Prob. 1.15 Av=50, Avs=33.33, Ai=1.25e+06, G=6.25e+07=78dB

Prob. 1.16: a. 1.515 V and 45.91 mW, b. 2.50 mV and 0.125 uW, c. Power to load with amp is 36728 times case with no amp. Use an amplifier with "good buffer" properties, i.e. high Zin and low Zout!

Prob. 1.18 Ro=1111 ohms

Prob. 1.21: a. Zin=3kohms, Zout=20ohms, Avo=49980, b. Zin=1Mohm, Zout=400ohm, Avo=49669

Prob. D1.24: Final Design: ACCB, Rin=10Mohm, Rout=1ohm, Avo=377

Prob. 8.23: As = -9/(1 + j1e-05w) , As = -99/(1 + j1e-04w)

Prob. 8.24: Rx = 16.6 kohm

Prob. 8.25: Rin=0.1ohms, Rout=0ohms, the corresponding model has these two values plus an Avo=-1e+05

In-Class Problem: Zin = Zin,miller = 909.1 ohms

Prob. 1.31: Zin=10kohm, Av=500=54dBV, Ai=500000=114dBA, G=0.25e+09=84dBW

Prob. 1.34: a. 3.162 V, b. 31.62 mV, c. 3.162 mV, d. 70.71 V

Prob. 1.35: a. 0.1 W, b. 1 uW, c. 10 W

Prob. 1.40 Voltage Amp Model: Zin=1Mohm, Avo=2e+07, Zout=500kohm (in series)

Prob. 1.40 Transconductance Amp Model: Zin=1Mohm, Gmsc=40, Zout=500kohm (in parallel)

Prob. 2.3: Ri=infinite, Aol=infinite, Acm=0, Ro=0, Bandwidth=infinite

Prob. 2.9: See Fig. 2.4, p.64 for circuit, Av=-Rf/Ri, Rin=Ri, Rout=0

Prob. 2.10: a. -2V, b. -1V, c. 3V, d. 0V, e. 3V

Prob. 2.11: -1.2mV < Vx < +1.2mV, -1.2V < Vin < +1.2V, Yes: since Vin=1000Vx, Yes: summing point assumption is valid!

Prob. 2.14: Av = -8

Prob. 2.15: +or-2%

Prob. 2.17: Ii=1mA, If=1mA, Iload=-10mA, op-amp sinks 11 mA

Prob. 2.18: A1 = -4/3, A2 = -8/3

Prob. 2.22: vo = (1+R2/R1)[(Rb/(Ra+Rb))*Va + (Ra/(Ra+Rb))*Vb]

Prob. 2.23: a. Io=(V1-V2)/R since Io is independent of the load have infinite output impedance seen by the load (assuming an independent current source with a parallel output resistance), b. Io=-Vin/Rf again with infinite output impedance seen by the load

Prob. D2.34: Ri=10kohms, Rf=91kohms (closest 5% resistor), 10kohm from non-inverting terminal to ground, Av=10.1

Prob. D2.35: If Rc=0, consider making R2=R4 then if R1=R3=1kOhm one gets R2=R4=9.05kOhm or 9.09kOhm as the closest 1% resistor. Total resistance is then 20.18kOhm.

Prob. D2.39: For stage one use a voltage follower; for stage two use an inverting amplifier with a gain of -20.

Prob. 2.42: (a) Av=(Aol+Ro/Rin)/(1+Aol+Ro/Rin)=0.99999, (b) Zin=(1+Aol)Rin+Ro=100.001Gohm, (c) (Ro/(1+Aol))//Rin=0.25mohm

Prob. 2.44: (a) 1.5 MHz, (b) 150 kHz

Prob. 2.46: (a) 9987.5ang(-87.14 deg), (b) 1000ang(-89.71 deg), (c) 1ang(-90 deg)

Prob. 2.47: (a) Acl = 100/(1+j(f/1e+04)) and BW(-3dB) = 10000 Hz, (b) Acl = 100/(1+j(f/1e+05))**2) and BW(-3dB) = 64358 Hz

Prob. 2.49: The slew-rate limitation is the maximum rate of change possible for the output voltage magnitude. The full-power bandwidth is the highest frequency for which the op amp can produce a sine wave of full amplitude without exceeding the slew-rate limit.

