Thevenin's Theorem_ Network Theorems

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THEVENINS THEOREM

INTRODUCTION

THEVENINS EQUIVALENT CIRCUIT

PRESENTATION OF THEVENINS THEOREM

PROOF OF THEVENINS THEOREM

WORKED EXAMPLE 2

WORKED EXAMPLE 3

WORKED EXAMPLE 4

SUMMARY

INTRODUCTION

Thevenins theorem is a popular theorem, used often for analysis of electronic circuits. Its theoretical

value is due to the insight it offers about the circuit. This theorem states that a linear circuit containing one

or more sources and other linear elements can be represented by a voltage source and a resistance. Using

this theorem, a model of the circuit can be developed based on its output characteristic. Let us try to find

out what Thevenins theorem is by using an investigative approach.

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THEVENINS EQUIVALENT CIRCUIT

In this section, the model of a circuit is derived based on its output characateristic. Let a circuit be

represented by a box, as shown in Figure 8. Its output characteristic is also displayed. As the load resistor

is varied, the load current varies. The load current is bounded between two limits, zero and Im, and the

load voltage is bounded between limits,EVolts and zero volts. When the load resistor is infinite, it is an

open circuit. In this case, the load voltage is at its highest, which isEvolts and the load current is zero.

This is the point at which the output characteristic intersects with the Y axis. When the load resistor is of

zero value, there is a short circuit across the output terminals of the circuit and in this instance, the load

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current is maximum, specified asIm and the load voltage is zero. It is the point at which the output

characteristic intersects with the X axis.

The circuit in Figure 9 reflects the output characteristic, displayed in Fig. 8. It has an output ofEvolts,

when the load current is zero. Hence the model of the circuit can have a voltage source ofEvolts. When

the output terminals are short-circuited, it can be stated that the internal resistance of circuit absorbsE

volts at a current ofIm. This means that the internal resistance of the circuit, called as RTh, has a value of

EoverIm, as shown by the equation displayed in Fig. 9. Hence the circuit model consists of a voltage

source of valueEvolts and a resistorRTh. This resistor is the resistance of the circuit, as viewed from the

load terminals.

Let us see how we can apply what we have learnt. A simple circuit is presented in Fig. 10. The task is to

get an expression for the load currentIL and express it in terms of Thevenins voltage and Thevenins

resistance. Thevenins voltage is the voltage obtained across the load terminals, with the load resistor

removed. In this case, the load resistor is named as R3.

At first, an expression for the load current is obtained without the use of Thevenins theorem. To get the

load current, the steps involved are as follows.

Get an expression for the equivalent resistanceReq, seen by the source, as shown in Fig. 11.

Divide the source voltage by the equivalent resistance to get currentIS supplied by the source. This




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source current flows through resistors,R2 andR3 connected in parallel.

Use the current division rule to get an expression for the load current.

As shown by equation (17), the equivalent resistance is obtained by adding resistorR1 to the parallel value

of resistors,R2 andR3 . The source current is the ratio of source voltage to the equivalent resistance, as

expressed by equation (18).. Then the load current through resistorR3 is obtained using the current

division rule, as shown by equation (19).

Now some mathematical manipulations are required to get Thevenins voltage and Thevenins resistance.

The expression for the load current is expressed by equation (20). Divide both the numerator and the

denominator of equation (20) by the sum of resistors,R1 andR2 , and then we get equation (21). The

numerator of equation (21) is Thevenins voltage. The first part of the denominator, containing resistorsR1 andR2 , is Thevenins resistance. It is the parallel value of resistors R1 andR2 . Once Thevenins

voltage and Thevenins resistance are known, the load current can be obtained as shown by equation (21)

Equation (22) defines the expressions for Thevenins voltage and Thevenins resistance. They are

obtained from equation (21).

From the expression for the load current, we can obtain a circuit and this circuit is presented in Figure 12.

We can now ask what Thevenins voltage and Thevenins resistance represent? How do we obtain them

in a simpler way? They can be obtained as shown next.




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Thevenins voltage is the voltage across the load terminals with the load resistor removed. In other words,

the load resistor is replaced by an open circuit. In this instance, the load resistor isR3 and it is replaced by

an open circuit. Then Thevenin,s voltage is the open circuit voltage, the voltage across resistorR2. This

voltage can easily be obtained by using the voltage division rule. The voltage division rule states the

division of source voltage is proportionate to resistance. Thevenins resistance is the resistance, as viewed

from the load terminals, with both the load resistor and the sources in the circuit removed. Here removal

of the voltage source means that it is replaced by a short circuit, and the load resistor is replaced by an

open circuit. Thevenins resistance is the parallel value of resistors R1 andR2 . Next Thevenins theorem

is presented in a formal manner.


