Experiment#6Thevenin Equivalent Circuits

1.0 Objectives:

- To study Thevenin Theorem.

- To study maximum power transfer to a load.

- To study the load line equation.

2.0 Theory:

- Thévenin's Theorem states that we can replace entire network, exclusive of the

load, by an equivalent circuit that contains only an independent voltage source in

series with an impedance (resistance) such that the current-voltage relationship at

the load is unchanged. Thevenin's Theorem is especially useful in analyzing

power systems and other circuits where one particular resistor in the circuit (called

the "load" resistor) is subject to change, and re-calculation of the circuit is

necessary with each trial value of load resistance, to determine voltage across it

and current through it. The advantage in performing the "Thevenin conversion" to

the simpler circuit, of course, is that it makes load voltage and load current so

much easier to solve than in the original network. Calculating the equivalent

Thevenin source voltage and series resistance is actually quite easy.

Fig.1 Circuit connection for Thevenin theory

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Thévenin Equivalent Circuit

Experiment#6Thevenin Equivalent Circuits

2.1 Thevenin's Theorem Procedures

1. Pick a good breaking point in the circuit (cannot split a dependent source and its control variable).

2. Compute the open circuit voltage, VOC.

3. Compute the Thevenin equivalent resistance, RTh (or impedance, ZTh).

a. If there are only independent sources, then short circuit all the voltage sources and open circuit the current sources (just like superposition).

b. If there are only dependent sources, then must use a test voltage or current source in order to calculate RTh= vTest/iTest (or ZTh=VTest/ITest).

c. If there are both independent and dependent sources, then compute RTh (or ZTh) from RTh= vOC/iSC (or ZTh=VOC/ISC).

4. Replace circuit with Thevenin/Norton equivalent. VOC in series with RTh (or ZTh).

Note: for 3(b) the equivalent network is merely RTh (or ZTh), that is, no current or voltage sources.

2.2The Maximum Power Theorem

Whenever we have a source of electrical energy we might pose the question of what circumstances produce maximum load power.

Consider a load connected to a source, and consider that the source is well represented by a Thevenin equivalent model. Then the situation is the one shown in the figure below.   

  First calculate the voltage across the load resistor, calling that voltage Vload.

Vload = Vo Rload/(Ro+ Rload)

Then the power dissipated in Rload is:

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=(Vload)2 / Rload

= (Vo)2  Rload/(Ro+ Rload)2

        For an extremum (maximum or minimum) we differentiate the load power with respect to Rload, and set that derivative to zero.  The derivative is:

(Vo)2(Ro- Rload)/(Ro+ Rload)

        The maximum power for positive load resistance occurs when:

Rload = Ro

        When the load resistance is equal to the internal resistance of the source, the load voltage is exactly half of the open circuit voltage, so the maximum power delivered to the load is:

Pmax = (Vo )2/(4Ro)

        Whenever a load resistance is equal to the internal resistance of a source, the two (source and resistance) are said to be matched.

3. Equipment:

1- DC Power supply. SN. = 097082

2- Resistor substitution box SN. = 8268012

3- Discrete Resistors.

4- Bread Board.

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Experiment#6Thevenin Equivalent Circuits

4. Procedure:

3.1. Part (A): Thevnin Theory:

Fig.1 Circuit connection for Thevenin theory experiment

1. Given R1=R3=2kΩ , R2=R4=1kΩ , RL=5kΩ

2. Measure the exact value of the resistors

3. Connect the circuit as shown in Fig.1.

4. Set the power supply Vs=10 Volts (measure it using DMM)

5. Measure the current passing through RL= IRL

6. Disconnect RL and measure the voltage between nodes a and b, Vth

7. Disconnect the supply voltage and replace it with awire (short circuit) and

use an ohmmeter to measure Thevenin resistance between terminals a and b,

Rth

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8. Fig.2 shows Thevenin equivalent circuit of the network shown in Fig.1. The

current flowing through the load resistance RL is given by:

IRL= Vth/ (Rth + RL)

Find the theoretical and the measured values of IRL.

