Previous Year Questions- Operational Amplifiers - 1 | Analog and Digital Electronics - Electrical Engineering (EE) PDF Download

Q1: A difference amplifier is shown in the figure. Assume the op-amp to be ideal. The CMRR (in dB) of the difference amplifier is _____(rounded off to 2 decimal places). (2024)
(a) 12.36
(b) 36.25
(c) 40.52
(d) 48.44
Ans: (c)
Sol: CMRR = A_d/A_c
A_c = A₁ + A₂
= 9.71 - 9.619 = 0.091
CMRR = (9.6645/0.091) = 106.20
CMRR = 20log(106.20) = 40.52 dB

Q2: The current gain (I_out/I_in) in the circuit with an ideal current amplifier given below is (2022)
(a) C_f/C_c
(b) -C_f/C_c
(c) C_c/C_f
(d) -C_c/C_f
Ans: (c)
Sol: Redraw the circuit:
From circuit,
V_o = V_c
Q3: The output impedance of a non-ideal operational amplifier is denoted by Z_out. The variation in the magnitude of Z_out with increasing frequency, f, in the circuit shown below, is best represented by (2022)
(a) (b) (c) (d) Ans: (c)
Sol: Bode plot of negative feedback amplifier:
Given amplifier is a voltage series feedback amplifier.
Therefore, Output impedance is given by
From Bode plot ; at low frequency, the open loop gain (A) is constant.
when ω↑, A↓, Z_out↑
At A = 0, Z_out = Z_o → constant
Therefore, z_out with frequency represented by

Q4: The steady state output (V_out), of the circuit shown below, will (2022)
(a) saturate to +V_DD
(b) saturate to −V_EE
(c) become equal to 0.1 V
(d) become equal to -0.1 V
Ans: (b)
Sol: Redraw the circuit:
From circuit,
Hence, V_out = −V_EE

Q5: A CMOS Schmitt-trigger inverter has a low output level of 0 V and a high output level of 5 V. It has input thresholds of 1.6 V and 2.4 V. The input capacitance and output resistance of the Schmitt-trigger are negligible. The frequency of the oscillator shown is ____________ Hz. (Round off to 2 decimal places.) (2021)
(a) 1245.23
(b) 3157.56
(c) 258.36
(d) 965.24
Ans: (b)
Sol: v_c(t)  = v_{c final} +[v_initial −v_c′ final] e^−t/RC 
Charging, v_c(t) = 5 + (1.6 − 5)e^−t/RC
= 5 − 3.4e^−t/RC  
t = t₁, v_c(t) = 2.4 V
2.4 = 5 - 3.4e ^-t/RC
3.4e^-11/RC = 2.6
Discharging,
v_c(t) = 0 + (2.4 − 0)e^−t/RC
= 2.4e^−t/RC  
t = t₂, v_c(t_a) = 1.6 V
1.6 = 2.4e^−12/AC

t₂ = In (2.4/1.6) RC = 0.405 × RC 

T = t₁ + t₂ = (0.268 + 0.405) RC
T = 0.673RC
f = 3157.46 Hz

Q6: The temperature of the coolant oil bath for a transformer is monitored using the circuit shown. It contains a thermistor with a temperature-dependent resistance, R_thermistor = 2(1 + αT)kΩ. Where T is the temperature in °C. The temperature coefficient α, is −(4 ± 0.25)%/°C. Circuit parameters: R₁ = 1kΩ, R₂ = 1.3kΩ, R₃ = 2.6kΩ. The error in the output signal (in V. rounded off lo 2 decimal places) at 150°C is ________. (2020)
(a) 0.01
(b) 0.08
(c) 0.04
(d) 0.06
Ans: (c)
Sol: As per GATE official answer key MTA (Marks to ALL)
Given data,
R_thermistor = R_th = 2(1 + αT)KΩ
α = −(4 + 0.25)%/°C = −(0.04 ± 0.0025)°C
α_max = −0.0424/°C, α_min = −0.375/°C
Temperature, T = 150°C
R₁ = 1KΩ, R₂ = 1.3KΩ, R₃ = 2.6KΩ
Considering, α = -0.04
R_th = 2[1 − 0.04 × 150] = −10KΩ
= −0.5V
Case-1:
Considering, α_max = −0.0425/°C
R_thmax = 2[1 + (-0.0425) x 150] = -10.75 kΩ
Case-2:
Considering, α_max = −0.0375/°C
R_thmax = 2[1 + (-0.0375) x 150] = -9.25 kΩ
 = -0.54 V
Output voltage, V₀ = 0.5 ± 0.04 ⇒ Error  = 0.04

