Analog circuits miscellaneous


Analog circuits miscellaneous

  1. In the given circuit the silicon transistor has β = 75 collector voltage VC = 9V. The ratio of RB and RC is ______.









  1. View Hint View Answer Discuss in Forum


    IC = BIB
    IC = 75 IB
    IC + IB = 75 IB + IB = 76IB

    IC + IB =
    15 - 9
    RC

    76 IB =
    6
    ...(i)
    RC

    IB =
    9 - 0.7
    =
    8.3
    ...(ii)
    RBRB

    Equation (i) / Equation (ii),
    76IB
    =
    6 / RC
    =
    6
    ×
    RB
    IB8.3 / RCRC8.3

    RB
    =
    76 × 8.3
    = 105.1
    RC6

    RB
    = 105.13
    RC

    Correct Option: B


    IC = BIB
    IC = 75 IB
    IC + IB = 75 IB + IB = 76IB

    IC + IB =
    15 - 9
    RC

    76 IB =
    6
    ...(i)
    RC

    IB =
    9 - 0.7
    =
    8.3
    ...(ii)
    RBRB

    Equation (i) / Equation (ii),
    76IB
    =
    6 / RC
    =
    6
    ×
    RB
    IB8.3 / RCRC8.3

    RB
    =
    76 × 8.3
    = 105.1
    RC6

    RB
    = 105.13
    RC


  1. Two perfectly matched silicon transistors are connected as shown in the figure. Assuming the β of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is










  1. View Hint View Answer Discuss in Forum

    Since Both thyristor are perfectly matched, so VBE1 = VBE2

    IC1
    = exp
    VBE1 - VBE2
    = e0 = 1
    IC2VT

    Since β for both are same, therefore IB1 = IB2 = IB
    Now by KVL in loop as shown,
    IR =
    0 - 0.7 - (-5)
    = 4.3 mA
    1 kΩ

    By KCL at point M
    IR = IC1 + 2 IB
    IR = IC1 + 2
    IC1
    β

    ∴ IC1 =
    β
    IR
    β + 2

    For large β ,
    β
    ≈ 1
    β + 2

    ∴ IC1 = IR = 4.3 mA
    ∴ I = IC2 = IC1 = 4.3 mA

    Correct Option: C

    Since Both thyristor are perfectly matched, so VBE1 = VBE2

    IC1
    = exp
    VBE1 - VBE2
    = e0 = 1
    IC2VT

    Since β for both are same, therefore IB1 = IB2 = IB
    Now by KVL in loop as shown,
    IR =
    0 - 0.7 - (-5)
    = 4.3 mA
    1 kΩ

    By KCL at point M
    IR = IC1 + 2 IB
    IR = IC1 + 2
    IC1
    β

    ∴ IC1 =
    β
    IR
    β + 2

    For large β ,
    β
    ≈ 1
    β + 2

    ∴ IC1 = IR = 4.3 mA
    ∴ I = IC2 = IC1 = 4.3 mA



  1. The following circuit has R = 10k&ohm:, C = 10&m;F. The input voltage is a sinusoid at 50Hz with an rms value of 10V under ideal conditions, the current is from the source is










  1. View Hint View Answer Discuss in Forum

    Xc =
    1
    =
    1
    =
    103
    ohm
    ω C2π × 50 × 10μFπ

    VA =
    - jXc
    . V0
    -jXc + R

    ∴ V0 =
    - jXc + R
    . VA
    -jXc


    Assuming virtual ground,
    Vin = VA = 10V
    ∴ V0 =
    - jXc + R
    . 10
    -jXc

    Current , is =
    Vin - V0
    R

    =
    1
    10 -
    -jXc + R
    .10
    104-jXc

    =
    1
    1 -
    -jXc + R
    103-jXc

    =
    1
    -jXc + jXc - R
    103-jXc

    =
    1
    R
    =
    1
    .
    1
    ×
    10 × 103
    103jXcj103(103 / π)

    =
    10π × 10-3
    = (10π mA) ∠ -90° lagging
    j

    Alternately
    V1 = Vs = R . is + V0
    V2 = Vo
    1
    jωC = Vo
    1
    1
    + R 1 + jωRC
    jωC

    But Vo = A(V1 – V2) = A Vs -
    V0
    1 + jωRC

    ⇒ Vo1 +
    A
    = AVs
    1 + jωRC

    V0
    =
    A(1 + jωRC)
    ≃ 1 + jωRC
    Vs1 + jωRC + A

    = 1 + j 100 π × 104 × 10 × 10–6 = 1 + j 10π
    ⇒ Vo = Vs (1 + j 10π) = 10 (1 + j 10π)
    But Vs = R . is + Vo
    or is =
    Vs - V0
    =
    10 - 10 - j100π
    R10 × 103

