Su perposition Ex a mple sThe following examples illustrate the proper use of superposition of dependent sources. All superpositionequations are written by inspection using voltage division, current division, series-parallel combinations, andOhm’s law. In each case, it is simpler not to use superposition if the dependent sources remain active.Example 1Theobjectistosolveforthecurrenti in the circuit of Fig. 1. By superposition, one can writei =243+2− 723+2−3i3+2=2−35iSolution for i yieldsi =21+3/5=54AFigure 1: Circuit for example 1.If superposition of the controlled source is not used, two solutions must be found. Let i = ia+ib,whereiais the current with the 7Asource zeroed and ibis the current with the 24 V source zeroed. By superposition,we can writeia=243+2−3ia3+2ib= −723+2−3ib3+2Solution for iaand ibyieldsia=243+21+33+2=3A ib=−723+21+33+2= −74AThe solution for i is thusi = ia+ ib=54AThis is the same answer obtained by using superposition of the controlled source.Example 2The object is to solve for the voltages v1and v2across the current sources in Fig. 2, where the datum nodeis the lower branch. By superposition, the current i is given b yi =277+15+5+37+15+5+4i7+157+15+5=1727+8827iSolution for i yieldsi =17/271 − 88/27= −1761AAlthough superposition can be used to solve for v1and v2, it is simpler to writev2=5i = −1.393 V v1= v2− (4i − i)15= 11.148 V1Figure 2: Circuit for example 2.Example 3Theobjectistosolveforthecurrenti1in the circuit of Fig. 3. By superposition, one can writei1=306+4+2+346+4+2− 8i166+4+2=4212− 4i1Solution for i1yieldsi1=42/121+4=0.7AFigure 3: Circuit for example 3.Example 4The object is to solve for the Thévenin equivalent circuit seen looking into the terminals A− A0in the circuitof Fig. 4. By superposition, the voltage vxis given byvx=(3− io)(2k40) + 5vx240 + 2=8042(3 − io)+1042vxwhere iois the current drawn by any external load and the symbol “k” denotes a parallel combination.Solution for vxyieldsvx=80/421 − 10/42(3 − io)=2.5(3− io)Although superposition can be used to solve for vo,itissimplertowritevo= vx− 5vx= −30 + 10ioIt follows that the Thévenin equivalent circuit consists of a −30 V source in series with a −10 Ω resistor. Thecircuit is shown in Fig. 5.2Figure 4: Circuit for example 4.Figure 5: Thévenin equivalent circuit.Example 5The object is to solve for the voltage voin the circuit of Fig. 6. By superposition, the current ibis given byib=704k20 + 2k10204+20+5010 + 4k20k220k24+20k2−2ib20k2+4k10104+10=353+2518−1136ibSolution for ibyieldsib=35/3+25/181+11/36=10AAlthough superposition can be used to solve for vo, it is simpler to writevo=70− 4ib=30VFigure 6: Circuit for example 5.3Example 6The object is to solve for the voltage voin the circuit of Fig. 7. By superposition, the voltage v∆is given byv∆= −0.4v∆× 10 + 5 × 10Thiscanbesolvedforv∆to obtainv∆=5 × 101+0.4 × 10=10VBy superposition, i∆is given byi∆=105+20− 0.4v∆2020 + 5=1025− 0.4v∆2025= −7025AThus vois given byvo=10− 5i∆=24VFigure 7: Circuit for example 6.Example 7The object is to solve for the voltage v as a function of vsand isin the circuit in Fig. 8. By superposition,the current i is given byi =vs5−25is−35× 3iThiscanbesolvedfori to obtaini =vs14−is7By superposition, the voltage v is given byv =vs5−25is+25× 3i=vs5−25is+25× 3µvs14−is7¶=27vs−47isExample 8This example illustrates the use of superposition in solving for the dc bias currents in a BJT. The object is tosolve for the collector current ICin the circuit of Fig. 9. Although no explicit dependent sources are shown,the three BJT currents are related b y IC= βIB= αIE,whereβ is the current gain and a = β/ (1 + β).Ifan y one of the currents is zero, the other t wo must also be zero. However, the currents can be treated asindependent variables in using superposition.4Figure 8: Circuit for Example 7.Figure 9: Circuit for example 8.5By superposition of V+, IB= IC/β,andIC, the voltage VBis given byVB= V+R2RC+ R1+ R2−ICβ[(RC+ R1) kR2]−ICRCR2RC+ R1+ R2A node-voltage solution for VBrequires the solution of two sim u ltaneous equations to obtain the same answerwhich superposition yields by inspection. This equation and the equationVB= VBE+ICαREcanbesolvedforICto obtainIC=V+R2RC+R1+R2− VBE(RC+R1)kR2β+RCR2RC+R1+R2+REαIn most contemporary electronics texts, the value VBE=0.7V is assumed in BJT bias calculations.Example 9This 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 REfromemittertogroundandaresistorRCfrom collector to ground. In the model, rπ= VT/IBand r0=(VA+ VCE) /IC,whereVTis the thermalvoltage, IBisthedcbasecurrent,VAis the Early voltage, VCEis the dc collector-emitter voltage, and ICisthedccollectorcurrent.Figure 10: Circuit for example 9.By superposition of iband βib, the base voltage vbis given byvb= ib[rπ+ REk (r0+ RC)] + βibr0RE+ r0+ RCREThis can be solved for the base input resistance rib= vb/ibto obtainrib= rπ+ REk (r0+ RC)+βr0RERE+ r0+ RCwhich simplifies torib= rπ+ RE(1 + β) r0+ RCRE+ r0+ RCA node-voltage solution for ribrequires the solution of three simultaneous equations to obtain the sameanswer which follows almost trivially by superposition.6Example 10This example illustrates the use of superposition with an op-amp circuit. The circuit is sho wn in Fig. 11.The object is to solve for vO.Withv2=0, it follows that vA= v1, vB=0,andvC=[1+R4/ (R3kR5)] v1.By superposition of vAand vC, vOcan be writtenvO= −R2R5vA−R2R1vC= −·R2R5+R2R1µ1+R4R3kR5¶¸v1With v1=0, it follows that vA=0, vB= v2,andvC= − (R4/R5) v2. By superposition of v2and vC, vOcan be writtenvO=µ1+R2R1kR5¶v2−R2R1vC=µ1+R2R1kR5+R2R1R4R5¶v2Th us the total expression for vOisvO= −·R2R5+R2R1µ1+R4R3kR5¶¸v1+µ1+R2R1kR5+R2R1R4R5¶v2Figure 11: Circuit for Example 10.Example 11Figure 12 shows a circuit that might be encountered in the noise analysis of amplifiers. The amplifier ismodeled b y a z-parameter model. The square sources represent noise sources. Vtsand ItA, respectively,model the thermal noise generated by Zxand ZA. Vnand Inmodel the noise generated by the amplifier.The amplifier load is an open circuit so that I2=0. The open-circuit output voltage is given byVo(oc)= z12I1+ IAZABy superposition, the currents I1and IAare given byI1=Vs+ Vts+ VnZS+ ZA+ z11+ InZS+ ZAZS+ ZA+ z11−ItAZAZS+