Thevenin equivalent circuits
We have seen the idea of equivalency same IS1 IS2 I +I used in several instances already. as S1 S2
R1 V + S1 – + same V +V R R = + – S1 S2 2 3 same as V + as S2 –
R1
V IS S + V same R same same – S IS 1 0 V 0 A as as as
=
EE 201 Thevenin – 1 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.
load
some two terminals some + R v circuit (two nodes) = circuit L RL port –
RTh RTh load
+ + Thevenin + V V R v Th – equivalent Th – L RL –
Need to determine VTh and RTh so that the model behaves just like the original.
EE 201 Thevenin – 2 Norton equivalent
load load
some + + I v circuit vRL N RN RL – –
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.
RTh
= + V IN R Th – N =
EE 201 Thevenin – 3 Example R 1.5 k! 1 RTh 1 k!
V R I R R S + 2 S L V + L 9V – 3 k! 6 mA Th – 12 V
Attach various load resistors to the R v i P original circuit. Do the same for 1 ! 11.99 mV 11.99 mA 0.144 mW the equivalent circuit. For each load resistance, calculate the load 10 ! 118.8 mV 11.88 mA 1.41 mW voltage (and current and power) for each of the circuits. The results 100 ! 1.091 V 10.91 mA 11.90 mW are identical. In terms of the load 1 k! 6.0 V 6.0 mA 36 mW that is attached at the port, the two circuits are indistinguishable. 10 k! 10.91 V 1.09 mA 11.9 mW
Check it yourself. 100 k! 11.88 V 0.119 mA 1.41 mW
EE 201 Thevenin – 4 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.
load RTh load
unknown + + + R v V R v = circuit 1 1 Th – 1 1 + – –
load RTh load
+ + unknown + = v V R v circuit R2 2 Th – 2 2 + – –
( ) ( ) 2 equations, 2 unknowns: = = EE 201 Thevenin – 5 More directly: open-circuit voltage, short-circuit current
1. Leave port open-circuited. (RL →∞, iL = 0) Measure open-circuit voltage.
RTh + + voc = VTh unknown + voc V v circuit Th – oc open-circuit voltage is a – – direct measure of VTh.
2. Short the output port. (RL = 0, vL = 0) Measure short-circuit current. RTh unknown = + isc V isc circuit Th – = =
Note, that isc can also be interpreted as a direct measurement of IN: isc = IN.
EE 201 Thevenin – 6 Calculating Thevenin equivalent The open-circuit voltage / short-circuit current approach can be used to calculate the Thevenin equivalent for a known circuit.
R1 1.5 k!
Consider the circuit from slide 4: VS + R2 IS 9V – 3 k! 6 mA
Open-circuit voltage – Use whatever method you prefer. We’ll use node voltage in this case.
R1 v a = + V + R2 I S S voc + = – – + +( . )( ) = = = 12 V + + . .
VTh = voc = 12 V. EE 201 Thevenin – 7 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.
R 1 va
a: Because of the short circuit, va = 0! V + R2 I isc S – S b: Because of the short, vR2 = 0 and iR2
= 0. So R2 plays no role and can be removed. R 1 va isc = iR1 + IS i VS + IS sc – = + = + = + = = = = . EE 201 Thevenin – 8 Thevenin Norton
RTh 1 k!
V + I Th – N RN 12 V 12 mA 1 k!
Alternate method for RTh 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. De-activate the sources
R1
R2 R = =( . ) ( )= Th
EE 201 Thevenin – 9 Summary
To measure VTh and RTh 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. 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 = EE 201 Thevenin – 10 Summary
To calculate VTh and RTh 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.
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 – 11 Example 1 R2 1 k! Find the Thevenin and Norton + R R equivalents of the circuit at I 1 3 voc S 6 k! right, with the port as shown. 3 k! 24 mA –
Find voc. Start with a current divider. + + = = ( )= . + + + +
= =( . )( )= . VTh = voc = 43.2 V
R2 Find isc. Note that R3 is shorted out. Use a current divider again. R1 IS isc = + = ( )= . IN = isc = 20.57 mA + EE 201 Thevenin – 12 . = = = . .
RTh 2.1 k!
V + I Th – N RN 43.2 V 20.57 mA 2.1 k!
Alternatively, we could use the short-cut method to find RTh. De-activating the current source:
R2
R1 R3 R = ( + ) eq =( ) ( + )= .
EE 201 Thevenin – 13 Example 2
R1 1.5 k! R4 3.3 k!
Find the Thevenin and Norton VS + equivalents of the circuit at R R 20 V – 2 5 left, with the port as shown. 4.7 k! 4.7 k!
R3 1.5 k! R6 3.3 k!
R1 R4 Find voc. Use mesh current method. + = = V S + voc ia R ib R – 2 5 ( ) = – ( ) =
R3 R6 ( + + ) = Insert values and solve: ib = 0.930 mA. ( + + + ) = EE 201 voc = ib R5 = 4.37 V. Thevenin – 14 R R Find isc. Note that R5 is shorted out 1 4 by the short circuit.
iS Use equivalent resistance to find iS. VS + R isc – 2 = + + ( + )= . = = . R3 R6
R 3.27 k! Use current divider to find isc. Th
+ + = V + Th – + 4.37 V
isc = 1.45 mA
. = = = . . I N RN 1.45 mA 3.27 k!
EE 201 Thevenin – 15 Alternatively, we can use the short-cut method to find RTh.
R1 R4
R2 R5 RTh
R3 R6
= [ + + ( + )] = 3.02 k!
EE 201 Thevenin – 16 R 15 ! Example 3 5
IS 4 A Find the Thevenin and Norton R R 30 ! equivalents of the circuit at 1 30 ! 3 right, with the port as shown. V + S R2 R4 30 V – 20 ! 15 !
Find voc. Use node voltage. = + = + R5 + + = I + + = S
R1 a R3 b + + = + V S + R R + + + = + – 2 4 voc – solve: va = 20 V, vb = 40 V. "
EE 201 voc = vb = 40 V Thevenin – 17 Find isc. Use node voltage, again. Note that R4 is shorted out (so ignore it)
and node b is shorted to ground, vb = 0.
= + + R5 IS = + +
R1 a R3 b = + +
VS + R i . – 2 sc = + + = . = = . . Need va.
= + RTh 6.36 ! = + 40 V + V IN R Th – N 6.28 A 6.36 ! = = 8.57 V + + EE 201 Thevenin – 18 Maximum power transfer Now 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.
RTh load = = + ( + ) V + v Th – RL RL –
In the usual way, find the max by setting the derivative to zero and solving. = = ( + ) ( + ) For maximum + = = power to the load
EE 201 See slide 4 for an example – look at power column in the table. Thevenin – 19