Analog electronics circuits miscellaneous
- The dissipation at the collector is zero in the quiescent state and increases with excitation in the case of:
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NA
Correct Option: A
NA
- The change in the value of the emitter resistance RE in a difference amplifier:
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ACM = –hfe · Rc hie + Rs + (1 + hfe)RE and ADM = 1 × –hfe·Rc 2 (Rs + hie)
Hence alternative (B) is the correct choice.Correct Option: B
ACM = –hfe · Rc hie + Rs + (1 + hfe)RE and ADM = 1 × –hfe·Rc 2 (Rs + hie)
Hence alternative (B) is the correct choice.
- A op-amp has an offset voltage of 1 mv and is ideal in all respects. If this op-amp is used in the circuit shown below the output voltage will be (select the nearest value)
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Output of the op-amp due to offset voltage of 1 mV
Vo = – 1 MΩ Vi 1 kΩ
or Vo = – (103) (± 1 × 10–3)
or Vo = ± 1 VCorrect Option: C
Output of the op-amp due to offset voltage of 1 mV
Vo = – 1 MΩ Vi 1 kΩ
or Vo = – (103) (± 1 × 10–3)
or Vo = ± 1 V
- Assume that the op-amp shown below is ideal. The current I through the 1 kΩ resistor is:
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Due to virtual ground voltage at terminal A is 0 V. KCL at node A
2 = VA – VB 2 2 = 0 – VB 2
or VB = – 4 V
Again KCL at BI = VB – VA + VB – 0 2 2 or I = (–4 – 0) + –4 2 2
or I = – 2 – 2 I = – 4 mA
Hence alternative (A) is the correct choice.
Correct Option: A
Due to virtual ground voltage at terminal A is 0 V. KCL at node A
2 = VA – VB 2 2 = 0 – VB 2
or VB = – 4 V
Again KCL at BI = VB – VA + VB – 0 2 2 or I = (–4 – 0) + –4 2 2
or I = – 2 – 2 I = – 4 mA
Hence alternative (A) is the correct choice.
- In the op-amp circuit V0 is given by:
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Due to virtual ground voltage at terminal A is Vs KCL at node A.
Vs + Vs – Vo = 0 R 2R Vs + Vs = Vo R 2R 2R Vs 3R = Vo 2R2 2R
or V0 = 3 Vs
Alternative methodV0 = Vs 1 + 2R = 3 Vs R
Correct Option: C
Due to virtual ground voltage at terminal A is Vs KCL at node A.
Vs + Vs – Vo = 0 R 2R Vs + Vs = Vo R 2R 2R Vs 3R = Vo 2R2 2R
or V0 = 3 Vs
Alternative methodV0 = Vs 1 + 2R = 3 Vs R