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Chapter 5 Solution Manual of Fundamentals of Microelectronics Bahzad Razavi Preview Edition.

### Transcript of Fundamental Of MIcroelectronics Bahzad Razavi Chapter 5 Solution Manual

5.3 (a) Looking into the base of Q1 we see an equivalent resistance of r1, so we can draw the followingequivalent circuit for nding Rin:RinR1R2 r1Rin = R1 +R2 r1(b) Looking into the emitter of Q1 we see an equivalent resistance of 1gm1 r1, so we can draw thefollowing equivalent circuit for nding Rin:RinR11gm1 r1Rin = R1 1gm1 r1(c) Looking down from the emitter of Q1 we see an equivalent resistance of 1gm2 r2, so we can drawthe following equivalent circuit for nding Rin:RinQ1VCC1gm2 r2Rin = r1 + (1 +1)

1gm2 r2

(d) Looking into the base of Q2 we see an equivalent resistance of r2, so we can draw the followingequivalent circuit for nding Rin:RinQ1VCCr2Rin = r1 + (1 +1)r25.4 (a) Looking into the collector of Q1 we see an equivalent resistance of ro1, so we can draw the followingequivalent circuit for nding Rout:RoutR1ro1Rout = ro1 R1(b) Lets draw the small-signal model and apply a test source at the output.RBr1+v1gm1v1 ro1vt+itit = gm1v1 + vtro1v1 = 0it = vtro1Rout = vtit= ro1(c) Looking down from the emitter of Q1 we see an equivalent resistance of 1gm2 r2 ro2, so wecan draw the following equivalent circuit for nding Rout:Q1Rout1gm2 r2 ro2Rout = ro1 + (1 + gm1ro1)

r1 1gm2 r2 ro2

(d) Looking into the base of Q2 we see an equivalent resistance of r2, so we can draw the followingequivalent circuit for nding Rout:Q1Routr2Rout = ro1 + (1 + gm1ro1) (r1 r2)5.5 (a) Looking into the base of Q1 we see an equivalent resistance of r1, so we can draw the followingequivalent circuit for nding Rin:RinR1R2 r1Rin = R1 + R2 r1(b) Lets draw the small-signal model and apply a test source at the input.r1+v1vt+itgm1v1 R1it = v1r1gm1v1v1 = vtit = vtr1+ gm1vtit = vt

gm1 + 1r1

Rin = vtit= 1gm1 r1(c) From our analysis in part (b), we know that looking into the emitter we see a resistance of1gm2 r2. Thus, we can draw the following equivalent circuit for nding Rin:RinQ1VCC1gm2 r2Rin = r1 + (1 + 1)

1gm2 r2

(d) Looking up from the emitter of Q1 we see an equivalent resistance of 1gm2 r2, so we can drawthe following equivalent circuit for nding Rin:RinQ11gm2 r2VCCRin = r1 + (1 + 1)

1gm2 r2

(e) We know that looking into the base of Q2 we see Rin = r2 if the emitter is grounded. Thus,transistor Q1 does not aect the input impedance of this circuit.5.6 (a) Looking into the collector of Q1 we see an equivalent resistance of ro1, so we can draw the followingequivalent circuit for nding Rout:RoutRCro1Rout = RC ro1(b) Looking into the emitter of Q2 we see an equivalent resistance of 1gm2 r2 ro2, so we can drawthe following equivalent circuit for nding Rout:Q1RoutRE = 1gm2 r2 ro2Rout = ro1 + (1 + gm1ro1)

