Superposition Examples

Superposition Examples The following examples illustrate the proper use of superposition of dependent sources. All superposition equations are written by inspection using voltage division, current division, series-parallel combinations, and Ohm s law. In each case, it is simpler not to use superposition if the dependent sources remain active. Example 1 Theobjectistosolveforthecurrenti in the circuit of Fig. 1. By superposition, one can write i = 4 7 3i = 3 5 i Solution for i yields i = 1+3/5 = 5 4 A Figure 1: Circuit for example 1. If superposition of the controlled source is not used, two solutions must be found. Let i = i a +i b,wherei a is the current with the 7Asource zeroed and i b is the current with the 4 V source zeroed. By superposition, we can write Solution for i a and i b yields The solution for i is thus i a = 4 3i a i a = 4 1+ 3 =3A i b = i = i a + i b = 5 4 A i b = 7 3i b 7 1+ 3 = 7 4 A This is the same answer obtained by using superposition of the controlled source. Example The object is to solve for the voltages v 1 and v across the current sources in Fig., where the datum node is the lower branch. By superposition, the current i is given by 7 i = 7+15+5 + 3 7+15+5 +4i 7+15 7+15+5 = 17 7 + 88 7 i Solution for i yields i = 17/7 1 88/7 = 17 61 A Although superposition can be used to solve for v 1 and v, it is simpler to write v =5i = 1.393 V v 1 = v (4i i)15=11.148 V 1

Figure : Circuit for example. Example 3 Theobjectistosolveforthecurrenti 1 in the circuit of Fig. 3. By superposition, one can write Solution for i 1 yields 30 i 1 = 6+4+ +3 4 6+4+ 8i 6 1 6+4+ = 4 1 4i 1 i 1 = 4/1 1+4 =0.7A Figure 3: Circuit for example 3. Example 4 The object is to solve for the Thévenin equivalent circuit seen looking into the terminals A A 0 in the circuit of Fig. 4. By superposition, the voltage v x is given by v x =(3 i o )(k40) + 5v x 40 + = 80 4 (3 i o)+ 10 4 v x where i o is the current drawn by any external load and the symbol k denotes a parallel combination. Solution for v x yields v x = 80/4 1 10/4 (3 i o)=.5(3 i o ) Although superposition can be used to solve for v o,itissimplertowrite v o = v x 5v x = 30 + 10i o It follows that the Thévenin equivalent circuit consists of a 30 V source in series with a 10 Ω resistor. The circuit is shown in Fig. 5.

Figure 4: Circuit for example 4. Figure 5: Thévenin equivalent circuit. Example 5 The object is to solve for the voltage v o in the circuit of Fig. 6. By superposition, the current i b is given by i b = 70 0 4k0 + k10 4+0 + 50 10 + 4k0k i b 10 0k+4k10 4+10 = 35 3 + 5 18 11 36 i b Solution for i b yields i b = 35/5/18 =10A 1+11/36 Although superposition can be used to solve for v o, it is simpler to write v o =70 4i b =30V 0k 4+0k Figure 6: Circuit for example 5. 3

Example 6 The object is to solve for the voltage v o in the circuit of Fig. 7. By superposition, the voltage v is given by v = 0.4v 10 + 5 10 Thiscanbesolvedforv to obtain 5 10 v = 1+0.4 10 =10V By superposition, i is given by i = 10 5+0 0.4v 0 0 + 5 = 10 5 0.4v 0 5 = 70 5 A Thus v o is given by v o =10 5i =4V Figure 7: Circuit for example 6. Example 7 The object is to solve for the voltage v as a function of v s and i s in the circuit in Fig. 8. By superposition, the current i is given by i = v s 5 5 i s 3 5 3i Thiscanbesolvedfori to obtain i = v s 14 i s 7 By superposition, the voltage v is given by v = v s 5 5 i s + 5 3i = v s 5 5 i s + 5 3 = 7 v s 4 7 i s vs 14 i s 7 Example 8 This example illustrates the use of superposition in solving for the dc bias currents in a BJT. The object is to solve for the collector current I C in the circuit of Fig. 9. Although no explicit dependent sources are shown, the three BJT currents are related by I C = βi B = αi E,whereβ is the current gain and a = β/(1 + β). If any one of the currents is zero, the other two must also be zero. However, the currents can be treated as independent variables in using superposition. 4

