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For common -emitter amplifier,

By Jessica Shaw,2014-05-06 14:46
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For common -emitter amplifier,

    UNION OF MYANMAR

    MINISTRY OF SCIENCE AND TECHNOLOGY

    DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION

    GOVERNMENT TECHNOLOGICAL COLLEGES AND INSTITUTES

    A.G.T.I Year 2 (2006)

    EcE-02011 Analog Electronics

    Attempt any SIX questions.

    Take assumption if necessary.

    1. Draw the circuit diagram of a darlington used as buffer between a common emitter

    amplifier and a low resistive load. The circuit parameters of common-emitter amplifier

    are V = 12V, R = 1kΩ and r′ = 5Ω. For the darlington emitter-follower: R1 = 10k Ω, ccce

    R2 = 22k Ω, RE = 100 Ω and βDC = βac = 200 for each transistor.

    (i) Determine the voltage gain of the common-emitter amplifier.

    (ii) Determine the voltage gain of the darlington emitter follower

    (iii) Determine the overall gain and compare the gain of the common-emitter

    amplifier driving the speaker directly without the darlington emitter follower.

     2. (i) Explain the working characteristics of Enhancement MOSFET (E-MOSFET) briefly.

    (ii) Draw the common E-MOSFET voltage-divider bias. Circuit parameters are R = 1

    100kΩ, R = 15k Ω, R = 470 Ω, V = 24V. The data sheet for this particular MOSFET 2DDD

    gives minimum value of I= 500mA at V= 10 V, V= 1V. D(on) GS GS(th)

    (iii) Sketch the common-source amplifier using E-MOSFET. Find V, IV and ac GS D, DS

    output voltage. I= 5mA at V= 10 V, V= 4V and g = 5.5mS, V = 50mV, D(on) GS GS(th) min

    V = 15V, R =47k Ω, R = 33k Ω, R = 3.3k Ω and R =33k Ω. DD12DL

    3. (i) Draw the op-amp integrator and derive the output voltage equation.

    (ii) Determine the time constant required in an integrator. Assume that voltage V across c

    the capacitor is zero at t = 0. The step input voltage of V = -1V is applied at t= 0. i

    Determine the time constant required such that output reaches +10V at t = 1ms.

    4. Calculate the overall small signal voltage gain of the op-amp circuit. Assume JFET

    -1transistor parameters of I= 0.4mA, V = 4V and λ = 0.01V and BJT transistor DSS p

    parameters of β(npn) = 150, V(npn) = 120V, β(pnp) = 75, V(pnp) = 80V. AA

     5.(i) Calculate the output resistance of the 741 op-amp. Assume the output current is Ir20 =

    2mA, VA = 50V, β= 200 and β = 50. n p

     (ii) Determine the dominant-pole frequency of the 741 op-amp using the appropriate results

    from part (i).

     6.(i) Draw the circuit diagram and equivalent circuit of one pole low pass filter.

     (ii) Design a one pole low pass filter has frequency gain is -1 and the control frequency is

    1kHz.

7. Draw the circuit diagram of Monostable Multivibrator and explain its operation briefly.

    ……………………………..END………………………………..

    (1) For common -emitter amplifier, V =12V CC

    R = 1kΩ C

    r' = 5Ω e

    B = B =200 DC ac

    For the darlington emitter -follower R =10kΩ 1

    R =22kΩ 2

    R = 100Ω E

    B = B =200 DC ac

    (a) A=? (2) A=? (3) A' (with EF) =? A' (without EF) =? C(CE) V(EF) VV

    VCC

    RCR1

    Vin

    R2RLRE

    DC analysis for darlington pair,

    VCCR1

    R2RE

     2 R= β?R IN(Base)DCE

    2 = 200? 100

     =4MΩ

    R= R// RIN(tot)2in(base)

    ?RR2in(base)= RR2in(base)

    22k?4M= 22k4M

     =21.88kΩ

    RIN(tot) ?VCCVBRR1IN(tot)

    21.88k ?1210k21.88k

     = 8.236V

    8.2360.7VVBBE 75.36mAIE100RE

    12VTr'= 159.2357Ωe75.36mIE

    AC analysis for darlington pair

    R2RE

    100k?10kR= 99Ωe100k10k

    2R= β(r'+ R) in(base)acee

    2 =200(159.2357+99)=10.33MΩ R= R//R//R in(tot)12in(base)

    1 111

    RRR12in(base)