PRESENTATION OF THEVENINS THEOREM

Thevenins theorem represents a linear network by an equivalent circuit. Let a network with one or more

sources supply power to a load resistor as shown in Fig. 14. Thevenins theorem states that the network

can be replaced by a single equivalent voltage source, marked as Thevenins Voltage or open-circuit

voltage and a resistor marked as Thevenins Resistance. Proof of this theorem is presented below.

Thevenins theorem can be applied to linear networks only. Thevenins voltage is the algebraic sum of

voltages across the load terminals, due to each of the independent sources in the circuit, acting alone. Itcan be seen that Thevenins theorem is an outcome of superposition theorem.




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Thevenins equivalent circuit consists of Thevenins voltage and Thevenins resistance. Thevenins

voltage is also referred to as the open-circuit voltage, meaning that it is obtained across the load terminals

without any load connected to them. The load is replaced by an open-circuit and hence Thevenins

voltage is called as the open-circuit voltage.

Figure 15 shows how Thevenins voltage is to be obtained. Here it is assumed that we have a resistive

circuit with one or more sources. As shown in Fig. 15, Thevenins voltage is the open-circuit voltage

across the load terminals. The voltage obtained across the load terminals without the load being connected

is the open-circuit voltage. This open-circuit voltage can be obtained as the algebraic sum of voltages, due

to each of the independent sources acting alone.Given a circuit, Thevenins voltage can be obtained as

outlined below.

Figure 16 shows how Thevenins resistance is to be obtained. Thevenins resistance is the resistance as

seen from the load terminals. To obtain this resistance, replace each independent ideal voltage source in

the network by a short circuit, and replace each independent ideal current source by an open circuit. If a

source is not ideal, only the ideal part of that source is replaced by either a short circuit or an open circuit

as the case may be. The internal resistance of the source, reflecting the non ideal aspect of the circuit, is

left in the circuit, as it is where it is. A voltage source is connected across the load terminals. Then

Thevenins resistance is the ratio of this source voltage to its current, as marked in Fig. 16. A few

examples are presented after this page to illustrate the use of Thevenins theorem.


PROOF OF THEVENINS THEOREM




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The circuit in Fig. 17 can be used to prove Thevenins theorem. Equation (1) in the diagaram expresses an

external voltage VY connected to the load terminals, as a function of currentIY and some constants. It is

valid to do so, since we are dealing with a linear circuit. Let us some that the internal independent sources

remain fixed. Then, as the external voltage VY is varied, currentIY will vary, and the variationIY with VY

is accounted for by provision of a coefficient , named as k1 in equation (1). It can be seen that k1 reflects

resistance of the circuit as seen by external voltage source VY. Coefficient k2 reflects the contribution to

terminal voltage by internal sources and components of the circuit. It is valid to do so, since we are

dealing with a linear circuit, and a linear circuit obeys the principle of superposition. Each independentinternal source within the circuit contributes its part to terminal voltage and constant k2 is the algebraic

sum of contributions of internal sources. Adjust external voltage source such that currentIY becomes

zero. As shown by equation (2), the coefficient k2 is Thevenins voltage. To determine Thevenins

resistance, set external source voltage to zero. If the internal sources are such as to yield positive

Thevenins voltage, currentIY will be negative and coefficient k1 is Thevenins resistance, as shown by

equation (3). This concludes the proof of Thevenins theorem.

The step involved in the application of Thevenins theorem are summarized below.


WORKED EXAMPLE 2




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A problem has been presented now. For the circuit in Fig. 18, you are asked to obtain the load current

using ThevEnins theorem. We have already looked at this circuit, but the purpose here is to show, how

to apply Thevenins theorem.

Solution:

It is a good practice to learn to apply a theorem in a systematic way. The solution is obtained in four

steps. The steps are as shown above.

The first step is to obtain Thevenins voltage as described now. Remove the load resistor, and represent

the circuit, as shown in Fig. 19 in order to get the value of Thevenins voltage, which is the voltage across

resistorR2. This voltage can be obtained is shown next by equation (23).

Equation (23) is obtained using the voltage division rule. The two resistors are connected in series and the

current through them is the same, and hence the voltage division rule can be applied.




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WORKED EXAMPLE 3

We take up another example now. Figure 22 contains the circuit. The source voltage is 10 Volts. The

circuit containing the small signal model of a bipolar junction transistor looks similar to this circuit in Fig.

22.

Solution:

You are asked to obtain the Thevenins equivalent of the circuit in Fig. 22. This problem is a bit moredifficult, since it has dependent sources. The Thevenins theorem can be applied to circuits containing

dependent sources also. The only constraint in applying Thevenins theorem to a circuit is that it should

be a linear circuit.

