Analog electronics circuits miscellaneous

Analog electronics circuits miscellaneous

Easy Questions

Analog Electronic Circuits

Analog electronics circuits miscellaneous
  1. In the circuit shown below the op-amp is ideal. If βF = 60, then the total current supplied by the 15 V source is:










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    Given
    V+ = 5 V
    V– = VE = 5 V (due to virtual ground)

    Ie =
    5
    		= 50 mA
    100

    since ideal op-amp draws no current, so
    Iz =
    15 – V+
    		, Ie = Ib + Ic
    47 kΩ

    =
    15 – 5
    		, or Ic = Ie – Ib
    47 kΩ

    = 0·213 mA , or βIb + Ib = Ie
    Ie =
    Vo
    		, or Ib =
    Ie
    		⇒ Ic =
    β·Ie
    1001 + β(1 + β)

    = 50 mA
    The total current supplied by the 15 V source
    = Iz + Ic = Iz +
    β·Ie
    (1 + β)

    =0·213 +
    60·(50)
    		mA = 49.39 mA
    61

    Hence alternative (C) is the correct choice.


    Correct Option: C

    Given
    V+ = 5 V
    V– = VE = 5 V (due to virtual ground)

    Ie =
    5
    		= 50 mA
    100

    since ideal op-amp draws no current, so
    Iz =
    15 – V+
    		, Ie = Ib + Ic
    47 kΩ

    =
    15 – 5
    		, or Ic = Ie – Ib
    47 kΩ

    = 0·213 mA , or βIb + Ib = Ie
    Ie =
    Vo
    		, or Ib =
    Ie
    		⇒ Ic =
    β·Ie
    1001 + β(1 + β)

    = 50 mA
    The total current supplied by the 15 V source
    = Iz + Ic = Iz +
    β·Ie
    (1 + β)

    =0·213 +
    60·(50)
    		mA = 49.39 mA
    61

    Hence alternative (C) is the correct choice.


  1. The value of C required for sinusoidal oscillation of frequency 1 kHz in the circuit shown below:










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    The given circuit represents the wien bridge oscillator.

    The ratio
    R2
    		must be greater than 2 for oscillation to start.
    R1

    Since here ratio
    R2
    		=
    2.1
    		= 2.1 > 2
    R11

    so there will oscillation and frequency of oscillation
    f =
    1
    2πRC

    1 × 103 =
    1
    2π × 1 × 103 × C

    or C =
    1
    		× 106 F
    2π

    or C =
    1
    		µF
    2π

    Correct Option: A

    The given circuit represents the wien bridge oscillator.

    The ratio
    R2
    		must be greater than 2 for oscillation to start.
    R1

    Since here ratio
    R2
    		=
    2.1
    		= 2.1 > 2
    R11

    so there will oscillation and frequency of oscillation
    f =
    1
    2πRC

    1 × 103 =
    1
    2π × 1 × 103 × C

    or C =
    1
    		× 106 F
    2π

    or C =
    1
    		µF
    2π

  1. The phase shift oscillator shown below operate at f = 80 kHz. The value of resistance RF is:










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    Frequency of oscillation in the phase shift oscillator,

    f =
    1
    2π 6 RC

    80 × 103 =
    1
    2π 6 × R × 100 × 10–12

    or R =
    106
    8 × 2π × 6

    or R = 8.12 kΩ
    and, gain =
    RF
    		= 29
    R

    or RF = 29 R = 29 × 8.12 kΩ = 235.65 kΩ
    or RF 236 kΩ

    Correct Option: B

    Frequency of oscillation in the phase shift oscillator,

    f =
    1
    2π 6 RC

    80 × 103 =
    1
    2π 6 × R × 100 × 10–12

    or R =
    106
    8 × 2π × 6

    or R = 8.12 kΩ
    and, gain =
    RF
    		= 29
    R

    or RF = 29 R = 29 × 8.12 kΩ = 235.65 kΩ
    or RF 236 kΩ

  1. In the filter circuit shown below the 3 dB cut-off frequency is:










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    f3dB =
    1
    2πRC

    Here R = R1 || R2 where R1 = 3 kΩ
    = 3 || 6 = 2 k R2 = 6 kΩ
    f3dB =
    1
    		= 1.59 kHz
    2π × 2 × 103 × 50 × 10–9

    Correct Option: B

    f3dB =
    1
    2πRC

    Here R = R1 || R2 where R1 = 3 kΩ
    = 3 || 6 = 2 k R2 = 6 kΩ
    f3dB =
    1
    		= 1.59 kHz
    2π × 2 × 103 × 50 × 10–9

  1. A 50 kHz square wave is to be amplified by an op-amp to have an output voltage swing ± 10 V. Two op-amp are available A has slew rate of 0.5 V/µs and B has slew rate of 13 V/µs. Then correct answer is:








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    Given that output voltage swing = ± 10 V and input wave is square wave with 50 kHz
    Slew rate for op-amp A = 0.5 V/µs.
    Slew rate for op-amp B = 13 V/µs.
    Maximum output voltage after passing a square wave can be calculated as first
    Check for op-amp A

    Slew rate of op-amp A =
    2πf·Vm
    106

    0.5 × 106 = 2π × 50 × 103. Vm
    or Vm =
    0·5 × 103
    314

    Vm = 1.59 V
    Since this voltage is less than the output voltage swing i.e. ± 10 V.
    Therefore op-amp A is not suitable for amplified a 50 kHz square wave.
    Now check for op-amp B
    Vm =
    Slew rate × 106
    2πf

    =
    13 × 106
    		= 41.4 V
    2 × 3·14 × 50 × 103

    Since this voltage is higher than the output voltage swing i.e. ± 10 V.
    Therefore op-amp B is suitable for amplifying 50 kHz square wave.
    Hence alternative (C) is the correct choice.

    Correct Option: C

    Given that output voltage swing = ± 10 V and input wave is square wave with 50 kHz
    Slew rate for op-amp A = 0.5 V/µs.
    Slew rate for op-amp B = 13 V/µs.
    Maximum output voltage after passing a square wave can be calculated as first
    Check for op-amp A

    Slew rate of op-amp A =
    2πf·Vm
    106

    0.5 × 106 = 2π × 50 × 103. Vm
    or Vm =
    0·5 × 103
    314

    Vm = 1.59 V
    Since this voltage is less than the output voltage swing i.e. ± 10 V.
    Therefore op-amp A is not suitable for amplified a 50 kHz square wave.
    Now check for op-amp B
    Vm =
    Slew rate × 106
    2πf

    =
    13 × 106
    		= 41.4 V
    2 × 3·14 × 50 × 103

    Since this voltage is higher than the output voltage swing i.e. ± 10 V.
    Therefore op-amp B is suitable for amplifying 50 kHz square wave.
    Hence alternative (C) is the correct choice.