r1 1gm2 r2 ro2

5.7 (a)VCC IB(100 k) = VBE = VT ln(IC/IS)VCC 1IC(100 k) = VT ln(IC/IS)IC = 1.754 mAVBE = VT ln(IC/IS) = 746 mVVCE = VCC IC(500 ) = 1.62 VQ1 is operating in forward active.(b)IE1 = IE2 VBE1 = VBE2VCC IB1(100 k) = 2VBE1VCC 1IC1(100 k) = 2VT ln(IC1/IS)IC1 = IC2 = 1.035 mAVBE1 = VBE2 = 733 mVVCE2 = VBE2 = 733 mVVCE1 = VCC IC(1 k) VCE2= 733 mVBoth Q1 and Q2 are at the edge of saturation.(c)VCC IB(100 k) = VBE + 0.5 VVCC 1IC(100 k) = VT ln(IC/IS) + 0.5 VIC = 1.262 mAVBE = 738 mVVCE = VCC IC(1 k) 0.5 V= 738 mVQ1 is operating at the edge of saturation.5.8 See Problem 7 for the derivation of IC for each part of this problem.(a)IC1 = 1.754 mAgm1 = IC1/VT = 67.5 mSr1 = /gm1 = 1.482 k100 kr1+v1gm1v1 500 (b)IC1 = IC2 = 1.034 mAgm1 = gm2 = IC1/VT = 39.8 mSr1 = r2 = /gm1 = 2.515 k100 kr1+v1gm1v1 1 kr2+v2gm2v2(c)IC1 = 1.26 mAgm1 = IC1/VT = 48.5 mSr1 = /gm1 = 2.063 k100 kr1+v1gm1v1 1 k5.9 (a)VCC VBE34 k VBE16 k = IB = ICIC = VCC VT ln(IC/IS)34 k VT ln(IC/IS)16 kIC = 677 AVBE = 726 mVVCE = VCC IC(3 k) = 468 mVQ1 is in soft saturation.(b)IE1 = IE2IC1 = IC2VBE1 = VBE2 = VBEVCC 2VBE9 k 2VBE16 k = IB1 = IC1IC1 = VCC 2VT ln(IC1/IS)9 k 2VT ln(IC1/IS)16 kIC1 = IC2 = 1.72 mAVBE1 = VBE2 = VCE2 = 751 mVVCE1 = VCC IC1(500 ) VCE2 = 890 mVQ1 is in forward active and Q2 is on the edge of saturation.(c)VCC VBE 0.5 V12 k VBE + 0.5 V13 k = IB = ICIC = VCC VT ln(IC/IS) 0.5 V12 k VT ln(IC/IS) + 0.5 V13 kIC = 1.01 mAVBE = 737 mVVCE = VCC IC(1 k) 0.5 V = 987 mVQ1 is in forward active.5.10 See Problem 9 for the derivation of IC for each part of this problem.(a)IC = 677 Agm = IC/VT = 26.0 mSr = /gm = 3.84 k(34 k) (16 k) r+vgmv 3 k(b)IC1 = IC2 = 1.72 mAgm1 = gm2 = IC1/VT = 66.2 mSr1 = r2 = /gm1 = 1.51 k(9 k) (16 k) r1+v1gm1v1 500 gm2v2r2v2+(c)IC = 1.01 mAgm = IC/VT = 38.8 mSr = /gm = 2.57 k(12 k) (13 k) r+vgmv 1 k5.11 (a)VCE VBE (in order to guarantee operation in the active mode)VCC IC(2 k) VBEVCC IC(2 k) VT ln(IC/IS)IC 886 AVCC VBERB VBE3 k = IB = ICVCC VT ln(IC/IS)RB VT ln(IC/IS)3 k = ICRB

IC + VT ln(IC/IS)3 k

= VCC VT ln(IC/IS)RB = VCC VT ln(IC/IS)IC + VT ln(IC/IS)3 kRB 7.04 k(b)VCC VBERB VBE3 k = IB = ICIC = VCC VT ln(IC/IS)RBVT ln(IC/IS)3 kIC = 1.14 mAVBE = 735 mVVCE = VCC IC(2 k) = 215 mVVBC = VBE VCE = 520 mV5.13 We know the input resistance is Rin = R1 R2 r. Since we want the minimum values of R1 andR2 such that Rin > 10 k, we should pick the maximum value allowable for r, which means pickingthe minimum value allowable for gm (since r 1/gm), which is gm = 1/260 S.gm = 1260 SIC = gmVT = 100 AVBE = VT ln(IC/IS) = 760 mVIB = IC = 1 AVCC VBER1 VBER2= IBR1 = VCC VBEIB + VBER2r = gm= 26 kRin = R1 R2 r=

VCC VBEIB + VBER2

R2 r> 10 kR2 > 23.57 kR1 > 52.32 k5.14gm = ICVT 126 Sr = gm= 2.6 kRin = R1 R2 r rAccording to the above analysis, Rin cannot be greater than 2.6 k. This means that the requirementthat Rin 10 k cannot be met. Qualitatively, the requirement for gm to be large forces r to besmall, and since Rin is bounded by r, it puts an upper bound on Rin that, in this case, is below therequired 10 k.5.15Rout = RC = R0Av = gmRC = gmR0 = ICVTR0 = A0IC = A0R0VTr = VTIC= R0A0VBE = VT ln(IC/IS) = VT ln