Figure 8: Circuit for Example 7. Figure 9: Circuit for example 8. 5

By superposition of V +, I B = I C /β, andi C, the voltage V B is given by V B = V + R R C + R 1 + R I C β [(R C + R 1 ) kr ] I C R C R R C + R 1 + R A node-voltage solution for V B requires the solution of two simultaneous equations to obtain the same answer which superposition yields by inspection. This equation and the equation canbesolvedfori C to obtain I C = V B = V BE + I C α R E V + R R C+R 1+R V BE (R C +R 1 )kr β + R CR R C+R 1+R + R E α In most contemporary electronics texts, the value V BE =0.7V is assumed in BJT bias calculations. Example 9 This example illustrates the use of superposition to solve for the small-signal base input resistance of a BJT. Fig. 10 shows the small-signal BJT hybrid-pi model with a resistor R E fromemittertogroundandaresistor R C from collector to ground. In the model, r π = V T /I B and r 0 =(V A + V CE ) /I C,whereV T is the thermal voltage, I B isthedcbasecurrent,v A is the Early voltage, V CE is the dc collector-emitter voltage, and I C isthedccollectorcurrent. Figure 10: Circuit for example 9. By superposition of i b and βi b, the base voltage v b is given by v b = i b [r π + R E k (r 0 + R C )] + βi b r 0 R E + r 0 + R C R E This can be solved for the base input resistance r ib = v b /i b to obtain βr 0 R E r ib = r π + R E k (r 0 + R C )+ R E + r 0 + R C which simplifies to (1 + β) r 0 + R C r ib = r π + R E R E + r 0 + R C A node-voltage solution for r ib requires the solution of three simultaneous equations to obtain the same answer which follows almost trivially by superposition. 6

Example 10 This example illustrates the use of superposition with an op-amp circuit. The circuit is shown in Fig. 11. The object is to solve for v O.Withv =0, it follows that v A = v 1, v B =0,andv C =[1+R 4 / (R 3 kr 5 )] v 1. By superposition of v A and v C, v O can be written v O = R R 5 v A R R 1 v C = R + R 1+ R 4 v 1 R 5 R 1 R 3 kr 5 With v 1 =0, it follows that v A =0, v B = v,andv C = (R 4 /R 5 ) v. By superposition of v and v C, v O can be written v O = 1+ R v R v C R 1 kr 5 R 1 = 1+ R + R R 4 v R 1 kr 5 R 1 R 5 Thus the total expression for v O is v O = R + R R 5 + 1+ R 4 R 1 R 3 kr 5 1+ R + R R 4 R 1 kr 5 R 1 R 5 v v 1 Figure 11: Circuit for Example 10. Example 11 Figure 1 shows a circuit that might be encountered in the noise analysis of amplifiers. The amplifier is modeled by a z-parameter model. The square sources represent noise sources. V ts and I ta, respectively, model the thermal noise generated by Z x and Z A. V n and I n model the noise generated by the amplifier. The amplifier load is an open circuit so that I =0. The open-circuit output voltage is given by V o(oc) = z 1 I 1 + I A Z A By superposition, the currents I 1 and I A are given by I 1 = V s + V ts + V n Z S + Z A + I n I ta Z A I A = V s + V ts + V n I n z 11 +I ta Z S + z 11 7

Note that when I n =0,thesourcesV s, V ts,andv n are in series and can be considered to be one source equal to the sum of the three. When these are substituted into the equation for V o(oc) and the equation is simplified, we obtain V o(oc) = z 1 + Z h A V s + V ts + V n +I n (Z S + Z A ) z 1 Z A z 11 z 1 + Z A I ta Z A z 1 (Z S + z 11 ) Z A z 1 + Z A i Figure 1: Circuit for Example 11. Example 1 It is commonly believed that superposition can only be used with circuits that have more than one source. This example illustrates how it can be use with a circuit having one. Consider the first-order all-pass filter shown in Fig. 13(a). An equivalent circuit is shown in Fig. 13(b) in which superposition can be used to write by inspection V o = 1+ R 1 RCs R 1 1+RCs V i R 1 V i = RCs 1 R 1 RCs +1 V i 8

Figure 13: Circuit for Example 1. 9