 k resistor has no effect so RTh = 2k + 1 k = 3 k.









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• Summary





RTh

= −2mA

Thévenin Resistance: RTh = 2k + 1 k = 3 kΩ

Thévenin Resistance:We set all the independent sources to zero (voltage sources → shortcircuit, current sources→ open circuit). Then we find the equivalentresistance between the two terminals.The 3 k resistor has no effect so RTh = 2k + 1 k = 3 k.









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• Summary





RTh

= −2mA

Thévenin Resistance: RTh = 2k + 1 k = 3 kΩ

Thévenin Resistance:We set all the independent sources to zero (voltage sources → shortcircuit, current sources→ open circuit). Then we find the equivalentresistance between the two terminals.The 3 k resistor has no effect so RTh = 2k + 1 k = 3 k.Any measurement gives the same result on an equivalent circuit.

Thévenin of Complicated Circuits








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• Summary


For a complicated circuit, you can usenodal analysis to find the Théveninequivalent directly in the form:

V = VTh + IRTh.









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• Summary



V = VTh + IRTh.

Step 1: Label ground as an output terminal + label other nodes.









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• Summary



V = VTh + IRTh.


Step 2: Write down the equations









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• Summary



V = VTh + IRTh.


Step 2: Write down the equations

X−V2 + X

1 + X−Y1 = 0









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• Summary



V = VTh + IRTh.


Step 2: Write down the equations (Y is a supernode)

X−V2 + X

1 + X−Y1 = 0

Y−3−V1 + Y−X

1 + Y−32 = 0









• Power Transfer




• Summary



V = VTh + IRTh.



X−V2 + X

1 + X−Y1 = 0

Y−3−V1 + Y−X

1 + Y−32 = 0

V−Y+31 + V−X

2 − I = 0









• Power Transfer




• Summary



V = VTh + IRTh.



X−V2 + X

1 + X−Y1 = 0

Y−3−V1 + Y−X

1 + Y−32 = 0

V−Y+31 + V−X

2 − I = 0

Step 3: Eliminate X and Y andsolve for V in terms of I :

V = 75I −

35 = RThI + VTh









• Power Transfer




• Summary



V = VTh + IRTh.



X−V2 + X

1 + X−Y1 = 0

Y−3−V1 + Y−X

1 + Y−32 = 0

V−Y+31 + V−X

2 − I = 0

Step 3: Eliminate X and Y andsolve for V in terms of I :

V = 75I −

35 = RThI + VTh

Norton Equivalent








• Power Transfer




• Summary


Norton Theorem: Any two-terminal network consisting of resistors, fixedvoltage/current sources and linear dependent sources is externallyequivalent to a circuit consisting of a resistor in parallel with a fixed currentsource.

KCL:−I − INo +

VRTh

= 0

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

Norton Equivalent








• Power Transfer




• Summary



KCL:−I − INo +

VRTh

= 0

⇔ I = 1RTh

V − INo

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

Norton Equivalent








• Power Transfer




• Summary



KCL:−I − INo +

VRTh

= 0

⇔ I = 1RTh

V − INo

c.f. Thévenin (slide 5-4):Same R and INo = VTh

RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

Norton Equivalent








• Power Transfer




• Summary



KCL:−I − INo +

VRTh

= 0

⇔ I = 1RTh

V − INo


RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)

Open Circuit Voltage: If I = 0 then VOC = INoRTh.

Norton Equivalent








• Power Transfer




• Summary



KCL:−I − INo +

VRTh

= 0

⇔ I = 1RTh

V − INo


RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)


Short Circuit Current: If V = 0 then ISC = −INo

Norton Equivalent








• Power Transfer




• Summary



KCL:−I − INo +

VRTh

= 0

⇔ I = 1RTh

V − INo


RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)



Thévenin Resistance: The slope of the characteristic is 1RTh

.

Norton Equivalent








• Power Transfer




• Summary



KCL:−I − INo +

VRTh

= 0

⇔ I = 1RTh

V − INo


RTh

-2 0 2 4 6 8-3

-2

-1

0

1

V (V)



Thévenin Resistance: The slope of the characteristic is 1RTh

.

Easy to change between Norton and Thévenin: VTh = INoRTh.Usually best to use Thévenin for small RTh and Norton for large RTh

compared to the other impedances in the circuit.

Power Transfer








• Power Transfer




• Summary


Suppose we connect a variable resistor, RL, across a two-terminalnetwork. From Thévenin’s theorem, even a complicated network isequivalent to a voltage source and a resistor.

Power Transfer








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• Summary



We know I = VTh

RTh+RL

Power Transfer








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• Summary



We know I = VTh

RTh+RL

⇒ power in RL is PL = I2RL =

V 2

ThRL

(RTh+RL)2

Power Transfer








• Power Transfer




• Summary



We know I = VTh

RTh+RL


V 2

ThRL

(RTh+RL)2

Power Transfer








• Power Transfer




• Summary



We know I = VTh

RTh+RL


V 2

ThRL

(RTh+RL)2

To find the RL that maximizes PL:

0 = dPL

dRL

=(RTh+RL)2V 2

Th−2V 2

ThRL(RTh+RL)

(RTh+RL)4

Power Transfer








• Power Transfer




• Summary



We know I = VTh

RTh+RL


V 2

ThRL

(RTh+RL)2


0 = dPL

dRL

=(RTh+RL)2V 2

Th−2V 2

ThRL(RTh+RL)

