Homework

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Note: Homework is due by 5 pm. on due date.

Please note that the answers given may be prone to errors. Please contact me if you question an answer.

Homework	Modifications	Answers:	Due

Problem Set #1 Solution	Semilog Graph Paper. Sketch the transfer function means to sketch the bode plots. #1. Assume gm1 and gm2 are known values.	1. Vth=(-gm1gm2(500)(500)2)/(5(1+500gm2)); Rth=(1/gm2\|\|500) 2. Vth=-15Vg/4(16); Rth=10ohm; 9. Vo/V1=10k(1+(2/10Meg))/11k(1+(s/1100Meg))	Fri, Jan 20
Problem Set #2(updated) Solution 2	#6. Use a Vsin source to create a sinusoidal signal. The Bode Plot for the PSpice circuit has been updated. Please see Problem Set #2.	9. VL/Vs=8V/V=18dB; 10. VL/Vs=80/3V/V=29dB	Fri, Jan 27
Problem Set #3 Solution 3	Skip problems 6 and 10. Problems 5 and 9 will be counted as 2 problems.	2. b) bandwith=80kHz, c) 105mVpeak; 5. ID1=24mA, Vo= -7.6V, ID2=0; 7. ID1=0, ID2=.1A, Vo=-12V, 9. Vo= -19.3V, ID2=34mA, ID1=0	Fri, Feb 3
Problem Set #4 Solution 4	Note that the transistor is "on" if equal to Vt.	10. I1=0, ID=IS=1m, VG=2V, Vs=-0.7V, Vo=6V, VGS=2.7	Tues, Feb. 21 (9am)
Problem Set #5 Solution 5		1. VS2=4V, ID2=4A, VD2=9V, VGS=3V; 2. VGS1=2V or ID1=1mA, VS1=0V, ID=4mA, VD2=4.5V; 3.a) I1=0A, IS=1mA; b) VG2=9V, VS2=3.75V, VS1=-4V; d) ID1=1mA or VGS1=4V; e) VS2_total=3.75V -8msin(20t)V; 5. gm=1.9mA/V^2	Fri, Feb 24
Problem Set #6 Solution 6		1.a. -20micros^3/(0.02s+1)(.2s+1)(500micro*s+1); 2. Vo/Vin=-0.019V/V, Rin=500k, Rout=10; 3. Vout/Vin=18m, Rout=19; 4. Rin=30k, Rout=238, Vo/Vsig=-6.35V/V; 5. Rin=5M, Rout=91, Vo/Vsig=-1V/V; 10.(a) Vs=4V, V5=-5V, R=3k; (b)V6=6V, V7=2V, R=1k	Fri, March 2
Problem Set #7 Solution 7	only do 9(b), ignore 9(a)	2. a) VB=0, IE=0.318m, IC=0.312mA, Vc=.81, IB=6.24uA; b) VB=0, VE=0.8, IE=0.7m, IC=0.69m, VC=-0.81, IB=14uA; c) VB=1, VE=1.8, IE=1m, IC=0.98m, VC=0.55, IB=20uA; d) VB=1.5V, VE=0.7, IE=1.5m, IC=1.47m, VC=1.52V, IB=29uA; 3.a) VC=1.4V, VE=-0.7; b) IE=1.05mA, VC=1.32; c) VB=0, VE=-0.7, VC=1.32; d) VE=1.4, VC=-0.55; e)VB=0.73, VE=1.43, VC=-0.55; 4.IE=1.29m, IB=12.8m, VB=4.57, IC=1.28m, VC=8.6; 5.a) VB=0, VE=-0.7, IE=1.95m, IC=1.93m, IB=19.3u, VC=4; b) IE=7.76m, IC=7.69m, IB=76.9u, VB=8.5, VE=7.8, VC=9.23; c) IE=17.6m, IC=17.5m, IB=174u, VB=-8.26, VE=-7.6, VC=-9.13; d) IE=0.4m, IB=4u, VB=0.96, VE=1.66, VC=-2.6; 6.IE=2m, IC=1.98m, IB=19.8u, VB=-0.099, VE=-.799, VC=13.02; 7. IE3=7m, IC3=7m, IB3=60u, IE1=683m, IB1=6.8n, VE3=13, VB3=13.7, VB2=14.4, VC115, VB1=15; 8. IE=9.4m, IC=9.2m, VB=1.16, VE=0.46, VC=8.08; 9. V2=1.36, V1=0.66, V4=-1.8, V3=-1.08, V5=3.8	Fri, March 23
Problem Set #8 Solution 8	#5. Change the emitter resistor from 40ohm to 4kohm. #10 Base R should be 4Meg	4. IB=70u, IE=7.09m, IC=7.02m, VB=1.44, VE=0.74, Vo=9.3; 5. IB=34u, IE=3.4m, IC=3.37m, VE=6.9, VB=7.5, Vo=9.3; 6. Rin=203,300, Rout=13333, Vo/Vsig=49V/V; 7. Rin=1.1k, Rout=2.7k, Vo/Vsig=9.09k; 9. No, 10. Vo/Vsig=-24V/V	Fri, March 30
Problem Set #9 Solution 9	Use RL=1k for Problem 7 and 10. Prob. 7 and 10 read Analyze circuit above to find .......	3. Vo/Vsig=(gm2)(RL)(rpi1)(gm1)(R3)(rpi2)/(rpi1+R1+Rsig)(R3+rpi2+RE(beta+1)); 4. (a) VBB=2.3V; (b) VBB=2.42; (c) VBB=10.3V; 5. IE=1.77m, IB=17.5micro; IC=1.74m; VB=2.3, VC=4.13, VE=1.6; 7. Rin=698, Rout=500, Vout/Vin=-0.16V/V; 8. IE=3.23m, IB=32micro, IE=3.20m; VB=1.7, VC=1.77, VE=1.0; 10. Rin=1,168; Rout=1,010; Vout/Vin=-1.2V/V	Fri, April 20