Jntu Hyd 2 2ece Eca Set 4

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    S.55Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad

    ( JNTU-Hyderabad ) B.Tech. II-Year II-Sem.

    S e t - 4S o l u t i o n s

    Code No: R09220402/R09

    II B.Tech. II Semester Examinations

    April/May - 2012ELECTRONIC CIRCUIT ANALYSIS

    (Common to ICE, E.COMP.E, ETM, EIE, ECE)

    Time: 3 Hours Max. Marks: 75

    Answer any FIVE questions

    All questions carry equal marks

    - - -

    1. (a) Define f and fT and also establish the relationship between f and fT. (Unit-III, Topic No. 3.5)

    (b) Derive the expression for the CE short-circuit current gain as a function of frequency. [7+8]

    (Unit-III, Topic No. 3.5)

    2. For the circuit shown in figure, show that,

    (a) (AVS

    )max

    =f

    i o r f

    h

    h h h hif R

    L= and R

    S= 0

    (b) Ri=

    i o r f

    o

    h h h h

    hif R

    L=.[8+7] (Unit-I, Topic No. 1.3)

    hr

    V2

    V1VS+

    +

    +

    RS hi

    V2

    IL

    ZL

    I2 2

    hfI1h

    o

    2'1'

    1

    hr

    V2

    V1VS+

    +

    +

    RS hi

    V2

    IL

    ZL

    I2 2

    hfI1h

    o

    2'1'

    1

    Figure

    3. (a) Sketch the circuit of a CS amplifier. Derive the expression for the voltage gain at low frequencies. What is the

    maximum value of voltage gain? (Unit-IV, Topic No. 4.5)

    (b) The FET shown in figure has the following parameters,

    IDSS

    = 5.6 mA and VP= 4 V. If V

    i= 10 V find V

    o. [8+7] (Unit-IV, Topic No. 4.3)

    +

    +24 V

    4.7 k

    10 kVi

    +

    Vo

    12 V

    +

    +24 V

    4.7 k

    10 kVi

    +

    Vo

    12 V

    Figure

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    4. (a) Draw the equivalent circuit of a single tuned capacitive coupled amplifier and derive the expression for gain at

    resonance. (Unit-VIII, Topic No. 8.3)

    (b) Draw the circuit diagram for tuned RF amplifier and explain its working. [7+8] (Unit-VIII, Topic No. 8.1)

    5. (a) Derive an expression for the output power of a class A large-signal amplifier in terms of Vmax

    , Vmin

    , Imax

    and Imin

    .

    (Unit-VII, Topic No. 7.2)

    (b) For a particular power amplifier, the optimum load impedance is 180 . Calculate the turns ratio required tomatch an 8 load to this transistor. If the amplifier takes a mean collector current of 2 A from a 15 V supply anddelivers an A.C load power of 2.5 W to the transformer coupled-load, calculate the efficiency and the collector

    dissipation (neglecting the losses). [7+8] (Unit-VII, Topic No. 7.2)

    6. (a) Discuss about the types of negative feedback amplifiers giving the effect of each type of feedback on the

    parameters of the amplifier. (Unit-V, Topic No. 5.2)

    (b) What sort of feedback is employed in a CE amplifier with unbypassed emitter resistor? Discuss its analysis in

    detail. [7+8] (Unit-V, Topic No. 5.5)

    7. Design a RC phase-shift oscillator to operate at a frequency of 5 kHz. Use a MOSFET with = 55 and rd= 5.5 k. The

    phase-shift network not load the amplifier.

    (a) Find the minimum value of the drain-circuit resistance for which the circuit will oscillate.

    (b) Choose reasonable values of R and find C. [8+7] (Unit-VI, Topic No. 6.2)

    8. (a) Write the expressions for overall voltage gain nV(A ) ,

    nLf and

    n

    Hf when n-non-identical amplifier stages are

    cascaded. (Unit-II, Topic No. 2.1)

    (b) Compute f H

    and fL

    a 2-stage amplifier if1H

    f = 3 kHz and2H

    f = 4 kHz,1L

    f = 300 Hz and2L

    f = 600 Hz. [15]

    (Unit-II, Topic No. 2.1)

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    S.57Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad

    ( JNTU-Hyderabad ) B.Tech. II-Year II-Sem.

