1. 1 ECE 221 Electric Circuit Analysis I Chapter 13 Thévenin & Norton Equivalent Herbert G. Mayer, PSU Status 3/5/2015

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

3. 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 the power source 

4. 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 model é One such model is the Thévenin equivalent, an imaginary CVS with identical behavior éé Named after 19 th century French telegraph engineer Léon Charles Thévenin, 1857-1926 é Such a Thévenin equivalent source consists of CVS with V Th Volt and an internal resistance of R Th Ω in series; nothing else!

5. 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. 6 Thought Experiment é 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 V Th with no load, i.e. load with resistance R L = ∞ Measuring the voltage yields V Th 2. Short-circuiting the terminals a and b allows us to measure the maximum load current i SC Measuring that short-circuit current yields i SC 3. Thus we can compute R Th R Th = V Th / i SC

7. 7 Conclusion: Thévenin Equivalent é A Thévenin Equivalent circuit is a simple, CVS circuit with a serial resistor, R TH...... electrically equivalent to an arbitrary circuit, whose key parameters, idle voltage and short- circuit current are known, as measured above Such a circuit’s equivalent resistance is: R Th = V Th / i SC é A sample Thévenin transformation follows, taken from [1], p. 113-115

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

9. 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 R Th and i SC equivalents First measure idle voltage V Th at the terminals, here called v ab, then measure the short-circuit current i SC Thus we can compute the resulting equivalent resistance R Th : R Th = V Th / i SC 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. 10 Sample Thévenin Problem First experiment: to measure voltage at open terminals = v ab = v1 identical to V TH

11. 11 Thévenin Problem: Open Terminals (v1 - 25)/5+ v1/20 - 3=0 4*v1 -100 -60 + v1=0 5*v1=160 v1= v ab = v Th = 32 V V Th = 32 V

12. 12 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 CCS We compute v2, the voltage along the 3 A CCS Once v2 is known all other units, specifically i SC can be computed

13. 13 Sample Thévenin Problem

14. 14 Thévenin Problem: Short-Circuit v2/20 + (v2-25)/5 - 3 + v2/4=0 v2 + 4*v2 + 5*v2=160 10*v2=160 v2 =16 V V TH = 32 V i SC = v2/4 = 16/4 = 4 A R Th = V Th /i SC = 32/4 = 8 Ω

15. 15 Final Thévenin Equivalent Circuit

16. 16 Short-Cut For Thévenin Resistance é 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 R EQ which is the final R TH

17. 17 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

18. 18 Norton Equivalent Method 1: Earlier we covered transforming some CVS into an equivalent CCS é 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 i N is 4 A, with the 8 Ω resistor in parallel: i N = i Th = 32/8=4 A i N =4 A

19. 19 Norton Equivalent Sample for R TH = R L VL CVS = 32 * 8 / ( 8 + 8 )= 16 V iL CVS = 32 / ( 8 + 8 )= 2 A iL CCS = 4 * 8 / ( 8 + 8 )= 2 A VL CCS = 4 * 8 / 2= 16 V

20. 20 Thévenin- vs. Source Transformation Thévenin Thévenin 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:

21. 21 Source Transformation 1 Transformation 1: as a result, have 2 resistors and 2 CCS in parallel; can be combined trivially

22. 22 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:

23. 23 Source Transformation 3 As a result, we have two 4 Ω resistors in series. Trivial transformation. See next simple replacement!

