1. 1 Unit Eight: Thevenin’s Theorem Maximimum Power Transfer Theorm John Elberfeld JElberfeld@itt-tech.edu ET115 DC Electronics

2. Schedule Unit Topic Chpt Labs 1.Quantities, Units, Safety12 (13) 2.Voltage, Current, Resistance23 + 16 3.Ohm’s Law35 (35) 4.Energy and Power36 (41) 5.Series CircuitsExam I47 (49) 6.Parallel Circuits59 (65) 7.Series-Parallel Circuits610 (75) 8.Thevenin’s, Power Exam 2619 (133) 9.Superposition Theorem 611 (81) 10.Magnetism & Magnetic Devices7Lab Final 11.Course Review and Final Exam 2

3. 3 Unit 8 Objectives - I Describe the Thevenin equivalent circuit. Reduce a resistive series/parallel circuit to its equivalent Thevenin form. Explain terminal equivalency in the context of Thevenin’s theorem. Calculate the load current and voltage in a Wheatstone bridge by applying Thevenin’s theorem. Determine the value of a load resistance for which maximum power is transferred from a given source.

4. 4 Unit 8 Objectives – II Calculate the load resistor for which maximum power is transferred for a given circuit. Construct basic DC circuits on a protoboard. Use a digital multimeter (DMM) to measure a predetermined low voltage on a power supply. Measure resistances and voltages in a DC circuit using a DMM.

5. Unit 8 Objectives – III Apply Ohm’s Law, Thevenin’s theorem, KVL and KCL to practical circuits. Construct and test a Wheatstone bridge on a protoboard. Test circuits by connecting simulated instruments in Multisim. Troubleshoot circuits constructed in Multisim exercises using simulated instruments. 5

6. Reading Assignment Read and study Chapter 6: Series-Parallel Circuits: Pages 237-247 (Second half of chapter) 6

7. Lab Assignment Experiment 19, “Thevenin’s Theorem,” beginning on page 133 of DC Electronics: Lab Manual and MultiSim Guide. Complete all measurements, graphs, and questions and turn in your lab before leaving the room 7

8. Written Assignments Complete the Unit 8 Homework sheet Show all your work! Be prepared for a quiz on questions similar to those on the homework. 8

9. 9 Thevenin’s Theorem This theorem is used to convert a complex linear network into a simple network consisting of a constant voltage source and resistors in series.

10. 10 Thevenin’s Theorem To solve a circuit using this theorem: –disconnect the load resistance from terminals –determine open circuit voltage between terminals –short circuit the voltage sources or open circuit the current source and then replace by its internal resistance, if any –Replace original circuit by the Thevenin’s circuit to analyze the total given circuit

11. 11 Thevenin Example In this example R L is the LOAD Resistor We want the value of ONE voltage source and ONE resistor that gives the same voltage to R L as this circuit V R2R2 R1R1 I1I1 I2I2 RLRL

12. 12 What is the Voltage across R L ? Using POS and Voltage Divider V L = V R 2 R L /(R 2 +R L ) / [R1+ R 2 R L /(R 2 +R L ) ] Substitute each value for R L and solve for the voltage Imagine a circuit with 10 resistors. Image 20 values for R L There must be an easier way! RLRL V R2R2 R1R1 A B

13. 13 Steps Note where the load resistor connects to the circuit, and remove it Calculate the voltage between the two points where R L used to be connected –This is V Th Short the voltage sources, open current sources Calculate the resistance between the two points where the load resistor used to be connected –The is R Th

14. 14 Step 1 Note where the load resistor connects to the circuit, and remove it V R2R2 R1R1 I1I1 I2I2 RLRL V R2R2 R1R1 A B

15. 15 Step 2 Calculate the voltage between the two points where R L used to be connected V AB = VR 2 /(R 1 +R 2 ) Voltage Divider V TH = VR 2 /(R 1 +R 2 ) V R2R2 R1R1 A B

16. 16 Steps 3 + 4 Short the voltage sources, open current sources Calculate the resistance between the two points where the load resistor used to be connected Parallel resistors here R TH = R 1 R 2 /(R 1 +R 2 ) POS V R2R2 R1R1 A B

