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GT ECE 3050 - Cascode Amplifier

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Cascode Amplifier Example - Spring 2002RPxy,()xy.xyFunction for calculating parallel resistors.R1390000 R2200000 R356000 R4100RC20000 RE4300 RS1000 RL10000Vplus35 Vminus30 VBE0.65 VT0.025β199α0.995rx20 r050000vs1With vs = 1, the voltage gain is equal to vo, that is Av = vo.1First, the dc bias solution Solve for the Thevenin equivalent circuit looking out of the base of Q1 VBB1VplusR3.VminusR1R2.R1R2R3αIE1.1βR1R1R2R3.R3.RBB1RPR1R2R3,The bias loop equation is VBB1VminusIE11βRBB1VBEIE1RE.It follows that the solution for IE1is IE1VplusR3.VminusR1R2.R1R2R3VminusVBERBB11βREα1βR1R3.R1R2R3.IE11.0552 103=2re1VTIE1re123.6922=IE2αIE1.IE21.0499 103=re2VTIE2re223.8113=Now, check to see that Q1 and Q2 are in the active mode VB1VBEIE1RE.VminusVB124.8126=VB2VplusR2R3.VminusR1.R1R2R3IE21βRPR1R2R3,.VB25.0528=VC1VB2VBEVC15.7028=VCB1VC1VB1VCB119.1098=Thus active mode for Q1.VC2VplusαIE2.RC.VC214.1065=VCB2VC2VB2VCB219.1594=Thus active mode for Q2.Now for the ac solution 3Make a Thevenin equivalent circuit looking out of the base of Q1 vtb1vsRPR2R3,RSRPR2R3,.vtb10.9777=Rtb1RPRSRPR2R3,,Rtb1977.6536=Rte1RPRER4,Rte197.7273=r'e1Rtb1rx1βre1r'e128.6805=r'e2rx1βre2r'e223.9113=4Next make a Thevenin equivalent circuit looking out of the emitter of Q2 ric1r0RPr'e1Rte1,1αRte1.r'e1Rte1ric12.1678 105=Gmb1αr'e1Rte1r0Rte1βr0RPr'e1Rte1,.ic1scvtb1Gmb1.ic1sc7.692 103=vte2ic1scric1.vte21.6674 103=Rte2ric1Rte22.1678 105=5Now make a Norton equivalent circuit looking into the collector of Q1.It follows that the circuit for the output voltage and output resistance is Rtc2RPRCRL,Rtc26.6667 103=Gme2αr'e2Rte2r0r'e2αr0RPr'e2Rte2,.ic2scGme2vte2.ic2sc7.6527 103=ric2r0RPr'e2Rte2,1αRte2.r'e2Rte2ric29.7899 106=voic2scRPric2Rtc2,.AvvoAv50.9832=This is the voltage gain.routRPRCric2,rout1.9959 104=6The circuit for the input resistance is rie2r'e2r0Rtc2r'e2r0Rtc21β.rie227.0685=Rtc1rie2Rtc127.0685=rib1rx1β()re1RPRte1r0Rtc1,.βRte1.Rtc1.Rtc1r0Rte1rib12.4255 104=rinRPrib1RPR2R3,,rin1.5604 104=7The following simpler solution is based on the r0 approximations for Q2 . ic2scαic1sc.ric1rie2ric1.ic2sc7.6526 103=The ric1rie2ric1 is a current divider.voic2scRPRtc2ric2,.AvvoAv50.9823=This is the voltage gain.routRPric2RC,rout1.9959 104=rib1rx1β()re1Rte1.rib12.4304 104=rinRPrib1RPR2R3,,rin1.5624


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