Prob. 2.50: (a) 159 kHz, (b) 10 V, (c) 2 V, (d) 1.59, (e) triangle wave with Vpp = 5 V

Prob. 2.51: SR = 3.14 V/us

Prob. 2.52: SR = 8 V/us

Prob. 2.53: Bcl=5e+4Hz, Tcl=7us and plotted function is vo(t)=100Vm[1-exp(-2pi(5e+4)t]. Slew-rate limited input is 31.83mV.

Prob. 2.54: (a) 15915 Hz, (b) 2.5 V, the op amp is output current limited!, (c) 10 V, the op amp is output voltage-saturation limited!, (d) 1.59 V, the op amp is slew-rate limited!

Prob. 2.55: (a) vo=4vs, (b) Note: In plot v2 is inverse of v1 and both are exactly half of vo., (c) Vo(max)= +-28V

Prob. 2.58: Vo(worst case) = 64 mV

Prob. 2.59: Add a 50-kohm resistor from the non-inverting input to circuit common.

Prob. 2.60: (a) 9.09 mV, (b) 1 uA, (c) 9.09 kOhm, (d) 1 uA

Prob. 2.75: Negative ramp from 0 V at 0 ms to -0.5 V at 2 ms then constant at -0.5 V to 5 ms then a negative ramp to -1.0 V at 7 ms then constant at -1.0 V to 10 ms etc., 20 pulses

Prob. 2.76: -5 V from 0 to 1 ms, 0 V from 1 to 2 ms, +10 V from 2 to 3 ms, then -5 V from 3 to 4 ms

Prob. 2.78: 1 V

Prob. 2.79: 6 V

Prob. 2.80: 1-V peak-to-peak triangle wave with a 0.5 ms period and a value of +0.5 V at t = 0

Prob. 2.81: 6-V peak-to-peak triangle wave with a 10 ms period and a value of +3 V at t = 0

Prob. 2.82: -2.35 V

Prob. 2.83: +7.0 V

Prob. 2.84: 12.8 Vpeak-to-peak

Prob. 2.85: 16 Vpeak-to-peak

Prob. 2.86: Square wave from -6.4 V to + 6.4 V with a 10 ms period, transition from low to high at 2.5 ms and back to low at 7.5 ms, etc.

Prob. 2.87: Square wave from -8.0 V to + 8.0 V with a 50 ms period, transition from low to high at 12.5 ms and back to low at 37.5 ms, etc.

Filter Handout: H(s)=(s^2+(3/RC2)s)/(s^2+(3/RC2)s+(2/R^2C1C2)), wo=sqrt(2/R^2C1C2), Q=sqrt(2C2/9C1), fo=22508Hz, Q=4.71, H(s)=(s^2+30000s)/(s^2+30000s+2e10), high-pass

Prob. 3.4: Vdrop = 0.300 V

Prob. 3.5: a. Forward drop at 0.6 V, reverse drop at -0.6 V, b. Forward drop at 7.4 V, no reverse drop

, c. Forward drop at 6.2 V, reverse drop at -6.2 V

Prob. 3.6: 50degC: 0.125 uA and 0.125 V, 70degC: 0.5 uA and 0.5 V, 100degC: 4 uA and 4 V

Prob. 3.8: 5 diodes and 3.33%

Prob. 3.9: a. Va=0.8V, Ia=2.1mA, b.Vb=0.67V, Ib=1.7mA, c. Vc=0.28V, Ic=1.15mA

Handout: vo(t) = 3.131 + 520e-06coswt V

Prob. 3.15: a. Diode on V=0, I=3.70mA, b. Diode off and V=10V, I=0, c. Diode on and V=0 , I=0, d. Assume diode on and get V=5V, I=5mA so ok.

Prob. 3.16: a. D1 on, D2 off, V=10V, I=0, b. D1 on, D2 off, V=6V, I=6mA, c. D1 on, D2 on, V=30V, I=33.63mA

Prob. D3.25: 833 uF and 15.6:1

Prob. 3.32: Have a clamp circuit followed by a peak detector so Va(t) = Vm + Vm sin(wt) and Vpeak(across C2) = 2Vm .

