The BJT Differential AmplifierECE 3050 Analog ElectronicsThe differential amplifier or diff ampisusedinapplicationswhereitisdesiredtohaveanoutputvo ltage that is proportional to the difference betw een two input voltages. Fig. 1(a) shows the basiccircuit diagram. The tail supply is modeled as a current source I0Qhaving an output resistance RQ.In the case of a n ideal current source, RQis an open circuit. Often a diff ampisdesignedwitharesistiv e tail supply. In this case, I0Q=0. There are two outputs sho wn. Either or both can beused. Often the difference voltage between the tw o outputs is used.Figure 1: (a) Circuit diagram of the differential amplifier. (b) First equivalent bias circuit for Q1.(b) Second equivalent bias circuit for Q1.The dc bias circuit is obtained by setting vi1= vi2=0. The tail supply can b e divided intotwo parallel current sources of value I0Q/2, each having an output resistance 2RQ. By symmetry,no dc current flows betw een the two sides of the circuit so that the t wo sides can be separated. Thecircuit obtained for Q1is shown in Fig. 1(b). The circ uit for Q2is identical. The circuit shown inFig. 1(c) is obtained by making a Thévenin equivalent of the tail supply in Fig. 1(b). The biasequation for IEis0 −¡V−− I0QRQ¢=IE1+βRB+ VBE+ IE(RE+2RQ)This can be solved IEto obtainIE=−V−+ I0QRQ− VBERB/ (1 + β)+RE+2RQThe dc collector-to-base voltage is given byVCB= VC− VB=¡V+− αIERC¢−µ−IE1+βRB¶= V+− αIERC+IE1+βRBThis must be greater than zero for the two BJTs to be biased in the activ e mode. The collector toemitter voltage is given byVCE= VC− VE= VC− (VB− VBE)=VCB+ VBE1It follows that re, r0e,andr0for each transistor are given byre=VTIEr0e=RB+ rx1+β+ rer0=VA+ VCEαIETo solve for the small-signal value of vo1, w e zero V+, V−,andI0Qto form the ac sign al circuit.Then the circuit seen looking out of the emitter of Q1is replaced by a Thévenin equiv alent circuit.To obtain this, we first replace the circuit seen looking into the emitter of Q2with a Théveninequiva len t circuit. This circuit is show n in Fig. 2(a), whereve2(oc)= vi2r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)rie2= r0er0+ RCr0e+ r0+ RC/ (1 + β)The Thév enin voltage and resistance seen looking out of the emitter of Q1are given byvte1= ve2(oc)RQRQ+ RE+ rie2Rte= RE+ RQk (RE+ rie2)The new circuit is shown in Fig. 2(b).Figure 2: (a) Ac signal circuit for Q1.(b) Circ uit for calc ulating vo1and rout.The circuit for calculating vo1and routis shown in Fig. 2(c). We can writevo1= −ic1(sc)× rickRC=(Gmbvi1− Gmevte1) × rickRCrout= rickRCwhereGmb=αr0e+ Rtekr0r0− Rte/βr0+ RteGme=1Rte+ r0ekr0αr0+ r0er0+ r0eric=r0+ r0ekRte1 − αRte/ (r0e+ Rte)When the abo ve results are combined, we obtainvo1= −Av1vi1+ Av2vi2= −Av1µvi1−Av2Av1vi2¶where Av1and Av2are the voltage gains given byAv1= Gmb× rickRCAv2= Gme× rickRC×RQRQ+ RE+ rie2r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)2By symmetry, vo2is given byvo2= −vo1= −Av1µvi2−Av2Av1vi1¶We see that the gains for vi1and vi2differ by the ratio