Analog circuits miscellaneous


Analog circuits miscellaneous

  1. A hysteresis type TTL inverter is used to realize an oscillator in the circuit shown in the figure.

    If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for which output is LOW, is __________.










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    Given LTP = 0.9
    UTP = 1.7
    VC (t) = Vmax + (Vinitial – Vmax )e – t / RC
    LTP = 0 + (1.7 – 0)e – t / RC = 0.9
    ⇒ t = 0.635 ms (Given R = 10k, C = 0.1 μF).

    Correct Option: A

    Given LTP = 0.9
    UTP = 1.7
    VC (t) = Vmax + (Vinitial – Vmax )e – t / RC
    LTP = 0 + (1.7 – 0)e – t / RC = 0.9
    ⇒ t = 0.635 ms (Given R = 10k, C = 0.1 μF).


  1. The sinusoidal ac source in the figure has an rms value of
    20
    V . Considering all
    2

    possible values of RL, the minimum value of Rs in Ω to avoid burnout of the Zener diode is _______.









  1. View Hint View Answer Discuss in Forum

    To avoid burn-out IS ≥ IZ + IL
    For RL = ∞ [considering all possible values]
    IL = 0

    then , IS ≥ IZ >
    20 - VZ
    RS

    Since , 5.IS =
    1
    4

    ⇒ IS =
    1
    A
    20

    Then ,
    1
    20 - 5
    20RS

    ⇒ RS ≥ 300 Ω
    Minimum value of RS = 300 Ω

    Correct Option: D

    To avoid burn-out IS ≥ IZ + IL
    For RL = ∞ [considering all possible values]
    IL = 0

    then , IS ≥ IZ >
    20 - VZ
    RS

    Since , 5.IS =
    1
    4

    ⇒ IS =
    1
    A
    20

    Then ,
    1
    20 - 5
    20RS

    ⇒ RS ≥ 300 Ω
    Minimum value of RS = 300 Ω



  1. The transistor in the given circuit should always be in active region. Take VCE (sat) = 0.2 V. VBE = 0.7V. The maximum value of Rc in Ω which can be used, is ______









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    In B-E loop, using KVL, we have
    5 – 2KIB – VBE = 0

    ⇒ IB =
    5 - 0.7
    = 2.15 mA
    2K

    and IC = βIB = 100 × 2.15 mA = 215 mA
    Using KVL in C-E loop, we get
    VCE = 5 – 0.215 RC
    For active region, VCE > 0.2 V
    ⇒ 0.215 RC < 5 – 0.2
    ⇒ RC =
    4.8
    ≅ 22.32 Ω
    0.215

    Correct Option: A


    In B-E loop, using KVL, we have
    5 – 2KIB – VBE = 0

    ⇒ IB =
    5 - 0.7
    = 2.15 mA
    2K

    and IC = βIB = 100 × 2.15 mA = 215 mA
    Using KVL in C-E loop, we get
    VCE = 5 – 0.215 RC
    For active region, VCE > 0.2 V
    ⇒ 0.215 RC < 5 – 0.2
    ⇒ RC =
    4.8
    ≅ 22.32 Ω
    0.215


  1. In the following circuit, the transistor is in active mode and VC = 2V. To get VC = 2V. To get VC = 4V, we replace RC with R'C. Then the ratio R' C / RC is ______.









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    we have Vc = 2V;
    Now, Ic Rc = 10 – 2 = 8 ...(i)
    We have Vc = 4V
    and Ic Rc = 10 – 4 = 6... (ii)

    Ic Rc
    =
    6
    Ic Rc8

    Rc
    =
    3
    = 0.75
    Rc4

    Correct Option: C

    we have Vc = 2V;
    Now, Ic Rc = 10 – 2 = 8 ...(i)
    We have Vc = 4V
    and Ic Rc = 10 – 4 = 6... (ii)

    Ic Rc
    =
    6
    Ic Rc8

    Rc
    =
    3
    = 0.75
    Rc4



  1. A(0– 50A) moving coil ammet er has a volt age dr op of 0.1V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0 – 500A) is ______.









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    Voltage drop across its terminals at full scale deflection = 0.1 volt

    450 × Rsh = 0.1

    Rsh =
    0.1
    = 0.22 mΩ
    450

    Required shunt resistance, Rsh = 0.22m Ω

    Correct Option: A


    Voltage drop across its terminals at full scale deflection = 0.1 volt

    450 × Rsh = 0.1

    Rsh =
    0.1
    = 0.22 mΩ
    450

    Required shunt resistance, Rsh = 0.22m Ω