Revision: June 11, E Main Suite D Pullman, WA (509) Voice and Fax

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1 Reision: June 11, E Main Suite D Pullman, WA (509) Voice and Fax Oeriew In preious chapters, we hae seen that it is possible to characterize a circuit consisting of sources and resistors by the oltage-current (or i-) characteristic seen at a pair of terminals of the circuit. When we do this, we hae essentially simplified our description of the circuit from a detailed model of the internal circuit parameters to a simpler model which describes the oerall behaior of the circuit as seen at the terminals of the circuit. This simpler model can then be used to simplify the analysis and/or design of the oerall system. In this chapter, we will formalize the aboe result as Théenin s and Norton s theorems. Using these theorems, we will be able to represent any linear circuit with an equialent circuit consisting of a single resistor and a source. Théenin s theorem replaces the linear circuit with a oltage source in series with a resistor, while Norton s theorem replaces the linear circuit with a current source in parallel with a resistor. In this chapter, we will apply Théenin s and Norton s theorems to purely resistie networks. Howeer, these theorems can be used to represent any circuit made up of linear elements. Before beginning this chapter, you should be able to: Represent a circuit in terms of its i- characteristic (Chapter 1.7.3) Represent a circuit as a two-terminal network (Chapter 1.7.3) After completing this chapter, you should be able to: Determine Théenin and Norton equialent circuits for circuits containing power sources and resistors Relate Théenin and Norton equialent circuits to i- characteristics of two-terminal networks This chapter requires: N/A Consider the two interconnected circuits shown in Figure 1 below. The circuits are interconnected at the two terminals a and b, as shown. Our goal is to replace circuit A in the system of Figure 1 with a simpler circuit which has the same current-oltage characteristic as circuit A. That is, if we replace circuit A with its simpler equialent circuit, the operation of circuit B will be unaffected. We will make the following assumptions about the oerall system: Circuit A is linear Circuit A has no dependent sources which are controlled by parameters within circuit B Circuit B has no dependent sources which are controlled by parameters within circuit A Doc: XXX-YYY page 1 of 12

2 Figure 1. Interconnected two-terminal circuits. In chapter 1.7.3, we determined i- characteristics for seeral example two-terminal circuits, using the superposition principle. We will follow the same basic approach here, except for a general linear twoterminal circuit, in order to deelop Théenin s and Norton s theorems. Théenin s Theorem: First, we will kill all sources in circuit A and determine the oltage resulting from an applied current, as shown in Figure 2 below. With the sources killed, circuit A will look strictly like an equialent resistance to any external circuitry. This equialent resistance is designated as R in Figure 2. The oltage resulting from an applied current, with circuit A dead is: 1 = R i (1) Figure 2. Circuit schematic with dead circuit. Now we will determine the oltage resulting from re-actiating circuit A s sources and open-circuiting terminals a and b. We open-circuit the terminals a-b here since we presented equation (1) as resulting from a current source, rather than a oltage source. The circuit being examined is as shown in Figure 3. The oltage is the open-circuit oltage. Figure 3. Open-circuit response. page 2 of 12

3 Superimposing the two oltages aboe results in: = 1 + (2) or = R i + (3) Equation (3) is Théenin s theorem. It indicates that the oltage-current characteristic of any linear circuit (with the exception noted below) can be duplicated by an independent oltage source in series with a resistance R, known as the Théenin resistance. The oltage source has the magnitude and the resistance is R, where is the oltage seen across the circuit s terminals if the terminals are open-circuited and R is the equialent resistance of the circuit seen from the two terminals, with all independent sources in the circuit killed. The equialent Théenin circuit is shown in Figure 4. R i + + V - - Equialent Circuit Figure 4. Théenin equialent circuit. Procedure for determining Théenin equialent circuit: 1. Identify the circuit and terminals for which the Théenin equialent circuit is desired. 2. Kill the independent sources (do nothing to any dependent sources) in circuit and determine the equialent resistance R of the circuit. If there are no dependent sources, R is simply the equialent resistance of the resulting resistie network. Otherwise, one can apply an independent current source at the terminals and determine the resulting oltage across the terminals; the oltage-to-current ratio is R. 3. Re-actiate the sources and determine the open-circuit oltage V across the circuit terminals. Use any analysis approach you choose to determine the open-circuit oltage. page 3 of 12

