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THEVENIN'S THEOREM

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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
varies. The load current is bounded between two limits, zero and Im, 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,
the load current is maximum, specified as Im 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
a current of Im. This means that the internal resistance of the circuit, called as RTh,
has a value of E over Im, as shown by the equation displayed in Fig. 9. Hence the
circuit model consists of a voltage source of value E volts and a resistor RTh. 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 current IL 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
case, the load resistor is named as R3.
At first, an expression for the load current is obtained without the use of
Thevenin’s theorem. To get the load current, the steps involved are as follows.

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

C     Divide the source voltage by the equivalent resistance to get current IS
supplied by the source. This source current flows through resistors, R2 and R3
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
resistor R1 to the parallel value of resistors, R2 and R3 . The source current is the
ratio of source voltage to the equivalent resistance, as expressed by equation (18)..
Then the load current through resistor R3 is obtained using the current division
rule, as shown by equation (19).




Now some mathematical manipulations are required to get Thevenin’s voltage and
Thevenin’s 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 and R2 , and then we get equation (21). The numerator
of equation (21) is Thevenin’s voltage. The first part of the denominator,
containing resistors R1 and R2 , is Thevenin’s resistance. It is the parallel value of
resistors R1 and R2 . Once Thevenin’s voltage and Thevenin’s resistance are
known, the load current can be obtained as shown by equation (21).
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 and
Thevenin’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
instance, the load resistor is R3 and it is replaced by an open circuit. Then
Thevenin,s voltage is the open circuit voltage, the voltage across resistor R2. 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 resistors
R1 and R2 . 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) in
the diagaram expresses an external voltage VY connected to the load terminals, as
a function of current IY 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, current IY will vary, and
the variation IY 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 independent internal 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 current IY becomes
zero. As shown by equation (2), the coefficient k2 is Thevenin’s voltage. To
determine Thevenin’s resistance, set external source voltage to zero. If the internal
sources are such as to yield positive Thevenin’s voltage, current IY will be negative
and coefficient k1 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
Thevenin’s voltage, which is the voltage across resistor R2. 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
source V1, has been replaced by a short circuit. From Fig. 20, it is seen that
Thevenin’s resistance is the equivalent of resistors, R1 and R2, in parallel. The
resultant 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 two
resistors.
Now the part of the circuit containing source V1 and resistors R1 and R2, 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 it
is the current that flows through the short circuit. This technique has been used
in the proof of Thevenin’s theorem.




Since there is no load connected to the output terminals, voltage V2 is the open
circuit voltage, which is the same as the Thevenin’s voltage. To obtain the open
circuit voltage, the following equations are obtained.
Equation (26) expresses the voltage across resistor R2. The current through
resistor R2 is ten times current I, and the value of resistor R2 is 100 W. Equation
(27) is written for the loop containing the independent source voltage. The
independent source voltage is 10 Volts. The value of resistor R1 is 10 W, 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 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
to resistor R1 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.




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 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
voltage is zero, the sum of voltage across resistor R1 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




It is possible to obtain an expression for the current Ix 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




Find the current through the load resistor RL.

Solution:

Thevenin’s theorem is used to get the solution. Remove RL. Find Thevenin’s
voltage. Replace RL 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-
circuit current. Then the current through the load resistor RL can be determined.
First let us obtain Thevenin’s voltage. The circuit without RL is shown below.




Let the resistance of the circuit in Fig. 26, as seen by the source be RA. The value
of RA can be obtained, as shown by equation (42).




Once RA is known, the current IA supplied by the source can be obtained, as shown
by equation (43).




From the circuit in Fig. 26, we can obtain currents I3 and I5, marked in Fig. 26, by
using the current division rule.



Once the values of currents I3 and I5 are known, Thevenin’s voltage can be
obtained as shown by equation (46).

To find the short-circuit current IN , we use the circuit in Fig. 27. Let the
resistance of the circuit in Fig. 27, as seen by the source be RB. The value of RB can
be obtained, as shown by equation (47).
Once RB is known, the current IB supplied by the source can be obtained, as shown
by equation (48).




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




The difference of currents I2 and Ic is the short-circuit current IN. 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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