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c Copyright 2010. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering.The BJT Diﬀerential 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 voltages 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 IQ divides 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.¶µ¡ IEIERB V αIE RC RBVCB VC VB V αIE RC 1 β1 β(e) Calculate the collector-emitter voltage.VCE VC VE VC (VB VBE ) VCB VBE1

Small-Signal AC Solution using the Emitter Equivalent CircuitThis solution uses the r0 approximations and assumes that the base spreading resistance rx is notzero.(a) Calculate gm , rπ , re , and rie .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 ie1 and ie2 . (b) Collector equivalent circuits.(c) Using Ohm’s Law, solve for ie1 and ie2 .ie1 vi1 vi22 (rie RE )ie2 ie1(d) The circuit for vo1 , vo2 , and rout is shown in Fig. 2(b).vo1 ic1(sc) ric kRC α ie1 ric kRC α ric kRC(vi1 vi2 )2 (rie RE )vo2 ic2(sc) ric kRC α iie2 ric kRC α ric kRC(vi2 vi1 )2 (rie RE )rout1 rout2 ric kRC β (2RE rie ) (RB rx rπ ) k (2RE rie )RB rπ rx 2RE rie(e) The resistance seen looking into either input with the other input zeroed is·ric r0 1 rin RB rx rπ (1 β) (2RE rie )The diﬀerential input resistance rind is the resistance between the two inputs for diﬀerential inputsignals. For an ideal current source tail supply, this is the same as the input resistance rin above.Diﬀ Amp with Non-Perfect Tail SupplyFig. 3 shows the circuit diagram of a diﬀerential amplifier. The tail supply is modeled as a current0 having a parallel resistance R . In the case of an ideal current source, R is an opensource IQQQ0 0. The solutionscircuit. Often a diﬀ amp is designed with a resistive tail supply. In this case, IQbelow are valid for each of these connections. The object is to solve for the small-signal outputvoltages and output resistances.2

Figure 3: BJT Diﬀerential amplifier.DC Solutions0 is known. If I is known, the solutions are the same as above.This solution assumes that IQQ(a) Zero both inputs. Divide the tail supply into two equal parallel current sources having a0 /2 in parallel with a resistor 2R . The circuit obtained for Q is shown on the left incurrent IQ1QFig. 4. The circuit for Q2 is 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 RB and V . To solve for IE , this equation is ¡IE0RB VBE IE (RE 2RQ )RQ 0 V IQ1 β(c) Solve the loop equation for the currents.IE 0 R V V IQQBEIC (1 β) IB αRB / (1 β) RE 2RQ(d) Verify that VCB 0 for the active mode.¶µ¡ IEIERB V αIE RC RBVCB VC VB V αIE RC 1 β1 β(e) Calculate the collector-emitter voltage.VCE VC VE VC (VB VBE ) VCB VBE0 /2. If the current source is replaced with a resistor(f) If RQ , it follows that IE1 IE2 IQRQ only, the currents are given byIE IC V VBE (1 β) IB αRB / (1 β) RE 2RQ3

Figure 4: DC bias circuits for Q1 .Small-Signal or AC SolutionsThis solutions use the r0 approximations.(a) Calculate gm , rπ , and rie .gm αIEVTrπ (1 β) VTIErie RB rx rπ1 βr0 VA VCEαIE0 0. Replace the two BJTs with the emitter(b) Redraw the circuit with V V 0 and IQequivalent 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, solve for ie1 and ie2 .ie1 RQvi2vi1 rie RE RQ k (rie RE ) rie RE RQ k (rie RE ) RQ rie RE4

ie2 RQvi1vi2 rie RE RQ k (rie RE ) rie RE RQ k (rie RE ) RQ rie REFor RQ , these becomeie1 vi1 vi22 (rie RE )ie2 vi2 vi12 (rie RE )(d) The circuit for vo1 , vo2 , rout1 , and rout2 is shown in Fig. 6.Figure 6: Circuits for calculating vo1 , vo2 , rout1 , and rout2 .vo1 i0c1 α ric kRC ric kRC α ie1 ric kRC rie RE RQ k (rie RE )vo2 i0c2 α ric kRC ric kRC α ie1 ric kRC rie RE RQ k (rie RE )·ric r0 1 µvi1 vi2µvi2 vi1RQRQ rie RERQRQ rie RE¶¶rout1 rout2 ric kRCβRte (RB rx rπ ) kRteRB rπ RteRte RE RQ k (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 ric kRC(vi1 vi2 )2 (rie RE )vo2 α ric kRC(vi2 vi1 )2 (rie RE )(g) The equivalent circuit seen looking into the two inputs is shown in Fig. 7. The resistorslabeled rπ0 are given byrπ0 rx rπ (1 β) REThe diﬀerential input resistance rid is defined the same way that it is defined for Fig. ?. Thatis, it is the resistance seen between the two inputs when vi1 vid /2 and vi2 vid /2, where vid isthe diﬀerential input voltage. In this case, the small-signal voltage at the upper node of the resistor(1 β) RQ is zero so that no current flows it. It follows that rid is given by¡ rid 2 RB rπ05

