ECE3050 Homework Set 21. A diode has the parameters IS=10fA, n =2,andVT=25mV.(a) Calculate rdfor VD=0.6V.rd=nVTID+ IS=nVTe−VD/nVTIS=30.72 MΩ(b) Calculate rdfor VD=0V.rd=nVTIS=5× 1012Ω(c)Atwhatvoltagedoesrdexceed 1015Ω?VD= nVTlnµnVTISrd¶= −0.265 V2. A diode current-controlled attenuator circuit is shown. It is given that R =20kΩ. The diodeparameters are n =2and VT=0.025 V.(a) Calculate the bias current which will provide a small-signal attenuation of 20 dB,i.e.vo/vs=0.1.I =2nVTR"µvovs¶−1− 1#=45µA(b) If the current is halved, what is the new attenuation? (−14.8dB). (c) If the current isdoubled, what is the new atten uation? (−25.6dB)3. A diode has the current ID1=1mAfor VD1=0.55 V and ID2=2mAfor VD2=0.58 V.IfIS¿ ID1, determine the ideality factor or emission coefficient n and the saturation currentIS.n =VD2− VD1VTln (ID2/ID1)=1.73IS=ID1exp (VD1/nVT)=ID2exp (VD2/nVT)=3.03 nA4. The diagram shows a zener diode regulator. It is given that V1=35V. The diode has thezener voltage VZ=24V. The load resistance varies bet ween the limits 500 Ω ≤ RL≤ 10 kΩ.1(a) Calculate R1if IZis to have a value that is no smaller than 10 mA.NotethatI1is aconstant once R1is determined and I1= IZ+ IL. Thus the minimum value of IZoccurswhen ILis a maximum (when RLis a minimum) because this makes IZhave the smallestvalue.R1=35 V − 24 V0.01 A + 24 V/500 Ω= 190 Ω(b) What is the po wer dissipation in R1and the maximum power dissipation in the zenerdiode? Note that I1is a constant and I1= IZ+ IL. Thus the maximum dissipation in thezener diode occurs when ILis its smallest value because this makes IZhave the largest value.P1=(35 V − 24 V)2190 Ω=0.637 W PZ max=24Vµ35 V − 24 V190 Ω−24 V10 kΩ¶=1.33 W5. Calculate the values of β and ISfor the transistor sho wn if VCB= VBE=0.7V, IB=0.2mA,and IE=10mA.Figure 1:β =10 mA − 0.2mA0.2mA=49 IS=9.8 × 10−3exp (0.7/0.025)=6.78 × 10−15A6. Calculate the values of β and ISfor the transistor shown if VEB= VBC=0.7V, IB=50µA,and IC=2.5mA.β =2.5mA50 µA=50 IS=2.5 × 10−3exp (0.7/0.025)=1.73 × 10−15A7. Calculate the collector, emitter, and base currents if V+=3.3V, VEE= −3.3V, VBE=0.7V,RE=47kΩ,andβ =90.IE=−0.7V− (−3.3V)47 kΩ=55.3 µA IB=55.3 µA91=0.608 µAIC= IE− IB=54.7 µA28. An npn transistor is operated in the active m ode with a base current of 3 µA. It is found thatIC= 240 µA for VCE=5Vand IC=265µA for VCE=10V. What are the values of β0andVAfor this transistor? [β0=71.7, VA=43.1V]9. A BJT has the parameters β0=75, VA= 100 V,andVCE=10V.(a) Calculate ICfor rπ=10kΩ.IB=VTrπ=2.5 µA IC= β0µ1+VCEVA¶IB=0.2063 mA(b) Calculate the values of gmand r0.gm=ICVT=1121.2r0=VA+ VCEIC= 533.3kΩ(c) Calculate α and re.α =β1+β=β0µ1+VCEVA¶1+β0µ1+VCEVA¶=0.9880re=VTIEor=VT(1 + β) IBor=rπ1+β0µ1+VCEVA¶=119.8 Ω10. The output characteristics of a BJT are shown. (a) Determine β0and VA.[β0= 120,VA=30V] (b) Calculate β at iB=4µA and VCE=5V.[135] (c) Calculate β at iB=8µAand
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