Course Syllabus - Equivalent Reading from 5th Edition
College of Engineering Guidelines - Fall 2011
Please select the exams you plan on taking through the following link
http://www.doodle.com/yy7kyxqmnce3iysz
Note: (each part of the exam can only be taken at the specified times)
Part 2 - 8 to 9am
Part 1 - 9 to 10am
Part 3 - 10 to 11am
MW, 8:05-9:25
WEB 2250
4 credit hours
Instructor: Professor
Office: MEB 3254
Office Hours: M, W
9:45-11:30
La
Grader: Farhana Masid
(801)-556-8497
TA: Dipanjan Bhadra
Text
Notes
| Notes | With Annotations | Notes | With Annotations | Notes | With Annotations | ||
| Aug 22 | Aug 22 (Review) | Oct 5 | Oct 5 | Nov 14 | Nov 14 | ||
| Aug 24 | Aug 24 (CS/CE/Cascode) | Oct 17 | Oct 17 | Nov 16 | Nov 16 | ||
| Aug 29 | Aug 29 | Oct 19 | Oct 19 | Nov 21 | Nov 21 | ||
| Aug 31 | Aug 31 | Oct 24 | Oct 24 | Nov 23 | Nov 23 | ||
| Sept 7 | Sept 7 | Oct 26 | Oct 26 | Nov 28 | Nov 28 | ||
| Sept 12(part of Sept. 7 notes) | Sept 12 | Oct 31 | Oct 31 | Nov 30 | Nov 30 | ||
| Sept 14 | Sept 14 | Nov 2 | Nov 2 | ||||
| Sept 19 | Sept 19 | Nov 7 | Nov 7 | ||||
| Sept 21 | Sept 21 | Review | Dec 7 | ||||
| Sept 26 | Sept 26 | ||||||
| Sept 28 | Sept 28 |
Quizes
| Quiz1 | Quiz2 | Quiz3 | Quiz4 |
Homework Assignments
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Homework |
Modifica |
Answers: |
Due |
| 1. IB=34micorA, IE=3.4mA, IC=3.37mA; Vo=9.3V, VE=6.9V, VB=7.6V; ic_total=[3.37m+12microsin(20t)]A; 2. Rin=1.1k, Rout=2.7k, Vo/Vsig=9.091k; 4.e)I1=0, ID=ID=1m; (f) VG=2, Vs=-0.7, Vo=6; (h)VGS=2.7 or ID=1mA; (i)Vototal=6-0.014sin(20t)V; 5.Rin=5Meg ohm; Rout=91ohm; Vo/Vsig=-1V/V; 7. (a) gm1=800microA/V; Ron=Rop=1.28Megohm; Rout=640kohm; Av=-512V/V; (b) Vov=0.3V; range is 0.7V to 1.8V; 8. L=0.4micrometer; I=0.2mA; (W/L)1,2=25; (W/L)3,4=100; 9. Rout=400k; 10. (a) Rin=25k; Rout=3.33Meg; Avo=-13.3kV/V; (b) Rin=25k; Rout=2.5Meg; Avo=-10kV/V; (c) Rin=infinity; Rout=2.5Meg; Avo=-2.5kV/V; (d) Rin=infinity; Rout=3.33Meg; Avo=-3.33kV/V |
Aug 31 Extended to Sept. 2, 9am |
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| #10 typo - should read 100microA (not 100A). Also note that for #10, you need to have the same voltage across the diode connected transistor as that across R. | 1. Io=1.02mA, Ro=100k; 2. IREF=0.497mA, R=8.65k, Vomin=0.3, Io=.53mA; 3. (a) Ro=10Meg, (b) Ro=54Meg; 4. IREF=0.994mA, R=4.3k, Max Vo=4.3, Io=1.09mA; 5. VB1=9.3V, VB2=-9.3, IR1=IC1=IC2=IC3=IC4=IC5=IC6=IC9=IC8=IC7=IC9, IC10, IC11=0.93mA, VC3=4.65, VC5=0.7, VC6=1.65, IR4=1.86mA, VC7=-3.72, VC9=4.3, VC11=2.79; 9. R=68.5k, Ro=115Meg; 10. R=6.42k | Sept. 8, 9pm | |
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Examples(includes
problems 4 and 8 from homework) Error on #1 kn'(W/L)=5mA/V^2 |
1. For kn'(W/L)=5m => (a) Vov=.2, VGS=0.7, (b) Vs=-0.7, ID=0.1mA, VD=0V; (c) Vs=-0.4, ID=0.1m, VD=0V; (d) Vs=-0.8, ID=0.1mA, VD=0V; (e) from -0.1 to 0.5V. For kn'(W/L)=0.4m => (a) Vov=0.7, VGS=1.2, (b,c,d) not possible, (e) from 0.4 to 0.5V. 2. (a) vid=0.28, (b) vid=-0.28, (c) -2 to 2. 3. Vov=0.2, gm=4m, ro=50k, Ad=18.2; 5. gm=4m, Rout=25k, Ad=100, Acm=5m, CMRR=86dB; 6. I=128microA; 7. (a) (W/L)1,2=12.5; (W/L)3,4=50; (b) from -0.5 to 0.8; (c) CMRR=74dB; 8. (a) I=0.4m, Rc =5k, VC=1.5; (b) Rid=25k; (c) VC1=-400m, VC2=+400m; (d) VCM=2.49; 9.Ad=200, Rid=20.2k, Acm=50m, CMRR=72dB, (e) 20.2k | Sept. 16 by 2pm | |
