GT ECE 3050 - The BJT Differential Amplifier

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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=IQ2IC1= 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.(a) Calculate gm, rπ, re,andr0e.gm=αIEVTrπ=(1 + β) VTIEre=VTIEr0e=RB+ rx1+β+ rer0=VA+ VCEαIE(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.Figure 2: Emitter equivalent circuit using the r0approximations.(c) Using Ohm’s Law, solve for i0e1and i0e2.i0e1=vi1− vi22(r0e+ RE)i0e2= −i0e1(d) The circuit for vo1, vo2, rout1,androut2isshowninFig.3.Figure 3: Circuits for calculating vo1, vo2, rout1,androut2.vo1= −i0c1rickRC= −αi0e1rickRC=−αrickRC2(r0e+ RE)(vi1− vi2)vo2= −i0c2rickRC= −αi0e1rickRC=−αrickRC2(r0e+ RE)(vi2− vi1)rout1= rout2= rickRCric=r0+ r0ekRte1 −αRter0e+ RteRte=2RE+ r0e2(e) The resistance seen looking into the vi1(vi2) input with vi2=0(vi1=0)isrin= RB+ rx+ rπ+(1+β) Rte= RB+ rx+(1+β)µ2RE+RB+ rx1+β+ re¶=2[RB+ rx+ rπ+(1+β) RE]=2¡RB+ r0π¢where rπ=(1+β) rehas been used andr0π= rx+ rπ+(1+β) REFigure 4: Base equivalent circuit for calculating ib1and ib2.The differential input resistance ridis the resistance seen between the two inputs when vi1=vid/2 and vi2= −vid/2,wherevidis the differential input voltage. It can be seen from the figurethat it is given by rid=2(RB+ r0π).The Diff AmpwithanActiveLoadFigure5showsaBJTdiff amp with an active load formed by a current mirror with base currentcompensation. The object is to solve for the open-circuit output voltage voc, the short-circuitoutput current isc, and the output resistance rout. By Thévenin’s theorem, these are related by theequation voc= iscrout. It will be assumed that the curren t mirror consisting of transistors Q3− Q5is perfect so that its output current is equal to its input current, i.e. ic4= ic1. In addition, the r0approximations will be used in solving for the currents. That is, the Early effect will be neglectedexcept in solving for rout. For the bias solution, it will be assumed that the tail current IQsplitsequally bet ween Q1and Q2so that IE1= IE2= IQ/2.Because the tail supply is assumed to be a curren t source, the common-mode gain of the circuit iszero when the r0approximations are used. In this case, it can be assumed that the two input signalsare pure differential signals that can be written vi1= vid/2 and vi2= −vid/2.Fordifferential inputsignals, it follows by symmetry that the signal voltage is zero at the node above the tail currentsupply IQ. Following the analysis abo ve, the small-signal collector current s in Q1and Q2are giv enbyi0c1=αr0e+ REvid2i0c2= −αr0e+ REvid2wherer0e=RB+ rx1+β+ reThe short-circuit output current is given byisc= i0c4− i0c23Figure 5: Diff amp with active current-mirror load.With i0c4= i0c3= i0c1and i0c2= −i0c1, this becomesisc=2i0c1=αr0e+ REvid=αr0e+ RE(vi1− vi2)The output resistance is given byrout= r04kric2where ric2is given byric2=r0+ r0ekRte21 −αRte2r0e+ Rte2Rte2=2RE+ rie1=2RE+ r0eBy Thévenin’s theorem, the small-signal open-circuit output voltage is given byvoc= iscrout=α × r04kric2r0e+ RE(vi1− vi2)Example 1 For IQ=2mA, RB= 100 Ω, RE=51Ω, V+=15V, V−= −15 V, VT=0.025 V,rx=50Ω, β =99, α =0.99 VBE1= VBE2=0.65 V, VEB3= VEB4= VEB5=0.65 V, VC2= VC4=13.7V,andVA=50V,calculateisc, rout,andvoc.Solution.re1= re2=2VTIQ=25Ω r0e1= r0e2=RB+ rx1+β+ re=26.5 ΩRte2=2RE+ rie1=2RE+ r0e= 128.5 Ω isc=αr0e+ RE(vi1− vi2)=0.0128 (vi1− vi2)4r02=VA+(VC2+ VBE)αIQ/2=65kΩ ric2=r02+ r0ekRte21 −αRte2r0e+ Rte2= 362.7kΩr04=VA+(V+− VC4)IQ=51.82 kΩ rout= r04kric2=45.34 kΩvoc= iscrout=579.2(vi1− vi2)This is a dB gain of 55.3dB.Diff Amp with Non-P erfect Tail SupplyFig. 6 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.Figure 6: BJT Differential amplifier.DC SolutionsThis solution assumes that I0Qis known. If IQis kno wn, the solutions are the same as above.(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. 7. The circuit for Q2is identical. Now make a Thévenin equivalent as shown in on the rightin Fig. 7. This is the basic bias circuit.5Figure 7: DC bias circuits for Q1.(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.IfI0Q=0, the currents are given byIE=ICα=(1+β) IB=−V−− VBERB/ (1 + β)+RE+2RQSmall-Signal or A C SolutionsEmitter Equivalent CircuitThis solution uses the r0approximations.6(a) Calculate gm, rπ, re,andr0e.gm=αIEVTrπ=(1 + β) VTIEre=VTIEr0e=RB+ rx1+β+ rer0=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


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