Analog Electronics - 6
10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Analog Electronics - 6
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The current flowing through the 3 Ω resistor R1 is _____ A.
Apply nodal at A:
Magnitude wise: 2 A
An op-amp has a voltage gain of 100 dB at dc and a unity gain frequency of 5 MHz. The lower 3-dB cut off frequency is _______ Hz.
100 dB voltage gain
20 log (Gain) = 100
Gain = 105
GBW = 5 MHz
Low frequency 3-dB = 
= 50 Hz.
An op-amp with slew rate 1 V/μ is used in the circuit below input step voltage Vi = V sin (105t) is given, the maximum value of V such that no raise time distortion occurs at the output is _____V.
The output Vo = -3V sin (ω0t)
For no distortion to occur
3Vω0 ≤ SR
The circuit shows an analog voltmeter of very high input impedance that uses an inexpensive moving coil-meter. The voltmeter measures the voltage ‘V’ applied between the op-amp’s positive-input terminal and ground. Assuming that the moving coil produces full-scale deflection when the current passing through it is 100 μA, the value of R (in kΩ) to obtain full-scale reading at +10 V is
VA = V (virtual ground)
Circuit shows an op-amp circuit used for generating the square waveform.
If the output frequency required in 1 kHz. then the possible values of R1 and R2 can be respectively
Frequency of oscillation
If R1 = 2k
R2 = 0.859 × 2
= 1.7 k
Option 1 is correct.
The circuit shows the series voltage regulator.
If reference voltage Vref is 2V and R1 = 3K, R2 = 1K, β = 99 for both transistors. Then the value of current I0 in micro-amperes is _____ μA.
V- = V+ = 2V
= 2 mA
I1 = (1 + β) (1 + β) I0
2 mA = 104 I0
I0 = 0.2 uA
The circuit shows an – op-amp in Schmidt Trigger configuration. If the hysteresis width is 3V. Then the value of R1 and R2 is
For the amplifier circuit shown, the op-amp can deliver a maximum current of 100 mA and is powered by ±15 V supply. If the input to the amplifier is a triangular waveform as shown. The peak value of the output waveform is ______.
The circuit is non-inverting.
Amplifier:
= 11V
However the current flowing through load when V0 = 11 V is
Which is more than the current op-amp can supply.
Thus peak value is limited by the maximum current:
50 Ω ≪ 110 kΩ
All the current flows through 50 Ω resistor.
V0 = (100 mA) (50 Ω)
= 5 V
An op-amp circuit is shown in the figure. If Vin is an AC source of frequency ω.
Consider the following statements:
a) For ω 
Circuit is an integrator
(b) For ω 
circuit is an Amplifier
Circuit acts like an Integrator
Circuit acts an Inverting Amplifier
The given conditions in question is opposite Hence None of a and b is correct
The average power delivered by the 3 V source is ________ mW.
V- = V+ = 3 V [Virtual ground]
Current through 1 k resistor
I = 1 mA
Voltage across 1k resistor from voltage division
Voltage across 
Vo = 15 V
KVL at V+ node:
3 – I'(8) = Vo
3 – I'(8) = 15
3 – 15 = I'(8)
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21 docs|263 tests
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21 docs|263 tests
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