| #3. (d) you can use up to 5 pnp transistors. |
All homework for the rest of the semester is on due on Friday by 2pm. Wed. Sept. 28(by 5pm) Extra Credit(optional) |
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| 1. (a) Beta=R1/(R1+R2); (c) R2/R1=100; (d) 40dB; (e) Vo=100, Vf=0.99, Vi=10mV; (f) Af=99.75V/V; 2. Af=90V/V; f_H=11.1kHz; 3. f_L=1Hz, f_H=1MHz; 4. Beta=-1; (5&6) R11=972, R22=36k, beta=(1/36), Af=31.9V/V, Rof=208, Rif=2879; (7&8) R11=3k, R22=9k, beta=(2/3), Af=0.5V/V, Rof=2.1k; (9&10) R11=2002, R22=21k, beta=(1/21), Af=19.9V/V, Rof=759, Rif=79.7k | Oct. 21 | ||
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Solution(updated) |
#2. Should read amount of feedback is 10. #6 RL=1k (do not include in feedback). #10 should read amount of feedback is 15. | 1. Rof=96, Rif=6.67, Rin=7.15; 2. Rif=3430, Rin=3,330, Rof=29820 3. Beta_feed=(1/3), R11=2.3k, R22=6k; 4.Af=0.57, 5. Rof=688, Rif=1.476k, Rin=1.276k, 6. R11=1.55k, R22=2.7k, Beta_feed=-(3/7); 8. Af=A/(1+ABeta_feed) where A=[-gm2ro2gm1(ro1||200k)R11rpi1]/(R22+ro2+200k)(R11+rpi1)(1+(R22gm2ro2)/(R22+ro1+200k)); Rof=(200k)/(1+ABeta_feed); Rif=(R11||rpi1)/(1+ABeta_feed) 10. Rif=6.7, Rof=22.5k, Beta_feed=-1/6k | Oct. 28 |
| 6. Note that to find PM=45 you have to subtract from the original 90 degrees to get -135 degrees. | 1. b) Beta= 0.09, c) Af=0.7; 2. R11=18.7k, R22=120k, Beta=0.17; 3.a)A=-112.5m, b)Af=-107m, c)Rif=108,794; 4. a) Beta=(1/2), b) Af=2, c)Rof=18.3, d)Rif=13.4M, e)Rin=13.4M; 5. b) Beta=0.1; 6. Beta=1.5x10^-2; 7. a) A=-90.9, b) Af=-90, c)Rif=1.1k, d)Rin=1k, e)Rof=1.24k, f) Rout=2.1k; 9. Rof=20k; | Mon (5pm) Nov. 7 | |
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#4 has VEE=-Vcc=-5V, #5. has +5 and -5 power supplies, #6 uses +5 and -5 power supplies, find RL when Vin=5V and Vout=2.5V, #7 note: IQ=1mA because VBE=0.7V (show that VBE=0.7V), skip 7b, 7c note that with Vin=6V (this is a bias) so Vout=6V constant with an additional max and min. | 2. a. VCC=61V, b. 3.54A, c. 137.4W, d. 73% 4. -4.3V to 4.7V for Vout, -3.6V to 5.4V for Vin, b. RL=1.09k, 24% 5. -2.416 to 6V for Vin and -3.416 to 4.131V for Vout, 6. Vout_peak=4.5V, RL=625, 7. a. IQ=1mA, b. 4V step (6V to max output) and 6V (6V to min output), 8. VGG=26V, 9. a. TA=70degreesC, b. theta_CA=30degreeC/W , c. Tc=28degrees, 10. a. Ts=65.1degrees C, b. theta_SA=1.17degreeC/W | Wed, Nov. 23(5pm) |
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Problem Set #9-reposted |
2. Use |VBE|=0.7, RL=10k,
HW Reposted to show tables (Table for 5 is given completed) |
3. 11.5%, 4.a. 88degrees, b. 45degrees, 5. I1-4=480micro, I5-8=960micro, VOV3-4=0.7, VGS3-4=1.5, ro1-4=83.3k, 6.a. fp2=1,512MHz, b. Cc=0.53pF, c. PM=34degrees; 7. I1-4=150micro, I5-8=300micro, Vov=0.24, VGS=0.94, gm1-4=1.25m, gm5-8=2.5m, ro1-4=100k, ro5-8=50k; VICM is -1.26 to 0.32; Vo is -1.26 to 1.26; 8. Aoverall=3,906V/V; 9. Anew=289V/V; 10.a. fp2=398MHz, b. Cc=1pF, c. PM=37degrees. | Dec. 2 |
La
Exams