    Q1. (a) Define f and fT and also establish the relationship between f and fT.

    Answer : April/May-12, Set-4, Q1(a) M[7]

    For answer refer Unit-III, Q21, Topics: cut-off frequency,f, Current Gain Bandwidth Product,fT.

    (b) Derive the expression for the CE short-circuit current gain as a function of frequency.

    Answer : April/May-12, Set-4, Q1(b) M[8]

    For answer refer Unit-III, Q17.

    Q2. (a) For the circuit shown in figure, show that,

    (a) (AVS

    )max

    = f

    i o r f

    h

    hh -h hif R

    L= and R

    S= 0

    (b) Ri=

    i o r f

    o

    h h -h h

    hR

    L= .

    hrV

    2V

    1V

    S

    +

    +

    +

    RS

    hi

    V2

    IL

    ZL

    I2

    2

    hfI

    1

    ho

    2'1'

    1

    hrV

    2V

    1V

    S

    +

    +

    +

    RS

    hi

    V2

    IL

    ZL

    I2

    2

    hfI

    1

    ho

    2'1'

    1

    Figure

    Answer : April/May-12, Set-4, Q2 M[8+7]

    Given that,

    hrV

    2V

    1V

    S

    +

    +

    +

    RS

    hi

    V2

    IL

    ZL

    I2

    2

    hfI

    1

    ho

    2'1'

    1

    hrV

    2V

    1V

    S

    +

    +

    +

    RS

    hi

    V2

    IL

    ZL

    I2

    2

    hfI

    1

    ho

    2'1'

    1

    Figure

    (a) (Avs

    )max

    =f

    i o r f

    h

    h h h h

    if RL = and RS = 0

    To prove that (AVs

    )max

    =froi

    f

    hhhh

    h

    at R

    L= andR

    S= 0, consider the overall voltage gain expression

    including source.

    The expression for voltage gain including source from figure is given as,

    AVS

    =V i

    i s

    A R

    Z R+... (1)

    SOLUTIONS TO APRIL/MAY-2012, SET-4, QP

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    By substitutingRS

    = 0 in equation (1), we get,

    AVS

    =V i

    i

    A R

    Z

    [Q Ri=Zi]

    =V i

    i

    A R

    R

    =AV

    ... (2)

    Since,AV

    =)( rfoiLi

    Lf

    hhhhRh

    Rh

    +

    ... (3)

    By substituting equation (3) in equation (2), we get,

    AVS

    =)( rfoiLi

    Lf

    hhhhRh

    Rh

    +

    =

    +

    )( rfoiL

    iL

    Lf

    hhhhR

    hR

    Rh

    =

    +

    )( rfoiL

    i

    f

    hhhhR

    h

    h... (4)

    SubstitutingRL

    = in equation (4),

    AVS =froi

    i

    f

    hhhhh

    h

    +

    = 01

    Q

    =froi

    f

    hhhh

    h

    +

    0

    max( )f

    VS

    i o r f

    hA

    h h h h

    =

    (b) Ri= i o r f

    o

    h h h h

    h

    if R

    L=

    The expression for input impedance from figure is

    given as,

    Ri= h

    i

    1

    f r

    oL

    h h

    hR

    +... (5)

    SubstitutingRL

    = in equation (5), we get,

    Ri= h

    i

    1

    f r

    o

    h h

    h +

    = hi

    0

    f r

    o

    h h

    h +

    = hi

    f r

    o

    h h

    h

    i o f r

    i

    o

    h h h hR

    h

    =

    Q3. (a) Sketch the circuit of a CS amplifier.Derive the expression for the voltagegain at low frequencies. What is themaximum value of voltage gain?

    Answer : April/May-12, Set-4, Q3(a) M[8]

    A common source amplifier is as shown in figure (1).

    +

    Vi V0

    +

    RG

    S

    RS C

    S

    D

    RD

    VDD

    G

    +

    Vi V0

    +

    RG

    S

    RS C

    S

    D

    RD

    VDD

    G

    Figure (1): Common Source Amplifier

    The small signal equivalent circuit of common source

    amplifier is shown in figure (2).

