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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.