Previous Year Questions- Operational Amplifiers - 3 | Analog and Digital Electronics - Electrical Engineering (EE) PDF Download

Q31: An ideal op-amp circuit and its input wave form as shown in the figures. The output waveform of this circuit will be (2009)
(a) (b) (c) (d) Ans: (d)
Sol: When V_i < V₁(upto t₂); V_o = +ve
When V_i > V₁(t₂ ≤ t ≤ t₄); V_o = −ve

Q32: Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power P_in and output power P_out in both the cases is (2009)
(a) $Pin=Pout$P_in = P_out for both transformer and emitter follower
(b) $Pin>Pout$P_in > P_out for both transformer and emitter follower
(c) P_in < P_out for transformer and P_in = P_out for emitter follower
(d) $Pin=Pout$P_in = P_out for transformer and P_in < P_out for emitter follower
Ans: (d)
Sol: For emitter follower
A_V ≅ 1; A_I is high
⇒ A_P = A_V⋅A_I is high ⇒ P_out > P_in
For transformer, P_in = P_out

Q33: The following circuit has R = 10kΩ, C = 10μF. The input voltage is a sinusoidal at 50 Hz with an rms value of 10 V. Under ideal conditions, the current i_s from the source is (2009)
(a) 10π mA leading by 90%
(b) 20π mA leading by 90%
(c) 10 mA leading by 90%
(d) 10π mA lagging by 90%
Ans: (d)
Sol: I_s = −V_sjωC
= −j(10 × 2π × 50 × 10 × 10⁻⁶)
= −j10π...mA
= 10π mA lagging by 90°

Q34: The nature of feedback in the op-amp circuit shown is (2009)
(a) Current-Current feedback
(b) Voltage-Voltage feedback
(c) Current-Voltage feedback
(d) Voltage-Current feedback
Ans: (b)
Sol: Voltage-series feedback arrangement or voltage-voltage feedback.

Q35:A general filter circuit is shown in the figure :
The output of the above circuit is given to the circuit shown below in figure :
The gain v/s frequency characteristic of the output (v_o) will be (2008)
(a) (b) (c) (d) Ans: (d)
Sol: Which is similar to equation of a band pass filter.

Q36: A general filter circuit is shown in the figure :
If R₁ = R₂ = R_A and R₃ = R₄ = R_B, the circuit acts as a (2008)
(a) all pass filter
(b) band pass filter
(c) high pass filter
(d) low pass filter
Ans: (c)
Sol: Let, V_i only on onverting terminal
Here,

Let, V_i only only on non-inverting terminal  

Putting the value of R_f, we get
So, it is high pass filter.

Q37: A waveform generator circuit using OPAMPs is shown in the figure. It produces a triangular wave at point 'P' with a peak to peak voltage of 5 V for v_i = 0V.
If the voltage v_i is made +2.5 V, the voltage waveform at point 'P' will become (2008)
(a)
 (b) (c) (d) Ans: (a)

Q38: The block diagrams of two of half wave rectifiers are shown in the figure. The transfer characteristics of the rectifiers are also shown within the block.
It is desired to make full wave rectifier using above two half-wave rectifiers. The resultants circuit will be (2008)
(a) (b) (c) (d) Ans: (b)
Sol: For (V_in > 0),
P → V₀₁ = negative
Q → V₀₂ = 0
V₀ will be positive due to inverting action.
For (V_in < 0),
P → V′₀₁ = 0
Q → V′₀₂ = positive
V₀ will be positive due to non-inverting action. So, output is always rectified.

Q39: The switch S in the circuit of the figure is initially closed, it is opened at time t = 0. You may neglect the zener diode forward voltage drops. What is the behavior of V_out for t > 0? (2007)
(a) It makes a transition from -5 V to +5 V at t = 12.98 μs
(b) It makes a transition from -5 V to +5 V at t = 2.57 μs
(c) It makes a transition from +5 V to -5 V at t = 12.98 μs
(d) It makes a transition from +5 V to -5 V at t = 2.57 μs
Ans: (c)
Sol: Let the voltage at the non-inverting terminal of the opamp be V_a volts and the voltage at inverting terminal be V_b volts.
At t = 0⁺ switch is open V_o = 5V and V_b = −10V
As t → ∞, V_b → 10 V
V_b = V_f+(V_i−V_f)e^−t/RC....(i)
Putiing , V_b = (50/11) at t = T₁
V_i = −10V and V_f = 10V in eq.(i)
We get, T₁ = 12.98 μsec
∴ At t = 12.98μsec, V_o changes from 5V to -5V.  