17. 17 Create the New Circuit and Test R TH = R 1 R 2 /(R 1 +R 2 ) V TH = VR 2 /(R 1 +R 2 ) V L = V TH R L /(R L +R TH ) The Claim: No matter what value R L is given, it will have the same voltage and current in this circuit that it would in the old, two resistor circuit with the other voltage source V TH R TH RLRL

18. 18 Test the Theorem R L2 = 3.9kΩ2.7kΩ/(3.9kΩ+2.7kΩ) = 1.60k Ω V L = 25V1.60k Ω /(1.6kΩ+18kΩ) = 2.04 V I L = 2.04 V / 2.7k Ω = 756 mA We should get the same values with the Thevenin Circuit R L = 2.7k V= 25V R 2 = 3.9k R 1 = 18k A B

19. 19 Step 1 Note where the load resistor connects to the circuit, and remove it R L = 2.7k V= 25V R 2 = 3.9k R 1 = 18k A B V= 25V R 2 = 3.9k R 1 = 18k A B

20. 20 Step 2 Calculate the voltage between the two points where R L used to be connected V TH = VR 2 /(R 1 +R 2 ) Voltage Divider V TH = 25v 3.9kΩ/(3.9kΩ +18kΩ) V TH = 4.45 V V= 25V R 2 = 3.9k R 1 = 18k A B

21. 21 Steps 3 + 4 Short the voltage sources, open current sources Calculate the resistance between the two points where the load resistor used to be connected R TH = R 1 R 2 /(R 1 +R 2 ) POS R TH = 18kΩ 3.9kΩ /(18kΩ +3.9kΩ ) R TH = 3.21k Ω R 2 = 3.9k R 1 = 18k A B

22. 22 Create the New Circuit and Test R TH = 3.21k Ω V TH = 4.45 V V L = V TH R L /(R L +R TH ) V L = 4.45 V 2.7k Ω /(3.21k Ω + 2.7k Ω ) V L = 2.04 V I L = 2.04V/ 2.7k Ω =756mA AGREES!!! V TH R TH R L = 2.7kΩ

23. 23 Usefulness Simplifying a two resistor circuit to a one resistor circuit does not save much effort Suppose you had to calculate the voltage and current for 10 load resistors in a complex circuit with 20 resistors and 2 power supplies? The time needed to find the Thevenin circuit will pay off handsomely

24. 24 Example

25. 25 Thevenin Practice R 1 = 8kΩ, R 2 = 12kΩ, R 3 = 6kΩ, R 4 = 15kΩ, R L = 5kΩ, V = 15V Find V TH and R TH V R2R2 R1R1 R3R3 RLRL R4R4

26. 26 Step 1 Note where the load resistor connects to the circuit, and remove it V R2R2 R1R1 R3R3 R4R4

27. 27 Step 2 Calculate the voltage between the two points where R L used to be connected V TH = V (R 2 +R 3 )/(R 1 +R 2 +R 3 +R 4 ) Voltage Divider V TH = 15v 18kΩ/(41kΩ) V TH = 6.59 V V= 15V 12k 8k 6k 15k

28. 28 Steps 3 + 4 Short the voltage sources, open current sources Calculate the resistance between the two points where the load resistor used to be connected R TH = 23k18k/(23k+18k) POS R TH = 10.1k Ω V 12k 8k 6k 15k

29. 29 Create the New Circuit and Test R TH = 10.1k Ω V TH = 6.59 V V L = V TH R L /(R L +R TH ) V L = 6.59 V 5k Ω /(5k Ω +10.1k Ω) V L = 2.18 V I L = 2.18/ 5k Ω =436μA V TH R TH R L = 5kΩ

30. 30 Check V L = 2.18 V, I L = 436 μA V R23 = 2.18 V, L R23 = 2.18 V/18k = 121μA I T = 436μA + 121μA = 557 μA V R1 = 557 μA 8k = 4.46V V R4 = 557 μA 15k = 8.36V V R1 + V R23 + V R4 = 15V Numbers CHECK!! 15V 12k 8k 6k 15k R L = 5kΩ

31. 31 Thevenin Practice R 1 = 5kΩ, R 2 = 2kΩ, R 3 = 1kΩ, R L = 5kΩ, V = 10V Find V TH and R TH Check your results V R2R2 R1R1 R3R3 RLRL

