THEVENIN’S THEOREM

INTRODUCTIONTHEVENIN’S EQUIVALENT CIRCUITILLUSTRATION OF THEVENIN’S THEOREMFORMAL PRESENTATION OF THEVENIN’S THEOREMPROOF OF THEVENIN’S THEOREMWORKED EXAMPLE 2WORKED EXAMPLE 3WORKED EXAMPLE 4SUMMARYINTRODUCTION

Thevenin’s theorem is a popular theorem, used often for analysis of electroniccircuits. Its theoretical value is due to the insight it offers about the circuit. Thistheorem states that a linear circuit containing one or more sources and other linearelements can be represented by a voltage source and a resistance. Using thistheorem, 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 investigativeapproach.

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 outputcharacteristic 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 loadvoltage is bounded between limits, E Volts and zero volts. When the load resistoris 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 outputcharacteristic 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 thepoint at which the output characteristic intersects with the X axis.

The circuit in Figure 9 reflects the output characteristic, displayed in Fig. 8. It hasan output of E volts, when the load current is zero. Hence the model of the circuitcan 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 the

Thcircuit model consists of a voltage source of value E volts and a resistor R . Thisresistor 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 interms of Thevenin’s voltage and Thevenin’s resistance. Thevenin’s voltage is thevoltage obtained across the load terminals, with the load resistor removed. In this

3case, the load resistor is named as R .

At first, an expression for the load current is obtained without the use ofThevenin’s theorem. To get the load current, the steps involved are as follows.

eqC Get an expression for the equivalent resistance R , seen by the source, asshown in Fig. 11.

SC Divide the source voltage by the equivalent resistance to get current I

2 3supplied by the source. This source current flows through resistors, R and Rconnected 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

1 2 3resistor R to the parallel value of resistors, R and R . The source current is theratio of source voltage to the equivalent resistance, as expressed by equation (18)..

3Then the load current through resistor R is obtained using the current divisionrule, as shown by equation (19).

Now some mathematical manipulations are required to get Thevenin’s voltage andThevenin’s resistance. The expression for the load current is expressed byequation (20). Divide both the numerator and the denominator of equation (20)

1 2by the sum of resistors, R and R , and then we get equation (21). The numeratorof equation (21) is Thevenin’s voltage. The first part of the denominator,

1 2containing resistors R and R , is Thevenin’s resistance. It is the parallel value of

1 2resistors R and R . Once Thevenin’s voltage and Thevenin’s resistance areknown, the load current can be obtained as shown by equation (21).

Equation (22) defines the expressions for Thevenin’s voltage and Thevenin’sresistance. They are obtained from equation (21).

From the expression for the load current, we can obtain a circuit and this circuitis 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? Theycan be obtained as shown next.

Thevenin’s voltage is the voltage across the load terminals with the load resistorremoved. 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 . Thisvoltage can easily be obtained by using the voltage division rule. The voltagedivision rule states the division of source voltage is proportionate to resistance.Thevenin’s resistance is the resistance, as viewed from the load terminals, withboth the load resistor and the sources in the circuit removed. Here removal of thevoltage source means that it is replaced by a short circuit, and the load resistor isreplaced by an open circuit. Thevenin’s resistance is the parallel value of resistors

1 2R 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. Leta network with one or more sources supply power to a load resistor as shown inFig. 14. Thevenin’s theorem states that the network can be replaced by a singleequivalent voltage source, marked as Thevenin’s Voltage or open-circuit voltageand a resistor marked as Thevenin’s Resistance. Proof of this theorem ispresented below. Thevenin’s theorem can be applied to linear networks only.Thevenin’s voltage is the algebraic sum of voltages across the load terminals, dueto each of the independent sources in the circuit, acting alone. It can be seen thatThevenin’s theorem is an outcome of superposition theorem.

Thevenin’s equivalent circuit consists of Thevenin’s voltage and Thevenin’sresistance. Thevenin’s voltage is also referred to as the open-circuit voltage,meaning that it is obtained across the load terminals without any load connectedto them. The load is replaced by an open-circuit and hence Thevenin’s voltage iscalled as the open-circuit voltage.

Figure 15 shows how Thevenin’s voltage is to be obtained. Here it is assumed thatwe 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. Thevoltage obtained across the load terminals without the load being connected is theopen-circuit voltage. This open-circuit voltage can be obtained as the algebraicsum of voltages, due to each of the independent sources acting alone. Given acircuit, Thevenin’s voltage can be obtained as outlined below.

