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NUMERICAL EXAMPLES FOR EE213 Part II
1) Determine the r’e of a transistor that is operating with IE = 5 mA.
r’e = 25 mV/ IE = 25 mV/ 5 mA = 5 W
2) For the CE amplifier, R1 = 33 kW, R2 = 8.2 kW, RC = 2.7 kW,
RE = 680 W, bDC = 100, and VCC = +15 V. Calculate: VB, IC, and VCE.
RIN(base) = bDCRE = 100 x 680 = 68 k (which is not >> R2)
VB = (R2//RIN(base))VCC /(R1+R2//RIN(base)) = (8.2//68)x15/(33+8.2//68)
= 7.32 x 15 / (33 + 7.32) = 2.72 V
VE = VB - VBE = 2.72 - 0.7 = 2.02 V
IC  IE = VE/RE = 2.02 / 680 = 2.97 mA
VC = VCC - ICRC = 15 - 2.97 mA x 2.7 k = 6.98 V
VCE = VC - VE = 6.98 - 2.02 = 4.96 V
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3) For the circuit in e.g. 2, what are Rin(tot), Rout, and Av if bac = 120,
and RL = 1 kW?
Rin(base) = bacr’e= 120 x 25 mV / IE = 120 x 25 / 2.97 = 1.01 kW
Rin(tot) = R1//R2//Rin(base) = 33k//8.2k//1.01k = 875.4 W
Rout = RC//RL = 2.7k // 1k = 729.7 W
Av = -Rc / r’e = -729.7 / (25/2.97) = -86.7
4) If a 50 mV, 200 W source is connected to the circuit in e.g. 2 & 3,
determine: Vb, A’v, and Vout
Vb = Rin(tot)Vs/(Rs+Rin(tot)) = 875.4 x 50 mV/(200+875.4) = 40.7 mV
A’v = Rin(tot)Av/(Rs+Rin(tot)) = 875.4 x (-86.7)/(200+875.4) = -70.6
Vout = A’vVin or AvVb = -70.6 x 50 mV = -3.53 V
5) What is the minimum value of CE in e.g. 2 if the amplifier
operates over a frequency range from 500 Hz to 100 kHz.
XC RE/10 = 68 W;  CE 1/(2pfminXC) = 4.68 mF
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6) For the swamped CE amplifier with R1 = 10 kW, R2 = 2 kW, RC
= 4.7 kW, RE1 = 150 W, RE2 = 470 W, VCC = 12 V, and bDC = 110,
bac = 125, calculate: Av, and Rin(tot)
RIN(base) = bDC(RE1+RE2) = 110 x (150+470) = 68.2 k >> R2
VB  R2VCC/(R1+R2) = 2 x 12 / (10 + 2) = 2 V
VE = VB - 0.7 = 1.3 V; IE = VE / (RE1+RE2) = 2.1 mA
r’e = 25 mV/ IE = 25 / 2.1 = 11.9 W (which is << RE1)
Av  -RC / RE1 = -4.7k / 150 = -31.3
Rin(base) = bac(r’e+RE1) = 125 x (11.9 + 150) = 20.24 kW
Rin(tot) = R1//R2//Rin(base) = 10k//2k//20.24k = 1.54 kW
7) If a 0.5 V, 300 W source is connected to the circuit in e.g. 6, and
RL = 1 kW, find A’v.
Av = -(RC//RL) / RE1 = -4.7k//1k / 150 = -5.5
A’v = Rin(tot)Av/(Rin(tot)+Rs) = 1.54k x (-5.5)/(1.54k+300) = -4.6
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8) For the emitter-follower amplifier, R1 = 27 kW, R2 = 10 kW, RE
= 1.2 kW, RL = 2 kW, bDC = 120, VCC = 15 V. Find: IC, and VCE.
