Thevenin equivalent circuits - Iowa State...

EE 201 Thevenin – 1

Thevenin equivalent circuitsWe have seen the idea of equivalency used in several instances already.

IS R1+– VS

R1

same as

96 = ,65�

IS0 A same

as

VS0 V

same as

+–

+–

VS1

VS2

+– VS1 +VS2

same as

same as

5HT = 5� + 5��5�

R1

R2 R3

IS1 IS2 IS1 +IS2same

as


The behavior of any circuit, with respect to a pair of terminals (port) can be represented with a Thevenin equivalent, which consists of a voltage source in series with a resistor.

somecircuit

two terminals (two nodes) = port

+–VTh

RTh

Thevenin equivalent

somecircuit RL

load

+

–vRL

+–VTh

RTh

RL

load

+

–vRL

Need to determine VTh and RTh so that the model behaves just like the original.


Norton equivalent

IN RNsomecircuit

load

+

–vRL

load

+

–vRL

Ideas developed independently (Thevenin in 1880’s and Norton in 1920’s). But we recognize the two forms as identical because they are source transformations of each other. In EE 201, we won’t make a distinction between the methods for finding Thevenin and Norton. Find one and we have the other.

+–VTh

RTh

IN RN 97K = ,157K

57K = 51


Example

+–

VS

R1

R2 IS +–VTh

RTh

9V 6 mA

1.5 k!

3 k! 12 V

1 k!

RL RL

R v i P

1 ! 11.99 mV 11.99 mA 0.144 mW

10 ! 118.8 mV 11.88 mA 1.41 mW

100 ! 1.091 V 10.91 mA 11.90 mW

1 k! 6.0 V 6.0 mA 36 mW

10 k! 10.91 V 1.09 mA 11.9 mW

100 k! 11.88 V 0.119 mA 1.41 mW

Attach various load resistors to the original circuit. Do the same for the equivalent circuit. For each load resistance, calculate the load voltage (and current and power) for each of the circuits. The results are identical. In terms of the load that is attached at the port, the two circuits are indistinguishable.

Check it yourself.


Determining the Thevenin (or Norton) components

How to find VTh and RTh?

Need two components, so two measurements or calculations should suffice. Use two different load resistors.

2 equations, 2 unknowns: 57K =5�5� (Y� � Y�)5�Y� � 5�Y�

97K =Y�Y� (5� � 5�)5�Y� � 5�Y�

Y� =5�

57K + 5�97K

Y� =5�

57K + 5�97K+

–VTh

RTh

R2

load

+

–v2

+–VTh

RTh

R1

load

+

–v1

unknowncircuit R2

load

+

–v2

unknowncircuit R1

load

+

–v1


More directly: open-circuit voltage, short-circuit current

unknowncircuit

+

–

voc

1. Leave port open-circuited. (RL →∞, iL = 0) Measure open-circuit voltage.

+–VTh

RTh+

–

voc

voc = VTh

open-circuit voltage is a direct measure of VTh.

iscunknowncircuit

2. Short the output port. (RL = 0, vL = 0) Measure short-circuit current.

+–VTh

RTh

iscLVF =

97K57K

57K =97KLVF

=YRFLVF

Note, that isc can also be interpreted as a direct measurement of IN: isc = IN.


Calculating Thevenin equivalentThe open-circuit voltage / short-circuit current approach can be used to calculate the Thevenin equivalent for a known circuit.

Consider the circuit from slide 4: +–

VS

R1

R2 IS9V 6 mA

1.5 k!

3 k!

Open-circuit voltage – Use whatever method you prefer. We’ll use node voltage in this case.

+–

VS

R1

R2 IS

va+

–

voc

YRF = YD � �

96 � YD5�

+ ,6 =YD5�

YD =96 + 5�,6�+ 5�

5�=�9+ (�.�Nż) (�P$)

�+ �.�Nż�.�Nż

= 12 V

VTh = voc = 12 V.


Short-circuit current – Use whatever method you prefer. We’ll use node voltage in this case. But proceed carefully – the short circuit introduces some unusual wrinkles into the circuit analysis.

R1 va

+–

VS R2 IS isca: Because of the short circuit, va = 0!

b: Because of the short, vR2 = 0 and iR2 = 0. So R2 plays no role and can be removed.

R1 va

+–

VS IS iscisc = iR1 + IS

=96 � YD

5�+ ,6

=965�

+ ,6

=�9

�.�Nż + �P$ = ��P$57K =97KLVF

=��9��P$ = �Nż


+–VTh

RTh

IN RN12 V

1 k!

1 k!12 mA

Thevenin Norton

If the circuit consists of independent sources and resistors only, then the Thevenin resistance can also be found by de-activating the independent sources and finding the equivalent resistance as seen from the port.

R1

R2 RTh

De-activate the sources

= 5��5� = (�.�Nż) � (�Nż) = �Nż

Alternate method for RTh


Summary

1. Use a voltmeter to measure the open-circuit voltage at the port of the circuit: voc = VTh.

2. Connect a short circuit across the output and use an ammeter to measure the short-circuit current: isc = IN.

3. Calculate RTh = VTh / IN.

To measure VTh and RTh

Note that shorting the output may not always be practical. For example, some devices may have over-current protection circuitry that prevents large short-circuit currents from flowing. Or the device might not be able to handle the large current that might flow when the output is shorted without being damaged. In those cases:

1. Use a voltmeter to measure the open-circuit at the port of the circuit: voc = VTh.

2. Attach a load resistance, RL that is small enough so that an appreciable current is flowing. Measure the resulting load voltage, vL.

