Analog electronics circuits miscellaneous

Analog electronics circuits miscellaneous

Easy Questions

Analog Electronic Circuits

Analog electronics circuits miscellaneous
  1. The circuit shown below:










  1. View Hint View Answer Discuss in Forum

    In the given circuit zener diode is connected at the non-inverting terminal. Hence circuit represents non-inverting time reference voltage.


    Correct Option: B

    In the given circuit zener diode is connected at the non-inverting terminal. Hence circuit represents non-inverting time reference voltage.


  1. A circuit shown below. The largest value of RL that can be used is:










  1. View Hint View Answer Discuss in Forum

    Since transistor in the saturation. op-amp having, Rf = 0

    Vo = 21 +
    Rf
    		= 2 V
    R1

    Negative voltage drop between base and collector terminal due to virtual ground,
    VA = 2 V
    so, current in 2 kΩ resistor =
    12 – 2
    		= 5 mA = I
    2 kΩ

    Now, since voltage at C is 2 V
    so, 2 = I × RL
    2 = 5 × 10–3. RL
    or RL = 400 Ω


    Correct Option: B

    Since transistor in the saturation. op-amp having, Rf = 0

    Vo = 21 +
    Rf
    		= 2 V
    R1

    Negative voltage drop between base and collector terminal due to virtual ground,
    VA = 2 V
    so, current in 2 kΩ resistor =
    12 – 2
    		= 5 mA = I
    2 kΩ

    Now, since voltage at C is 2 V
    so, 2 = I × RL
    2 = 5 × 10–3. RL
    or RL = 400 Ω


  1. For the circuit shown below, the output voltage is ν0 = 2.5 V in response to input voltage νi = 5 V. The finite open-loop differential gain of the op-amp is:










  1. View Hint View Answer Discuss in Forum

    From figure

    VA = Vi ×
    1
    1 + 500

    VA =
    Vi
    501

    and Vo = Aod. VA = Aod.
    Vi
    501

    or Aod = 501.
    V0
    		= 501 ×
    2·5
    		= 250.5
    V15


    Correct Option: B

    From figure

    VA = Vi ×
    1
    1 + 500

    VA =
    Vi
    501

    and Vo = Aod. VA = Aod.
    Vi
    501

    or Aod = 501.
    V0
    		= 501 ×
    2·5
    		= 250.5
    V15


  1. In the circuit shown below the output voltage ν0 is:










  1. View Hint View Answer Discuss in Forum

    Vo = – 0.5
    20
    		+ 1
    20
    		– 2
    20
    204060

    = – 0.5 + 0.5 – 0.667
    = – 0.667 V


    Correct Option: C

    Vo = – 0.5
    20
    		+ 1
    20
    		– 2
    20
    204060

    = – 0.5 + 0.5 – 0.667
    = – 0.667 V


  1. For the op-amp shown in fig below open loop differential gain is Aod = 103. The output voltage ν0 for νi = 2 V is:










  1. View Hint View Answer Discuss in Forum

    KCL at node A.
    I1 = I0

    I1 =
    Vi – VA
    100 kΩ

    I0 =
    VA – V0
    100 kΩ

    given, Aod = 103 = –
    V0
    		and Vi = 2 V
    VA

    or VA =
    –V0
    103

    Now Vi – VA = VA – Vo
    or Vi + Vo = 2 VA
    or Vi + Vo = 2 ·
    –V0
    103

    or 2 + Vo =
    –2V0
    103

    or Vo +
    2V0
    		= – 2
    103

    or Vo =
    –2 × 103
    		= – 1.996 V
    (103 + 2)

    Hence alternative (A) is the correct choice


    Correct Option: A

    KCL at node A.
    I1 = I0

    I1 =
    Vi – VA
    100 kΩ

    I0 =
    VA – V0
    100 kΩ

    given, Aod = 103 = –
    V0
    		and Vi = 2 V
    VA

    or VA =
    –V0
    103

    Now Vi – VA = VA – Vo
    or Vi + Vo = 2 VA
    or Vi + Vo = 2 ·
    –V0
    103

    or 2 + Vo =
    –2V0
    103

    or Vo +
    2V0
    		= – 2
    103

    or Vo =
    –2 × 103
    		= – 1.996 V
    (103 + 2)

    Hence alternative (A) is the correct choice