The steps involved are the same, as outlined before for the previous problem. The steps involved are

shown above.:

Given a circuit with dependent sources, it may at times be preferable to obtain the open circuit voltage and

the short circuit current, and then obtain Thevenins resistance as the ratio of open circuit voltage to the

short circuit current. The short circuit current is obtained by replacing the load resistor by a short circuit,

and it is the current that flows through the short circuit. This technique has been used in the proof of




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Thevenins theorem.

Since there is no load connected to the output terminals, voltage V2 is the open circuit voltage, which is

the same as the Thevenins voltage. To obtain the open circuit voltage, the following equations are

obtained.

Equation (26) expresses the voltage across resistorR2. The current through resistorR2 is ten times

currentI, and the value of resistorR2 is 100 . Equation (27) is written for the loop containing the

independent source voltage. The independent source voltage is 10 Volts. The value of resistorR1 is 10

and the current through it can be obtained as shown by equation (27). Equation (28) is obtained by

replacing voltage V2 in equation (27) by its corresponding expression in equation (26).

On simplifying, we can obtain the value of currentI, and the Thevenins voltage, as illustrated by equation

(29).

The second step is to obtain Thevenins resistance. The circuit in Fig. 23 is used for this purpose.

To obtain Thevenins resistance of a circuit with dependent source, it is preferable to obtain the short

circuit current and then obtain Thevenins resistance as the ratio of Thevenins voltage to short circuit




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current. The circuit in Fig. 23 is used to obtain the short circuit current.

Equations (30) to (33) are obtained from the circuit in Fig. 23. When the output terminals are shorted, the

short circuit current, known also as the Nortons current, is ten times current I, as shown by equation (30).

Note that the source voltage is 10 Volts. When the output voltage is zero, current Iis the ratio of source

voltage to resistorR1 and it equals one Ampere, as displayed by equation (31). Equations (32) and (33)

show how Nortons current and Thevenins resistance can be obtained.

Now it is shown how the Thevenins resistancecan be obtained by another way. The circuit in Fig. 24 is

presented for this purpose.

Alternate Method to obtain RTh

Remove the independent voltage source and replace it by a short circuit. Connect a source at the output

as shown in Fig. 24. Then Thevenins resistance is obtained as follows.

Thevenins resistanceis expressed by equation (34). It is obtained with the independent source voltage,

contained in the circuit, being replaced by a short circuit, as shown in Fig. 24.




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Equations (35) to (38) are obtained from the circuit in Fig. 24. Since the source voltage is zero, the sum ofvoltage across resistorR1 and the voltage across the dependent voltage source is zero and we get

equation (35). Equation (36) is obtained by using KVL at node a. The expression obtained for currentIin

equation (35) is used to replace the currentIin equation (36) and this leads to equation (37). On

simplifying, we get equation (38) and the value ofThevenins resistanceis 2 Ohms.

Another View

It is possible to obtain an expression for the currentIx marked in Fig. 24. Equation (39) shows how this

current is obtained. Since we know the voltage across the dependent current source and the current

through it, we can replace it by a resistor, as shown in Fig. 24. The parallel value of two resistors is the

Thevenins resistance.

Equations (40) and (41) illustrate how Thevenins resistance is obtained. Since Thevenins voltage and

Thevenins resistance are known, the equivalent circuit can be drawn.


WORKED EXAMPLE 4




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Find the current through the load resistorRL.

Solution:

Thevenins theorem is used to get the solution. RemoveRL. Find Thevenins voltage. ReplaceRL by a

short-circuit. Find the current through the short-circuit. Then Thevenins resistance is the ratio of theopen-circuit voltage and the short-circuit current. Then the current through the load resistorRL can be

determined.

First let us obtain Thevenins voltage. The circuit withoutRL is shown below.

Let the resistance of the circuit in Fig. 26, as seen by the source beRA.The value ofRA can be obtained,

as shown by equation (42).

OnceRA is known, the currentIA supplied by the source can be obtained, as shown by equation (43).




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From the circuit in Fig. 26, we can obtain currents I3 andI5, marked in Fig. 26, by using the current

division rule.

Once the values of currentsI3 andI5 are known,Thevenins voltage can be obtained as shown by

equation (46).

To find the short-circuit currentIN , we use the circuit in Fig. 27. Let the resistance of the circuit in Fig.

27, as seen by the source beR

B.The value ofR

B can be obtained, as shown by equation (47).

OnceRB is known, the currentIB supplied by the source can be obtained, as shown by equation (48).

Use the current division rule. Find currents I2 andIc, marked in Fig. 27.




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Thevenin's Theorem_ Network Theorems