32. 32 Step 1 Note where the load resistor connects to the circuit, and remove it V R2R2 R1R1 R3R3 RLRL

33. 33 Step 2 Calculate the voltage between the two points where R L used to be connected V TH = V (R 2 )/(R 1 +R 2 ) Voltage Divider V TH = 10v 2 kΩ/(5 kΩ + 2 kΩ) V TH = 2.857 V No current flows through R 3, so it has no effect on the output voltage V R2R2 R1R1 R3R3

34. 34 Steps 3 + 4 Short the voltage sources, open current sources Calculate the resistance between the two points where the load resistor used to be connected R TH = R 3 + R 1 R 2 /(R 1 +R 2 ) R TH =1kΩ+5kΩ 2kΩ /(5kΩ+2kΩ ) R TH = 2.429 kΩ R2R2 R1R1 R3R3

35. 35 Create the New Circuit and Test R TH = 2.429 kΩ V TH = 2.857 V V L = V TH R L /(R L +R TH ) V L = 2.86 V 5kΩ /(5kΩ +2.43k Ω) V L = 1.923 V I L = 1.923 V/ 5k Ω = 384.6 μA V TH R TH R L = 5kΩ

36. 36 Check V L = 1.923 V, I L = 384.6 μA R T = 5kΩ + 2kΩ 6kΩ (2kΩ + 6kΩ ) = 6.5kΩ I T = 10 V/ 6.5kΩ = 1.538 mA V 1 = 1.538 mA 5kΩ = 7.690 V V 2 = V 3L = 10V-7.69V=2.31V I 3L =2.31V/6k = 385 μA V L = 385 μA 5kΩ = 1.925V Numbers CHECK!! 10V 5kΩ 2kΩ 1kΩ 5kΩ

37. Wheatstone Bridge The Wheatstone bridge is a complex circuit that can’t be simplified using series and parallel combinations 37

38. Thevenin Theorem We can apply Thevenin’s Theorem to find the voltage across the center load resistor in an unbalanced Wheatstone bridge 38

39. 39 Thevenin Practice R 1 = 330 Ω, R 2 = 680 Ω, R 3 = 680 Ω, R 4 = 560 Ω, R L = 1kΩ, V = 24 V Find V TH and R TH There is no easy way to check your results!

40. Step 1 Note where the load resistor connects to the circuit, and remove it R L used to go from A to B 40 24 V 330 Ω 680 Ω 560 Ω A B

41. Step 2 Calculate the voltage between the two points where R L used to be connected V TH = V A - V B V A =24 V 680Ω/(330Ω + 680Ω) = 16.16V V B = 24 V 560Ω/(680Ω + 560Ω) = 10.84V V TH = V A - V B V TH = 5.32 V 41 24 V 330 Ω 680 Ω 560 Ω A B

42. 42 Steps 3 + 4 Short the voltage sources, open current sources Calculate the resistance between the two points where the load resistor used to be connected This needs some imagination! 330 Ω 680 Ω 560 Ω A B

43. Analysis Point A connects the 330 Ω and 680 Ω at one end, and ground connects them at the other end, so they are parallel Point B connects the 680 Ω and 560 Ω at one end, and ground connects them at the other end, so they are parallel 43 330 Ω 680 Ω 560 Ω A B

44. Result All four resistors are still connected to ground A and B both contact the same resistors as the original circuit 44 330 Ω 680 Ω 560 Ω A B 330 Ω 680 Ω 560 Ω A B

45. Calculate R TH R Top = 330 Ω 680 Ω/(330 Ω+ 680 Ω) =222 Ω R Bot = 560 Ω 680 Ω/(560 Ω+ 680 Ω) =307 Ω R Th = 222 Ω + 307 Ω = 529 Ω 45 330 Ω 680 Ω A B 560 Ω

46. 46 Create the New Circuit and Test R TH = 529 Ω V TH = 5.32 V V L = V TH R L /(R L +R TH ) V L = 5.32 V 1kΩ /(529 Ω +1 k Ω) V L = 3.48 V I L = 3.48 V/ 1k Ω = 3.48 mA V TH R TH R L = 1kΩ

47. Balanced Bridge Analysis What if the bridge is balanced? 330 Ω/680 Ω = 271.8 Ω/560 Ω 47 24 V 330 Ω 271.8 Ω 680 Ω 560 Ω A B

48. Step 1 Note where the load resistor connects to the circuit, and remove it R L used to go from A to B 48 24 V 330 Ω 271.8 Ω 680 Ω 560 Ω A B

49. Step 2 Calculate the voltage between the two points where R L used to be connected V TH = V A - V B V A =24 V 680Ω/(330Ω + 680Ω) = 16.16V V B = 24 V 560Ω/(271.8Ω + 560Ω) = 16.16V V TH = V A - V B V TH = 0 V 49 24 V 330 Ω 271.8 Ω 680 Ω 560 Ω A B

50. 50 Steps 3 + 4 Short the voltage sources, open current sources Calculate the resistance between the two points where the load resistor used to be connected This needs some imagination! 330 Ω 680 Ω 560 Ω A B 271.8 Ω

51. Analysis Point A connects the 330 Ω and 680 Ω at one end, and ground connects them at the other end, so they are parallel Point B connects the 271.8 Ω and 560 Ω at one end, and ground connects them at the other end, so they are parallel 51 330 Ω 680 Ω 560 Ω A B 271.8 Ω

52. Result All four resistors are still connected to ground A and B both contact the same resistors as the original circuit 52 330 Ω 680 Ω 560 Ω A B 330 Ω 680 Ω 560 Ω A B 271.8 Ω

53. Calculate R TH R Top = 330 Ω 680 Ω/(330 Ω+ 680 Ω) =222 Ω R Bot = 560 Ω 271.8 Ω/(560 Ω+ 271.8 Ω) =183 Ω R Th = 222 Ω + 183 Ω = 405 Ω 53 330 Ω A B 680 Ω 560 Ω 271.8 Ω

54. 54 Create the New Circuit and Test R TH = 405 Ω (Really doesn’t matter!) V TH = 0 V V L = V TH R L /(R L +R TH ) V L = 0 V 1kΩ /(529 Ω +1 k Ω) V L = 0 V I L = 0 V/ 1k Ω = 0 mA V TH R TH R L = 1kΩ

55. Balanced Bridge In a balanced bridge, there is NO CURRENT and there in NO VOLTAGE DROP across the center load resistor, no matter what value resistor is used as a load 55

56. Maximum Power Transfer For a given voltage, maximum power is transferred from a source to a load when the load resistance is equal to the internal source resistance 56

57. General Theory P = V I = I 2 R = V 2 / R For a simple circuit: 57 VSVS RSRS RLRL

58. Non-linear Relationship This is a non-linear relationship As R L increases, both the top and bottom of the fraction increase, but not at equal rates because the denominator is squared 58

59. Let Excel Do the Work P = = $A$2^2*A5/(A5+$B$2)^2 59

60. Logic Since P = I 2 R, it is logical that increasing R would increase power However, as R increases, the current decreases as well, reducing the power The “Tipping Point” occurs when R L = R S 60

61. Check Theory Find P L when R L = 50 Ω, 100 Ω, 150 Ω V L = 15 V 50 Ω / (50 Ω + 100 Ω) = 5V P = V 2 /R = (5V) 2 / 50 Ω =.5 W (50 Ω Load) V L = 15 V 100 Ω / (100 Ω + 100 Ω) = 7.5V P = V 2 /R = (7.5V) 2 / 100 Ω =.563 W (100 Ω ) V L = 15 V 150 Ω / (150 Ω + 100 Ω) = 9V P = V 2 /R = (9V) 2 / 150 Ω =.54 W (150 Ω Load) 61 V S = 15V R S =100 Ω RLRL

62. Maximum Transfer The highest power transfer occurs when the load resistor is equal to the internal or source resistance A little calculus can confirm this – which is why advanced math is important in electronics 62

63. Unit 8 Summary Thevenin’s theorem and terminal equivalency Applying Thevenin’s theorem to resistive circuits Applying Thevenin’s theorem to the Wheatstone bridge Calculating the load required for maximum power transfer from a specific source 63