24. 24 Source Transformation 4 Simple transformation, two 4 Ω become one 8 Ω Now convert CVS back into equivalent CCS Happens to be Thévenin Equivalent

25. 25 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: V TH = 32 V, i SC = 4 A, and R TH = 8 Ω With CCS, happens to be Norton Equivalent

26. 26 Thévenin Equivalent With Dependent Sources (Taken from [1])

27. 27 Find Thévenin: Dependent Sources é 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 i x = 0 ? é To answer that we exercise the Thévenin transformation: open terminals and short-circuited terminals (a) and (b)

28. 28 Find Thévenin: Dependent Sources Can circuit C1 be simplified by omitting left part, since i x = 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

29. 29 Find Thévenin: Dependent Sources Is C3 equivalent to C1? Yes, since ix = 0 A, but circuit analysis needs both parts due to mutual dependencies é Goal to find Thévenin Equivalent of circuit C1 Step 1: Compute v = v ab = v TH of C1 with terminals (a) and (b) open Step 2: Compute current i SC with terminals (a) and (b) short- circuited

30. 30 Thévenin Equivalent Step 1 Step 1: Compute v = v ab = v TH of C1 with terminals (a) and (b) open: i = ( 5 – 3 * v ) / 2000=( 5 – 3 * v TH ) / 2000

31. 31 Thévenin Equivalent Step 1 Step 1: Compute v = v ab = v TH of C1 with terminals (a) and (b) open: i = ( 5 – 3 * v ) / 2000AKAi =( 5 – 3 * v TH ) / 2000 v = -20 * 25 * iAKAv TH = -500 * i v TH = -500 * ( 5 – 3 * v TH ) / 2000= -5 * ( 5 – 3 * v TH ) / 20 v TH = -25 / 20 + v TH * 15 / 20-- multiply by * 20 20 * v TH = -25 + v TH * 15 5 * v TH = -25 v TH = -5 V

32. 32 Thévenin Equivalent Step 2 Step 2: Compute i = i SC 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 i SC run in opposite directions! Sign!!

33. 33 Thévenin Equivalent Step 2 Step 2: Compute i = i SC with terminals (a) and (b) short-circuited Note that the 20 * i and i SC run opposite directions! i=5 / 2000=2.5 mA i SC =- 20 * i i SC =-50 mA With i SC = -50 mA, v TH = -5 V we get R TH = - 5 / -50 * 1000 [ V / A ] R TH = 1/10 k Ω

34. 34 Thévenin Equivalent With Dependent é Below we have C5, the Thévenin Equivalent circuit of the original C1 circuit With dependent sources the convenient short-cut (p. 15) for computing R TH does not work! Note reversed polarity of 5 V CVS in C5!

35. 35 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!

36. 36 Exercise 1

37. 37 Exercise 1: Thévenin Equivalent Given the circuit below, generate the Thevénin equivalent First we leave terminals (a) and (b) open and compute V TH Then we short-circuit the terminals, and compute i SC Yielding the Thevénin equivalent V TH, i SC, including the resistance R TH

38. 38 Exercise 1 Open: Thévenin Equivalent With open plugs, the voltage drop V TH 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: V TH = 24 + 12 = 36 V

39. 39 Exercise 1 SC: Thévenin Equivalent With terminals (a) and (b) short-circuited, i SC runs through the 14 Ω resistor, and i 6 through the 6 Ω resistor KCL: i SC – 4 + i 6 =0 KVL:14 * i SC – 12 – 6 * i 6 =0 i SC =1.8 A R TH = V TH / i SC =36 / 1.8= 20 Ω

40. 40 Exercise 1: Thévenin Equivalent V TH Easier to use the Node Voltage Methodology to compute i SC: KCL: v / 14 – 4 + ( v – 12 ) / 6=0 v=14*36 / 20 Yields i SC : i SC = V / 14=14 * 36 / ( 20 * 14 ) i SC =36 / 20= 1.8 A R TH = V TH / i SC =36 / 1.8= 20 Ω

41. 41 Exercise 1: A Simpler Way To identify R TH, all we need to do is short-circuit all CVSs, eliminate all CCSs, and compute the equivalent R EQ, which is R TH This just leaves the 2 resistors in series, resulting R TH = 20 Ω

42. 42 Finally: Thévenin Equivalent The circuit below shows the equivalent Thévenin arrangement

43. 43 References 1. 1.Electric Circuits, James W. Nielsson and Susan A. Riedel, Pearson Education Inc, publishing as as Prentice Hall, © 2015, ISBN- 13: 978-0-13-376003-3