Figure 7: Equivalent circuits for calculating ib1 and ib2 .Diﬀerential and Common-Mode Gains(a) Define the common-mode and diﬀerential input voltages as follows:vid vi1 vi2vicm vi1 vi22With these definitions, vi1 and vi2 can be writtenvi1 vicm vid2vi2 vicm vid2By linearity, it follows that superposition of vicm and vid can be used to solve for the currents andvoltages.(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 follows by superposition that va 0 andie1 vid /2rie REie2 vid /2rie REvo1 α ric kRC α ric kRC vid α ie1 ric kRC rie RE2rie REvo2 α ric kRC α ric kRC vid α ie2 ric kRC rie RE2rie RE6µµvi1 vi22vi1 vi22¶¶

The diﬀerential voltage gain is given byAd vo1vo21 αric kRC vidvid2 rie RE(d) For vi1 vi2 vicm , it follows by superposition that ia 0 andie1 vicmrie RE 2RQie2 vicmrie RE 2RQvo1 α ric kRC α ric kRC α ie1 ric kRC vicm rie RE 2RQrie RE 2RQvo2 α ric kRC α ric kRC α ie2 ric kRC vicm rie RE 2RQrie RE 2RQµµvi1 vi22vi1 vi22¶¶The common-mode voltage gain is given byAcm vo1vo2α ric kRC vicmvicmrie RE 2RQ(e) If the output is taken from the collector of Q1 or Q2 , the common-mode rejection ratio isgiven by vo1 /vid vo2 /vid 1 rie RE 2RQRQ1 CM RR vo1 /vicm vo2 /vicm 2rie RE2 rie REThis can be expressed in dB.CM RRdB 20 logµRQ1 2 rie RE¶0 2 mA, R 50 kΩ, R 1 kΩ, R 100 Ω, R 10 kΩ, V 20 V,Example 1 For IQQBECV 20 V, VT 0.025 V, rx 20 Ω, β 99, VBE 0.65 V, and VA 50 V, calculate vo1 , vo2 ,vod , rout , and CM RR.Solution.IE VCB ³0 R0 V IQQ VBE 1.192 mARB / (1 β) RE 2RQµ¶ ¡ IERB 8.209 V VC VB V αIE RC 1 βαIE(1 β) VT 0.0472 Srπ 2.097 kΩVTIEVTRB rx re 31.17 Ω 20.97 Ωrie re IE1 βgm VA VCE 49.869 kΩRte RE RQ k (rie RE ) 230.83 ΩIC· β (2RE rie ) (RB rπ ) k (2RE rie ) 390.5 kΩric r0 1 RB rπ 2RE rieµ¶RQ α ric kRCvi1 vi2 36.84vi1 36.75vi2rie RE RQ k (rie RE )RQ rie REr0 vo1vo2 36.84vi2 36.75vi17

rout ric kRC 9.75 kΩAvd 1 α ric kRC 36.802 rie RE α ric kRC 0.0964rie RE 2RQ Avd 51.63 dBCM RRdB 20 log Avcm Avcm The Diﬀ Amp with an Active LoadFigure 9 shows a BJT diﬀ amp with an active load formed by a current mirror with base currentcompensation. Similar circuits are commonly seen as the input stages of operational amplifiers andaudio amplifiers. 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 isc rout . It will be assumed that the current 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 bias current IQsplits equally between Q1 and Q2 so that IE1 IE2 IQ /2.Figure 9: Diﬀ amp with active current-mirror load.Because the tail supply is assumed to be a current source, the common-mode gain of the circuit iszero when the r0 approximations 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. For diﬀerential inputsignals, it follows by symmetry that the signal voltage is zero at the node above the tail current8

supply IQ . Following the analysis above, the small-signal collector currents in Q1 and Q2 are givenbyvidvidααic2(sc) ic1(sc) rie RE 2rie RE 2whererie RB rπ1 βThe short-circuit output current is given byisc ic4 ic2With ic4 ic3 ic1 and ic2 ic1 , this becomesisc 2ic1(sc) ααvid (vi1 vi2 )rie RErie REThe output resistance is given byrout r04 kric2where ric2 is given byric2· r0 1 β (2RE rie ) (RB rπ ) k (2RE rie )RB rπ 2RE rieBy Thévenin’s theorem, the small-signal open-circuit output voltage is given byvoc isc rout α (r04 kric2 )(vi1 vi2 )rie REExample 2 For IQ 2 mA, RB 100 Ω, RE 51 Ω, V 15 V, V 15 V, VT 0.025 V,rx 50 Ω, β 99, α 0.99 VBE1 VBE2 0.65 V, VEB3 VEB4 VEB5 0.65 V, VC2 VC4 13.7 V, and VA 50 V, calculate isc , rout , and voc . Because rx 0, we add it to RB in theequations above.Solution.re1 re2 2VT 25 ΩIQRte2 2RE rie1 128.5 Ωr02 VA (VC2 VBE ) 65 kΩαIQ /2r04 ric2rie1 rie2 RB rx re 26.5 Ω1 βα(vi1 vi2 ) 0.0128 (vi1 vi2 )rie RE· βRte (RB rπ ) kRte 362.7 kΩ r0 1 RB rπ Rteisc VA (V VC4 ) 51.82 kΩIQrout r04 kric2 45.34 kΩvoc isc rout 579.2 (vi1 vi2 )This is a dB gain of 55.3 dB.9