Homework
Homework is due by 5 pm. on due date.
Please note that the answers given may be prone to errors.
| Homework | Modifications | Answers | Due Date |
|
Semilog Graph Paper(4 to a page) Semilog Graph Paper (2 to a page) |
1. Vo/Vg=5.9, Rth=40, Vth=5.9Vg; 2.a. 11.8, b. 29.5, c. 5.9sin(10t); 3.Vo/V1=0.833/(1+0.05s) | Jan. 13 | |
| 1. Low pass filter with bandwidth up to 10 rad/sec. 4. Bandwidth from 10 to 1000 rad/sec. 5. Bandwidth from 1 to 1k rad/sec. 6. ID1=1mA, ID2=0, Vo=2.7V 7. ID1=18mA, ID2=0, Vo=-9.3V 8. Id_total=18m+-1mA 9. ID1=0, ID2=34mA, Vo=-19.3, Id2_total=34m+-989microA | Jan. 27 | ||
| Assume all capacitors act as an "open" for DC analysis. | 2. VB=0, VE=-0.7, VC=4, IB=19.3micro, IC=1.93m, IE=1.95m; 3. VB=8.5, VE=7.8, VC=9.2, IB=77micro, IC=7.69m, IE=7.76m; 4. VB=-8.26, VE=-7.6, VC=-9.13, IB=176micro, IC=17.4m, IE=17.6m; 5. VB=0.96, VE=1.66, VC=-2.6, IB=4micro, IE=0.4m, IC=0.396m; 6. VB=-0.099, VE=-.799, VC=13.02, IB=19.8micro, IC=1.98m, IE=2m; 7. IB1=6.8n, IB2=IE1=683n, IB3=IE2=69micro, IC3=IE3=7m, VE3=13, VB3=13.7, VB2=14.4, VC1=15, VB1=15; 8. VC=2.85, VB=0.25, VE=-0.45, IE=1.55m, IB=15.35micro, IC=1.53m; 9. VB=-0.3, VC=2.3, VE=-1, IE=7.46m, IB=73.8micro, IC=7.39m; 10. VE=-6.8, VB=-6.1, VC= 8.4, IE=1.59m, IB=15.7micro, IC=1.57m | Feb. 10 | |
| 1. Rc=302 ohms, 4.b. beta_forced=10, c. Rc<2030; 5.b. beta_forced=9; 6. beta_forced=27.3; 7.c. Saturation | Feb. 19 | ||
|
All AC Sweeps
are on a logarithm scale. 6. add 1pF to the value list. |
3. Vout/Vin=1.51; Rin=9630, Rout=2k; 4.gain=1.5(similar to 3a); Rin=10k; Rout=2k; 5. b. C1 effects the starting value for the bandwidth of the circuit. c. The trace where the -3dB is less than 20Hz; 6.b. The 1pF value creates the first pole too high for the circuit to be able to amplify. c. The value that has the starting value less than 20Hz will be acceptable; 7. They are the same. The capacitors are blocking any DC effect on the biasing of the circuit. gain=18V/V. 8. hand analysis: gain=17.3V/V, Rin=3400, Rout=2k; 9. Vout/Vin=1.5V/V, Rin=440, Rout=2k; 10. gain=877mV/V, Rin=9.5k, Rout=36. | March 3 | |
| 1. IE1=1.84m, IB1=18.2u, IC1=1.82m, VB1=-2.46, VE1=-3.16, VC1=1.34, IB2=9.9u, IE2=1m, IC2=0.99m, VB2=2.98, VE2=2.28, VC2=4; VIMAX=0.198V; 2. IE1=1m, IB1=9.9u, IC1=0.99m, VB1=3.09, VE1=2.4, VC1=4.6, IE2=10m, IB2=99.9u, IC2=9.99m, VB2=1.05, VE2=0.35, VC2=2.01; VIMAX=0.192V; 3. Rin=23,793, Rout=20k, Vo/Vsig=0.86; 4. Rin=337,000, Rout=313, Vo/Vsig=-4; 7. IB=10m, IC=105m, Bforced=10.5, RB<10k; 8. IB=1m, IC=9.8m, Bforced=9.8, RB<46k; 9. Rin=1732, Rout=118, Vo/Vsig=-243; 10. Rin=105,100, Rout=10k, Vo/Vsig=-480.5 | March 17 | ||
| 2.a.
uCox(W/L)=200micro, Vt=1; b. uCox(W/L)=400micro,
Vt=-1.5; c. uCox(W/L)=400micro, Vt=-1; d.
uCox(W/L)=100micro, Vt=0.8 (the bottom case is triode.)
6 and 7. V1=-4, V2=2, V3=3.41, V4=4, V5=-5, V6=6, V7=2
a. 3.01k, V2=2.04, V1=-4.01; b. 6.65k, V3=3.41; c.
3.01k, V4=4.01, V5=-5.03; d. 1k, V6=6, V7=2; 8.
VGS=3, VS2=4, ID=9, VD2=9; 9. VGS1=2 or ID1=1m, Vs1=0,
ID=4m, VD2=4.5; 10. gm=1.9mA/V^2 |
March 31 | ||
| 3. Rin=5M, Rout=91, Vo/Vsig=-1, 4. Rin=30k, Rout=238, Vo/Vsig=-6.35; 5. Rin=1M, Rout=4k, Vo/VI=-4; 6. I1=0, IS=1m, VG2=9V, VS2=3.75, VS1=-4, Vs2_total= 3.75-8msin(20t); 7. I1=0, IS=1.4m, Vs1=-2.7, VG2=3.6, VS2=0.6; 8. Rin=60k, Rout=20,548, Vo/Vsig= -389(gm1=80m) or -1231(gm1=.253) | April 7 | ||
| 4.b. R3=100k, R1=R2=10k | 1 and 2 R placed at + terminal. 3. a. 4, b. 3MHz, c. 63.7kHz, d. 8mV, e. 16mV, f. R at + terminal=2.4K; 4. b. 80KHz, c. 105mVpeak, d. R at + terminal; 5. a. 20, b. 4MHz, 6. a. max=83mV, min=43mV, b. R at + terminal. 7. -(Vs+isR1)R4/(R1+R3), 8. I1=0, Is=16m, VG2=4V, VS2=-1V, VS1=-2V; 9. Won't be 0.1V ever. 10. Rin=30,500, Vo/Vsig=-.97V/V | April 14 |