thevenin theoram

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THEVENIN’S THEOREM

INTRODUCTION

THEVENIN’S EQUIVALENT CIRCUIT

ILLUSTRATION OF THEVENIN’S THEOREM

FORMAL PRESENTATION OF THEVENIN’S THEOREM

PROOF OF THEVENIN’S THEOREM

WORKED EXAMPLE 2

WORKED EXAMPLE 3

WORKED EXAMPLE 4

SUMMARY

INTRODUCTION

Thevenin’s 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 Thevenin’s theorem is by using an investigative

approach.

THEVENIN’S 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

mvaries. The load current is bounded between two limits, zero and I , and the load

voltage is bounded between limits, E Volts 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 is E volts 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,

mthe load current is maximum, specified as I 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 of E volts, when the load current is zero. Hence the model of the circuit

can have a voltage source of E volts. When the output terminals are short-

circuited, it can be stated that the internal resistance of circuit absorbs E volts at

m Tha current of I . This means that the internal resistance of the circuit, called as R ,

mhas a value of E

over I

, as shown by the equation displayed in Fig. 9. Hence theThcircuit model consists of a voltage source of value E volts and a resistor R . 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

LFig. 10. The task is to get an expression for the load current I and express it in

terms of Thevenin’s voltage and Thevenin’s resistance. Thevenin’s voltage is the

voltage obtained across the load terminals, with the load resistor removed. In this

3case, the load resistor is named as R .





Equation (22) defines the expressions for Thevenin’s voltage and Thevenin’s

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 Thevenin’s voltage andThevenin’s resistance represent? How do we obtain them in a simpler way? They

can be obtained as shown next.

Thevenin’s 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

3instance, the load resistor is R and it is replaced by an open circuit. Then

2Thevenin,s voltage is the open circuit voltage, the voltage across resistor R . 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.

Thevenin’s 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. Thevenin’s resistance is the parallel value of resistors1 2 R and R . Next Thevenin’s theorem is presented in a formal manner.

FORMAL PRESENTATION OF THEVENIN’S THEOREM



Thevenin’s 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. Thevenin’s theorem states that the network can be replaced by a single

equivalent voltage source, marked as Thevenin’s Voltage or open-circuit voltage

and a resistor marked as Thevenin’s Resistance. Proof of this theorem is presented below. Thevenin’s theorem can be applied to linear networks only.

Thevenin’s voltage is the algebraic sum of voltages across the load terminals, due

to each of the independent sources in the circuit, acting alone. It can be seen that

Thevenin’s theorem is an outcome of superposition theorem.

Thevenin’s equivalent circuit consists of Thevenin’s voltage and Thevenin’s

resistance. Thevenin’s 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 Thevenin’s voltage is

called as the open-circuit voltage.

Figure 15 shows how Thevenin’s 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,Thevenin’s 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, Thevenin’s voltage can be obtained as outlined below.



Figure 16 shows how Thevenin’s resistance is to be obtained. Thevenin’s

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

Thevenin’s 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

Thevenin’s theorem.

PROOF OF THEVENIN’S THEOREM

The circuit in Fig. 17 can be used to prove Thevenin’s theorem. Equation (1) inYthe diagaram expresses an external voltage V connected to the load terminals, as

Ya function of current I 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

Y Yremain fixed. Then, as the external voltage V is varied, current I will vary, and

Y Ythe variation I with V is accounted for by provision of a coefficient , named as

1 1k in equation (1). It can be seen that k reflects resistance of the circuit as seen by

Y 2external voltage source V . Coefficient k 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 principleof superposition. Each independent internal source within the circuit contributes



2its part to terminal voltage and constant k is the algebraic sum of contributions of

Yinternal sources. Adjust external voltage source such that current I becomes

2zero. As shown by equation (2), the coefficient k is Thevenin’s voltage. To

determine Thevenin’s resistance, set external source voltage to zero. If the internal

Ysources are such as to yield positive Thevenin’s voltage, current I will be negative1and coefficient k is Thevenin’s resistance, as shown by equation (3). This

concludes the proof of Thevenin’s theorem.

The step involved in the application of Thevenin’s theorem are summarized below.

WORKED EXAMPLE 2

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

obtain the load current using ThevEnin’s theorem. We have already looked at this

circuit, but the purpose here is to show, how to apply Thevenin’s 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 Thevenin’s voltage as described now. Remove the load

resistor, and represent the circuit, as shown in Fig. 19 in order to get the value of

2Thevenin’s voltage, which is the voltage across resistor R . 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.

You can obtain Thevenin’s resistance from the circuit shown in Fig. 20. Here

1source V , has been replaced by a short circuit. From Fig. 20, it is seen that

1 2Thevenin’s resistance is the equivalent of resistors, R

and R

, in parallel. Theresultant value of Thevenins resistance is obtained as shown by equation (24).

When two resistors are connected in parallel, the equivalent conductance is the

sum of conductances of the resistors. As shown by equation (24), Thevenin’s

resistance is obtained as the reciprocal of the sum of conductances of the tworesistors.