    = – j 10π mA = 10π ∠– 90° mA

    Correct Option: D

    Xc =
    1
    =
    1
    =
    103
    ohm
    ω C2π × 50 × 10μFπ

    VA =
    - jXc
    . V0
    -jXc + R

    ∴ V0 =
    - jXc + R
    . VA
    -jXc


    Assuming virtual ground,
    Vin = VA = 10V
    ∴ V0 =
    - jXc + R
    . 10
    -jXc

    Current , is =
    Vin - V0
    R

    =
    1
    10 -
    -jXc + R
    .10
    104-jXc

    =
    1
    1 -
    -jXc + R
    103-jXc

    =
    1
    -jXc + jXc - R
    103-jXc

    =
    1
    R
    =
    1
    .
    1
    ×
    10 × 103
    103jXcj103(103 / π)

    =
    10π × 10-3
    = (10π mA) ∠ -90° lagging
    j

    Alternately
    V1 = Vs = R . is + V0
    V2 = Vo
    1
    jωC = Vo
    1
    1
    + R 1 + jωRC
    jωC

    But Vo = A(V1 – V2) = A Vs -
    V0
    1 + jωRC

    ⇒ Vo1 +
    A
    = AVs
    1 + jωRC

    V0
    =
    A(1 + jωRC)
    ≃ 1 + jωRC
    Vs1 + jωRC + A

    = 1 + j 100 π × 104 × 10 × 10–6 = 1 + j 10π
    ⇒ Vo = Vs (1 + j 10π) = 10 (1 + j 10π)
    But Vs = R . is + Vo
    or is =
    Vs - V0
    =
    10 - 10 - j100π
    R10 × 103

    = – j 10π mA = 10π ∠– 90° mA


  1. For the circuit shown below,

    the CORRECT transfer characteristic is









  1. View Hint View Answer Discuss in Forum

    Redrawing the first stage,

    Assuming ideal OPAMP’s

    V- = V+ =
    V2
    2

    V+ =
    V2 R
    =
    V2
    R + R2

    Using KCL at inverting terminal, we get
    V1 - V-
    =
    V- - V01
    RR

    ⇒ V01 = 2V – V1 = V2 – V1
    = – Vi [as V1 – V2 = Vi]
    For the second-stage schmitt-trigger.

    For V01 < V+ or Vi > V+, V0 = 12 Volts
    v+ =
    R
    VR +
    R
    V0 ≡ V1
    R + RR + R

    As VR = 0 [VR is grounded]
    v+ =
    V0
    ≡ V1 .......(A)
    2

    For v01 > v+
    or vi < v+ , v0 = – 12 volts
    The voltage at the non-inverting terminal is,
    v+ =
    R VR
    -
    R
    V0 ≡ V2
    R + RR + R

    v+ = -
    V0
    ≡ V2
    2



    Correct Option: D

    Redrawing the first stage,

    Assuming ideal OPAMP’s

    V- = V+ =
    V2
    2

    V+ =
    V2 R
    =
    V2
    R + R2

    Using KCL at inverting terminal, we get
    V1 - V-
    =
    V- - V01
    RR

    ⇒ V01 = 2V – V1 = V2 – V1
    = – Vi [as V1 – V2 = Vi]
    For the second-stage schmitt-trigger.

    For V01 < V+ or Vi > V+, V0 = 12 Volts
    v+ =
    R
    VR +
    R
    V0 ≡ V1
    R + RR + R

    As VR = 0 [VR is grounded]
    v+ =
    V0
    ≡ V1 .......(A)
    2

    For v01 > v+
    or vi < v+ , v0 = – 12 volts
    The voltage at the non-inverting terminal is,
    v+ =
    R VR
    -
    R
    V0 ≡ V2
    R + RR + R

    v+ = -
    V0
    ≡ V2
    2





  1. Given that the op-amps in the figure are ideal, the output voltage V0 is










  1. View Hint View Answer Discuss in Forum


    V2 - V1
    +
    V2 - V02
    = 0
    2RR

    ∴ V02 =
    3V2 - V1
    .....(i)
    2

    V1 - V2
    +
    V1 - V01
    = 0
    2RR

    ∴ V01 =
    3V1 - V2
    ......(ii)
    2

    ∵ I1 = If
    V02 -
    V01
    =
    V01
    - V0
    22
      R  R

    From equation (i) and (ii)
    ∴ V0 = V01 – V02
    =
    3V1 - V2
    -
    3V2 - V1
    22

    V0 = 2(V1 – V2)

    Correct Option: B


    V2 - V1
    +
    V2 - V02
    = 0
    2RR

    ∴ V02 =
    3V2 - V1
    .....(i)
    2

    V1 - V2
    +
    V1 - V01
    = 0
    2RR

    ∴ V01 =
    3V1 - V2
    ......(ii)
    2

    ∵ I1 = If
    V02 -
    V01
    =
    V01
    - V0
    22
      R  R

    From equation (i) and (ii)
    ∴ V0 = V01 – V02
    =
    3V1 - V2
    -
    3V2 - V1
    22

    V0 = 2(V1 – V2)