Prob. 4.8: beta=30, alpha=0.968, iE=9.3mA

Prob. 4.9: alpha=0.980

Prob. 4.10: vBE=0.6585V, vBC=-9.341V, iC=9.90mA, iB=99.0uA, alpha=0.990

Prob. 4.14: a. beta=38.0, b. beta=19.6

Prob. 4.20: IBQ=5uA, VBEQ=0.58V, ICQ=2mA, VCEQ=16V, 2uA<=iB<=10uA, 12.5V<=vCE<=19V, 0.5mA<=iC<=4.0mA

Prob. 4.20: Av=Delta vCE/Delta vin=16.25 and Ai=Delta iC/Delta iB=440

Prob. 4.28: a. Active, b. Cutoff, c. Cutoff, d. Saturation

Prob. 4.33: a. IC=1.93mA, VCE=10.93V and IC=4.21mA, VCE=0.2V b. IC=1.47mA, VCE=4.98V and IC=2.18mA, VCE=0.2V, c. IC=0, VCE=15V, and IC=0, VCE=15V, d. IC=6.5mA, VEC=8.5V and IC=14.8mA, VEC=0.2V

Prob. 4.35: RE = 753 ohms, RB = 31492 ohms

Prob. 4.38: ICQ = 1.79 mA, VCEQ = 6.06 V

Prob. D4.39: RE = 1 kohm, R1 = 13 kohms, R2 = 5.6 kohms (There are many possible answers!)

Prob. 4.45: ICQ=3.97mA, Rpi=655ohms, Av=-76.3, Avo=-152.7, Zin=544ohms, Zout=1kohm, Ai=-41.5, G=3167=35.0dB

Prob. 4.46: ICQ=39.7uA, Rpi=65.4kohms, Av=-76.5, Avo=-152.7, Zin=54.4kohm, Zout=100kohm, Ai=-41.5, G=3167=35.0dB

Prob. 4.49: a. Diagram: Don't forget RE between emitter and ground.

Prob. 4.49: b. Diagram: Here have RL between emitter and ground, while collector is grounded.

Prob. 4.49: c. Diagram: Here RE is between input and ground, emitter is at input, while base is grounded.

Prob. 4.51: ICQ=6.415mA, rpi=405ohms, Zin=4360ohms, Zout=12.1ohms, Av=0.988, Avo=0.996, Ai=8.61, G=8.51

Prob. 4.52: ICQ=64.15uA, rpi=40529ohms, Zin=436kohms, Zout=1212ohms, Av=0.988, Avo=0.996, Ai=8.61, G=8.51

Prob. 4.53: Av=-(beta(RC\\RL))/(rpi+(1+beta)RE), Zin=RB(rpi+(1+beta)RE)/(RB+rpi+(1+beta)RE)=RB\\(rpi+(1+beta)RE)

Prob. 4.54: RE=100ohms: ICQ=5.105mA, rpi=509.3ohms, Zin=10208ohms, Av=-4.71

Prob. 4.54: RE=0ohms: ICQ=5.30mA, rpi=490.6ohms, Zin=489.7ohms, Av=-101.9

Prob. 4.55: Zout=RC

Prob. 4.56: Av=(rpi/RB+(1+beta))(R1\\R2\\RE\\RL)/(rpi+(rpi/RB+(1+beta))(R1\\R2\\RE\\RL))

Prob. 4.56: Zin=(rpi+(rpi/RB+(1+beta))(R1\\R2\\RE\\RL))/(1+rpi/RB)

Prob. 4.57: ICQ=0.643mA, rpi=8087ohms, Av=0.980, Zin=370252ohms

Prob. 4.58: With vhum=0 both circuits are the same and Av=-beta(Rc/rpi) for both.

Prob. 4.58: With vin=0: a. Ahum=beta(Rc/rpi), b. Ahum=1 (Note: Here rpi is shorted so ib=0!)

Prob. 4.58: ICQ=1.43mA so rpi=1818ohms, thus a. Av=-258, Ahum=+258, b. Av=-258, Ahum=1, so circuit b is preferable since output is much less affected by hum!

