Homework

Modifications

Answers:

Due

Problem Set #1

Solution

Semilog Graph Paper. Sketch the transfer function means to sketch the bode plots.

1.  Vo/Vg=5.9, Vo=59mV, Vth=5.9Vg, Rth=40, 2.  Vth=(Vsig(500*gm2)(-gm1*500)(2))/((1+gm2*500)5), Rth=(1/gm2)||500,

1/18/2013

Problem Set #2

Solution 2

3.  10k(s/1e6+1)/11k(s/1.1e6+1), 4.  R4(R1+R2+R5)/(R1+R2)(R3+R4); 5.  -R4(Vs+isR1)/(R1+R3); 6.  -Vs-isR1-isR2R3/(R2+R3); 7.  8V/V, 18dB; 8.  19V/V, 25.6dB, c)  36/190; 9.  (a)  51, (b) 78,431Hz, (c) 53,052Hz, (d) 2*(Vmax-2) (e) 2Mohm (f) 0.102V; 10.  (a) (R3/R2||R1+1), (b) 80kHz, (c)105mVpeak

Monday

1/28/2013

Problem Set #3

Solution 3

 1.  First amp, 900 ohms; second amp 1.8k; 3.  ID1=1mA, ID2=0, Vo=2.7V; 4.  ID1=18mA, ID2=0A, Vo=-9.3V; 5.  ID1_total=18m+-1mA; 6.  ID1=0, ID2=34m, Vo=-19.3V, 7.  ID2_total=34m+-989uA;  8.  b)  40dB, c) 10 to 1k; 9. a) 23dB, b)No clipping, c) yes clipping; 10.  a) 4V/V (inverting), b) 3MHz, c) 63.7kHz, d) 8mV, d) 16mV, e) add 2.4k ohm to positive terminal.

Monday

2/4/2013

Problem Set #4

Solution 4

6-7 V1=-4, V2=2, V3=3.41, V4=4, V5=-5, V6=6, V7=2, (a) R=3.01k; (b) R=6.65k; (c)R=3.01k; (d)R=1k; 8.Qpoint=3V, Vs2=4V, ID=9A, VD2=9V; 9.  Qpoint=2V or 1mA, Vs1=0V, ID=4mA, VD2=4.5V; 10.  gm=1.9mA/V^2

Wed.  2/20/2013

Problem Set #5

Solution 5

3.  Rin=5M, Rout=91, Vo/Vsig=-1V/V; 4.  Rin=30k, Rout=238, Vo/Vsig=-6.35V/V; 5.  Vo/VI=-4V/V, Rin=1M, Rout=4k; 6.  I1= 0, Is=1mA , VG2=9V , VS2=3.75V , VS1=-4V , 7.  VS2_total=3.75-8msin(20t) 8.  No, 9.  No, 10.  No.

2/25/2013

Problem Set #6

Solution 6

1.  I1=0, Is=1.4m; 2.  VG2=3.6V, VS2=0.6V, Vs1=-2.7V; 6.  Rin=60k, 7.  20,548; 8.  For VGS1=9V => Vo/vsig=-389V/V; For ID1=3.2A => Vo/vsig=-1,231V/V;

3/4/2013

Problem Set #7

Solution 7

 #2a.  The resistor located in the collector is 500 Ohms.  The emitter resistor is 2.2k.

#10.  Also find Rin, Rout, Vo/Vs gain.

2.a.  VB=0, VE=-0.7, VC=4; IB=19.3micro, IE=1.95m, IC=1.93m; b.  VB=8.5, VE=7.8, VC=9.2; IB=76.9micro, IE=7.76m, IC=7.69m; c.  VB=-8.26, VE=-7.6, VC=-9.13; IB=174micro, IE=17.6m, IC=17.4m; d.  VB=0.96, VE=1.66, VC=-2.6; IB=4micro, IE=0.4m.  3.  IB4=19.8micro, IE4=2m, IC4=1.98m, VB4=-0.99, VE4=-.799, VC4=13; 4.  IB1=6.8n, IB2=683n, IB3=69micro, IC3=7m, IE3=7m, VB1=15, VC1=15,VB2=14.4,VB3=13.7, VE3=13; 5.  IE=1.55m, VE=-0.45, IB=15.4mico, VB=0.25, IC=1.53m, VC=2.85; 6.  IB=,10micro,  IE=1m, IC=990micro, VB=1.7, VE=1, Vo=2.03, active; 7.  IB=15.6micro, IE=1.58m,  IC=1.56m, VB=-11.08, VE=-11.78, Vo=11.8, active; 9.  V1=421.9m, V2=1.122, V3=-1.225, V4=-1.925, V5=1.39; 10.  IE=1.59m, VE=-6.8, IB=15.7micro, VB=-6.1, IC=1.57m, VC=8.4, Rin=7,698, Rout=91, Vout/Vin=0.99.

3/25/2013

Problem Set #8

Solution 8

#2.  The emitter R=4k NOT 40.  Uploaded new file. Note that IC is going through only the 100 ohm R.

#3.  IC is going through only the 200 ohm R’s.

2. IB=70u, IE=7.09m, IC=7.02m, VB=1.44, VE=0.74, Vo=9.3, ICtotal=7.02m-50micro*sin(20t); 3. IB=34u, IE=3.4m, IC=3.37m, VE=6.9, VB=7.6, Vo=9.3, ic_total=3.37m-10micro*sin(20t); 4.  Rin=203,300, Rout=1,333, Vo/Vsig=49V/V; 5.  Rin=1.1k, Rout=2.7k, Vo/Vsig=9.09k; 7.  No; 8.  Yes, Vo/Vsig=-24V/V

 4/1/2013

Problem Set #9

Solution 9

 2.  Assume that the input VIN=3+0.5sin(20t) is connected through a capacitor to the junction of the 2k,1mA,2k, and 3k.  All current sources for this problem are not ideal and have a voltage drop across them. 

1.  IB=70.9micro, IE=7.16m, IC=7.09m, VB=2.86, VC=4.29, VE=2.16, Active, Rmax=302; 2.  IB=21.9micro, IE=2.21m, IC=2.19m, VB=-0.09, VE=-.79, VC=7.78, active; 3.  Rin=105,100, Rout=10k, Vo/Vsig=-480.5; 4.  Rin=1732, Rout=118, Vo/Vsig=-243; 7.  (b) 10, © Rc<2,030; 8.  beta_forced=infinity; 9.  (b) 9; 10. 27.3

4/8/2013