VB R2VCC/(R1+R2) = 10 x 15 /(27+10) = 4.05 V
VE = VB - 0.7 = 4.05 - 0.7 = 3.35 V; IC  VE/RE = 3.35/1.2k = 2.8 mA
VCE = VC - VE = VCC - VE = 15 - 3.35 = 11.65 V
9) If bac = 85 for e.g. 8, and a 2 V, 300 W source is connected to the
input, determine: Rin(tot), Rout, Av, and Ai.
r’e = 25 mV/ IE = 25 / 2.8 = 8.93 W; Re = RE//RL = 1.2k//2k = 750 W
Rin(base) = bac(r’e+Re) = 85 x (8.93+750) = 64.5 kW
Rin(tot) = R1//R2//Rin(base) = 27k//10k//64.5k = 6.56 kW
Rout  (Rs/bac) //Re = (300/85) // 750 = 3.5 W
Av = Re/(r’e+Re) = 750/(8.93+750) = 0.988
Ie = Vout/Re = AvVb/Re = AvRin(tot)Vin/((Rs+Rin(tot))Re) = 2.52 mA
Iin = Vin/Rin(tot) = 2/6.56k = 0.3 mA;  Ai = Ie/Iin = 2.52 / 0.3 = 8.4
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10) For e.g. 8 & 9, if the transistor is replaced by a Darlington pair
where bDC = bac = 100, and RE = RL = 8 W, find: Av, and Rin(tot).
RIN(base) = b2DCRE = 1002 x 8 = 80 kW
VB = (R2//RIN(base)VCC) / (R1+R2//RIN(base)) = 3.72 V
VE2 = VB-2VBE = 3.72 - 2x0.7 = 3.02V; IE = VE2/RE = 377.5 mA
r’e = 25 mV/IE = 25 / 377.5 = 66.2 mW; Re = RE//RL = 8//8 = 4 W
Av = Re/(Re+r’e) = 0.984
Rin(base) = b2ac(r’e+ Re) = 1002 x (0.0662 + 4) = 40.66 kW
11) Find Rin, Av, and Ap for a CB amplifier with R1 = 68 kW, R2
= 15 kW, RC = 3.3 kW, RE = 1.5 kW, RL = 5 kW, VCC = 24 V, bDC
= 180.
VB  R2VCC / (R1+R2) = 4.34 V; VE = VB - 0.7 = 3.64 V
IE = VE/RE = 2.43 mA; Rin  r’e = 25 mV/IE = 10.3 W
Av = Rc/r’e = (RC//RL)/r’e = (3.3k//5k) / 10.3 = 193
Ap  Av = 193
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12) A 30 dB amplifier is cascaded with a 20 dB amplifier. Calculate
their overall gain in dB. What would be the output voltage at the 2nd
amplifier if the input voltage to the 1st amplifier is 20 mV?
AvT = 30 + 20 = 50 dB; the gain in non-dB value is 316.2
Vout = AvTVin = 316.2 x 20 mV = 6.32 V
13) Two capacitively-coupled CE amplifier stages have the following
components: R1 = R5 = 56 kW, R2 = R6 = 12 kW, R3 = R7 = 3.6 kW,
R4 = R8 = 1 kW, RL = 2 kW, VCC = 20 V, bDC = bac = 140 for Q1 and
Q2. Find AvT
DC voltages and currents for both stages are the same:
VB  R2VCC /(R1+R2) = 3.53 V; VE = VB - 0.7 = 2.83 V
IE = VE/R4 = 2.83 mA; r’e = 25 mV/IE = 8.83 W
Rc1 = R3//R5//R6//Rin(base2) = 3.6k//56k//12k//(140x8.83) = 841.8 W
Av1 = Rc1/r’e = 95.3; Av2 = Rc2/r’e = (R7//RL)/r’e = 145.6
AvT = Av1Av2 = 13,876 or 82.85 dB
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14) For a CE amplifier with R1 = 12 kW, R2 = 5.6 kW, RC = 1.2 kW,
RE = 560 W, RL = 1.8 kW, VCC = 12 V, and bDC = bac = 110, draw the
dc and ac load lines. Obtain all the important parameters for the lines.