Figure 16 shows how Thevenin’s resistance is to be obtained. Thevenin’sresistance is the resistance as seen from the load terminals. To obtain thisresistance, replace each independent ideal voltage source in the network by ashort 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 ashort circuit or an open circuit, as the case may be. The internal resistance of thesource, reflecting the non ideal aspect of the circuit, is left in the circuit, as it iswhere it is. A voltage source is connected across the load terminals. ThenThevenin’s resistance is the ratio of this source voltage to its current, as markedin Fig. 16. A few examples are presented after this page to illustrate the use ofThevenin’s theorem.

PROOF OF THEVENIN’S THEOREM

The circuit in Fig. 17 can be used to prove Thevenin’s theorem. Equation (1) in

Ythe 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 aredealing 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 terminalvoltage 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. Todetermine 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 negative

1and coefficient k is Thevenin’s resistance, as shown by equation (3). Thisconcludes 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 toobtain the load current using ThevEnin’s theorem. We have already looked at thiscircuit, 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 solutionis obtained in four steps. The steps are as shown above.

The first step is to obtain Thevenin’s voltage as described now. Remove the loadresistor, 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 beobtained is shown next by equation (23).

Equation (23) is obtained using the voltage division rule. The two resistors areconnected in series and the current through them is the same, and hence thevoltage division rule can be applied.

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

1 source 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 thesum of conductances of the resistors. As shown by equation (24), Thevenin’sresistance is obtained as the reciprocal of the sum of conductances of the tworesistors.

1 1 2Now the part of the circuit containing source V and resistors R and R , can bereplaced by the Thevenin’s equivalent circuit as shown in Fig. 21. Thevenin’sequivalent 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 obtainthe load current. The circuit in Fig. 21 shows the load resistor connected to theThevenin’s equivalent circuit. From this circuit, the load current can becalculated.

Equation (25) shows how the load current can be obtained. Another workedexample is presented next.

WORKED EXAMPLE 3

We take up another example now. Figure 22 contains the circuit. The sourcevoltage is 10 Volts. The circuit containing the small signal model of a bipolarjunction 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. Thisproblem is a bit more difficult, since it has dependent sources. The Thevenin’stheorem can be applied to circuits containing dependent sources also. The onlyconstraint in applying Thevenin’s theorem to a circuit is that it should be a linearcircuit.

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 theopen circuit voltage and the short circuit current, and then obtain Thevenin’sresistance as the ratio of open circuit voltage to the short circuit current. The shortcircuit 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 usedin the proof of Thevenin’s theorem.

2Since there is no load connected to the output terminals, voltage V is the opencircuit 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 100 W. 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 10 W, and thecurrent through it can be obtained as shown by equation (27). Equation (28) is

2obtained by replacing voltage V in equation (27) by its corresponding expressionin 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 usedfor this purpose.

To obtain Thevenin’s resistance of a circuit with dependent source, it is preferableto obtain the short circuit current and then obtain Thevenin’s resistance as the ratioof Thevenin’s voltage to short circuit current. The circuit in Fig. 23 is used toobtain the short circuit current.

Equations (30) to (33) are obtained from the circuit in Fig. 23. When the outputterminals 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 is10 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 beobtained.

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. Connecta source at the output as shown in Fig. 24. Then Thevenin’s resistance is obtainedas follows.

Thevenin’s resistance is expressed by equation (34). It is obtained with theindependent source voltage, contained in the circuit, being replaced by a shortcircuit, 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 thedependent voltage source is zero and we get equation (35). Equation (36) isobtained by using KVL at node a. The expression obtained for current I inequation (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’sresistance 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 voltageacross the dependent current source and the current through it, we can replace itby a resistor, as shown in Fig. 24. The parallel value of two resistors is theThevenin’s resistance.

Equations (40) and (41) illustrate how Thevenin’s resistance is obtained. SinceThevenin’s voltage and Thevenin’s resistance are known, the equivalent circuitcan 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 shownby equation (43).

3 5From the circuit in Fig. 26, we can obtain currents I and I , marked in Fig. 26, byusing the current division rule.

3 5Once the values of currents I and I are known, Thevenin’s voltage can beobtained 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 canbe obtained, as shown by equation (47).

B BOnce R is known, the current I supplied by the source can be obtained, as shownby 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 theThevenin’s voltage and the short-circuit current, we can obtain the Thevenin’sresistance. 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 theThevenin’s resistance. Equation (53) expresses the load current. It is somewhatmore difficult to solve using either mesh or nodal analysis.

SUMMARY

This page has described the Thevenin’s theorem. Its use has been illustrated byusing a few examples. The next page is on Norton’s theorem.