Q40: The circuit shown in the figure is (2007)
(a) a voltage source with voltage

(b) a voltage source with voltage
(c) a current source with current
(d) a current source with current
Ans: (d)
Sol: It behaves as current source because the output current I_o depends upon (V_in) and resistance only.
Where,

Q41: A relaxation oscillator is made using OPAMP as shown in figure. The supply voltages of the OPAMP are ±12 V. The voltage waveform at point P will be (2006)
(a) (b) (c) (d) Ans: (a)
Sol: Output will be either at +V_sat or −V_sat. When output will be at +V_sat diodde connected to 10kΩ resistance will be on making voltage at point P equal to 6V.
When output will be at −V_sat diodde connected to 2kΩ resistance will be on making voltage at point P equal to -10V.

Q42: For a given sinusoidal input voltage, the voltage waveform at point P of the clamper circuit shown in figure will be (2006)
(a) (b) (c) (d) Ans: (d)
Sol: For -ve half of the input +ve terminal of OPAMP is at higher potential than -ve terminal and output goes to +V_sat but due to this high potential diode gets on and restricts the output to 0.7V only.
And for +ve half of the input +ve terminal of OPAMP is at lower potential than -ve terminal's potential and output goes to −V_sat and remains at −V_sat.

Q43: In the given figure, if the input is a sinusoidal signal, the output will appear as shown (2005)
 (a) (b) (c) (d) Ans: (c)
Sol: Output will be at its saturation values and it is having a phase difference of (180°).
For +ve half of the input, first diode will be on making -ve terminal of op-amp to 0.7V larger than the voltage at +ve input, so output will be −V_sat.
For -ve half of the input reverse of the above will happen and output will go +V_sat.

Q44: Consider the inverting amplifier, using an ideal operational amplifier shown in the figure. The designer wishes to realize the input resistance seen by the smallsignal source to be as large as possible, while keeping the voltage gain between -10 and -25. The upper limit on R_F is 1 MΩ. The value of R₁ should be (2005)
(a) Infinity
(b) 1MΩ
(c) 100kΩ
(d) 40kΩ
Ans: (d)
Sol: If gain = -25 then 

If gain = -10 then

So if we keep R₁ to be 100kΩ then we never get the gain -25 for any R_F so we can keep R₁ to be 40kΩ.

Q45: The input resistance   of the circuit in figure is (2004)
(a) +100kΩ
(b) $-100k\Omega$−100kΩ
(c) +1MΩ
(d) −1MΩ
Ans: (b)
Sol: Here,

But,

Q46: Assuming the operational amplifier to be ideal, the gain V_out/V_in for the circuit shown in figure is (2004)
(a) -1
(b) -20
(c) -100
(d) -120
Ans: (d)
Sol: Using KCL at node 1, we have,
12V₁ = V_out   ...(i)
Also, using KCL at inverting node, we get
V₁ = −V_in × 10 ...(ii)  
From equation (i) and (ii), we get

Q47: For the circuit of figure with an ideal operational amplifier, the maximum phase shift of the output V_out with reference to the input V_in is (2003)
(a) 0°
(b) -90°
(c) +90°
(d) ±180°
Ans: (d)
Sol:where, V₍₋₎ = V₍₊₎ [For ideal Op-amp]
For −90° ≤ θ ≤ 90°, maximum phase shift occurs (+180°).  

Q48: An op-amp, having a slew rate of 62.8 V/μsec, is connected in a voltage follower configuration. If the maximum amplitude of the input sinusoid is 10V, then the minimum frequency at which the slew rate limited distortion would set in at the output is (2001)
(a) 1 MHz
(b) 6.28 MHz
(c) 10 MHz
(d) 62.8 MHz
Ans: (a)
Sol: S.R. = ωV_m
62.8V/μs = 2πf × 10