1. ECE 221 Electric Circuit Analysis I Chapter 13 Thévenin & Norton Equivalent
Herbert G. Mayer, PSU
Status 11/3/2015

2. Syllabus Motivation Thought Experiment Purpose Thévenin Problem
Norton Equivalent
A Sample
Thévenin vs. Source Transformations
Thévenin With Dependent Sources
Exercise 1
References

3. Motivation
When working with real electric sources, such as typical household power supplies, the actual circuit behind the terminals is unknown
It should be unknown! Unreasonable to expect users to know what exactly is behind the electric terminal!
We only know 1.) that there is a source of constant voltage at terminals a and b with a complex but finite resistance internally
And 2.) that the current supplied depends on external load, up to some practical limits
When the limit is exceeded, a fuse flips and we lose power 

4. Motivation
It is desirable to understand the physical limits of such a CVS
Once we know the limits, we can model the whole complex CVS source with an equivalent, but simpler alternative
One such model is the Thévenin equivalent, an imaginary CVS with identical behavior
Named after 19th century French telegraph engineer Léon Charles Thévenin,
Such a Thévenin equivalent source consists of CVS with vTh Volt and an internal resistance of RTh Ω in series; and nothing else!

5. Thought Experiment
We only know 1.) that the black (green) box contains constant voltage- and current sources; 2.) its voltage at the terminals; 3.) that the current is finite, limited by some practical maximum
How to model this situation? Answered by the Thévenin Equivalent

6. RTh = vTh / iSC Thought Experiment
We find the Thévenin parameters, modeling the green box, by practicing the following experiments:
1. Leaving the circuit open at terminals a and b, allows measuring the voltage vTh with no load, i.e. load with resistance RL = ∞
Measuring the voltage yields vTh
2. Short-circuiting the terminals a and b allows us to measure the maximum load current iSC
Measuring that short-circuit current yields iSC
3. Thus we can compute RTh
RTh = vTh / iSC

7. Conclusion: Thévenin Equivalent
A Thévenin Equivalent circuit is a simple, CVS circuit with a serial resistor, RTH . . .
. . . electrically equivalent to an arbitrary linear circuit, whose key parameters, idle voltage and short-circuit current are measurable
Such a circuit’s equivalent resistance is:
RTh = vTh / iSC
A sample Thévenin transformation follows, taken from [1], p

8. Sample Thévenin Problem
Looking at the terminals (a) and (b) from the right, we pretend not to know what’s behind them; only that vab is delivered. Measure 2 key parameters; in the end replace what’s there with the Thévenin equivalent:

9. Sample Thévenin Problem
The following circuit has various internal sources and resistances, 2 external terminals (a) and (b)
Goal is to model the exact behavior of this circuit via the RTh and iSC equivalents
First measure idle voltage VTh at the terminals, here called vab, then measure the short-circuit current iSC
Thus we can compute the resulting equivalent resistance RTh: RTh = vTh / iSC
With terminals open, no current flows in 4 Ω resistor
We compute v1, parallel to the CCV and the 20 Ω resistor
Note: with open terminals, the 4 Ω resistor in this circuit might as well be missing! The voltage drop along the 4 Ω resistor is 0 V

10. Thévenin Problem: Open Terminals
First experiment: to measure voltage at open terminals = vab = v1 identical to vTH

11. Thévenin Problem: Open Terminals
(v1 - 25)/5 + v1/ = 0
Students compute v1 = vTH!

12. Thévenin Problem: Open Terminals
(v1 - 25)/5 + v1/ = 0
4*v v1 = 0
5*v = 160
v1 = vab = vTh = 32 V
vTh = 32 V

13. Thévenin Problem: Short-Circuit
In the next experiment terminals (a) and (b) are short-circuited
A real current flows through the 4 Ω R, now parallel to the 3 A CCS
We compute v2, the voltage along the 3 A CCS, using the NoVoMo
Once v2 is known, then all other units, specifically iSC can be computed

14. Sample Thévenin Problem

15. Thévenin Problem: Short-Circuit
v2/20 + (v2-25)/ v2/4 = 0
Students Compute v2

16. Thévenin Problem: Short-Circuit
v2/20 + (v2-25)/ v2/4 = 0
v2 + 4*v2 + 5*v = 160
10*v = 160
v = 16 V
vTH = 32 V
iSC = v2/4 = 16/4 = 4 A
RTh = vTh/iSC = 32/4 = 8 Ω

