# Linear Integrated Circuit Questions and Answers – Slew Rate – 2

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This set of Linear Integrated Circuit Assessment Questions and Answers focuses on “Slew Rate – 2”.

1. 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

Explanation: The slew rate is given as, SR =dVc/dt|max = I/C
Therefore the higher current should be given a small compensating capacitor is used internally or outside an op-amp.

2. Find the expression for full power response.
a) fmax(Hz) =(slew rate×106)/(6.28×Vm)
b) fmax(Hz) =slew rate /(628×106×Vm)
c) fmax(Hz) =(slew rate×Vm×106)/6.28
d) fmax(Hz) =(6.28×Vm× 106)/ slew rate

Explanation: The maximum input frequency at which an undistorted output voltage can be obtained for a peak value of Vm is given as fmax(Hz) =(slew rate×106)/(6.28×Vm)
where fmax is called as full power response.

3. 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

Explanation: Slew rate = dv/dt
=> Time taken = 14-(-14)/ 0.5V/µs = 28v/0.5V/µs = 56µs.
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4. 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

Explanation: 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.

5. 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 maximum output amplitude obtained is 4.2V sine wave?
a) 159mv
b) 0.168mv
c) 207mv
d) 111mv

Explanation: Given, Vm= 4.2Vpeak
∴ the output voltage = 4.2+4.2 =8.4 V peak to peak.
Hence for the output to be undistorted sine wave, the maximum input signal should be less than => 8.4/50= 0.168 = 168mVpeak to peak.

6. What happens if the frequency or amplitude of the input signal is increased to exceed slew rate of the op-amp?
a) All of the mentioned
b) High frequency output
c) Distorted output
d) Large amplitude output

Explanation: Slew rate determines the maximum frequency of operation for a desired output swing. If the slew rate is greater than 2πfVm /106 then the output is distorted, whereas an increase in the frequency /amplitude of input signal distort the output.

7. Compute the peak output amplitude, when the voltage gain verses frequency curve of 741C is flat upto 25Hz.
a) 4Vpeak
b) 9Vpeak
c) 20Vpeak
d) None of the mentioned

Explanation: The slew rate of 741C op-amp = 0.5V/µs. So, the maximum output voltage at 25kHz is SR= (2πfVm)/ 106 V/µs
=> Vm = (SR×106)/(2πf ) = (0.5×106)/(2π×25kHz)
Vm = 3.18Vpeak.

8. Calculate the maximum input frequency at which the output will be distorted from the given specifications
Vo = 30 Vpp ; Slew rate = 0.6v/µs.
a) 1000Hz
b) 10kHz
c) 1kHz
d) 10kHz

Explanation: The minimum time between the two zero crossing is given as
=> 30v/(0.6v/µs) =50µs. Hence the maximize input frequency fmax at which the output get distorted is fmax = 1/(2×50µs) =10000 =10kHz.

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

Explanation: Analysis of difference between three AC parameter; bandwidth, transient response and slew rate.

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