    1 6.87kΩ11110k22k10.33MAC analysis for common emitter amplifier

    RC

    r`e

     R=R// RcCin(tot)

    ?1k?6.87kRRCin(tot)?= 872.94Ω1k6.87kRRCin(tot)

    R872.94e A174.588()VEF'5re

    R100e(b) . A= 0.386V(EF);;;;R'159.2357100ere

    ( c) A'= AA=174.5880.386=67.39 ??vV(CE)V(CF)(with darlington emitter follower)

    With out emitter follower, AC analysis

    RC

    r`e

    1k?10k R909.09C1k10k

    RCA'==181.82 v're

    ( A' ( without emitter follower) A' ( with emitter follower) vv

(2) (b)

    RDR1

    Vin

    R2

     V=1V GS(th)

     I=500 mA at V=10V D(on)GS

    2 I=k(V-V) DGSGS(th)

    I500m2D k6.1728mA/V22;;;;VV107GSGSth()

    R15k2 V?V?243.13V2DDRR15k100k121

     V=V-IR=3.13V GSGDS

(c)

    RDR1

    VinRL

    R2

     V=? I=? V=? Gm =5.5mS GSDDS

    R33k2V?V?156.1875VGDDRR33k47k 12

    AC analysis

    RR21RRLE

    3.3k?33k R3kd3.3k33k

    ?A=gR=5.5m 3k=16.5 Vmd

    Vout 16.5Vin

    ?V=16.550m=0.825V out

    V= 0.825 V DS

    I=0.825/R=0.25mA DD

    V=6.1875V GS

3( a) The op-amp integrator

    1 XCSC

    t1? since V=0 at t=o VVVdtc?0()CIt?RC120

    t1? VVdt?0()It?RC120

     (b) time constant , RC=? 12

     V= -1 V I

    V=0 at t=0 C

    V=10V at t=1ms 0

    t1? VVdt?0()It?RC120

    t1? 10(1)dt?RC120

    1t?RC= - ,;t12010

    1tt0 = ,;1010At t=1ms, Time constant RC= 0.1ms 12

(4) A?v

    I=0.4mA DSS

    I=0.2mA D

    -1V=4V, λ=0.01V, β=150 , β=75 ,V=120V ,V=80V P(npn)(pnp)A(npn)A(pnp)

    I=0.54mA C8

    I=0.675mA C9

    R=50kΩ , R= 1kΩ 45

AAAAVd23

    A?13

    (;;ββ1Rn4,;AR//R//R2)09010i3RRr(1β)R??i24π9S

    ;;,; Rr(1β)R//r(1β)Ri2π8n4π9S

    1;;Agg(R//R//R)dmm20204i22

    VA(pnp)RR0409ID

    1R02λID

    2g=g= IIm1m2DDSSVp

    VnT r8ICS

    VnT r9ICS

    R= negligible i3

    Sol;

    2g=g= 0.2m?0.4m141.42sm1m24

    80R= 400k0100.2m

    R=R=120/0.2m=600kc 0409

    150?0.026 r7.222k80.54m

    150?0.026 r5.78k90.675m

    1R= 500k020.01?0.2m

    R=7.222k+(1+150)(50k//(5078k+(1+50)1k)=5073MΩ i2

    (150(151)50k ,;A600k//400k227.882)5073M50k5078k151(1k)??

     AAA369.82?227.888380.5vd2

(5)

    (a) R=? ,V=0.026V 0T

    I=2mA , I=0.25 , I= 0.25*7.2m=1.8mA r20C13AREF

    V=50V , R= 50k , , 50k50A7nP

    //rRR20e22C!3 Re20(1P)

    V500.026?PT r20I2mr20

    //rRR22C17C!3 Re22(1P)

    VT r65022IC22

    V50A R27.78kC13I1.8mC13A

    R=50 0

    (b) f=? PD

    1 = 2RCi2i

     CC(1Ai2?

     =10p(1+679)=6.7nF

     R=R=26kΩ i213

    1 f913063HzPD2?26k?607n

(6) (a) One- pole Lowpass filter

    RF

    CF øøVin12-

    V+out

    C 2

     øø21

     -VoV+ in

    C 1

     fCc2f3dB2C F

    T(j)= - C/C~21 (b)T=- C/C=-1 21

     C/C=1 21

     f =F=1kHz 3dBC

    fCc2 Let f=10kΩ fcdB32CF

    C2?1k2 0.628C10kf

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