Analog electronics circuits miscellaneous
- If the input to the circuit given below is a sine wave the output will be a:
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Since the given circuit arrangement represent open loop system, so the output will be ± Vsat, which resembles a square wave.
Correct Option: D
Since the given circuit arrangement represent open loop system, so the output will be ± Vsat, which resembles a square wave.
- The op-amp of figure has a very poor open loop voltage gain of 45 but is otherwise ideal.
The gain of the amplifier equals:
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A closed loop gain,
ACL = V0 = AOL Vi 1 + AOL
where, AOL = 45 (given)β = 2k (from given figure) 2k + 8k
or β = 0.2Now, ACL = 45 1 + 45 × ·2 = 45 = 4.5 10
Correct Option: D
A closed loop gain,
ACL = V0 = AOL Vi 1 + AOL
where, AOL = 45 (given)β = 2k (from given figure) 2k + 8k
or β = 0.2Now, ACL = 45 1 + 45 × ·2 = 45 = 4.5 10
- For the op-amp circuit shown below the voltage gain Aν = ν0/νi is:
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Apply KCL at node A
VA – VC + VA + VA – VB = 0 ............(i) R R R
KCL at node BVB – VA + VB – V0 + VB = 0 ............(ii) R R R
KCL at node CVC – Vi + VC – VA = 0 .............(iii) R R
on putting VC = 0 (due to virtual ground) and eliminating VA and VB from equation (i), (ii) and (iii)V0 = – 8 Vi
Correct Option: A
Apply KCL at node A
VA – VC + VA + VA – VB = 0 ............(i) R R R
KCL at node BVB – VA + VB – V0 + VB = 0 ............(ii) R R R
KCL at node CVC – Vi + VC – VA = 0 .............(iii) R R
on putting VC = 0 (due to virtual ground) and eliminating VA and VB from equation (i), (ii) and (iii)V0 = – 8 Vi
- For the circuit shown below gain is Aν = ν0/νi = – 10. The value of R is:
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Given
V0 = – 10 Vi
KCL at node AVA – Vi + VA – VB = 0 ...........(1) 100 R
KCL at node BVB – VA + VB + VB – V0 = 0 ...........(2) R 100 100
on manipulating
VA = 0. (due to virtual ground)
V0 = – 10 Vi
in equation (1) and (2) we get0 – Vi + 0 – VB = 0 100 R or Vi = – VB. 100 .............(3) R VB – 0 + VB + VB = V0 R 100 100 100 or VB 1 + 1 + 1 = –10Vi R 100 100 100 or VB 1 + 2 = –16 · –VB 100 R 100 100 R or 1 + 2 = 10 R 100 R or 10 – 1 = 2 R R 100 or 10R – R = 2 R2 100 or 9 = 2 R 100
or R = 450 kΩ
Correct Option: B
Given
V0 = – 10 Vi
KCL at node AVA – Vi + VA – VB = 0 ...........(1) 100 R
KCL at node BVB – VA + VB + VB – V0 = 0 ...........(2) R 100 100
on manipulating
VA = 0. (due to virtual ground)
V0 = – 10 Vi
in equation (1) and (2) we get0 – Vi + 0 – VB = 0 100 R or Vi = – VB. 100 .............(3) R VB – 0 + VB + VB = V0 R 100 100 100 or VB 1 + 1 + 1 = –10Vi R 100 100 100 or VB 1 + 2 = –16 · –VB 100 R 100 100 R or 1 + 2 = 10 R 100 R or 10 – 1 = 2 R R 100 or 10R – R = 2 R2 100 or 9 = 2 R 100
or R = 450 kΩ
- In circuit shown below, the input voltage νi is 0.2 V. The output voltage ν0 is:
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From figure.
VA = – 50 . Vi 10
or VA = – 5 Viand V0 = – 150 (VA) 25
V0 = – 6. (– 5 Vi)
V0 = 30 Vi
V0 = 30 × 0.2
V0 = 6 V.
Correct Option: A
From figure.
VA = – 50 . Vi 10
or VA = – 5 Viand V0 = – 150 (VA) 25
V0 = – 6. (– 5 Vi)
V0 = 30 Vi
V0 = 30 × 0.2
V0 = 6 V.