thevenin equivalent circuits - iowa state...
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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
EE 201 Thevenin – 2
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.
EE 201 Thevenin – 3
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
EE 201 Thevenin – 4
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.
EE 201 Thevenin – 5
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
EE 201 Thevenin – 6
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.
EE 201 Thevenin – 7
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$)
�+ �.�Nż�.�Nż
= 12 V
VTh = voc = 12 V.
EE 201 Thevenin – 8
Short-circuit current – Use whatever method you prefer. We’ll use node voltage in this case. But proceed carefully – the short circuit introduces some unusual wrinkles into the circuit analysis.
R1 va
+–
VS R2 IS isca: Because of the short circuit, va = 0!
b: Because of the short, vR2 = 0 and iR2 = 0. So R2 plays no role and can be removed.
R1 va
+–
VS IS iscisc = iR1 + IS
=96 � YD
5�+ ,6
=965�
+ ,6
=�9
�.�Nż + �P$ = ��P$57K =97KLVF
=��9��P$ = �Nż
EE 201 Thevenin – 9
+–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
EE 201 Thevenin – 10
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/
� ��
EE 201 Thevenin – 11
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.)
EE 201 Thevenin – 12
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
EE 201 Thevenin – 13
+–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
EE 201 Thevenin – 14
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.
EE 201 Thevenin – 15
+–
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
EE 201 Thevenin – 16
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!
EE 201 Thevenin – 17
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
EE 201 Thevenin – 18
+–
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
EE 201 Thevenin – 19
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.