Analog electronics circuits miscellaneous

Analog electronics circuits miscellaneous

Easy Questions

Analog Electronic Circuits

Analog electronics circuits miscellaneous
  1. A zener diode in the circuit shown below, has a knee current of 5 mA, and a maximum allowed power dissipation of 300 mW. What are the minimum and maximum load current that can be drawn safely from the circuit, keeping the output voltage V0 at 6 V?










  1. View Hint View Answer Discuss in Forum

    Given, IZ (min) = 5 mA
    V0 = 6 V = Vz
    (according to given condition).
    Maximum power dissipation means maximum zener current
    i.e. Iz (max) = Pmax V = 300 mW 6 V = 50 mA
    from figure
    I = Vin – Vz 50 = 9 – 6 50 = 60 mA
    and I = IZ + IL
    IL (min) = I – Iz (max) = 60 – 50 = 10 mA
    IL (max) = I – Iz (min) = 60 – 5 = 55 mA
    Hence alternative (C) is the correct choice.


    Correct Option: C

    Given, IZ (min) = 5 mA
    V0 = 6 V = Vz
    (according to given condition).
    Maximum power dissipation means maximum zener current
    i.e. Iz (max) = Pmax V = 300 mW 6 V = 50 mA
    from figure
    I = Vin – Vz 50 = 9 – 6 50 = 60 mA
    and I = IZ + IL
    IL (min) = I – Iz (max) = 60 – 50 = 10 mA
    IL (max) = I – Iz (min) = 60 – 5 = 55 mA
    Hence alternative (C) is the correct choice.


  1. In order to obtain a 12-V stabilized supply from the current shown in figure below, the input to terminals A and B should be:










  1. View Hint View Answer Discuss in Forum

    Input voltage should be greater than 12 V to allow for drop across R. Since Zener diode always work in reverse - biased mode. Terminal A should obviously, be positive.


    Correct Option: D

    Input voltage should be greater than 12 V to allow for drop across R. Since Zener diode always work in reverse - biased mode. Terminal A should obviously, be positive.


  1. The current flowing through the zener diode of figure below is:










  1. View Hint View Answer Discuss in Forum

    From given figure

    I =
    50 – Vz
    		=
    50 – 30
    R4 × 103

    or I = 5 × 10–3
    or I = 5 mA.


    Correct Option: A

    From given figure

    I =
    50 – Vz
    		=
    50 – 30
    R4 × 103

    or I = 5 × 10–3
    or I = 5 mA.


  1. In the case of self Biasing or potential divider biasing, for good stability factor:








  1. View Hint View Answer Discuss in Forum

    The stability factor,

    S =
    β + 1(1 + RB/RE)
    		→ (A)
    1 + β + RB/RE

    where, RB =
    R1.R2
    R1 + R2

    RE = emitter resistance
    β = transistor gain
    from relation we conclude that for good bias stability i.e. lower value of stability factor S, RB << RE.
    Therefore alternative (B) is the correct choice.

    Correct Option: B

    The stability factor,

    S =
    β + 1(1 + RB/RE)
    		→ (A)
    1 + β + RB/RE

    where, RB =
    R1.R2
    R1 + R2

    RE = emitter resistance
    β = transistor gain
    from relation we conclude that for good bias stability i.e. lower value of stability factor S, RB << RE.
    Therefore alternative (B) is the correct choice.

  1. The cutin voltage for each diode shown below Vγ = 0.6 V. Each diode current is 0.5










  1. View Hint View Answer Discuss in Forum

    From figure
    VA = 5 – 0.6 = 4.4 V
    VB = 0 – 0.6 = – 0.6 V

    R1 =
    10 – 0.6 – 4.4
    		= 10 kΩ
    0.5 × 10–3

    R2 =
    VA – VB
    		=
    4·4 – (0·6)
    		= 5 kΩ
    1 × 10–31 × 10–3

    R3 =
    VB – (–5)
    		=
    –0.6 + 5
    		=
    4.6
    		= 2.93 kΩ
    1.5 × 10–31.5 × 10–31.5 × 10–3


    Correct Option: A

    From figure
    VA = 5 – 0.6 = 4.4 V
    VB = 0 – 0.6 = – 0.6 V

    R1 =
    10 – 0.6 – 4.4
    		= 10 kΩ
    0.5 × 10–3

    R2 =
    VA – VB
    		=
    4·4 – (0·6)
    		= 5 kΩ
    1 × 10–31 × 10–3

    R3 =
    VB – (–5)
    		=
    –0.6 + 5
    		=
    4.6
    		= 2.93 kΩ
    1.5 × 10–31.5 × 10–31.5 × 10–3