Linear Integrated Circuit MCQ [Free PDF] - Objective Question Answer for Linear Integrated Circuit Quiz

81. To obtain a faster slew rate the op-amp should have

A. High current and large compensating capacitor
B. Small compensating capacitor
C. High current or small compensating capacitor
D. Low current or large compensating capacitor

Answer: C

The slew rate is given as,

SR =dV_c/dt|_max = I/C

Therefore the higher current should be given a small compensating capacitor is used internally or outside an op-amp.

82. Find the expression for full power response.

A. f_max(Hz) =(slew rate×10⁶)/(6.28×V_m)
B. f_max(Hz) =slew rate /(628×10⁶×V_m)
C. f_max(Hz) =(slew rate×V_m×10⁶)/6.28
D. f_max(Hz) =(6.28×V_m× 10⁶)/ slew rate

Answer: A

The maximum input frequency at which an undistorted output voltage can be obtained for a peak value of V_m is given as

f_max(Hz) =(slew rate×10⁶)/(6.28×V_m)

where

f_max is called a full power response.

83. Calculate the time taken by the output to swing from +14v to -14v for a 741C op-amp having a slew rate of 0.5V/µs?

A. 22µs
B. 42µs
C. 56µs
D. 70µs

Answer: C

Slew rate = dv/dt

=> Time taken = 14-(-14)/ 0.5V/µs

= 28v/0.5V/µs = 56µs.

84. Consider a square wave having a peak to peak amplitude of 275mv and it is amplified to a peak to peak amplitude of 4v, with rise time of 5.2µs. Calculate the slew rate?

A. 0.615 v/µs
B. 0.712 v/µs
C. 0.325 v/µs
D. None of the mentioned

Answer: A

From the definition of rise time, the change in the output voltage is 5.2µs

△v= (90%-10%)×4v

= (0.9-0.1)×4v =3.2v.

Therefore, slew rate = 3.2v/5.2µs =0.615v/µs.

85. Determine the maximum input signal to be applied to an op-amp to get distortion-free output. If the op-amp used is an inverting amplifier with a gain of 50 and the maximum output amplitude obtained is 4.2V sine wave?

A. 159mv
B. 0.168mv
C. 207mv
D. 111mv

Answer: B

Given, V_m= 4.2V_peak

∴ the output voltage = 4.2+4.2 =8.4 V peak to peak.

Hence for the output to be an undistorted sine wave, the maximum input signal should be less than => 8.4/50= 0.168 = 168mVpeak to the peak.

86. What happens if the frequency or amplitude of the input signal is increased to exceed the slew rate of the op-amp?

A. All of the mentioned
B. High-frequency output
C. Distorted output
D. Large amplitude output

Answer: C

The slew rate determines the maximum frequency of operation for the desired output swing. If the slew rate is greater than 2πfV_m /10⁶ then the output is distorted, whereas an increase in the frequency /amplitude of the input signal distorts the output.

87. Compute the peak output amplitude, when the voltage gain versus frequency curve of 741C is flat upto 25Hz.

A. 4V_peak
B. 9V_peak
C. 20V_peak
D. None of the mentioned

Answer: D

The slew rate of 741C op-amp = 0.5V/µs. So, the maximum output voltage at 25kHz is SR= (2πfV_m)/ 10⁶ V/µs

=> V_m = (SR×10⁶)/(2πf )

= (0.5×10⁶)/(2π×25kHz)

V_m = 3.18V_peak.

88. Calculate the maximum input frequency at which the output will be distorted from the given specifications
V_o = 30 V_pp ; Slew rate = 0.6v/µs.

A. 1000Hz
B. 10kHz
C. 1kHz
D. 10kHz

Answer: D

The minimum time between the two zero crossings is given as

=> 30v/(0.6v/µs) =50µs.

Hence the maximize input frequency f_max at which the output get distorted is

f_max = 1/(2×50µs)

=10000 =10kHz.

89. Match the AC parameter of the op-amp in column 1 with column 2.

Column 1	Column-2
1. Bandwidth	i . a large signal phenomenon
2. Transient response	ii. Rise time is related to bandwidth and overshoots measure stability
3. Slew rate	iii. Depends on compensating components and closed-loop gain

A. 1-i 2-iii 3-ii
B. 1-ii 2-iii 3-i
C. 1-iii 2-ii 3-i
D. 1-iii 2-i 3-ii

Answer: C

Analysis of difference between three AC parameters; bandwidth, transient response, and slew rate.

Column 1	Column-2
1. Bandwidth	Depends on compensating components and closed-loop gain
2. Transient response	Rise time is related to bandwidth and overshoots measure stability
3. Slew rate	a large signal phenomenon

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