VBQ  R2VCC/(R1+R2) = 3.82 V; VEQ = VBQ - 0.7 = 3.12 V
ICQ  IEQ = VEQ/RE = 3.12 / 560 = 5.57 mA
VCQ = VCC - ICQRC = 12 - 5.57 mA x 1.2 k = 5.32 V
IC(mA)
VCEQ = VCQ - VEQ = 2.2 V
IC(sat) = VCC/(RC+RE) = 6.82 mA 8.63
VCE(cutoff) = VCC = 12 V
6.82
Q
Rc = RC//RL = 720 W
Vce(cutoff) = VCEQ + ICQRc = 6.21 V
Ic(sat) = ICQ+VCEQ/Rc = 8.63 mA
0
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6.21 V
VCE
12 V
7
15) Determine the value of RE so that the Q-point in e.g. 14 is
approximately centred on the ac load line
At mid-point of the ac load line, ICQ = Ic(sat)/2 = 8.63/2 = 4.32 mA
and VCEQ = Vce(cutoff)/2 = 6.21/2 = 3.11 V
Since VCEQ = VCC - ICQ(RC+RE)
 RE = (VCC - VCEQ - ICQRC) / ICQ = 858 W
16) Calculate a) the min. transistor power rating, b) the max.ac
output power without distortion, c) the efficiency for e.g. 14.
PD(min) = PDQ = VCEQICQ = 2.2 x 5.57 mA = 12.3 mW
Since the Q-point is closer to saturation, the max. voltage swing is
VCEQ, and the corresponding max. current swing is VCEQ/Rc
Max. Pout = Vout(rms)Iout(rms) = 0.707VCEQ(0.707VCEQ/Rc) = 3.36 mW
PDC = VCCICQ = 12 x 5.57 mA = 66.8 mA
So, h = Pout / PDC = 3.36 / 66.8 = 0.05 or 5%
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17) Determine the dc voltages at the bases and emitters of the
matched complementary transistors Q1 and Q2 of the class AB
amplifier with R1 = R2 = 120 W, and VCC = 24 V. Also determine
VCEQ for each transistor. Assume VD1 = VD2 = VBE = 0.7 V
The total current through R1, D1, D2, and R2 is
IT = (VCC-VD1-VD2)/(R1+R2) = (24-0.7-0.7)/(120+120) = 94.2 mA
VB1 = VCC - ITR1 = 24 - 94.2 mA x 120 = 12.7 V
VB2 = VB1 - VD1 - VD2 = 12.7 - 0.7 - 0.7 = 11.3 V
VE1 = VE2 = VB1 - VBE = 12.7 - 0.7 = 12 V
VCEQ1 = VCEQ2 = VCC / 2 = 24/2 = 12 V
18) If RL = 8 W, bac = 120, and r’e = 5 W for e.g. 17, find: Iout(pk)
and Rin.
Iout(pk) = Ic(sat) = VCEQ / RL = 24/8 = 3 A
Rin = bac(r’e + RL) = 120 (5 + 8) = 1.56 kW
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19) Find the max. Pout, the dc input power, and h for e.g. 17 & 18.
Max. Pout = 0.25VCCIc(sat) = 0.25 x 24 x 3 = 18 W
PDC = VCCIc(sat) / p = 24 x 3 / p = 22.92 W
h = Pout / PDC = 18 / 22.92 = 0.785 or 78.5%
20) If the circuit in e.g. 17 & 18 is replaced by a Darlington class
AB push-pull amplifier with bac1 = bac2 = 55, find Rin.
Rin = btot (r’e + RL) = 552 (5 + 8) = 39.3 kW
21) A tuned class C amplifier has a VCC = 12 V, VCE(sat) = 0.2 V,
IC(sat) = 120 mA, Rc = 80 W, and a duty cycle of 15%, determine:
PD(avg), and efficiency assuming max. output power operation.
PD(avg) = (tON/T)VCE(sat)IC(sat) = 0.15 x 0.2 x 120 mA = 3.6 mW
Max. Pout = 0.5V2CC / Rc = 0.5 x 122 / 80 = 900 mW
h = Pout / (Pout + PD(avg) ) = 900 / (900 + 3.6) = 0.996 or 99.6 %
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22) If VGS(off) = -5 V, IDSS = 10 mA, and RD = 1 kW,
what is the min. value of VDD to put the JFET in
constant-current region of operation.
Since VGS = 0, VGS(off) = -VP = -5, I.e. VP = 5 V,
and ID = IDSS = 10 mA
So, min. VDS = VP = 5 V
Min. VDD = VD(min) + IDRD = 5 + 10 mA x 1k = 15 V
RD
ID
VDD
23) What is the value of gm0 for the JFET in e.g. 22?
gm0 = 2IDSS / |VGS(off)| = 2 x 10 mA / 5 = 4 mS
24) Determine ID, gm, and RIN for e.g. 22 when VGS = -2 V, and
IGSS = -0.2 nA.
ID = IDSS (1 - VGS/VGS(off))2 = 10 (1 - (-2/-5))2 = 3.6 mA
gm = gm0(1-VGS/VGS(off)) = 4 mS (1 - (-2/-5)) = 2.4 mS
RIN = |VGS/IGSS| = 2/0.2 nA = 10 GW
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25) Find VDS and VGS for the circuit when RG = 10 MW,
+VDD
RD = 1.5 kW, RS = 330 W, VDD = 20 V, and ID = 4 mA.
RD
VS = IDRS = 4 mA x 330 = 1.32 V
VG = 0
VD = VDD - IDRD = 20 - 4 mA x 1.5k = 14 V
VDS = VD - VS = 14 - 1.32 = 12.68 V
R
G
VGS = VG - VS = 0 - 1,32 = -1.32 V
RS
26) Find RS to self-bias the JFET circuit where
IDSS = 15 mA, VGS(off) = -7 V, and VGS is to be -2.5 V.
ID = IDSS(1-VGS/VGS(off))2 = 15 mA(1-(-2.5/-7))2 = 6.2 mA
RS = |VGS/ID| = 2.5 / 6.2 mA = 403 W
27) Determine RD and RS for midpoint bias where IDSS = 15 mA,
VDD = 20 V, and VGS(off) = -7 V.
For midpoint bias, VD = VDD /2 = 10 V, and ID = IDSS /2 = 7.5 mA
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27) cont’d
VGS = VGS(off)/3.4 = - 7/3.4 = -2.06 V
So, RS = |VGS/ID| = 2.06 / 7.5 mA = 275 W
RD = (VDD - VD) / ID = (20 - 10) / 7.5 mA = 1.33 kW
28) Find ID and VGS for the JFET circuit with voltage-divider bias
given R1 = 5.6 MW, R2 = 1 MW, RD = 4.7 kW, RS = 2.7 kW, VDD
= 15 V, and VD = 8 V.
ID = (VDD - VD) / RD = (15 - 8) / 2.7k = 2.6 mA
VS = IDRS = 2.6 mA x 2.7 k = 7 V
VG = R2VDD/(R!+R2) = 1 x 15/(5.6 + 1) = 2.27 V
VGS = VG - VS = 2.27 - 7 = - 4.63 V
29) A D-MOSFET has VGS(off) = -6 V, IDSS = 14 mA and VGS = 2 V.
Find ID. What mode is it operating in?
ID = IDSS(1-VGS/VGS(off))2 = 14(1- 2/(-6))2 = 24.9 mA
Enhancement mode.
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30) For an E-MOSFET, ID = 400 mA at VGS = 8 V, and VGS(th)
1 V. Determine ID for VGS = 4 V.
K = ID / (VGS - VGS(th))2 = 400 mA /(8 - 1)2 = 8.16 mA/V2
ID = K(VGS - VGS(th))2 = 8.16 mA (4 - 1)2 = 73.4 mA
31) Determine VDS for a D-MOSFET with zero-bias, RD = 910 W,
VDD = 20 V, VGS(of) = -6 V, and IDSS = 15 mA.
For zero-bias, ID = IDSS = 15 mA
VDS = VDD - IDRD = 20 - 15 mA x 910 = 6.35 V
32) If the E-MOSFET from e.g. 30 is used in a circuit with voltagedivider bias where R1 = 200 kW, R2 = 40 kW, RD = 180 W, and VDD
= 24 V, determine VGS and VDS.
VGS = R2VDD / (R1+R2) = 40 x 24 / (200 + 40) = 4 V
As found in e.g 30, when VGS = 4 V, ID = 73.4 mA
Therefore, VDS = VDD - IDRD = 24 - 73.4 mA x 180 = 10.8 V
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33) Determine ID for an E-MOSFET in a circuit with drain-feedback
bias, RG = 10 MW, RD = 2.2 kW, VGS = 5 V, and VDD = 20 V.
With drain-feedback bias, VDS = VGS = 5 V
ID = (VDD - VDS) / RD = (20 - 5) / 2.2 k = 6.82 mA
34) The values for a CS self-biased JFET amplifier are: RG = 10 MW,
RD = 1.2 kW, RS = 680 W, IDSS = 10 mA, VGS(off) = -6 V, VDD = 15 V,
and Vin = 2 V. Assuming midpoint biasing, find: VDS and Vout.
For midpoint biasing, ID = IDSS / 2 = 10 /2 = 5 mA
VDS = VDD - ID(RD+RS) = 15 - 5 mA(1.2k+680) = 5.6 V
VGS = - IDRS = - 5 mA x 680 = - 3.4 V
gm0 = 2IDSS / |VGS(off)| = 2 x 10 mA/6 = 3333 mS
gm = gm0(1-VGS/VGS(off)) = 3333 mS(1- 3.4/6) = 1444.3 mS
Av = -gmRD = -1444.3 mS x 1.2 k = -1.733
Vout = AvVin = -1.733 x 2 = -3.5 V
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35) Determine the output voltage for e.g. 34 if the output is connected
to a load, RL = 1 kW
Rd = RD // RL = 1.2k // 1k = 545.5W
Vout = -gmRdVin = - 1444.3 mS x 545.5 x 2 = -1.58 V
36) With reference to the zero-biased CS D-MOSFET amplifier, RG
= 10 MW, RD = 100 W, VDD = 20 V, IDSS = 100 mA, gm = 100 mS,
RL = 1 kW, and Vin = 600 mV. Determine: VD and Vout.
Since it is zero-biased operation, ID = IDSS = 100 mA
VD = VDD - IDRD = 20 - 100 mA x 100 = 10 V
Vout = -gmRdVin = -100 mS x 100//1k x 600 mV = -5.45 V
37) For the amplifier in e.g. 34, if IGSS = -10 nA at VGS = -4 V , what
is Rin?
RIN(gate) = |VGS/IDSS| = 4/10nA = 400 MW
Rin = RG // RIN(gate) = 10 M // 400M = 9.76 MW
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38) A voltage-divider biased CS E-MOSFET amplifier has R1 = 45 k,
R2 = 10 k, RD = 2.7 k, VDD = 12 V, RL = 10 k, ID(on) = 150 mA at VGS
= 3.5 V, VGS(th) = 2 V, gm = 20 mS, and Vin = 100 mV. Find: VDS and
Vout.
K = ID(on) / (VGS - VGS(th) )2 = 150 mA / (3.5 - 2)2 = 66.7 mA/V2
VGS = R2VDD/(R1+R2) = 10 x 12 / (45 + 10) = 2.18 V
ID = K(VGS - VGS(th))2 = 66.7 (2.18 - 2)2 = 2.2 mA
VDS = VDD - IDRD = 12 - 2.2 mA x 2.7k = 6.05 V
Vout = -gmRdVin = -20 mS x 2.7k//10k x 100 mV = -4.25 V
39) A JFET is connected as a CD amplifier with self bias, RG = 10 M,
RS = 5 k, RL = 10 k, VDD = 15 V, Vin = 5 V, and gm = 2 mS. Find:
Av and Vout.
Av = gmRs / (1+gmRs) = 2 mS x 5k//10k (1+2 mS x 5k//10k) = 0.87
Vout = AvVin = 0.87 x 5 = 4.35 V
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40) A JFET connected as a zero-biased CG amplifier has RD = 5 k, RS
= 2.7 k, VDD = 15 V, RL = 5 k, Vin = 0.2 V and gm = 3 mS. Find: Vout
and Rin.
Vout = gmRdVin = 3 mS x 5k//5k x 0.2 = 1.5 V
Rin = Rin(source) // RS = (1/gm) // 2.7k = 297 W
41) An amplifier has an output of 3 V across a load of 100 W when the
input is 0.2 V. Express the gain in dB, and the output in dBm.
Av = 20log(Vout/Vin) = 20 x log (3/0.2) = 23.5 dB
Pout = V2out / RL = 32 / 100 = 90 mW = 10log(90/1) = 19.5 dBm
42) An input of 0.1 V is applied to a 25 dB amplifier. Find Vout.
Av = 25 dB = antilog (25/20) = 17.78
Vout = AvVin = 17.78 x 0.1 = 1.78 V
43) What is fcl for an amplifier with Rin = 3 k, and C1 = 1 mF?
fcl = 1/(2pRinC1) = 1/(2p x 3k x 1m) = 53.1 Hz
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44) Av(mid) of an amplifier is 50 and the input RC circuit has an fcl
= 500 Hz. Determine the voltage gain and phase shift at 50 Hz.
Since 50 Hz is one decade below fcl, Av is 20 dB less than Av(mid)
i.e. Av = Av(mid) (dB) - 20 dB or Av(mid) / (antilog (20/20)) = 5
A decade below fc, Xc = 10Rin, so q = tan-1(Xc/Rin) = tan -1(10) = 84.3o
45) The output RC circuit of a CE amplifier consists of RC = 5 k, RL
= 2 k, and C3 = 1 mF. Determine: fcl, and Av at fcl when Av(mid) = 20.
fcl = 1/(2p(RC+RL) C3) = 1/(2p(5k+2k) 1m) = 22.7 Hz
At fcl, Av = 0.707 Av(mid) = 0.707 x 20 = 14.1
46) Find fcl of the bypass RC circuit where RE = 560 W, CE = 5 mF,
R1 = 47 k, R2 = 10 k, Rs = 60 W, r’e = 15 W, and bac = 120.
Rth = R1//R2//Rs = 47k//10k//60 = 59.6 W
Rin(emitter) = r’e+Rth / bac = 15 + 59.6 / 120 = 15.5 W
fcl = 1/(2p(Rin(emitter) // RE) CE) = 1/(2p (15.5//560) 5m) = 2.1 kHz
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47) For the circuit on the left, R1 = 33k, R2
= 6.8 k, RC = 2 k, RE = 680 W, RL = 5 k,
Cbc = 4 pF, Cbe = 8 pF, VCC = 12 V, bac = 100.
Find: Cin(Miller) and Cout(Miller).
C1
VB = R2VCC/(R1+R2) = 2.05
Vin
IE = VE/RE = (VB-0.7)/RE = 1.99 mA
r’e = 25 mV/IE = 12.6 W
Av = -Rc/r’e = -(RC//RL)/r’e = -113.4
Cin(Miller) = Cbc( |Av| +1) = 457.6 pF
+VCC
R1
RC C3
Vout
R2
RE
C2
RL
Cout(Mileer) = Cbc( |Av| +1) / |Av|  4 pF
48) Determine the upper critical frequency of the input RC circuit for
e.g. 47 if the input source resistance is Rs = 100 W .
Rin(tot) = Rs//R1//R2//bacr’e = 100//33k//6.8k//(100x12.6) = 91.1W
Cin(tot) = Cin(Miller) + Cbe = 457.6 + 8 = 465.6 pF
fcu = 1/(2pRin(tot)Cin(tot)) = 1/(2p x 91.1 x 465.6 pF) = 3.75 MHz
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49) Determine the fcu of the output RC circuit in e.g. 47.
fcu = 1/(2pRcCout(Miller)) 1/(2p x 2k//5k x 4 pF) = 27.9 MHz
50) What are the bandwidth and gain bandwidth product for the
amplifier circuit in e.g. 47 assuming fcl << fcu?
BW  fcu = 3.75 MHz
fT = |Av(mid)| x BW = 113.4 x 3.75 MHz = 425.3 MHz
51) What would be the approximate gain of the amplifier in e.g. 47
at a frequency of 50 MHz?
|A’v| = fT / BW’ = 425.3 / 50 = 85.1
52) Determine the total low-frequency response of a JFET
amplifier with RG = 10 MW, RD = RL = 10 k, VGS = -8 V, C1 = C2
= 0.05 mF,and IGSS = 40 nA.
Rin(gate) = |VGS / IDSS| = 8 / 40 nA = 200 MW
fcl(input) = 1/(2p(Rin(gate)//RG)C1) = 0.33 Hz
fcl(output) = 1/(2p(RD+RL)C2) = 159.2
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53) A self-biased CS JFET amplifier has RG = 10 MW, RD = 5 k, RS
= 1k, RL = 10 k, Rs = 50 W, Ciss = 6 pF, Crss = 2 pF, gm = 3 mS.
Find the upper critical frequency for the input and output RC circuits.
Cgd = Crss = 2 pF; Cgs = Ciss - Crss = 6 - 2 = 4 pF
Av = gm(RD//RL) = -3 mS x (5k//10k) = -10
Cin(Miller) = Cgd( |Av| + 1) = 2 (10 + 1 ) = 22 pF
Cin(tot) = Cgs + Cin(Miller) = 4 + 22 = 26 pF
fcu(input) = 1/(2pRsCin(tot)) = 1/(2p x 50 x 26 pF) = 122.4 MHz
Cout(Miller) = Cgd (( |Av| +1 ) / |Av| ) = 2((10+1)/10) = 2.2 pF
fcu(output) = 1/(2pRdCout(Miller)) = 1/(2p x 5k//10k x 2.2 pf) = 21.7 MHz
54) An amplifier with fcl = 1 kHz and fcu = 700 kHz is cascaded
with another amplifier with fcl = 80 Hz and fcu = 250 kHz. What
is the overall bandwidth?
BW = f’cu - f’cl = 250 kHz - 1 kHz = 249 kHz
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55) 3 amplifiers with the same fcl = 250 Hz and fcu = 50 kHz are
cascaded together. Find the overall bandwidth.
1
1
f 'cu  f cu 2  1  50k 2 3  1 = 25.48 kHz
f 'cl 
f cl
1
n

250
1
= 490.6 Hz
2 n 1
2 3 1
Overall BW = f’cu - f’cl = 25.48 - 490.6 = 24.99 kHz
56) The rise and fall times for an amplifier in response to a step
voltage input are 10 ns and 1.2 ms respectively. Find fcl and fcu.
fcl = 0.35 / tf = 0.35 / 1.2 ms = 291.7 Hz
fcu = 0.35 / tr = 0.35 / 10 ns = 35 MHz
57) Calculate the line regulation in %/V of a regulator whose output
at 20 V increases by 0.2 V when its input increases by 6 V.
Line regulation = (DVout x 100)/(DVin x Vout) = 0.17 %/V
H. Chan; Mohawk College
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58) A voltage regulator has an output of 24 V at no load. The output
drops to 23.5 V when a current of 30 mA is drawn. What is the load
regulation as a percentage and as %/mA?
Load regulation = (VNL - VFL) x 100 / VFL = 2.13 %
Load regulation = (VNL - VFL) x 100 / (VFL x IFL) = 0.071 %/mA
Q1
59) Given: VZ = 15 V, RL = R = 100 W,
hFE = 100, and Vin = 18 V. Find IL, and IZT.
Vin
R
IL = (VZ - VBE) / RL = (15 - 0.7) / 100 = 143 mA
RL VL
IR = (Vin - VZ) / R = (18 - 15) / 100 = 30 mA
VZ
IZT = IR - IB = IR - IL/hFE = 30 - 1.43 = 28.57 mA
60) If the same component values are used for a basic
shunt voltage regulator, except RS = 10 W, find IL, and IE.
IL = (VZ + VBE) / RL = (15 + 0.7) / 100 = 157 mA
IRs = [Vin - (VZ +VBE)] / RS = [18 - 15.7] / 10 = 230 mA
So, IE = Irs - IL = 230 - 157 = 73 mA
H. Chan; Mohawk College
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