Previous Year Questions- Operational Amplifiers - 2 | Analog and Digital Electronics - Electrical Engineering (EE) PDF Download

Q16: The operational amplifier shown in the figure is ideal. The input voltage (in Volt) is V_i = 2sin(2π × 2000t). The amplitude of the output voltage V_o (in Volt) is _______. (SET-2  (2015))

(a) 0.14
(b) 1.24
(c) 2.25
(d) 2.44
Ans: (b)
Sol: V_i = 2sin(2π × 2000t)
The transfer function of the system is
The input is 2sin⁡(2π × 2000t)
ω = 4000π
V₀ = 2∣H(jω)∣_{ω = 4000π} × sin(4000πt − ϕ)  
Output = 1.245sin(4000πt − 51.46°)
So, amplitude of output is 1.245.

Q17: The op-amp shown in the figure has a finite gain A = 1000 and an infinite input resistance. A step-voltage V_i = 1mV is applied at the input at time t = 0 as shown. Assuming that the operational amplifier is not saturated, the time constant (in millisecond) of the output voltage  V_o is (SET-1 (2015))

(a) 1001
(b) 101
(c) 11
(d) 1
Ans: (a)
Sol: The circuit is
Now, the gain of op-amp is 1000
So, V_o = 1000(V₊ − V₋)
Since V₊ = 0 (grounded)
V_- = V_A
The above circuit can be redrawn as
So, V_o = 1000VA
Now, KCL at node 'A'
V_A − V_i = (1001) × 1000V_AsC
V_A − V_i = s(1001000 × 10⁻⁶)V_A
V_A − s1.001V_A = V_i
Since, V_o = −1000V_A
Since, pole is at (1/1.001), so time constant is approax '1001'.

Q18: Consider the circuit shown in the figure. In this circuit R = 1kΩ, and C = 1μF. The input voltage is sinusoidal with a frequency of 50 Hz, represented as a phasor with magnitude V_i and phase angle 0 radian as shown in the figure. The output voltage is represented as a phasor with magnitude V_o and phase angle δ radian. What is the value of the output phase angle δ (in radian) relative to the phase angle of the input voltage? (SET-1 (2015))
(a) 0
(b) π
(c) $\pi /2$π/2
(d) −π/2
Ans: (d)
Sol: The circuit is
Since circuit has negative feedback, so with help of virtual short V_A = V_B
So, KVL in the loop from A to B given,
So, V_A = −IR
and V_A = V_B
So,

Now,

So, V_o = V_B − IR

= −V_i(sCR)
So, V_o = −jωRCV_i
So, V_o lag V_i by 90° or phase of V_o w.r.t V_i is −90°.  

Q19: Of the four characteristics given below, which are the major requirements for an instrumentation amplifier?
P. High common mode rejection ratio
Q. High input impedance
R. High linearity
S. High output impedance (SET-1(2015))
(a) P, Q and R only
(b) P and R only
(c) P, Q and S only
(d) Q, R and S only
Ans: (a)

Q20: The transfer characteristic of the Op-amp circuit shown in figure is  (SET-3(2014))
(a) (b) (c) (d) Ans: (c)
Sol: CASE-I: V_i > 0, the circuit will look like
Hence V_o = 0
CASE-II: V_i < 0, the circuit will look like  
V_o = −V₀₁ ...(iii)
From equation (ii) and (iii),
V_o = V_i
 
Q21: An operational amplifier circuit is shown in the figure.
The output of the circuit for a given input v_i is (SET-3 (2014))
(a)

(b)

(c)

(d) +V_sat or −V_sat

Ans: (d)
Sol: The circuit of op-amp '1' is a Schmitt trigger, therefore
V₀₁ = ±V_sat
and the circuit of op-amp '2' is a non-inverting amplifier
⇒ V₀ = 2V₀₁
where, V₀₁ = ±V_sat
Therefore, the answer is V₀ = ±V_sat

Q22: A 10 kHz even-symmetric square ware is passed through a bandpass filter with centre frequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is (SET-2 (2014))
(a) a highly attenuated square wave at 10 kHz
(b) nearly zero
(c) a nearly perfect cosine wave at 30 kHz
(d) a nearly perfect sine wave at 30 kHz
Ans: (c)

Q23: In the figure shown, assume the op-amp to be ideal. Which of the alternatives gives the correct Bode plote for the transfer function  (SET-1 (2014))
(a) (b) (c) (d) Ans: (a)
Sol: This filter is considered as low-pass filter (LPF). The transfer function for LPF ω_c (corner frequency)= 10³ rad/sec.
tha gain at low frequencies,

Q24: Given that the op-amps in the figure are ideal, the output voltage V_o is (SET-1 (2014))
(a) (V₁ − V₂)
(b) 2(V₁ − V₂)
(c) (V₁ − V₂)/2
(d) $(V1+V2)$(V₁ + V₂)
Ans: (b)
Sol: Op-amp '3' circuit is a differential amplifier so,
V_o = V₀₁ − V₀₂ ...(i)
Now, apply KCL at node '2'
and apply KCL at node '1'
From equation (i), (ii) and (iii), we get
V₀ = 2(V₁ − V₂)

Q25: In the circuit shown below the op-amps are ideal. Then, V_out in Volts is (2013)
(a) 4
(b) 6
(c) 8
(d) 10
Ans: (c)
Sol: Output of the first op-amp

Q26: In the circuit shown below what is the output voltage (V_out) if a silicon transistor Q and an ideal op-amp are used? (2013)
(a) -15 V
(b) -0.7V
(c) +0.7V
(d) +15V
Ans: (b)
Sol: Using the concept of vitual ground, V = 0
V_out = −0.7V

Q27: The circuit shown is a (2012)
(a) low pass filter with f_3dB =

(b) high pass filter with f_3dB =
(c) low pass filter with f_3dB =
(d) high pass filter with f_3dB =
Ans: (b)
Sol: It is a high pass filter with

Q28: For the circuit shown below,
The CORRECT transfer characteristic is (2011)
(a) (b) (c) (d) Ans: (d)
Sol: First section is differential amplifier having gain-1.
Output is
Second stage-schmitt trigger

Q29: A low-pass filter with a cut-off frequency of 30 Hz is cascaded with a high pass filter with a cut-off frequency of 20 Hz. The resultant system of filters will function as (2011)
(a) an all-pass filter
(b) an all-stop filter
(c) an band stop (band-reject) filter
(d) a band-pass filter
Ans: (d)
Sol: It is band pass filter.

Q30: Given that the op-amp is ideal, the output voltage V₀ is (2010)
(a) 4V
(b) 6V
(c) 7.5V
(d) 12.12V
Ans: (b)
Sol: = (1 + 2)(2) = 6V