Analog circuits miscellaneous
- For the circuit shown in figure, true relation is
-
View Hint View Answer Discuss in Forum
At second stage input to both op-amp circuit is same. Upper op-amp circuit is buffer having gain, AV = 1. Lower op-amp circuit is inverting amplifier having gain
Aυ = - R = - 1 R
∴ υ01 = – υ02Correct Option: B
At second stage input to both op-amp circuit is same. Upper op-amp circuit is buffer having gain, AV = 1. Lower op-amp circuit is inverting amplifier having gain
Aυ = - R = - 1 R
∴ υ01 = – υ02
- Assuming operational amplifier to be ideal, gain Vout / Vin for the circuit shown in the given figure is
-
View Hint View Answer Discuss in Forum
Using KCL at the node 1 we haveV - 0 + V + V - Vout = 0 10 1 10
⇒ 12V = Vout ...(i)
Also, using KCL at inverting node, we getVin - 0 = 0 – V 1 10
⇒ V = – Vin. 10 ...(ii)
From equations (i) and (ii), we getVout = -120 Vin
Correct Option: D
Using KCL at the node 1 we haveV - 0 + V + V - Vout = 0 10 1 10
⇒ 12V = Vout ...(i)
Also, using KCL at inverting node, we getVin - 0 = 0 – V 1 10
⇒ V = – Vin. 10 ...(ii)
From equations (i) and (ii), we getVout = -120 Vin
- For the circuit of the given figure with an ideal operational amplifier, maximum phase shift of the output Vout with reference to the input Vin is
-
View Hint View Answer Discuss in Forum
From the circuit,
V+ = Vin 1 + jωRC
and V– =V+ (Ideal OPAMP) NowNow Vin - V– = V– – V0 R1 R1
⇒ V0 = 2V– – Vin = 2V+ – Vin= 2 - 1 Vin = 1 - jωRC V0 1 + jωRC 1 + jωRC
∴ ∠ (V0 / Vi) = - 2 tan-1 ωRC
For – 90 ≤ θ ≤ 90°
Phase-shift ∠ (V0 / Vi) = ± 180°
Correct Option: D
From the circuit,
V+ = Vin 1 + jωRC
and V– =V+ (Ideal OPAMP) NowNow Vin - V– = V– – V0 R1 R1
⇒ V0 = 2V– – Vin = 2V+ – Vin= 2 - 1 Vin = 1 - jωRC V0 1 + jωRC 1 + jωRC
∴ ∠ (V0 / Vi) = - 2 tan-1 ωRC
For – 90 ≤ θ ≤ 90°
Phase-shift ∠ (V0 / Vi) = ± 180°
- In the circuit of shown in the figure, LED will be ON if vi is
-
View Hint View Answer Discuss in Forum
υ_ = (10)(10k) = 5 V 10k + 10k
When υ+ > 5 V, output will be positive and LED will be ON Hence (c) is correct answer.Correct Option: C
υ_ = (10)(10k) = 5 V 10k + 10k
When υ+ > 5 V, output will be positive and LED will be ON Hence (c) is correct answer.
- Assuming that operational amplifier shown in the figure is ideal, the current, I, through 1 kΩ resistor is
-
View Hint View Answer Discuss in Forum
In this case, V0 = 4V
Is = - 4 = - 2 mA 2
But Is = I + 2
∴ I = – 4 mA.Correct Option: A
In this case, V0 = 4V
Is = - 4 = - 2 mA 2
But Is = I + 2
∴ I = – 4 mA.