    +

    Vi Vgs

    G

    RG

    + Vgs

    RD

    +

    V0

    D id

    S

    +

    Vi Vgs

    G

    RG

    + Vgs

    RD

    +

    V0

    D id

    S

    Figure (2)

    Voltage Gain

    The capacitor Cs

    is used as a bypass for source

    current. Because ofCs,R

    sis removed.

    Applying KVL to source-drain loop, we get,

    D

    o

    R

    V+

    Dd

    gs

    Rr

    V

    +

    = 0

    Vo =dD

    D

    rR

    R

    +

    Vgs

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    IfVgs

    = Vi

    o D

    V

    i D d

    V RA

    V R r

    = =

    +The maximum value of voltage gain is nearly equal to

    maximum theoretical gain (i.e.,) from FET amplifier circuits.

    (b) The FET shown in figure has thefollowing parameters,

    IDSS

    = 5.6 mA and VP

    = 4 V. If Vi= 10 V

    find Vo.

    +

    +24 V

    4.7 k

    10 kVi

    +

    Vo

    12 V

    +

    +24 V

    4.7 k

    10 kVi

    +

    Vo

    12 V

    Figure

    Answer : April/May-12, Set-4, Q3(b) M[7]

    Given that,

    For a Field Effect Transistor (FET),

    IDSS

    = 5.6 mA

    VP

    = 4 V

    Vi

    = 10 V

    Vo

    = ?

    +

    +24 V

    4.7 k

    10 kVi

    +

    Vo or VDS

    12 V

    VDD

    Rd

    Id

    D

    S

    G

    RsVGS

    VSS

    +

    +24 V

    4.7 k

    10 kVi

    +

    Vo or VDS

    12 V

    VDD

    Rd

    Id

    D

    S

    G

    RsVGS

    VSS

    Figure

    From figure,

    VDD

    = + 24 V

    VSS = 12 V

    Rd

    = 4.7 k

    RS

    = 10 k

    Vi

    = VGS

    = 10 V

    From figure, VGS

    is expressed as,

    VGS

    = ID

    RS

    ID

    =S

    GS

    R

    V

    = 31010

    10

    = 1 103

    1mADI =

    The output voltage Vo

    can be obtained by applying

    KVL (Kirchoffs Voltage Law) at the output terminal,

    i.e., VDS

    Vo= V

    DDI

    D(R

    S+R

    D) V

    SS

    = 24 ( 1) 103(4.7 103+ 10 103) ( 12)

    = 24 + 103(4.7 + 10) 103 + 12

    = 36 + 103+3 (14.7)

    = 36 + (1) (14.7)

    VDS

    = 50.7 V

    50.7VoV =

    Q4. (a) Draw the equivalent circuit of a singletuned capacitive coupled amplifier andderive the expression for gain atresonance.

    Answer : April/May-12, Set-4, Q4(a) M[7]

    For answer refer Unit-VIII, Q6.

    (b) Draw the circuit diagram for tuned RFamplifier and explain its working.

    Answer : April/May-12, Set-4, Q4(b) M[8]

    Tuned RF Amplifier

    Amplifier which are used to select and amplify a

    particular range of high frequencies or radio frequencies are

    known as tuned RF amplifier.

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    The circuit diagram for tuned RF amplifier is as shown

    in figure,

    L OutputR1

    Cin

    L

    R2 REInput

    CE

    +VCC

    C L OutputR1

    Cin

    L

    R2 REInput

    CE

    +VCC

    L OutputR1

    Cin

    L

    R2 REInput

    CE

    +VCC

    C

    Figure: Tuned RF Amplifier

    It consists of a parallel tuned circuit at the collector

    end and two resistorsR1

    andR2

    connected in parallel with

    parallel tuned circuit and emitter resistance respectively. The

    input and output capacitors are used as a tuning dial which

    tune to a desired frequency by rejecting all the unwanted

    frequencies.

    Tunning can be accomplished by varying the

    capacitance and inductance values in the parallel tunedcircuit. The impedance of the tuned circuit varies with

    varying frequency. At resonant frequency, the circuit offers

    a very high impedance whereas low impedance is achieved

    at other frequencies. If the frequency of signal applied to

    the tuned circuit is identical to the frequency of LC circuit,

    larger amplification of signal takes place in the amplifiers. In

    contrast to this, when various signals with different

    frequencies are applied, the circuit selects signals with

    frequencies lying between the range of resonant frequency

    and rejects all the other frequencies.