3. Calculate 57K = 5/

�YRFY/

� ��


Summary

1. Using whatever techniques are appropriate, calculate the open-circuit voltage at the port of the circuit: voc = VTh.

2. Connect a short circuit across the output. Using whatever techniques are appropriate, calculate the short-circuit current: isc = IN.

3. Calculate RTh = VTh / IN.

To calculate VTh and RTh

Alternate method (for circuits that consist only of independent sources and resistors).

1. Using whatever techniques are appropriate, calculate the open-circuit voltage at the port of the circuit: voc = VTh.

2. De-activate all independent sources. Calculate the equivalent resistance as seen from the port. (If dependent sources are present in the circuit, the test generator method can be used to find equivalent resistance. See the equivalent resistance notes to review the test generator technique.)


Example 1

R1

R2

R3ISFind the Thevenin and Norton equivalents of the circuit at right, with the port as shown.

voc

+

–6 k!

1 k!

3 k!24 mA

Find voc. Start with a current divider.

L5� =�

5�+5��5� + �

5�+5�,6 =

��+�

�� + �

�+�(��P$) = ��.�P$

YRF = L5�5� = (��.�P$) (�Nż) = ��.�9 VTh = voc = 43.2 V

Find isc. Note that R3 is shorted out. Use a current divider again. R1

R2

IS iscLVF =

�5�

�5� + �

5�,6

=��

�� + �

�(��P$) = ��.��P$ IN = isc = 20.57 mA


+–VTh

RTh

IN RN

2.1 k!

43.2 V

57K =97KLVF

=��.�9

��.��P$ = �.�Nż

2.1 k!20.57 mA

Alternatively, we could use the short-cut method to find RTh. De-activating the current source:

R1

R2

R3 Req 5HT = 5�� (5� + 5�)

= (�Nż) � (�Nż + �Nż) = �.�Nż


Example 2

20 V

1.5 k!

1.5 k!

3.3 k!

3.3 k!

4.7 k! 4.7 k!

+–

R1

R2

R3

R4

R5

R6

VSFind the Thevenin and Norton equivalents of the circuit at left, with the port as shown.

Find voc. Use mesh current method.

+–

R1

R2

R3

R4

R5

R6

VS

+

–

vocia ib

96 � Y5� � Y5� � Y5� = �Y5� � Y5� � Y5� � Y5� = �

96 � 5�LD � 5� (LD � LE) � 5�LD = �5� (LD � LE) � 5�LE � 5�LE � 5�LE = �

(5� + 5� + 5�) LD � 5�LE = 96

5�LD � (5� + 5� + 5� + 5�) LE = �Insert values and solve: ib = 0.930 mA.

voc = ib R5 = 4.37 V.


+–

R1

R2

R3

R4

R6

VS isc

Find isc. Note that R5 is shorted out by the short circuit.

iS Use equivalent resistance to find iS.

56 = 5� + 5� + 5�� (5� + 5�) = �.��Nż

L6 =9656

= �.��P$

Use current divider to find isc.

LVF =�

5�+5��

5�+5� + �5�

L6

isc = 1.45 mA

+–VTh

RTh

IN RN

3.27 k!

3.27 k!

4.37 V

1.45 mA

57K =97KLVF

=�.��9�.��P$ = �.��Nż


Alternatively, we can use the short-cut method to find RTh.

R1

R2

R3

R4

R5

R6

RTh

57K = 5�� [5� + 5� + 5�� (5� + 5�)]

= 3.02 k!


Example 3

+–

R1

R2

R3

R4

R5

VS

ISFind the Thevenin and Norton equivalents of the circuit at right, with the port as shown.

30 V

15 !

4 A

30 !

20 !

30 !

15 !

+–

R1

R2

R3

R4

R5

VS

IS

–

+

voc

a b

Find voc. Use node voltage.L5� + L5� + ,6 = L5�

96 � YD5�

=YD5�

+YD � YE5�

L5� = L5� + L5�

YD � YE5�

+96 � YE

5�+ ,6 =

YE5�

��+

5�5�

+5�5�

�YD � 5�

5�YE = 96

�5�5�

YD +

��+

5�5�

+5�5�

�YE = 96 + 5�,6

solve: va = 20 V, vb = 40 V. " voc = vb = 40 V


+–

R1

R2

R3

R5

VS

IS

isc

Find isc. Use node voltage, again. Note that R4 is shorted out (so ignore it) and node b is shorted to ground, vb = 0.

a b

LVF = L5� + L5� + ,6

=YD � YE5�

+96 � YE

5�+ ,6

=YD5�

+965�

+ ,6

Need va.

L5� = L5� + L5�

96 � YD5�

=YD5�

+YD5�

YD =96

�+ 5�5� + 5�

5�= 8.57 V

LVF =�.��9��ż +

��9��ż + �$ = �.��$

+–VTh

RTh

IN RN

6.36 !

6.36 !

40 V

6.28 A

57K =��9�.��$ = �.��ż


Maximum power transferNow that we have the ability to model any circuit using a simple Thevenin (or Norton) equivalent, we can answer another important question: How much power can a given circuit supply to an attached load?

Start with the Thevenin equivalent and determine the load resistance that would lead to the maximum amount of power being dissipated in the load.

+–VTh

RTh

RL

load

+

–vRL

3/ =Y�5/5/

=5/9�7K

(5/ + 57K)�

In the usual way, find the max by setting the derivative to zero and solving.

G3/G5/

= � 9�7K(5/ + 57K)�

� � 5/9�7K(5/ + 57K)�

= �

5/ + 57K � �5/ = � 5/ = 57KFor maximum power to the load

See slide 4 for an example – look at power column in the table.

Thevenin equivalent circuits - Iowa State...

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