Prob. 4.60: VCEQ=7.25V, ICEQ=3.96mA, rpi=656.6ohms, Zo=(rpi+R1(1+rpi/R2))/(1+beta+rpi/R2), Zo=126.2ohms

Prob. 4.61: a. Vo=5.1V, Zo=R\\rd, Zo=99ohms, b. Vo=4.9V, ICQ=4.9mA, rpi=1061ohms, Zo=RE\\((rpi+R\\rd)/(1+beta)), Zo=5.74ohms

Prob. 4.66: 22

Prob. 5.56: VGSQ = -1 V

Prob. 5.57: a. VDS > 2 V, b. VDS > 1 V

Prob. 5.58: a. IDQ = 3 mA, b. IDQ = 4 mA

Prob. 5.59: VGS < -3 V

Prob. 5.3: Note: Easiest to plot output characteristics then: a. Saturation, iD=2.25mA, b. Triode, iD=2.00mA, c. Cutoff, iD=0mA

Prob. 5.4: Saturation: vDS > 2V, Triode: 0 <= vDS <= 2V, Plot: iD=0.5e-03(vGS-3)**2

Prob. 5.18: a. vGS=sin((2000pi)t)+3V, b. Sketch-Be careful with scale!, c. Sketch, d. VDSQ=16V, VDSmin=11V, VDSmax=19V

Prob. 5.65: a. I1=IDSS=8mA, b. I2=IDSS=8mA, c. I3=IDSS=8mA, V1=7V, V2=0V, d. I5=IDSS=8mA, I4=4mA, V4=7V, V5=0.586V

Prob. 5.23: a. VGSQ=6V, IDQ=4mA, VDSQ=12V, b. VGSQ=3.77V, IDQ=6.23mA, VDSQ=7.54V

Prob. 5.66: a. IDQ=IDSS=8mA, VDSQ=7V, b. IDSQ=4.54mA, VDSQ=1.37V (Triode region!), c. IDQ=2.0mA, VDSQ=7.6V, d. IDQ=2mA, VDSQ=8V

Prob. 5.67: RS=396ohms for VGSQ=-1.58V so RDmax=4250ohms

Prob. 5.68: Have VGSQ=1V so need R2=1Mohm; also, RDmax=889ohms

Prob. 5.26: IDQ = 0.5625 mA, VDSQ = 4.375 V

Prob. 5.27: IDQ = 4.675 mA, VDSQ = 5.325 V

Prob. 5.40: VGSQ = 0 V, IDQ = 9 mA, VDSQ = 9.2 V, Av = -7.2, Zin = 3.3 Mohms, Zout = 1.2 kohm

Prob. 5.72: a. diagram, b. Av=(RD\\RL)gm, Zin=1/(gm+1/RS), Zout=RD, c. VGSQ=-1.22V, IDQ=1.22mA, VDSQ=10.49V, gm=3.12mS, d. Av=12.63, Zin=243ohms, Zout=6.8kohms, e. vo=0.895sin((2000pi)t)V, f. noninverting, low

Prob. 5.60: IDSS = 13 mA and Vto = -3 V

Prob. 5.73: rd = infinity, infinity, 250 ohms, 125 ohms, 83.3 ohms

Homework Assignment #24: PSpice will give you all the answers!!!

Homework Assignment #25: PSpice will give you all the answers!!!

Homework Assignment #26: PSpice will give you all the answers!!!

Prob. 8.40: a. RB=1.43Mohms, RC= 7kohms, b. fH=6.48MHz, Avs=-31.03 (Note: Av=-32.2)

Prob. 8.56: fH=(1/(2piRs'(Cgd+Cgs/(1+gmRL')))) where Rs'=(Rsig\\RG) and RL'=(rd\\Rs\\RL) and for given parameters fH = 56.60 MHz

Homework Assignment #29: 2N4401 Hybrid-Pi Estimated Model Parameters: gm=0.770S, hfe=250, rpi=325ohms, rmu=812500ohms, ro=40kohms, Cmu=6.5pF, Cpi=712pF (30pF from sheet makes no sense!)

Prob. 4.66: beta(min) = 22.3

Homework Assignment #30: Rc = 160 ohms, 5%, Rb = 3.9 kohms, 5%