Q7: A common-source amplifier with a drain resistance, R_D = 4.7kΩ, is powered using a 10 V power supply. Assuming that the transconductance, g_m, is 520μA/V, the voltage gain of the amplifier is closest to:  (2020)
(a) -2.44
(b) -1.22
(c) 1.22
(d) 2.44
Ans: (a)
Sol: Given data:
R_D = 4.7KΩ, g_m = 520μA/V
Voltage gain of CS amplifier
= −g_mR_D = −520μA/V × 4.7kΩ = −2.44

Q8: In the circuit below, the operational amplifier is ideal. If V₁ = 10 mV and  V₂ = 50 mV, the output voltage (V_out) is (2019)
(a) 100 mV
(b) 400 mV
(c) 500 mV
(d) 600 mV
Ans: (b)
Sol: V₀ = R₂/R₁ (V₂ - V₁)
= 100k/10k (50mV - 10mV)
= 10(40 mV) = 400 mV

Q9: The op-amp shown in the figure is ideal. The input impedance v_in/i_in is given by (2018)
(a) Z(R₁/R₂)
(b) -Z(R₁/R₂)
(c) Z
(d) Z(R₁/(R₁+R₂))
Ans: (b)
Sol: According to virtual ground,
V_A = V_B = V_in
At node A,
Equation (ii) in euation (i),

Q10: For the circuit shown below, assume that the OPAMP is ideal.
Which one of the following is TRUE? (SET-2(2017))
(a) $vo=vs$v_o = v_s 
(b) $vo=1.5vs$v_o = 1.5v_{s
(c) $vo=2.5vs$}v_o = 2.5v_s  
(d) v_o = 5v_s 
Ans: (c)
Sol: 
Q11: The approximate transfer characteristic for the circuit shown below with an ideal operational amplifier and diode will be (SET-1 (2017))
(a) (b) (c) (d) Ans: (a)
Sol: V_in > 0 V_A = +V₅₅, D on, V_o = V_in
V_in < 0 V_A = −V₅₅,  D off, V_o = 0

Q12: For the circuit shown below, taking the opamp as ideal, the output voltage V_out in terms of the input voltages V₁, V₂ and V₃ is (SET-2(2016))
(a) 1.8V₁ + 7.2V₂ − V₃
(b) 2V₁ + 8V₂ − 9V₃
(c) 7.2V₁ + 1.8V₂ − V₃
(d) 8V₁ + 2V₂ − 9V₃ 
Ans: (d)
Sol: V_out = −9V₃ + 10V_A
= −9V₃ + 8V₁ + 2V₂

Q13: The circuit shown below is an example of a (SET-2(2016))
(a) low pass filter.
(b) band pass filter
(c) high pass filter.
(d) notch filter.
Ans: (a)
Sol: So the system is a low pass filter.

Q14: The saturation voltage of the ideal op-amp shown below is ±10 V. The output voltage v₀ of the following circuit in the steady-state is (SET-2 (2015))
(a) square wave of period 0.55 ms.
(b) triangular wave of period 0.55 ms
(c) square wave of period 0.25 ms.
(d) triangular wave of period 0.25 ms.
Ans: (a)
Sol: The given circuit in a astable multivibrator, so output will be a periodic square wave and from the circuit β = 0.5.
So, time period will be 2τ ln ⁡(1+β)/(1−β)

Q15: The filters F1 and F2 having characteristics as shown in Figures (a) and (b) are connected as shown in Figure (c).
The cut-off frequencies of F1 and F2 are f₁ and f₂ respectively. If f₁ < f₂ , the resultant circuit exhibits the characteristic of a  (SET-2(2015))
(a) Band-pass filter
(b) Band-stop filter
(c) All pass filter
(d) High-Q filter
Ans: (b)
Sol: To check the type of system:
we apply a delta function at input V_i = δ(t)
V(f) = 1∀f
So, voltage at A will be same as voltage at output V_o. V_A will be equal to voltage due to F₁+ voltage due to F₂. Since, f₁ < f₂, so V_A/V_i will be
so, the system works as a band stop filter.