4 Example: Determine the Théenin equialent of the circuit below, as seen by the load, R L. We want to create a Théenin equialent circuit of the circuit to the left of the terminals a-b. The load resistor, R L, takes the place of circuit B in Figure 1. The circuit has no dependent sources, so we kill the independent sources and determine the equialent resistance seen by the load. The resulting circuit is shown below. From the aboe figure, it can be seen that the Théenin resistance R is a parallel combination of (6Ω)(3Ω) a 3Ω resistor and a 6Ω resistor, in series with a 2Ω resistor. Thus, R = + 2Ω = 4Ω. 6Ω + 3Ω The open-circuit oltage is determined from the circuit below. We (arbitrarily) choose nodal analysis to determine the open-circuit oltage. There is one independent oltage in the circuit; it is labeled as 0 in the circuit below. Since there is no current through the 2Ω resistor, = V 0 Applying KCL at 0, we obtain: 2A + + = 0 0 = = 6V. Thus, the Théenin 6Ω 3Ω equialent circuit is on the left below. Re-introducing the load resistance, as shown on the right below, allows us to easily analyze the oerall circuit. page 4 of 12

5 Norton s Theorem: The approach toward generating Norton s theorem is almost identical to the deelopment of Théenin s theorem, except that we apply superposition slightly differently. In Théenin s theorem, we looked at the oltage response to an input current; to deelop Norton s theorem, we look at the current response to an applied oltage. The procedure is proided below. Once again, we kill all sources in circuit A, as shown in Figure 2 aboe but this time we determine the current resulting from an applied oltage. With the sources killed, circuit A still looks like an equialent resistance to any external circuitry. This equialent resistance is designated as R in Figure 2. The current resulting from an applied oltage, with circuit A dead is: i 1 = (4) R Notice that equation (4) can be obtained by rearranging equation (1) Now we will determine the current resulting from re-actiating circuit A s sources and short-circuiting terminals a and b. We short-circuit the terminals a-b here since we presented equation (4) as resulting from a oltage source. The circuit being examined is as shown in Figure 5. The current i SC is the short-circuit current. It is typical to assume that under short-circuit conditions the short-circuit current enters the node at a; this is consistent with an assumption that circuit A is generating power under short-circuit conditions. Figure 5. Short-circuit response. Employing superposition, the current into the circuit is (notice the negatie sign on the short-circuit current, resulting from the definition of the direction of the short-circuit current opposite to the direction of the current i) i = i 1 (5) i SC so i = isc (6) R Equation (6) is Norton s theorem. It indicates that the oltage-current characteristic of any linear circuit (with the exception noted below) can be duplicated by an independent current source in parallel page 5 of 12

6 with a resistance. The current source has the magnitude i SC and the resistance is R, where i SC is the current seen at the circuit s terminals if the terminals are short-circuited and R is the equialent resistance of the circuit seen from the two terminals, with all independent sources in the circuit killed. The equialent Norton circuit is shown in Figure 6. Figure 6. Norton equialent circuit. Procedure for determining Norton equialent circuit: 1. Identify the circuit and terminals for which the Norton equialent circuit is desired. 2. Determine the equialent resistance R of the circuit. The approach for determining R is the same for Norton circuits as Théenin circuits. 3. Re-actiate the sources and determine the short-circuit current i SC across the circuit terminals. Use any analysis approach you choose to determine the short-circuit current. page 6 of 12

7 Example: Determine the Norton equialent of the circuit seen by the load, R L, in the circuit below. This is the same circuit as our preious example. The Théenin resistance, R, is thus the same as calculated preiously: R = 4Ω. Remoing the load resistance and placing a short-circuit between the nodes a and b, as shown below, allows us to calculate the short-circuit current, i SC. Performing KCL at the node 0, results in: 0 0 6V + 2Ω 6Ω 0 + = 2A 3Ω so = 0 3V Ohms law can then be used to determine i SC : 3V i SC = = 1. 5A 2Ω and the Norton equialent circuit is shown on the left below. Replacing the load resistance results in the equialent oerall circuit shown to the right below. page 7 of 12