A0VTR0IS

VCC VBER1 VBER2= IB = ICR1 = VCC VBEIC + VBER2Rin = R1 R2 r=VCC VT ln

A0VTR0IS

IC + VTR2ln

A0VTR0IS

R2 R0A0In order to maximize Rin, we can let R2 . This gives usRin,max =VCC VT ln

A0VTR0IS

IC R0A05.16 (a)IC = 0.25 mAVBE = 696 mVVCC VBE IERER1 VBE + IERER2= IB = ICR1 =VCC VBE 1+ ICREIC + VBE+1+ ICRER2= 22.74 k(b) First, consider a 5 % increase in RE.RE = 210 VCC VBE IERER1 VBE + IERER2= IB = ICVCC VT ln(IC/IS) 1+ ICRER1VT ln(IC/IS) + 1+ ICRER2= IB = ICIC = 243 AIC IC,nomIC,nom100 = 2.6 %Now, consider a 5 % decrease in RE.RE = 190 IC = 257 AIC IC,nomIC,nom100 = +2.8 %5.17VCE VBE (in order to guarantee operation in the active mode)VCC ICRC VT ln(IC/IS)IC 833 AVCC VBE IERE30 k VBE + IERER2= IB = ICR2 = VBE + IEREVCCVBEIERE30 k IC=VT ln(IC/IS) + 1+ ICREVCCVT ln(IC/IS)1+ ICRE30 k ICR2 20.66 k5.18 (a) First, note that VBE1 = VBE2 = VBE, but since IS1 = 2IS2, IC1 = 2IC2. Also note that1 = 2 = = 100.IB1 = IC1 = VCC VBE (IE1 + IE2)RER1 VBE + (IE1 + IE2)RER2IC1 = VCC VT ln(IC1/IS1) 321+ IC1RER1VT ln(IC1/IS1) + 321+ IC1RER2IC1 = 707 AIC2 = IC12 = 354 A(b) The small-signal model is shown below.R1 R2 r1+v1r2+v2gm1v1RCgm2v2REWe can simplify the small-signal model as follows:R1 R2 r1 r2+vgm1vRCgm2v2REgm1 = IC1/VT = 27.2 mSr1 = 1/gm1 = 3.677 kgm2 = IC2/VT = 13.6 mSr2 = 2/gm2 = 7.355 k5.19 (a)IE1 = IE2 VBE1 = VBE2VCC 2VBE19 k 2VBE116 k = IB1 = IC11IC1 = 1VCC 2VT ln(IC1/IS1)9 k 12VT ln(IC1/IS1)16 kIC1 = IC2 = 1.588 mAVBE1 = VBE2 = VT ln(IC1/IS1) = 754 mVVCE2 = VBE2 = 754 mVVCE1 = VCC IC1(100 ) VCE2 = 1.587 V(b) The small-signal model is shown below.(9 k) (16 k) r1+v1gm1v1r2+v2gm2v2100 gm1 = gm2 = IC1VT= 61.1 mSr1 = r2 = 1gm1= 1.637 k5.22VCC IE(500 ) IB(20 k) IE(400 ) = VBEVCC 1 + IC(500 + 400 ) 1IC(20 k) = VT ln(IC/IS)IC = 1.584 mAVBE = VT ln(IC/IS) = 754 mVVCE = VCC IE(500 ) IE(400 )= VCC 1 + IC(500 + 400 ) = 1.060 VQ1 is operating in forward active.5.23VBC 200 mVVCC IE(1 k) IBRB (VCC IE(1 k) IC(500 )) 200 mVIC(500 ) IBRB 200 mVIBRB IC(500 ) 200 mVVCC IE(1 k) IBRB = VBE = VT ln(IC/IS)VCC 1 + IC(1 k) IC(500 ) + 200 mV VT ln(IC/IS)IC 1.29 mARB IC(500 ) 200 mVIC 34.46 k5.25 (a)IC1 = 1 mAVCC (IE1 + IE2)(500 ) = VT ln(IC2/IS2)VCC

1 + IC1 + 1 + IC2

(500 ) = VT ln(IC2/IS2)IC2 = 2.42 mAVB (IE1 + IE2)(500 ) = VT ln(IC1/IS1)VB

1 + IC1 + 1 + IC2

(500 ) = VT ln(IC1/IS1)VB = 2.68 V(b) The small-signal model is shown below.r1+v1r2+v2200 gm1v1 gm2v2500 gm1 = IC1/VT = 38.5 mSr1 = 1/gm1 = 2.6 kgm2 = IC2/VT = 93.1 mSr2 = 2/gm2 = 1.074 k5.26 (a)VCC IB(60 k) = VEBVCC 1pnpIC(60 k) = VT ln(IC/IS)IC = 1.474 mAVEB = VT ln(IC/IS) = 731 mVVEC = VCC IC(200 ) = 2.205 VQ1 is operating in forward active.(b)VCC VBE1 IB2(80 k) = VEB2VCC VT ln(IC1/IS) IB2(80 k) = VT ln(IC2/IS)IC1 = npn1 + npnIE1= npn1 + npnIE2=