    Applications

    It has wide range of applications in various areas.

    For example, in electronic devices such as television

    receivers, radio receivers and radar systems.

    Q5. (a) Derive an expression for the outputpower of a class A large-signal amplifierin terms of V

    max, V

    min, I

    maxand I

    min.

    Answer : April/May-12, Set-4, Q5(a) M[7]

    For answer refer Unit-VII, Q2.

    (b) For a particular power amplifier, theoptimum load impedance is 180 .Calculate the turns ratio required to

    match an 8 load to this transistor. Ifthe amplifier takes a mean collectorcurrent of 2 A from a 15 V supply anddelivers an A.C load power of 2.5 W tothe transformer coupled-load, calculatethe efficiency and the collectordissipation (neglecting the losses).

    Answer : April/May-12, Set-4, Q5(b) M[8]

    Given that,

    For a power amplifier,

    Optimum load impedance, 'LR = 180

    Load resistance,RL

    = 8Mean collector current,I

    CQ= 2 A

    Power supply, VCC

    = 15 V

    A.C load power, Pac

    = 2.5 W

    (i) Turns ratio, n = ?

    (ii) Efficiency, = ?(iii) Power dissipation, P

    D= ?

    (i) The expression for turns ratio of a transformer is given as,

    '

    2

    L

    L

    R

    Rn =

    ... (1)

    By substituting given values in equation (1),

    We get ,

    n2 =180

    8

    n2 = 0.044

    n = 044.0

    21.0=n

    (ii) The efficiency of a power amplifier is expressed as,

    DS

    ac

    P

    P= ... (2)

    Where,

    D.Cinput power PD.C

    is given as,

    PDC

    =ICQ

    . VCC

    PDC

    = 2 15 [Q ICQ= 2 A, VCC= 15 A]

    30 WDCP = ... (3)

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    On substituting PDC

    and Pac

    in equation (2), we get,

    =30

    5.2= 0.083

    %3.8=

    (iii) Collector power dissipation, PD

    is obtained as,

    PD

    = PDC

    Pac

    = 30 2.5

    = 27.5

    27.5 WDP =

    Q6. (a) Discuss about the types of negative feedback amplifiers giving the effect of each type of

    feedback on the parameters of the amplifier.Answer : April/May-12, Set-4, Q6(a) M[7]

    Types of Negative Feedback Amplifier

    For answer refer Unit-V, Q5.

    Effect of Each Type of Feedback on the Parameters of the Amplifier

    The effect of each type of feedback on the parameters of the amplifier is illustrated in table (1),

    Types of Parameters

    Negative Feedback Improved Desensitises Bandwidth Non Linear Noise Overall

    Amplifier Characteristics Distortion Transmission Gain

    1. Voltage- Voltage AV Increases Decreases Decreases Decreasesseries amplifier

    feedback

    amplifier

    2. Current Transcon-

    -series ductance Gm

    Increases Decreases Decreases Decreases

    feedback amplifier

    amplifier

    3. Current Current

    shunt amplifier Ai

    Increases Decreases Decreases Decreases

    feedback

    amplifier

    4. Voltage- Transresis-

    shunt -tance Rm

    Increases Decreases Decreases Decreases

    feedback amplifier

    amplifier

    (b) What sort of feedback is employed in a CE amplifier with unbypassed emitter resistor? Discussits analysis in detail.

    Answer : April/May-12, Set-4, Q6(b) M[8]

    The type of feedback which is employed in common emitter amplifier with unbypassed emitter resistor is current-

    series feedback.

    For remaining answer refer Unit-V, Q25.

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    Q7. Design a RC phase-shift oscillator to operate at a frequency of 5 kHz. Use a MOSFET with = 55and r

    d= 5.5 k. The phase-shift network not load the amplifier.

    (a) Find the minimum value of the drain-circuit resistance for which the circuit will oscillate.

    (b) Choose reasonable values of R and find C.

    Answer : April/May-12, Set-4, Q7 M[8+7]

    Given that,

    For a phase shift oscillator using MOSFET,

    Frequency of oscillation,fr= 5 kHz

    Amplification factor,= 55

    Drain resistance, rd= 5.5 k

    (a) Minimum value ofRD

    for which circuit oscillates = ?