1 1 2 Now the part of the circuit containing source V and resistors R and R , can be

replaced by the Thevenin’s equivalent circuit as shown in Fig. 21. Thevenin’s

equivalent circuit contains only the Thevenin’s voltage and Thevenin’s resistance.

The last two steps are to draw the Thevenin’s equivalent circuit and then to obtain

the load current. The circuit in Fig. 21 shows the load resistor connected to the

Thevenin’s equivalent circuit. From this circuit, the load current can be

calculated.

Equation (25) shows how the load current can be obtained. Another worked

example is presented next.

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 Thevenin’s equivalent of the circuit in Fig. 22. This

problem is a bit more difficult, since it has dependent sources. The Thevenin’s

theorem can be applied to circuits containing dependent sources also. The only

constraint in applying Thevenin’s theorem to a circuit is that it should be a linear

circuit.

Steps involved can be listed as follows:

C Obtain the Thevenin’s Voltage.

C Obtain the Thevenin’s Resistance.

C Draw the Thevenin’s equivalent circuit.

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 Thevenin’s

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 itis the current that flows through the short circuit. This technique has been used

in the proof of Thevenin’s theorem.

2Since there is no load connected to the output terminals, voltage V is the open

circuit voltage, which is the same as the Thevenin’s voltage. To obtain the opencircuit voltage, the following equations are obtained.



2Equation (26) expresses the voltage across resistor R . The current through

2 2resistor R is ten times current I , and the value of resistor R is 100W. Equation

(27) is written for the loop containing the independent source voltage. The

1independent source voltage is 10 Volts. The value of resistor R is 10W, and the

current through it can be obtained as shown by equation (27). Equation (28) is2obtained by replacing voltage V in equation (27) by its corresponding expression

in equation (26).

On simplifying, we can obtain the value of current I , and the Thevenin’s voltage,

as illustrated by equation (29).

The second step is to obtain Thevenin’s resistance. The circuit in Fig. 23 is used

for this purpose.

To obtain Thevenin’s resistance of a circuit with dependent source, it is preferable

to obtain the short circuit current and then obtain Thevenin’s resistance as the ratio

of Thevenin’s voltage to short circuit 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 I is the ratio of source voltage

1to resistor R and it equals one Ampere, as displayed by equation (31). Equations



(32) and (33) show how Norton’s current and Thevenin’s resistance can be

obtained.

Now it is shown how the Thevenin’s resistance can be obtained by another way.

The circuit in Fig. 24 is presented for this purpose.

ThAlternate Method to obtain R

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

a source at the output as shown in Fig. 24. Then Thevenin’s resistance is obtained

as follows.

Thevenin’s resistance is 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.

Equations (35) to (38) are obtained from the circuit in Fig. 24. Since the source

1voltage is zero, the sum of voltage across resistor R 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 current I in

equation (35) is used to replace the current I in equation (36) and this leads to



equation (37). On simplifying, we get equation (38) and the value of Thevenin’s

resistance is 2 Ohms.

Another View

xIt is possible to obtain an expression for the current I 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

Thevenin’s resistance.

Equations (40) and (41) illustrate how Thevenin’s resistance is obtained. Since

Thevenin’s voltage and Thevenin’s resistance are known, the equivalent circuit

can be drawn.

WORKED EXAMPLE 4

LFind the current through the load resistor R .

Solution:

LThevenin’s theorem is used to get the solution. Remove R . Find Thevenin’s

Lvoltage. Replace R by a short-circuit. Find the current through the short-circuit.

Then Thevenin’s resistance is the ratio of the open-circuit voltage and the short-

Lcircuit current. Then the current through the load resistor R can be determined.



LFirst let us obtain Thevenin’s voltage. The circuit without R is shown below.

ALet the resistance of the circuit in Fig. 26, as seen by the source be R . The value

Aof R can be obtained, as shown by equation (42).

A AOnce R is known, the current I supplied by the source can be obtained, as shown

by equation (43).

3 5From the circuit in Fig. 26, we can obtain currents I and I , marked in Fig. 26, by

using the current division rule.

3 5Once the values of currents I and I are known, Thevenin’s voltage can be

obtained as shown by equation (46).

NTo find the short-circuit current I , we use the circuit in Fig. 27. Let the

B Bresistance of the circuit in Fig. 27, as seen by the source be R . The value of R can

be obtained, as shown by equation (47).



B BOnce R is known, the current I supplied by the source can be obtained, as shown

by equation (48).

2 cUse the current division rule. Find currents I and I , marked in Fig. 27.

2 c NThe difference of currents I and I is the short-circuit current I . From the

Thevenin’s voltage and the short-circuit current, we can obtain the Thevenin’s

resistance. Once the Thevenin’s voltage and the Thevenin’s resistance are known,the load current can be determined.



Equation (51) expresses the short-circuit current. Equation (52) expresses the

Thevenin’s resistance. Equation (53) expresses the load current. It is somewhat

more difficult to solve using either mesh or nodal analysis.

SUMMARY

This page has described the Thevenin’s theorem. Its use has been illustrated by

using a few examples. The next page is on Norton’s theorem.

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