1 Unit Eight: Thevenin’s Theorem Maximimum Power Transfer Theorm John Elberfeld...

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Unit Eight:Thevenin’s Theorem

Maximimum Power Transfer Theorm

John Elberfeld

[email protected]

ET115 DC Electronics

mailto:[email protected]

Schedule

Unit Topic Chpt Labs1. Quantities, Units, Safety 1 2 (13)2. Voltage, Current, Resistance 2 3 + 163. Ohm’s Law 3 5 (35)4. Energy and Power 3 6 (41)

5. Series Circuits Exam I 4 7 (49)

6. Parallel Circuits 5 9 (65)

7. Series-Parallel Circuits 6 10 (75)

8. Thevenin’s, Power Exam 2 6 19 (133)

9. Superposition Theorem 6 11 (81)

10. Magnetism & Magnetic Devices7 Lab Final 11. Course Review and Final Exam

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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.

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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.

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.

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Reading Assignment

• Read and study

• Chapter 6: Series-Parallel Circuits:Pages 237-247 (Second half of chapter)

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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

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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.

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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.

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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

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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

R2

R1

I1

I2

RL

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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! RL

V

R2

R1

A

B

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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

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Step 1


V

R2

R1

I1

I2RL

V

R2

R1

A

B

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Step 2


• VAB = VR2/(R1+R2) Voltage Divider

• VTH = VR2/(R1+R2)

V

R2

R1

A

B

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Steps 3 + 4


• Calculate the resistance between the two points where the load resistor used to be connected

• Parallel resistors here

• RTH = R1R2/(R1+R2) POSV

R2

R1

A

B

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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

VTHRTH RL

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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

RL=2.7k

V=25V

R2=3.9k

R1=18k

A

B

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Step 1


RL=2.7k

V=25V

R2=3.9k

R1=18k

A

B

V=25V

R2=3.9k

R1=18k

A

B

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Step 2


• VTH = VR2/(R1+R2) Voltage Divider

• VTH = 25v 3.9kΩ/(3.9kΩ +18kΩ)

• VTH = 4.45 V

V=25V

R2=3.9k

R1=18k

A

B

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Steps 3 + 4



• RTH = R1R2/(R1+R2) POS

• RTH = 18kΩ 3.9kΩ /(18kΩ +3.9kΩ )

• RTH = 3.21k Ω R2=3.9k

R1=18k

A

B

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• RTH = 3.21k Ω

• VTH = 4.45 V


• VL = 4.45 V 2.7k Ω /(3.21k Ω + 2.7k Ω )

• VL = 2.04 V

• IL = 2.04V/ 2.7k Ω =756mA

• AGREES!!!VTH

RTH RL=2.7kΩ

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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

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Example

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Thevenin Practice

• R1 = 8kΩ, R2 = 12kΩ, R3 = 6kΩ, R4 = 15kΩ, RL = 5kΩ, V = 15V

• Find VTH and RTH

V R2

R1

R3 RLR4

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Step 1


V R2

R1

R3R4

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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

V=15V

12k

8k

6k15k

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Steps 3 + 4



• RTH = 23k18k/(23k+18k) POS

• RTH = 10.1k Ω V 12k

8k

6k15k

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• RTH = 10.1k Ω

• VTH = 6.59 V


• VL = 6.59 V 5k Ω /(5k Ω +10.1k Ω)

• VL = 2.18 V

• IL = 2.18/ 5k Ω =436μA

VTHRTH RL=

5kΩ

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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!!15V

12k

8k

6k15k

RL=5kΩ

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Thevenin Practice

• R1 = 5kΩ, R2 = 2kΩ, R3 = 1kΩ, RL = 5kΩ, V = 10V


• Check your results

V

R2

R1

R3

RL

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Step 1


V

R2

R1

R3

RL

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• 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

V

R2

R1

R3

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Steps 3 + 4



• RTH = R3 + R1R2/(R1+R2)

• RTH=1kΩ+5kΩ 2kΩ /(5kΩ+2kΩ )

• RTH = 2.429 kΩR2

R1

R3

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• RTH = 2.429 kΩ

• VTH = 2.857 V


• VL = 2.86 V 5kΩ /(5kΩ +2.43k Ω)

• VL = 1.923 V

• IL = 1.923 V/ 5k Ω = 384.6 μA

VTHRTH RL=

5kΩ

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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!!