Av2/Av1.ThisisgivenbyAv2Av1=GmeGmbRQRQ+ RE+ rie2r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)If this ratio is unity, vo1and vo2are proportional to the difference betw een the two input voltages.It can be seen that the ratio is exactly unity only if RQ→∞and r0→∞.Let vobe the differential output voltage. This is given byvo= vo1− vo2= − (Av1+ Av2)(vi1− vi2)=−Av1µ1+Av2Av1¶(vi1− vi2)This is proportional to the difference betw een the tw o input voltages even if RQ< ∞ and r0< ∞.The r0approximations to the gains are obtained b y letting r0→∞except in the expressionfor ric. We obtainAv1' Gm× rickRCAv2= Gm× rickRC×RQRQ+ RE+ r0ewhereGm=αr0e+ RteRte= RE+ RQk¡RE+ r0e¢The expression for routis the same except ricis calculated with Rte= RE+ RQk (RE+ r0e).The e quivalent circuit seen looking into the base of Q1consists of the resistor ribin series withthe voltage vb1(oc).Thesearegivenbyrib= rx+(1+β) re+ Rte(1 + β) r0+ RCr0+ Rte+ RCvb1(oc)= vte1r0+ RCRte+ r0+ RC= vi2r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)r0+ RCRte+ r0+ RCThe equivalent circuit is shown in Fig. 3(a). The circuit for Q2is shown in Fig. 3(b), where vb2(oc)is given byvb2(oc)= vte2r0+ RCRte+ r0+ RC= vi1r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)r0+ RCRte+ r0+ RCThetwobaseequivalentcircuitscanbeusedtocalculatethebasecurrentsib1and ib2.Figure 3: (a) Equivalent circuit for ib1. (b) Equivalent circuit for ib2.3Example 1 It is given that I0Q=2mA, RQ=50kΩ, RB=1kΩ, RE= 100 Ω, RC=10kΩ,V+=20V, V−= −20 V, VT=0.025 V, rx=20Ω, β =99, VBE=0.65 V ,andVA=50V.Calculate the small-signal values for vo1and vo2, the output resistance rout, an d the equivalentinput circuit seen by each source.Solution. First, we solve for the dc emitter current in each transistor. It isIE=−V−+ I0QRQ− VBERB/ (1 + β)+RE+2RQ=1.192 mATo verify that th e transistors are in the active mode, we calculate the collector-to-base voltageVCB= V+− αIERC+IE1+βRB=8.209 VBecause this is greater than zero, both BJTs are in the active mode. The collector-to-emittervoltage is VCE= VCB+ VBE=8.859 V. Thus the resistances re, r0e,andr0arere=VTIE=20.97 Ω r0e=RB+ rx1+β+ re=31.17 Ω r0=VA+ VCEαIE=49.87 kΩThe resistances rieand Rtearerie= r0er0+ RCr0e+ r0+ RC/ (1 + β)=37.32 Ω Rte= RE+ RQk (RE+ rie) = 236.9 ΩThe transconductances Gmband Gmeand the resistance ricareGmb=αr0e+ Rtekr0r0− Rte/βr0+ Rte=1271S Gme=1Rte+ r0ekr0αr0+ r0er0+ r0e=1270.8Sric=r0+ r0ekRte1 − αRte/ (r0e+ Rte)= 398.9 ΩIt follows that the gains Av1and Av2areAv1= Gmb× rickRC=36Av2= Gme× rickRC×RQRQ+ RE+ rie2r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)=35.91Thus vo1and vo2arevo1= −36vi1+35.91vi2vo2= −36vi2+35.91vi1The differential output voltage isvo= vo1− vo2=71.91 (vi1− vi2)The output resistance isrout= rickRC=9.755 kΩIn the equiv a len t input circuits, rib, vb1(oc),andvb2(oc)arerib= rx+(1+β) re+ Rte(1 + β) r0+ RCr0+ Rte+ RC=19.82 kΩvb1(oc)= vi2r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)r0+ RCRte+ r0+ RC=0.9953vi2vb2(oc)= vi1r0+ RC/ (1 + β)r0e+ r0+ RC/ (1 + β)r0+ RCRte+ r0+