BJT Differential Amplifier ExampleRPxy,()xy.xyFunction for calculating parallel resistors.RC20000 RB1000 RE100 IQ0.001Vp20 Vm20 VBE0.65 VT0.025β199αβ1βrx20 r050000There are two ac solutions, one for the second input zeroed and one for the first input zeroed. By superposition, the total solution would the be sum of these two. To keep Mathcad happy, all source voltages are taken to be equal to 1 V so that the output voltage is equal to the voltage gain. In general, the output voltage is equal to the voltage gain multiplied by the source voltage.1DC Bias SolutionAssume the dc value of the sources is zero.IE1IQ2IE15 104=IE2IE1VC1VpαIE1.RC.VC110.05=VB1IE11βRB.VB12.5 103=VCB1VC1VB1VCB110.0525=Thus active mode. Same for Q2.re1VTIE1re150=re2re1r'e1RBrx1βre1r'e155.1=r'e2r'e12AC Solutions Circuit for the first output.vi11 vi21With the input equal to 1, the voltage gain is equal to the output voltage.ve2ocvi2r0RC1βr'e2r0RC1β.ve2oc0.9989=rie2r'e2r0RCr'e2r0RC1β.rie276.9015=vtb1vi1Rtb1RBvte1ve2ocRte12RE.rie2Rte1276.9015=3Gmb1αr'e1Rte1r0Rte1βr0RPr'e1Rte1,.Gmb12.9941 103=Gme1αr'e1Rte1r0r'e1αr0RPr'e1Rte1,.Gme12.9975 103=ric1r0RPr'e1Rte1,1αRte1.r'e1Rte1ric12.9416 105=Voltage gain from first input to first output:ic1scGmb1vtb1.ic1sc2.9941 103=vo1ic1scRPric1RC,.Av1vo1Av156.0705=This is the voltage gain from the first input to the first output. The gain from the second input to the second output is the same.4Voltage gain from the second input to the first output.ic1scGme1vte1.ic1sc2.9942 103=vo1ic1scRPric1RC,.Av2vo1Av256.0725=This is the voltage gain from the second input to the first output. The gain from the first input to the second output is the same.vo156.0705 vi1.56.0725 vi2.This is the sum ac output from Q1.vo256.0705 vi2.56.0725 vi1.This is the sum ac output from Q2.Differential input resistance.rib1rx1β()re1RPRte1r0RC,.βRte1.RC.Rte1r0RCrib14.95 104=rib2rib1rid2RB.rib1rib2rid1.01 105=5Common-Mode Rejection RatioAv156.0705=Av256.0725=Let us take the output from the collector of the first transistor. Because neither β nor r0 is infinity, the two voltage gains are not equal. This causes the CMRR to be non infinite. We calculate it below.vid1 vi1vid2vi2vid2vo1Av1vi1.Av2vi2.Advo1Ad56.0715=This is the differential voltage gain.vicm1 vi1vicmvi2vicmvo1Av1vi1.Av2vi2.Acmvo1Acm1.9938 103=This is the common mode voltage gain.CMRRAdAcmCMRR 2.8123 104=CMRRdB20 log CMRR().CMRRdB88.9811=If RQ (the ac resistance of the current source) is not infinity, the CMRR would be lower. 6Solution with the r0 approximations. We neglect r0 except in calculating ric. Thus we can use the emitter equivalent circuit to solve for ie1 and ie2, then multiply by α to solve for the collector currents. Because the common mode gain is zero if we neglect r0, we will assume a differential input signal.vid1 vi1vid2vi2vid2Differential input signal of 1 V.ie1vi1vi2r'e12RE.rie2ie13.012 103=ie2ie1vo1αie1.RPRCric1,.vo156.1236=This is the differential voltage gain to the first output.vo2vo1vo256.1236=This is the differential voltage gain to the second output.rib1rx1β()re1Rte1.rib16.54 104=rib2rib1rid2RB.rib1rib2rid1.328 105=This is the differential input resistance.There is more error using the r0 approximations than I had expected for this problem. Usually the answers are much closer. The major cause of the error here is the effect of r0 on rie. If r0 is infinity, then rie is equal to r'e. There is a fairly big difference between these two resistances in this problem.
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