Numerical -Source resistance for given h-parameters

Q.  Consider a single stage CE amplifier is estimated to possess the bandwidth of about 2MHz in addition to the resistive load of 500 ohm. What should be the value of source resistance in order to get the required bandwidth for the hybrid π equivalent circuit in accordance to the transistor assumptions given below?

hfe = 100, gm = 30 mA , r'bb = 80Ω, Cc = 3pF, fT = 200MHz, Ce = 20pF, fH = 5MHz, r' be = 2kΩ
- Published on 24 Sep 15
Hybrid


a. 497.4 Ω
b. 531.15 Ω
c. 731.04 Ω
d. 900 Ω

ANSWER: 531.15 Ω

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    Discussion

  • Satyawati   -Posted on 30 Sep 15
    Given data: From the given hybrid- π equivalent circuit,

    hfe = 100,
    gm = 30 mA,
    r'bb = 80 Ω,
    Cc =3pF,
    fT = 200 MHz ,
    Ce = 20 pF,
    fH = 5MHz,
    r'be =2 kΩ
    Bandwidth = 2 MHz,
    RL = 500 Ω

    To determine: Source resistance (Rs)

    Formula: R = ( Rs + r'bb) r'be / Rs + r'bb + r'be

    Method of computation: From the given schematic of hybrid- π equivalent circuit of single stage CE amplifier, the resistance can be given by,

    R = ( Rs + r'bb) r'be / Rs + r'bb + r'be

    But, R = 1 / 2π fH C & C = Ce + Cc [1 + gmRL ]

    Therefore, C = Ce + Cc [1 + gmRL ]

    = 20 x 10-12 + 3 x 10-12 [ 1 + 30 x 10-3 x 500 ]

    = 68 x 10-12 F

    = 68 pF


    R = 1 / 2π fH C

    = 1 / 2π x 5 x 106 x 68 x 10-12

    = 468.10 Ω


    In order to compute the source resistance,

    R = (Rs + r'bb) r'be / Rs + r'bb + r'be

    Substituting all the given as well as evaluated values in above expression,

    468.10 = ( Rs + 80 ) 2 x 103 / Rs + 80 + 2 x 103

    Therefore,

    Rs + 2080 = (Rs +80) 4.2725

    Rs + 2080 = 4.2725 Rs + 341.8

    2080 = [ 4.2725 Rs – Rs ] + 341.8

    2080 – 341.8 = 3.2725 Rs

    1738.2 = 3.2725 Rs

    Thus,

    Rs = 531.15 Ω

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