10V

5kΩ

2kΩ

1kΩ

5kΩ

Wheatstone Bridge

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

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Thevenin Theorem

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

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Thevenin Practice

• R1 = 330 Ω, R2 = 680 Ω, R3 = 680 Ω, R4 = 560 Ω, RL = 1kΩ, V = 24 V


• There is no easy way to check your results!

Step 1


• RL used to go from A to B

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24 V

330 Ω

680 Ω

680 Ω 560 Ω

A B



• 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

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24 V

330 Ω

680 Ω

680 Ω 560 Ω

A B

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Steps 3 + 4



• This needs some imagination!

330 Ω

680 Ω

680 Ω 560 Ω

A B

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

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330 Ω

680 Ω

680 Ω 560 Ω

A B

Result

• All four resistors are still connected to ground

• A and B both contact the same resistors as the original circuit

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330 Ω

680 Ω

680 Ω

560 Ω

A

B

330 Ω

680 Ω

680 Ω 560 Ω

A B

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

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330 Ω

680 Ω

A

B

680 Ω

560 Ω

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• RTH = 529 Ω

• VTH = 5.32 V


• VL = 5.32 V 1kΩ /(529 Ω +1 k Ω)

• VL = 3.48 V

• IL = 3.48 V/ 1k Ω = 3.48 mA

VTHRTH RL=

1kΩ

Balanced Bridge Analysis

• What if the bridge is balanced?• 330 Ω/680 Ω = 271.8 Ω/560 Ω

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24 V

330 Ω

271.8 Ω

680 Ω 560 Ω

A B

Step 1


• RL used to go from A to B

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24 V

330 Ω

271.8 Ω

680 Ω 560 Ω

A B



• 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

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24 V

330 Ω

271.8 Ω

680 Ω 560 Ω

A B

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Steps 3 + 4



• This needs some imagination!

330 Ω

680 Ω 560 Ω

A B

271.8 Ω

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

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330 Ω

680 Ω 560 Ω

A B

271.8 Ω

Result

• All four resistors are still connected to ground

• A and B both contact the same resistors as the original circuit

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330 Ω 680 Ω

560 Ω

A

B

330 Ω

680 Ω 560 Ω

A B

271.8 Ω

271.8 Ω

Calculate RTH

• RTop = 330 Ω 680 Ω/(330 Ω+ 680 Ω) =222 Ω

• RBot = 560 Ω 271.8 Ω/(560 Ω+ 271.8 Ω) =183 Ω

• RTh = 222 Ω + 183 Ω = 405 Ω

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330 Ω

A

B

680 Ω

560 Ω271.8 Ω

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• RTH = 405 Ω (Really doesn’t matter!)

• VTH = 0 V


• VL = 0 V 1kΩ /(529 Ω +1 k Ω)

• VL = 0 V

• IL = 0 V/ 1k Ω = 0 mA

VTHRTH RL=

1kΩ

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

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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

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General Theory

• P = V I = I2 R = V2 / R• For a simple circuit:

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VS

RS RL

T S L

S ST L

T S L

SL L L

S L

S LL

S L

2S S L S L

L L L 2S L S L S L

2S L

L 2S L

R = R + R

V VI = = = I

R (R +R )

VV =I R = x RL

(R +R )

V RV =

(R +R )

V V R V RP = I V =

(R +R ) (R +R ) (R +R )

V RP =

(R +R )

x

Non-linear Relationship

• This is a non-linear relationship

• As RL increases, both the top and bottom of the fraction increase, but not at equal rates because the denominator is squared

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2S L

L 2S L

V RP =

(R +R )

Let Excel Do the Work

• P = = $A$2^2*A5/(A5+$B$2)^2

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Logic

• Since P = I2R, 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 RL = RS

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Check Theory• Find PL when RL = 50 Ω, 100 Ω, 150 Ω

• VL = 15 V 50 Ω / (50 Ω + 100 Ω) = 5V

• P = V2/R = (5V)2 / 50 Ω = .5 W (50 Ω Load)

• VL = 15 V 100 Ω / (100 Ω + 100 Ω) = 7.5V

• P = V2/R = (7.5V)2 / 100 Ω = .563 W (100 Ω )

• VL = 15 V 150 Ω / (150 Ω + 100 Ω) = 9V

• P = V2/R = (9V)2 / 150 Ω = .54 W (150 Ω Load)

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VS = 15V

RS=100 Ω

RL

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

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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

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