17. Final Thévenin Equivalent Circuit

18. Short-Cut For Thévenin Resistance
A simpler way to compute the Thévenin Equivalence resistance for a linear circuit with only CCS, CVS, and resistances --does not work with dependent sources!:
Open all CCS, and short-circuit all CVS, compute the resulting resistance REQ which is the final RTH

19. Short-Cut For Thévenin Resistance
All CCS are left open, all CVS are short-circuited, resulting equivalent R = 8 Ω
20 Ω and 5 Ω in parallel = 4 Ω, plus 4 Ω in series = 8 Ω

20. Norton Equivalent iN = 4 A
Method 1: Earlier we covered transforming some CVS into an equivalent CCS
Method 2: The Thévenin transformation, introduced here, yields a CVS with R in series
We can combine the 2 methods and generate a CCS equivalent to any CVS-circuit by just measuring idle voltage and short-circuit current
The equivalent Norton current iN is 4 A, with the 8 Ω resistor in parallel:
iN = iTh = 32/8 = 4 A
iN = 4 A

21. Norton Equivalent Sample for RTH = RL
VLCVS = 32 * 8 / ( ) = 16 V
iLCVS = 32 / ( ) = 2 A
iLCCS = 4 * 8 / ( ) = 2 A
VLCCS = 4 * 8 / 2 = 16 V

22. Thévenin- vs. Source Transformation
Instead of using Thévenin equivalent, use successive source-to- source transformations to reduce a complex circuit into one as simple as Thévenin equivalent. What will it look like?
Transformation 1: starting on the left side, substitute 25 V CVS with R in series into equivalent CCS with R in parallel:

23. Thévenin- vs. Source Transformation
Students, transform 25 V CVS with 5 Ohm in series into equivalent CCS with 5 Ohm in parallel

24. Source Transformation 1
Transformation 1: as a result, we have 2 resistors or 5 Ohm and 20 Ohm and 2 CCS in parallel; can be combined trivially

25. Source Transformation 2
Combined 20 Ω and 5 Ω into an equivalent 4 Ω. Combined 5 A and parallel 3 A CCS into an equivalent 8 A CCS.
Now convert back into CVS:

26. Source Transformation 3
As a result, we have two 4 Ω resistors in series. Trivial transformation. See next simple replacement!

27. Source Transformation 4
Simple transformation, two 4 Ω become one 8 Ω
Now convert CVS back into equivalent CCS
Happens  to be Thévenin Equivalent

28. Source Transformation 4
Students, transform 32 V CVS and 8 Ohm in series into equivalent CCS with 8 Ohm in parallel

29. Source Transformation 5
Final transformation shows that Thévenin equivalent at terminals (a) and (b) is the same as source-to-source transformation down to minimal number of components!
Thévenin equivalent is: vTH = 32 V, iSC = 4 A, and RTH = 8 Ω
With CCS, happens  to be the Norton Equivalent

30. Thévenin Equivalent With Dependent Sources
(Taken from [1])

31. Find Thévenin: Dependent Sources
Consider circuit C1 below: what is the Thévenin Equivalent at terminals (a) and (b)?
Can it be simplified by totally omitting the left part, since ix = 0 ?
To answer that we exercise the Thévenin transformation: open terminals and short-circuited terminals (a) and (b)

32. Find Thévenin: Dependent Sources
Can circuit C1 be simplified by omitting left part, since ix = 0 A?
Clearly not! See C2, which is NOT an equivalent option, since both parts, left and right, depend on electric units of their mutual other half
The dependent current source is a function of i namely 20 * i, flowing through the 2 kΩ resistor on the left part
And the dependent voltage source is a function of voltage vab along the 25 Ω resistor, namely 3*v = 3*vab

33. Find Thévenin: Dependent Sources
Is C3 equivalent to C1? Yes, since ix = 0 A, but circuit analysis needs both parts due to mutual dependencies
Goal to find Thévenin Equivalent of circuit C1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b) open
Step 2: Compute current iSC with terminals (a) and (b) short- circuited

