c° Copyright 2010. W. Marshall Leach, Jr., Professor, Georgia Institute of Tec hnology, School ofElectrical and Computer Engineering.The BJT Differen tial AmplifierBasic CircuitFigure 1 shows the circuit diagram of a differential amplifier. The tail supply is modeled as a currentsource IQ. The object is to solve for the small-signal output v oltages and output resistances. Itwill be assumed that the transistors are identical.Figure 1: Circuit diagram of the differential amplifier.DC SolutionZero both base inputs. For identical transistors, the current IQdivides equally between the twoemitters.(a) The dc currents are given byIE1= IE2=IQ2IB1= IB2=IQ2(1+β)IC1= IC2=αIQ2(b) Verify that VCB> 0 for the active mode.VCB= VC− VB=¡V+− αIERC¢−µ−IE1+βRB¶= V+− αIERC+IE1+βRB(e) Calculate the collector-emitter voltage.VCE= VC− VE= VC− (VB− VBE)=VCB+ VBE1Small-Signal A C Solution using the Emitter Equivalent CircuitThis solution uses the r0approximations and assumes that the base spreading resistance rxis notzero.(a) Calculate gm, rπ, re,andrie.gm=ICVTrπ=VTIBre=VTIErie=RB+ rx+ rπ1+βr0=VA+ VCEIE(b) Redraw the circuit with V+= V−=0. Replace the two BJTs with the emitter equivalentcircuit. The emitter part of the circuit obtained is shown in Fig. 2(a).Figure 2: (a) Emitter equivalent circuit for ie1and ie2. (b) Collector equivalent circuits.(c) Using Ohm’s Law, solve for ie1and ie2.ie1=vi1− vi22(rie+ RE)ie2= −ie1(d) The circuit for vo1, vo2,androutisshowninFig. 2(b).vo1= −ic1(sc)× rickRC= −α × ie1× rickRC=−α × rickRC2(rie+ RE)(vi1− vi2)vo2= −ic2(sc)× rickRC= −α × iie2× rickRC=−α × rickRC2(rie+ RE)(vi2− vi1)rout1= rout2= rickRCric= r0·1+β (2RE+ rie)RB+ rπ+ rx+2RE+ rie¸+(RB+ rx+ rπ) k (2RE+ rie)(e) The resistance seen looking into either input with the other input zeroed isrin= RB+ rx+ rπ+(1+β)(2RE+ rie)The differential input resistance rindis the resistance between the two inputs for differential inputsignals. For an ideal current source tail supply, this is the same as the input resistance rinabove.Diff Amp with Non-P erfect Tail SupplyFig. 3 shows the circuit diagram of a differential amplifier. The tail supply is modeled as a currentsource I0Qhaving a parallel resistance RQ. In the case of an ideal current source, RQis an opencircuit. Often a diff amp is designed with a resistive tail supply. In this case, I0Q=0.Thesolutionsbelow are valid for each of these connections. The object is to solve for the small-signal outputvoltages and output resistances.2Figure 3: BJT Differential amplifier.DC SolutionsThis solution assumes that I0Qis known. If IQis kno wn, the solutions are the same as abo ve.(a) Zero both inputs. Divide the tail supply into two equal parallel current sources hav ing acurrent I0Q/2 in parallel with a resistor 2RQ. The circuit obtained for Q1isshownontheleftinFig. 4. The circuit for Q2is identical. Now make a Thévenin equivalent as shown in on the rightin Fig. 4. This is the basic bias circuit.