1 / 63

Unit Eight: Thevenin’s Theorem Maximimum Power Transfer Theorm

ET115 DC Electronics. Unit Eight: Thevenin’s Theorem Maximimum Power Transfer Theorm. John Elberfeld JElberfeld@itt-tech.edu. Schedule. Unit Topic Chpt Labs Quantities, Units, Safety 1 2 (13) Voltage, Current, Resistance 2 3 + 16 Ohm’s Law 3 5 (35) Energy and Power 3 6 (41)

tania
Download Presentation

Unit Eight: Thevenin’s Theorem Maximimum Power Transfer Theorm

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Schedule Unit Topic Chpt Labs • Quantities, Units, Safety 1 2 (13) • Voltage, Current, Resistance 2 3 + 16 • Ohm’s Law 3 5 (35) • Energy and Power 3 6 (41) • Series Circuits Exam I 4 7 (49) • Parallel Circuits 5 9 (65) • Series-Parallel Circuits 6 10 (75) • Thevenin’s, Power Exam 2 6 19 (133) • Superposition Theorem 6 11 (81) • Magnetism & Magnetic Devices 7 Lab Final • Course Review and Final Exam

  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. 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.

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

  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

  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.

  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. 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. Thevenin Example • In this example RL is the LOAD Resistor • We want the value of ONE voltage source and ONE resistor that gives the same voltage to RL as this circuit V I1 R1 RL R2 I2

  12. What is the Voltage across RL? • Using POS and Voltage Divider • VL = V R2RL/(R2+RL) / [R1+ R2RL/(R2+RL) ] • Substitute each value for RL and solve for the voltage • Imagine a circuit with 10 resistors. • Image 20 values for RL • There must be an easier way! V R1 A R2 RL B

  13. Steps • Note where the load resistor connects to the circuit, and remove it • Calculate the voltage between the two points where RL used to be connected • This is VTh • Short the voltage sources, open current sources • Calculate the resistance between the two points where the load resistor used to be connected • The is RTh

  14. Step 1 • Note where the load resistor connects to the circuit, and remove it I1 R1 V V R1 A R2 R2 RL I2 B

  15. Step 2 • Calculate the voltage between the two points where RL used to be connected • VAB = VR2/(R1+R2) Voltage Divider • VTH = VR2/(R1+R2) V R1 A R2 B

  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 • RTH = R1R2/(R1+R2) POS V R1 A R2 B

  17. Create the New Circuit and Test • RTH = R1R2/(R1+R2) • VTH = VR2/(R1+R2) • VL = VTHRL/(RL+RTH) • The Claim: No matter what value RL 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 RTH VTH RL

  18. Test the Theorem • RL2 = 3.9kΩ2.7kΩ/(3.9kΩ+2.7kΩ) = 1.60k Ω • VL = 25V1.60k Ω /(1.6kΩ+18kΩ) = 2.04 V • IL = 2.04 V / 2.7k Ω = 756 mA • We should get the same values with the Thevenin Circuit R1=18k V=25V A R2=3.9k RL=2.7k B

  19. Step 1 • Note where the load resistor connects to the circuit, and remove it R1=18k V=25V R1=18k V=25V A A R2=3.9k RL=2.7k R2=3.9k B B

  20. Step 2 • Calculate the voltage between the two points where RL used to be connected • VTH = VR2/(R1+R2) Voltage Divider • VTH = 25v 3.9kΩ/(3.9kΩ+18kΩ) • VTH = 4.45 V R1=18k V=25V A R2=3.9k B

  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 • RTH = R1R2/(R1+R2) POS • RTH = 18kΩ 3.9kΩ/(18kΩ +3.9kΩ) • RTH = 3.21k Ω R1=18k A R2=3.9k B

  22. Create the New Circuit and Test • RTH = 3.21k Ω • VTH = 4.45 V • VL = VTHRL/(RL+RTH) • VL = 4.45 V2.7k Ω /(3.21k Ω+ 2.7k Ω ) • VL = 2.04 V • IL = 2.04V/ 2.7k Ω =756mA • AGREES!!! RTH VTH RL=2.7kΩ

  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. Example

  25. Thevenin Practice • R1 = 8kΩ, R2 = 12kΩ, R3 = 6kΩ, R4 = 15kΩ, RL = 5kΩ, V = 15V • Find VTH and RTH R1 R2 V R3 RL R4

  26. Step 1 • Note where the load resistor connects to the circuit, and remove it R1 R2 V R3 R4

  27. Step 2 • Calculate the voltage between the two points where RL used to be connected • VTH = V (R2 +R3)/(R1+R2+R3+R4) Voltage Divider • VTH = 15v 18kΩ/(41kΩ) • VTH = 6.59 V 8k 12k V=15V 6k 15k

  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 • RTH = 23k18k/(23k+18k) POS • RTH = 10.1k Ω 8k 12k V 6k 15k

  29. Create the New Circuit and Test • RTH = 10.1k Ω • VTH = 6.59 V • VL = VTHRL/(RL+RTH) • VL = 6.59 V5k Ω /(5k Ω +10.1k Ω) • VL = 2.18 V • IL = 2.18/ 5k Ω =436μA RTH VTH RL=5kΩ

  30. Check • VL = 2.18 V, IL = 436 μA • VR23 = 2.18 V, LR23 = 2.18 V/18k = 121μA • IT = 436μA + 121μA = 557 μA • VR1= 557 μA 8k = 4.46V • VR4= 557 μA 15k = 8.36V • VR1+ VR23 + VR4 = 15V • Numbers CHECK!! 8k 12k RL=5kΩ 15V 6k 15k

  31. Thevenin Practice • R1 = 5kΩ, R2 = 2kΩ, R3 = 1kΩ, RL = 5kΩ, V = 10V • Find VTH and RTH • Check your results R1 V R3 R2 RL

  32. Step 1 • Note where the load resistor connects to the circuit, and remove it R1 V R3 R2 RL

  33. Step 2 • Calculate the voltage between the two points where RL used to be connected • VTH = V (R2)/(R1+R2) Voltage Divider • VTH = 10v 2 kΩ/(5 kΩ + 2 kΩ) • VTH = 2.857 V • No current flows through R3, so it has no effect onthe output voltage R1 V R3 R2

  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 • RTH = R3 + R1R2/(R1+R2) • RTH=1kΩ+5kΩ 2kΩ /(5kΩ+2kΩ ) • RTH = 2.429 kΩ R1 R3 R2

  35. Create the New Circuit and Test • RTH = 2.429 kΩ • VTH = 2.857 V • VL = VTHRL/(RL+RTH) • VL = 2.86 V5kΩ /(5kΩ +2.43k Ω) • VL = 1.923 V • IL = 1.923 V/ 5k Ω = 384.6 μA RTH VTH RL=5kΩ

  36. Check • VL = 1.923 V, IL = 384.6 μA • RT = 5kΩ + 2kΩ 6kΩ (2kΩ + 6kΩ ) = 6.5kΩ • IT = 10 V/ 6.5kΩ = 1.538 mA • V1 = 1.538 mA 5kΩ = 7.690 V • V2 = V3L= 10V-7.69V=2.31V • I3L=2.31V/6k = 385 μA • VL = 385 μA 5kΩ = 1.925V • Numbers CHECK!! 5kΩ 1kΩ 10V 2kΩ 5kΩ

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

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

  39. Thevenin Practice • R1 = 330 Ω, R2 = 680 Ω, R3 = 680 Ω, R4 = 560 Ω, RL = 1kΩ, V = 24 V • Find VTH and RTH • There is no easy way to check your results!

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

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

  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 A B 560 Ω 680 Ω

  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 330 Ω 680 Ω A B 560 Ω 680 Ω

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

  45. Calculate RTH • RTop = 330 Ω 680 Ω/(330 Ω+ 680 Ω) =222 Ω • RBot = 560 Ω 680 Ω/(560 Ω+ 680 Ω) =307 Ω • RTh = 222 Ω + 307 Ω = 529 Ω A 330 Ω 680 Ω 560 Ω 680 Ω B

  46. Create the New Circuit and Test • RTH = 529 Ω • VTH = 5.32 V • VL = VTHRL/(RL+RTH) • VL = 5.32 V1kΩ /(529 Ω +1 k Ω) • VL = 3.48 V • IL = 3.48 V/ 1k Ω = 3.48 mA RTH VTH RL=1kΩ

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

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

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

  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 271.8 Ω A B 560 Ω 680 Ω

More Related