34. Thévenin Equivalent Step 1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b) open:
i = ( 5 – 3 * v ) / 2000 = ( 5 – 3 * vTH ) / 2000

35. Thévenin Equivalent Step 1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b) open:
i = ( 5 – 3 * v ) / AKA i = ( 5 – 3 * vTH ) / 2000
v = -20 * 25 * i AKA vTH = -500 * i
Students, compute v = vTH

36. Thévenin Equivalent Step 1
Step 1: Compute v = vab = vTH of C1 with terminals (a) and (b) open:
i = ( 5 – 3 * v ) / AKA i = ( 5 – 3 * vTH ) / 2000
v = -20 * 25 * i AKA vTH = -500 * i
vTH = -500 * ( 5 – 3 * vTH ) / 2000 = -5 * ( 5 – 3 * vTH ) / 20
vTH = -25 / 20 + vTH * 15 / multiply by * 20
20 * vTH = vTH * 15
5 * vTH = -25
vTH = -5 V

37. Thévenin Equivalent Step 2
Step 2: Compute i = iSC with terminals (a) and (b) short-circuited
Short-circuit bypasses 25 Ω resistor; that means the voltage controlling the dependent voltage source is also short- circuited, i.e. the voltage drop is 0 V. That dependent source can be removed.
Note that the 20 * i and iSC run in opposite directions! Sign!!

38. Thévenin Equivalent Step 2
Step 2: Compute i = iSC with terminals (a) and (b) short-circuited
Note that the 20 * i and iSC run opposite directions!
i = 5 / 2000 = 2.5 mA
iSC = - 20 * i
iSC = -50 mA
With iSC = -50 mA, vTH = -5 V we get
RTH = - 5 / -50 * [ V / A ]
RTH = 1/10 k Ω = 100 Ω

39. Thévenin Equivalent With Dependent
Below we have C5, the Thévenin Equivalent circuit of the original C1 circuit
With dependent sources the convenient short-cut (p. 19 above) for computing RTH does not work!
Note reversed polarity of 5 V CVS in C5!

40. Norton Equivalent of C1
The Norton Equivalent has the 100 Ω resistor parallel to the constant current source of 5 V / 100 Ω = 50 mA
Note the direction of the CCS, the tip pointing to the + sign of the equivalent CVS!

41. Exercise 1

42. Exercise 1: Thévenin Equivalent
Given the circuit below, generate the Thevénin equivalent
First leave terminals (a) and (b) open and compute vTH
Then short-circuit terminals (a) and (b), and compute iSC
Yielding the Thevénin equivalent vTH, iSC, including resistance RTH

43. Exercise 1 Open: Thévenin Equivalent
With open plugs, the voltage drop vTH at terminals (a) and (b) is the same as along the CCS with 4 A
CCS pushes 4 A through 6 Ω resistor, creating 4 * 6 = 24 V voltage drop
Ohms Law: yields 6 * 4 = 24 V: vTH = = 36 V

44. Exercise 1 SC: Thévenin Equivalent
With terminals (a) and (b) short-circuited, iSC runs through the 14 Ω resistor, and i6 through the 6 Ω resistor
KCL: iSC – 4 + i6 = 0
KVL: 14 * iSC – 12 – 6 * i6 = 0
iSC = 1.8 A
RTH = vTH / iSC = 36 / 1.8 = 20 Ω

45. Exercise 1: Thévenin Equivalent VTH
Easier to use the Node Voltage Methodology to compute iSC:
KCL: v / 14 – 4 + ( v – 12 ) / 6 = 0
v = 14*36 / 20
Yields iSC: iSC = v / = 14 * 36 / ( 20 * 14 )
iSC = 36 / 20 = 1.8 A
RTH = vTH / iSC = 36 / 1.8 = 20 Ω

46. Exercise 1: A Simpler Way
To identify RTH, all we need to do is short-circuit all CVSs, eliminate all CCSs, and compute the equivalent REQ, which is RTH
This just leaves the 2 resistors in series, resulting RTH = 20 Ω

47. Finally: Thévenin Equivalent
The circuit below shows the equivalent Thévenin arrangement

48. References
Electric Circuits, James W. Nielsson and Susan A. Riedel, Pearson Education Inc., publishing as as Prentice Hall, © 2015, ISBN- 13: