h. chan; mohawk college1 numerical examples for ee213 part ii 1) determine the r’ e of a...

H. Chan; Mohawk College 1

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

2) For the CE amplifier, R1 = 33 k, R2 = 8.2 k, RC = 2.7 k, RE = 680 , DC = 100, and VCC = +15 V. Calculate: VB, IC, and VCE.

RIN(base) = DCRE = 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 VVE = 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


3) For the circuit in e.g. 2, what are Rin(tot), Rout, and Av if ac = 120,and RL = 1 k?Rin(base) = acr’e= 120 x 25 mV / IE = 120 x 25 / 2.97 = 1.01 kRin(tot) = R1//R2//Rin(base) = 33k//8.2k//1.01k = 875.4 Rout = RC//RL = 2.7k // 1k = 729.7 Av = -Rc / r’e = -729.7 / (25/2.97) = -86.7

4) If a 50 mV, 200 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 ; CE 1/(2fminXC) = 4.68 F


6) For the swamped CE amplifier with R1 = 10 k, R2 = 2 k, RC

= 4.7 k, RE1 = 150 , RE2 = 470 , VCC = 12 V, and DC = 110,ac = 125, calculate: Av, and Rin(tot)

RIN(base) = DC(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 mAr’e = 25 mV/ IE = 25 / 2.1 = 11.9 (which is << RE1)Av -RC / RE1 = -4.7k / 150 = -31.3

Rin(base) = ac(r’e+RE1) = 125 x (11.9 + 150) = 20.24 kRin(tot) = R1//R2//Rin(base) = 10k//2k//20.24k = 1.54 k

7) If a 0.5 V, 300 source is connected to the circuit in e.g. 6, andRL = 1 k, 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


8) For the emitter-follower amplifier, R1 = 27 k, R2 = 10 k, RE

= 1.2 k, RL = 2 k, DC = 120, VCC = 15 V. Find: IC, and VCE.

VB R2VCC/(R1+R2) = 10 x 15 /(27+10) = 4.05 VVE = 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 ac = 85 for e.g. 8, and a 2 V, 300 source is connected to theinput, determine: Rin(tot), Rout, Av, and Ai.r’e = 25 mV/ IE = 25 / 2.8 = 8.93 ; Re = RE//RL = 1.2k//2k = 750 Rin(base) = ac(r’e+Re) = 85 x (8.93+750) = 64.5 kRin(tot) = R1//R2//Rin(base) = 27k//10k//64.5k = 6.56 kRout (Rs/ac) //Re = (300/85) // 750 = 3.5 Av = Re/(r’e+Re) = 750/(8.93+750) = 0.988Ie = 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


10) For e.g. 8 & 9, if the transistor is replaced by a Darlington pair where DC = ac = 100, and RE = RL = 8 , find: Av, and Rin(tot).

RIN(base) = 2DCRE = 1002 x 8 = 80 k

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 mAr’e = 25 mV/IE = 25 / 377.5 = 66.2 m; Re = RE//RL = 8//8 = 4 Av = Re/(Re+r’e) = 0.984Rin(base) = 2

ac(r’e+Re) = 1002 x (0.0662 + 4) = 40.66 k

11) Find Rin, Av, and Ap for a CB amplifier with R1 = 68 k, R2

= 15 k, RC = 3.3 k, RE = 1.5 k, RL = 5 k, VCC = 24 V, DC

= 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 Av = Rc/r’e = (RC//RL)/r’e = (3.3k//5k) / 10.3 = 193Ap Av = 193


12) A 30 dB amplifier is cascaded with a 20 dB amplifier. Calculatetheir overall gain in dB. What would be the output voltage at the 2ndamplifier if the input voltage to the 1st amplifier is 20 mV?

AvT = 30 + 20 = 50 dB; the gain in non-dB value is 316.2Vout = AvTVin = 316.2 x 20 mV = 6.32 V

13) Two capacitively-coupled CE amplifier stages have the followingcomponents: R1 = R5 = 56 k, R2 = R6 = 12 k, R3 = R7 = 3.6 k,R4 = R8 = 1 k, RL = 2 k, VCC = 20 V, DC = ac = 140 for Q1 andQ2. Find AvTDC voltages and currents for both stages are the same:VB R2VCC /(R1+R2) = 3.53 V; VE = VB - 0.7 = 2.83 VIE = VE/R4 = 2.83 mA; r’e = 25 mV/IE = 8.83 Rc1 = R3//R5//R6//Rin(base2) = 3.6k//56k//12k//(140x8.83) = 841.8 Av1 = Rc1/r’e = 95.3; Av2 = Rc2/r’e = (R7//RL)/r’e = 145.6AvT = Av1Av2 = 13,876 or 82.85 dB


14) For a CE amplifier with R1 = 12 k, R2 = 5.6 k, RC = 1.2 k,RE = 560 , RL = 1.8 k, VCC = 12 V, and DC = ac = 110, draw thedc 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

VCEQ = VCQ - VEQ = 2.2 VIC(sat) = VCC/(RC+RE) = 6.82 mAVCE(cutoff) = VCC = 12 VRc = RC//RL = 720 Vce(cutoff) = VCEQ + ICQRc = 6.21 V

Ic(sat) = ICQ+VCEQ/Rc = 8.63 mA

Q

IC(mA)8.63

6.82

VCE

12 V6.21 V0


15) Determine the value of RE so that the Q-point in e.g. 14 isapproximately centred on the ac load line At mid-point of the ac load line, ICQ = Ic(sat)/2 = 8.63/2 = 4.32 mAand VCEQ = Vce(cutoff)/2 = 6.21/2 = 3.11 V

Since VCEQ = VCC - ICQ(RC+RE)

RE = (VCC - VCEQ - ICQRC) / ICQ = 858 16) Calculate a) the min. transistor power rating, b) the max.acoutput 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 isVCEQ, 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, = Pout / PDC = 3.36 / 66.8 = 0.05 or 5%


17) Determine the dc voltages at the bases and emitters of thematched complementary transistors Q1 and Q2 of the class ABamplifier with R1 = R2 = 120 , and VCC = 24 V. Also determineVCEQ for each transistor. Assume VD1 = VD2 = VBE = 0.7 VThe total current through R1, D1, D2, and R2 isIT = (VCC-VD1-VD2)/(R1+R2) = (24-0.7-0.7)/(120+120) = 94.2 mAVB1 = 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 VVCEQ1 = VCEQ2 = VCC / 2 = 24/2 = 12 V

18) If RL = 8 , ac = 120, and r’e = 5 for e.g. 17, find: Iout(pk)

and Rin.Iout(pk) = Ic(sat) = VCEQ / RL = 24/8 = 3 A

Rin = ac(r’e + RL) = 120 (5 + 8) = 1.56 k


19) Find the max. Pout, the dc input power, and for e.g. 17 & 18.

Max. Pout = 0.25VCCIc(sat) = 0.25 x 24 x 3 = 18 WPDC = VCCIc(sat) / = 24 x 3 / = 22.92 W

= Pout / PDC = 18 / 22.92 = 0.785 or 78.5%

20) If the circuit in e.g. 17 & 18 is replaced by a Darlington classAB push-pull amplifier with ac1 = ac2 = 55, find Rin.

Rin = tot (r’e + RL) = 552 (5 + 8) = 39.3 k

21) A tuned class C amplifier has a VCC = 12 V, VCE(sat) = 0.2 V,IC(sat) = 120 mA, Rc = 80 , 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

= Pout / (Pout + PD(avg) ) = 900 / (900 + 3.6) = 0.996 or 99.6 %


22) If VGS(off) = -5 V, IDSS = 10 mA, and RD = 1 kwhat is the min. value of VDD to put the JFET inconstant-current region of operation. VDD

RD

ID

Since VGS = 0, VGS(off) = -VP = -5, I.e. VP = 5 V,and ID = IDSS = 10 mASo, min. VDS = VP = 5 VMin. VDD = VD(min) + IDRD = 5 + 10 mA x 1k = 15 V

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, andIGSS = -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 G


25) Find VDS and VGS for the circuit when RG = 10 M,RD = 1.5 k, RS = 330 , VDD = 20 V, and ID = 4 mA.

+VDD

RD

RS

RG

VG = 0VS = IDRS = 4 mA x 330 = 1.32 V

VD = VDD - IDRD = 20 - 4 mA x 1.5k = 14 V

VDS = VD - VS = 14 - 1.32 = 12.68 V

VGS = VG - VS = 0 - 1,32 = -1.32 V

26) Find RS to self-bias the JFET circuit whereIDSS = 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

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


27) cont’d

VGS = VGS(off)/3.4 = - 7/3.4 = -2.06 V

So, RS = |VGS/ID| = 2.06 / 7.5 mA = 275 RD = (VDD - VD) / ID = (20 - 10) / 7.5 mA = 1.33 k28) Find ID and VGS for the JFET circuit with voltage-divider biasgiven R1 = 5.6 M, R2 = 1 M, RD = 4.7 k, RS = 2.7 k, 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.


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 ,VDD = 20 V, VGS(of) = -6 V, and IDSS = 15 mA.For zero-bias, ID = IDSS = 15 mAVDS = 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 voltage-divider bias where R1 = 200 k, R2 =40 k, RD = 180 , 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


33) Determine ID for an E-MOSFET in a circuit with drain-feedbackbias, RG = 10 M, RD = 2.2 k, VGS = 5 V, and VDD = 20 V.With drain-feedback bias, VDS = VGS = 5 VID = (VDD - VDS) / RD = (20 - 5) / 2.2 k = 6.82 mA

34) The values for a CS self-biased JFET amplifier are: RG = 10 MRD = 1.2 k, RS = 680 , 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 Vgm0 = 2IDSS / |VGS(off)| = 2 x 10 mA/6 = 3333 S

gm = gm0(1-VGS/VGS(off)) = 3333 S(1- 3.4/6) = 1444.3 S

Av = -gmRD = -1444.3 S x 1.2 k = -1.733

Vout = AvVin = -1.733 x 2 = -3.5 V


35) Determine the output voltage for e.g. 34 if the output is connectedto a load, RL = 1 kRd = RD // RL = 1.2k // 1k = 545.5Vout = -gmRdVin = - 1444.3 S x 545.5 x 2 = -1.58 V

36) With reference to the zero-biased CS D-MOSFET amplifier, RG

= 10 M, RD = 100 , VDD = 20 V, IDSS = 100 mA, gm = 100 mS,RL = 1 k, 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 , whatis Rin?RIN(gate) = |VGS/IDSS| = 4/10nA = 400 MRin = RG // RIN(gate) = 10 M // 400M = 9.76 M


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 andVout.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 mAVDS = 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


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

41) An amplifier has an output of 3 V across a load of 100 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 dBPout = V2

out / 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 F?

fcl = 1/(2RinC1) = 1/(2 x 3k x 1) = 53.1 Hz


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)) = 5A decade below fc, Xc = 10Rin, so = 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 F. Determine: fcl, and Av at fcl when Av(mid) = 20.fcl = 1/(2(RC+RL) C3) = 1/(2(5k+2k) 1) = 22.7 HzAt fcl, Av = 0.707 Av(mid) = 0.707 x 20 = 14.1

46) Find fcl of the bypass RC circuit where RE = 560 , CE = 5 F,R1 = 47 k, R2 = 10 k, Rs = 60 , r’e = 15 , and ac = 120.Rth = R1//R2//Rs = 47k//10k//60 = 59.6 Rin(emitter) = r’e+Rth / ac = 15 + 59.6 / 120 = 15.5 fcl = 1/(2(Rin(emitter) // RE) CE) = 1/(2(15.5//560) 5) = 2.1 kHz


47) For the circuit on the left, R1 = 33k, R2

= 6.8 k, RC = 2 k, RE = 680 , RL = 5 k,Cbc = 4 pF, Cbe = 8 pF, VCC = 12 V, ac = 100. Find: Cin(Miller) and Cout(Miller).

+VCC

C1

R1

R2

RC

RE

C3

RLC2

Vin

VoutVB = R2VCC/(R1+R2) = 2.05IE = VE/RE = (VB-0.7)/RE = 1.99 mAr’e = 25 mV/IE = 12.6 Av = -Rc/r’e = -(RC//RL)/r’e = -113.4Cin(Miller) = Cbc( |Av| +1) = 457.6 pF

Cout(Mileer) = Cbc( |Av| +1) / |Av| 4 pF

48) Determine the upper critical frequency of the input RC circuit fore.g. 47 if the input source resistance is Rs = 100 .

Rin(tot) = Rs//R1//R2//acr’e = 100//33k//6.8k//(100x12.6) = 91.1

fcu = 1/(2Rin(tot)Cin(tot)) = 1/(2 x 91.1 x 465.6 pF) = 3.75 MHzCin(tot) = Cin(Miller) + Cbe = 457.6 + 8 = 465.6 pF


49) Determine the fcu of the output RC circuit in e.g. 47.fcu = 1/(2RcCout(Miller)) 1/(2 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 MHzfT = |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. 47at a frequency of 50 MHz?|A’v| = fT / BW’ = 425.3 / 50 = 85.1

52) Determine the total low-frequency response of a JFETamplifier with RG = 10 M, RD = RL = 10 k, VGS = -8 V, C1 = C2

= 0.05 F,and IGSS = 40 nA.Rin(gate) = |VGS / IDSS| = 8 / 40 nA = 200 Mfcl(input) = 1/(2Rin(gate)//RG)C1) = 0.33 Hz

fcl(output) = 1/(2(RD+RL)C2) = 159.2


53) A self-biased CS JFET amplifier has RG = 10 M, RD = 5 k, RS

= 1k, RL = 10 k, Rs = 50 , 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) = -10Cin(Miller) = Cgd( |Av| + 1) = 2 (10 + 1 ) = 22 pFCin(tot) = Cgs + Cin(Miller) = 4 + 22 = 26 pFfcu(input) = 1/(2RsCin(tot)) = 1/(2 x 50 x 26 pF) = 122.4 MHz

Cout(Miller) = Cgd (( |Av| +1 ) / |Av| ) = 2((10+1)/10) = 2.2 pF

fcu(output) = 1/(2RdCout(Miller)) = 1/(2 x 5k//10k x 2.2 pf) = 21.7 MHz

54) An amplifier with fcl = 1 kHz and fcu = 700 kHz is cascadedwith 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


55) 3 amplifiers with the same fcl = 250 Hz and fcu = 50 kHz are cascaded together. Find the overall bandwidth.

125012' 311

kff ncucu

= 25.48 kHz

12

250

12'

311

n

clcl

ff = 490.6 Hz

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 stepvoltage 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 outputat 20 V increases by 0.2 V when its input increases by 6 V.

Line regulation = (Vout x 100)/(Vin x Vout) = 0.17 %/V


58) A voltage regulator has an output of 24 V at no load. The outputdrops to 23.5 V when a current of 30 mA is drawn. What is the loadregulation 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 %/mA59) Given: VZ = 15 V, RL = R = 100 , hFE = 100, and Vin = 18 V. Find IL, and IZT.

Q1

Vin

RRL

VZ

VL

IL = (VZ - VBE) / RL = (15 - 0.7) / 100 = 143 mAIR = (Vin - VZ) / R = (18 - 15) / 100 = 30 mA

IZT = IR - IB = IR - IL/hFE = 30 - 1.43 = 28.57 mA60) If the same component values are used for a basicshunt voltage regulator, except RS = 10 , 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 college1 numerical examples for ee213 part ii 1) determine the r’ e of a...

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