(RTh+RL)4

=V 2

Th(RTh+RL)−2V 2

ThRL

(RTh+RL)3

Power Transfer








• Power Transfer




• Summary



We know I = VTh

RTh+RL


V 2

ThRL

(RTh+RL)2


0 = dPL

dRL

=(RTh+RL)2V 2

Th−2V 2

ThRL(RTh+RL)

(RTh+RL)4

=V 2

Th(RTh+RL)−2V 2

ThRL

(RTh+RL)3

⇒ V2Th ((RTh +RL)− 2RL) = 0

Power Transfer








• Power Transfer




• Summary



We know I = VTh

RTh+RL


V 2

ThRL

(RTh+RL)2


0 = dPL

dRL

=(RTh+RL)2V 2

Th−2V 2

ThRL(RTh+RL)

(RTh+RL)4

=V 2

Th(RTh+RL)−2V 2

ThRL

(RTh+RL)3

⇒ V2Th ((RTh +RL)− 2RL) = 0

⇒ RL = RTh ⇒ P(max) =V 2

Th

4RTh

Power Transfer








• Power Transfer




• Summary



We know I = VTh

RTh+RL


V 2

ThRL

(RTh+RL)2


0 = dPL

dRL

=(RTh+RL)2V 2

Th−2V 2

ThRL(RTh+RL)

(RTh+RL)4

=V 2

Th(RTh+RL)−2V 2

ThRL

(RTh+RL)3

⇒ V2Th ((RTh +RL)− 2RL) = 0

⇒ RL = RTh ⇒ P(max) =V 2

Th

4RTh

For fixed RTh, the maximum power transferis when RL = RTh (“matched load”).

Source Transformation








• Power Transfer




• Summary


Sometimes changing between Thévenin and Norton can simplify a circuit.Suppose we want to calculate I.









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• Summary



Norton → Thévenin on current source: I = 18−(−10)5 = 5.6 A









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• Summary




If you can’t spot any clever tricks, you can always find out everything withnodal analysis.

−6 + X3 + X−(−10)

2 = 0









• Power Transfer




• Summary





−6 + X3 + X−(−10)

2 = 0

⇒ 5X = 36− 30 = 6









• Power Transfer




• Summary





−6 + X3 + X−(−10)

2 = 0

⇒ 5X = 36− 30 = 6

⇒ X = 1.2









• Power Transfer




• Summary





−6 + X3 + X−(−10)

2 = 0

⇒ 5X = 36− 30 = 6

⇒ X = 1.2

⇒ I = X−(−10)2 = 5.6

Source Rearrangement








• Power Transfer




• Summary


If all but one branches connecting to a node are voltage sources or arecurrent sources, you can choose any of the branches to be the sourcelessone.









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• Summary



Voltage Sources:









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• Summary



Voltage Sources:

We can use the leftnode as the reference









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• Summary



Voltage Sources:

We can use the leftnode as the reference =









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• Summary



Voltage Sources:


Current Sources:









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• Summary



Voltage Sources:


Current Sources:

KCL gives current intorightmost node









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• Summary



Voltage Sources:


Current Sources:

KCL gives current intorightmost node

=

Series Rearrangement








• Power Transfer




• Summary


If we have any number of voltage sources and resistors in series we cancalculate the total voltage across the chain as:

V = 8I − 2 + 7I + 5 + 9I









• Power Transfer




• Summary



V = 8I − 2 + 7I + 5 + 9I = (−2 + 5) + (8 + 7 + 9)I









• Power Transfer




• Summary



V = 8I − 2 + 7I + 5 + 9I = (−2 + 5) + (8 + 7 + 9)I

= 3 + 24I









• Power Transfer




• Summary



V = 8I − 2 + 7I + 5 + 9I = (−2 + 5) + (8 + 7 + 9)I

= 3 + 24I

We can arbitrarilyrearrange the order ofthe componentswithout affectingV = 3 + 24I .









• Power Transfer




• Summary



V = 8I − 2 + 7I + 5 + 9I = (−2 + 5) + (8 + 7 + 9)I

= 3 + 24I

We can arbitrarilyrearrange the order ofthe componentswithout affectingV = 3 + 24I .

If we move all the voltage sources together and all the resistors together wecan merge them and then we get the Thévenin equivalent.

Summary








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• Summary


• Thévenin and Norton Equivalent Circuits

Summary








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• Summary



A network has Thévenin and Norton equivalents if:

⊲ only 2 terminals connect it to the outside world

⊲ it is made of resistors + sources + linear dependent sources

Summary








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• Summary






How to determine VTh, INo and RTh

Summary








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• Summary







⊲ Method 1: Connect current source → Nodal analysis

Summary








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• Summary








⊲ Method 2: Find any two of:

(a) VOC = VTh, the open-circuit voltage

(b) ISC = −INo, the short-circuit current

(c) RTh, equivalent resistance with all sources set to zero

Summary








• Power Transfer




• Summary












⊲ Related by Ohm’s law: VTh = INoRTh

Summary








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• Summary













• Load resistor for maximum power transfer = RTh

Summary








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• Summary














• Source Transformation and Rearrangement

Summary








• Power Transfer




• Summary














• Source Transformation and Rearrangement

For further details see Hayt Ch 5 & A3 or Irwin Ch 5.

Date post:	24-Feb-2023
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5: Thévenin and Norton Equivalents

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