    (b) For the oscillator to work satisfactorily, reasonable values ofR = ?, C= ?

    The circuit arrangement of a phase shift oscillator using MOSFET is as shown in figure,

    G

    R RVi

    +

    R

    Vo

    C C C

    CS

    VDD

    RD

    D

    S

    RS

    G

    R RVi

    +

    R

    Vo

    C C C

    CS

    VDD

    RD

    D

    S

    RS

    Figure

    (a) Minimum Value of RD

    In the phase shift oscillator, the FET is required to have a gain greater than 29 i.e.,A > 29 for obtaining |AB| = 1

    Consider |A| = 40

    Then, the expression for open loop gain of phase shift oscillator using MOSFET is given as,

    |A| =dD

    Ddm

    rR

    Rrg

    +

    |A| =

    dD

    D

    rR

    R

    +

    [Q = gmrd]

    |A|RD

    + |A|rd

    =RD

    ( |A|) RD

    = |A|rd

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    RD

    =A

    rA d

    =4055

    105.5403

    =15

    102203

    = k6.14DR

    (b) R and C Values

    For the proper and satisfactory operation of the circuit,

    Consider, C= 0.01F

    Since, the frequency of oscillation of a phase shift oscillator using MOS is expressed as,

    fr =62

    1

    RC

    R =62

    1

    rCf... (1)

    On substituting values offr

    and Cin equation (1), we get,

    R =61051001.014.32

    136

    = 310769.0

    1

    = 1300

    = k3.1R

    Q8. (a) Write the expressions for overall voltage gain )(A ''V ,''

    Lf and''

    Hf when n-non-identical amplifier

    stages are cascaded.

    Answer : April/May-12, Set-4, Q8(a)

    Overall Voltage Gain )(nvA

    The expression for overall voltage gain ( )( )nVA , when n-non-identical amplifier stages are cascaded is given as,

    ( )nVA =Av1Av2 ... (1)

    It can be observed from equation (1) that, over all voltage gain is the product of voltage gain of each amplifier stage.

    Over All Lower 3dB Frequency (n)Lf

    The lower 3 dB frequency )(nLf can be expressed as,

    2)(

    f

    f nL=

    2

    1

    f

    fL+

    2

    2

    f

    fL+ ...

    ...22)(

    21++= LL

    nL fff ... (2)

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    Over All Upper 3dB Frequency ( )(n)HfThe upper 3 dB frequency )(n

    Hf can be expressed as,

    ...11

    1

    22

    )(

    21

    ++=

    HH

    nH

    ff

    f

    ... (3)

    (b) Compute fH

    and fL

    a 2-stage amplifier if1H

    f = 3 kHz and2H

    f = 4 kH;1L

    f = 300 Hz and2L

    f = 600

    Hz.

    Answer : April/May-12, Set-4, Q8(b)

    Given that,

    For a multistage amplifier,

    n = 2

    1Hf = 3 kHz

    2Hf = 4 kHz

    1L

    f = 300 Hz

    2L

    f = 600 Hz

    fH

    = ?

    fL

    = ?

    The expression for lower frequencyfL

    of a multistage amplifier is given as,

    fL

    = )(nLf 12

    1

    n ... (1)

    Where,

    )(nLf =

    22

    21 LLff +

    = 22 )600()300( + = 36000090000+

    )(nLf = 670.82 Hz. ... (2)

    On substituting )(nLf and n values in equation (1), we get,

    fL

    = 670.82 1122 1

    = 670.82 414.0 = 670.82 0.64

    = 429.3 Hz

    Hz429= Lf

    The higher frequency (fH

    ) can be expressed as,

    fH

    =

    121

    )(

    n

    nHf

    ... (3)

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    Where,

    )(n

    Hf =22

    21

    11

    1

    HH ff+

    =

    2323)104(

    1

    )103(

    1

    1

    +

    =

    16000000

    1

    9000000

    1

    1

    +

    = 41015.4

    1

    )(nHf = 2409.6 Hz

    On substituting values of )(nHf and n in equation (3), we get,

    fH

    =

    ( )

    1

    2 1

    nH

    n

    f

    =

    12

    6.2409

    21

    =

    41.0

    2404

    = 5863.4 Hz

    kHz8.5= Hf