(b) Make an “educated guess” for VBE. Write the loop equation between the ground node tothe left of RBand V−.TosolveforIE,thisequationis0 −¡V−− I0QRQ¢=IE1+βRB+ VBE+ IE(RE+2RQ)(c) Solv e the loop equation for the currents.IE=ICα=(1+β) IB=−V−+ I0QRQ− VBERB/ (1 + β)+RE+2RQ(d) Verify that VCB> 0 for the active mode.VCB= VC− VB=¡V+− αIERC¢−µ−IE1+βRB¶= V+− αIERC+IE1+βRB(e) Calculate the collector-emitter voltage.VCE= VC− VE= VC− (VB− VBE)=VCB+ VBE(f) If RQ= ∞, it follows that IE1= IE2= I0Q/2. If the curren t source is replaced with a resistorRQonly, the curren ts are giv en byIE=ICα=(1+β) IB=−V−− VBERB/ (1 + β)+RE+2RQ3Figure 4: DC bias circuits for Q1.Small-Signal or A C SolutionsThis solutions use the r0approximations.(a) Calculate gm, rπ,andrie.gm=αIEVTrπ=(1 + β) VTIErie=RB+ rx+ rπ1+βr0=VA+ VCEαIE(b) Redraw the circuit with V+= V−=0and I0Q=0. Replace the two BJTs with the emitterequivalent circuit. The emitter part of the circuit obtained is shown in 5(a).Figure 5: (a) Emitter equivalent circuit. (b) Collector equivalent circuits.(c) Using superposition, Ohm’s Law, and current division, solv e for ie1and ie2.ie1=vi1rie+ RE+ RQk (rie+ RE)−vi2rie+ RE+ RQk (rie+ RE)×RQRQ+ rie+ RE4ie2=vi2rie+ RE+ RQk (rie+ RE)−vi1rie+ RE+ RQk (rie+ RE)×RQRQ+ rie+ REFor RQ= ∞, these becomeie1=vi1− vi22(rie+ RE)ie2=vi2− vi12(rie+ RE)(d) The circuit for vo1, vo2, rout1,androut2isshowninFig.6.Figure 6: Circuits for calculating vo1, vo2, rout1,androut2.vo1= −i0c1× rickRC= −α × ie1× rickRC=−α × rickRCrie+ RE+ RQk (rie+ RE)µvi1− vi2RQRQ+ rie+ RE¶vo2= −i0c2× rickRC= −α × ie1× rickRC=−α × rickRCrie+ RE+ RQk (rie+ RE)µvi2− vi1RQRQ+ rie+ RE¶rout1= rout2= rickRCric= r0·1+βRteRB+ rπ+ Rte¸+(RB+ rx+ rπ) kRteRte= RE+ RQk (rie+ RE)(e) The resistance seen looking into the vi1(vi2) input with vi2=0(vi1=0)isrib= RB+ rx+ rπ+(1+β) Rte(f) Special case for RQ= ∞.vo1=−α × rickRC2(rie+ RE)(vi1− vi2) vo2=−α × rickRC2(rie+ RE)(vi2− vi1)(g) The equivalent circuit seen looking into the two inputs is shown in Fig. 7. The resistorslabeled r0πare giv en byr0π= rx+ rπ+(1+β) REThe differential input resistance ridis defined the same w ay that it is defined for Fig. ??.Thatis, it is the resistance seen between the tw o inputs when vi1= vid/2 and vi2= −vid/2,wherevidisthe differential input voltage. In this case, the small-signal voltage at the upper node of the resistor(1 + β) RQiszerosothatnocurrentflows it. It follows that ridis given byrid=2¡RB+ r0π¢5Figure 7: Equivalent circuits for calculating ib1and ib2.Diffe rential and Com m on-M ode G ains(a) Define the common-mode and differential input voltages as follows:vid= vi1− vi2vicm=vi1+ vi22With these definitions, vi1and vi2can be writtenvi1= vicm+vid2vi2= vicm−vid2By linearity, it follows that superposition of vicmand vidcanbeusedtosolveforthecurrentsandvoltages.(b) Redraw the emitter equivalent circuit as shown in Fig. 8.Figure 8: Emitter equivalent circuit.(c) For vi1= vid/2 and vi2= −vid/2, it follo ws by superposition that va=0andie1=vid/2rie+ REie2=−vid/2rie+ REvo1= −α × ie1× rickRC=−α × rickRCrie+ REvid2=−α × rickRCrie+ REµvi1− vi22¶vo2= −α × ie2× rickRC=+α × rickRCrie+ REvid2=+α × rickRCrie+ REµvi1− vi22¶6The differential voltage gain is giv en byAd=vo1vid= −vo2vid= −12αrickRCrie+ RE(d) For vi1= vi2= vicm, it follows by