Fig.2 Thevenin equivalent of the network shown in Fig.1

Experiment#6Thevenin Equivalent Circuits

4.2 Part (B): load Line and Maximum Power Transformation:

Fig.3 Circuit connection for the load line and maximum power transfer

1. Given R1= 2K

2. Set the power supply Vs to 10 Volts (measure it using DMM)

3. Vary the value of RL as shown in Table 2 and measure the current through

and the voltage across.

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Experiment#6Thevenin Equivalent Circuits

5.1 Data sheet (theoretical)

Table 1 resistor valuesR1 = 2KΩ R2=1 KΩ R3=2 KΩ R4=1 KΩ RL=5 KΩ Table 2 measured data

IRL= 0.2941 mA Vth= 2 V Rth = 1.8 kΩ

Table 3 results for the current through and voltage across RLRL (Ω) VRL (Volts) IRL (mA) Power=VxI (mW)

0 0 5 0

500 2 4 8

1000 3.333 3.333 11.109

1500 4.286 2.857 12.245

2000 5 2.5 12.5

2500 5.556 2.222 12.345

3000 6 2 12

5000 7.143 1.429 10.207

8000 8 1 8

12000 8.571 0.714 6.1197

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Experiment#6Thevenin Equivalent Circuits

5.2 Data sheet ( Practical )

Table 4 resistor valuesR1 = KΩ R2= KΩ R3= KΩ R4=KΩ RL= KΩ

Table 5 measured data IRL= 0.291 mA Vth= 2 V Rth = 1.77 kΩ

Table 6 results of current through and voltage across RLRL (Ω) VRL (Volts) IRL (mA) Power=VxI (mW)

0 0 5.025 0

500 2 4.019 8.1

1000 3.35 3.346 11.2

1500 4.3 2.868 12.3

2000 5.01 2.508 12.6

2500 5.6 2.229 12.4

3000 6.02 2.006 12.1

5000 7.2 1.432 10.3

8000 8.02 1.003 8.05

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12000 8.6 0.717 6.2

Experiment#6Thevenin Equivalent Circuits

6.0 Simulation:

6.1 DC Voltage:

6.2 Current:

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Experiment#6Thevenin Equivalent Circuits

7.0 Questions:

Q1: Find theoretically the Thevenin equivalent for the circuit shown in Fig.1. Vth: Open circuit: Req = R1 + R4 + R3 = 2 + 1 + 2 = 5 kΩ

I = V/ R = 10 / 5 = 2 mA V4 = Vth = I x R =2 x 1 = 2 V Rth: Short circuit: -10 + (2 x I1) + [1 x ( I1 - Isc)] + (2 x I1) = 0 → (1) [1 x ( Isc – I1)] + (1 x Isc) = 0 → (2) Solving (1) & (2): I1 = 2.22 mA Isc = 1.11 mA Rth = Vth / Isc

= 2 / 1.11 = 1.80 kΩ

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Experiment#6Thevenin Equivalent Circuits

Q2: Calculate the power dissipated in RL.

It is Solved in table 6.

Q3: On a graph paper plot the power dissipated versus RL.

It is in figure 4.

Q4: What is the value of RL for max power transfer?

From figure 4 and table 6: Pmax = 12.6 mW at RL = 2000 Ω

Q5: On graph paper plot IRL versus VRL.

It is in figure 5.

Q6: From the plot of (Q5) if VRL = 6.6 v find the value of RL.

If VRL = 6.6 v then IRL = 1.7 mA

RL = VRL / IRL = (6.6/1.7) = 3.88 kΩ = 3880 Ω

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Experiment#6Thevenin Equivalent Circuits

8. Conclusion:

1. Thevenin's theorem for electrical networks states that any combination of voltage sources and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R.

2. The power dissipated is maximum when Rth equals RL.

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Experiment#6Thevenin Equivalent Circuits

9. References:

http://en.wikipedia.org/wiki/Ohm's_law

http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.combo.html

Lab manual ENG207 fundamentals of electric circuits

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