8 Exceptions: Not all circuits hae Théenin and Norton equialent circuits. Exceptions are: 1. An ideal current source does not hae a Théenin equialent circuit. (It cannot be represented as a oltage source in series with a resistance.) It is, howeer, its own Norton equialent circuit. 2. An ideal oltage source does not hae a Norton equialent circuit. (It cannot be represented as a current source in parallel with a resistance.) It is, howeer, its own Théenin equialent circuit. Source Transformations: Circuit analysis can sometimes be simplified by the use of source transformations. Source transformations are performed by noting that Théenin s and Norton s theorems proide two different circuits which proide essentially the same terminal characteristics. Thus, we can write a oltage source which is in series with a resistance as a current source in parallel with the same resistance, and ice-ersa. This is done as follows. Equations (3) and (6) are both representations of the i- characteristic of the same circuit. Rearranging equation (3) to sole for the current i results in: i R = (7) R Equating equations (6) and (7) leads to the conclusion that i SC = (8) R Likewise, rearranging equation (6) to obtain an expression for gies: = i R + i R (9) SC Equating equations (9) and (3) results in: = i R (10) SC which is the same result as equation (8). Equations (8) and (10) lead us to the conclusion that any circuit consisting of a oltage source in series with a resistor can be transformed into a current source in parallel with the same resistance. Likewise, a current source in parallel with a resistance can be transformed into a oltage source in page 8 of 12

9 series with the same resistance. The alues of the transformed sources must be scaled by the resistance alue according to equations (8) and (10). The transformations are depicted in Figure 7. V S R I S R Figure 7. Source transformations. Source transformations can simplify the analysis of some circuits significantly, especially circuits which consist of series and parallel combinations of resistors and independent sources. An example is proided below. page 9 of 12

10 Example: Determine the current i in the circuit shown below. We can use a source transformation to replace the 9V source and 3Ω resistor series combination with a 3A source in parallel with a 3Ω resistor. Likewise, the 2A source and 2Ω resistor parallel combination can be replaced with a 4V source in series with a 2Ω resistor. After these transformations hae been made, the parallel resistors can be combined as shown in the figure below. The 3A source and 2Ω resistor parallel combination can be combined to a 6V source in series with a 2Ω resistor, as shown below. The current i can now be determined by direct application of Ohm s law to the three series 6V 4V resistors, so that i = = 0. 25A. 2Ω + 4Ω + 2Ω page 10 of 12

11 Voltage Current characteristics of Théenin and Norton Circuits: Preiously, in Chapter 1.7.3, we noted that the i- characteristics of linear two-terminal networks containing only sources and resistors are straight lines. We now look at the oltage-current characteristics in terms of Théenin and Norton equialent circuits. Equations (3) and (6) both proide a linear oltage-current characteristic as shown in Figure 8. When the current into the circuit is zero (open-circuited conditions), the oltage across the terminals is the open-circuit oltage,. This is consistent with equation (3), ealuated at i = 0: = R i + = R 0 + =. Likewise, under short-circuited conditions, the oltage differential across the terminals is zero and equation (6) readily proides: i = R SC i sc = 0 R i sc = i sc which is consistent with Figure 8. Figure 8. Voltage-current characteristic for Théenin and Norton equialent circuits. Figure 8 is also consistent with equations (8) and (10) aboe, since graphically the slope of the line is obiously R =. i SC Figure 8 also indicates that there are three simple ways to create Théenin and Norton equialent circuits: 1. Determine R and. This proides the slope and y-intercept of the i- characteristic. This approach is outlined aboe as the method for creating a Théenin equialent circuit. page 11 of 12

12 2. Determine R and i SC. This proides the slope and x-intercept of the i- characteristic. This approach is outlined aboe as the method for creating a Norton equialent circuit. 3. Determine and i SC. The equialent resistance R can then be calculated from R = isc to determine the slope of the i- characteristic. Either a Théenin or Norton equialent circuit can then be created. This approach is not commonly used, since determining R the equialent resistance of the circuit is usually easier than determining either or i SC. Note: It should be emphasized that the Théenin and Norton circuits are not independent entities. One can always be determined from the other ia a source transformation. Théenin and Norton circuits are simply two different ways of expressing the same oltage-current characteristic. page 12 of 12