Practical Limitations of Op Amp Circuit MCQ [Free PDF] – Objective Question Answer for Practical Limitations of Op Amp Circuit Quiz

31. Which type of op-amp offers relatively broader open-loop bandwidth?

A. Compensated op-amp
B. Uncompensated op-amp
C. Tailored frequency response op-amp
D. Non-compensated op-amp

Answer: B

The uncompensated op-amps offer broader open-loop bandwidth whereas; the internally compensated op-amps have very small open-loop bandwidth.

 

32. How the performance of an op-amp circuit can be improved?

A. By using a non-compensating network
B. By using frequency network
C. By using compensating network
D. None of the mentioned

Answer: C

The compensating networks are used to improve /modify the performance of an op-amp circuit over the desired frequency range by controlling its gain and phase shift.

 

33. Which op-amp requires an external compensating network?

A. Op-amp 771
B. Op-amp 351
C. Op-amp 709
D. Op-amp 741

Answer: C

Op-amp 709 is the first-generation op-amp. Generally, a first-generation op-amp is required for an external compensating network.

 

34. IC 741c op-amp belongs to

A. Compensated op-amp
B. Uncompensated op-amp
C. Non-compensated op-amp
D. None of the mentioned

Answer: A

IC 741c belongs to later generation op-amp and it has an internal compensating network. In an internally compensated op-amp, the compensating network is designed into the circuit to control the gain and phase shift of the op-amp and they are called compensating op-amp.

 

35. Determine the output voltage for an op-amp with single break frequency.

A. VO ={ jXC /[Ro+(iXC)]} × AVid
B. VO = = AVid / [1+ j2πfRoC]
C. VO = AVid /(Ro+j2πfC)
D. VO = Vid / [Ro-(j2πfRoC]

Answer: B

The output voltage for an op-amp with single break frequency,

VO = {(-jXC) / [(Ro)-(jXC)} ×AVid

∵ -j=1/j & XC =1/2πfC

=> VO = {(1/j2ΠfC./[Ro + (1/ j2πfC.] } × AVid

= AVid / [1+ j2πfRoC].

 

36. Compute the break frequency of an op-amp, if the output resistance=10kΩ and the capacitor connected to the output =0.1µF.

A. 159.2Hz
B. 6.28Hz
C. 318.4Hz
D. 1000Hz

Answer: A

Break frequency of the op-amp is given as

fo = 1/(2πRoC.= 1/ (2π×10kΩ×0.1µF)

= 1/ (6.28×10-3) = 159.2Hz.

 

37. The open-loop voltage gain as a function of frequency is defined as

A. AOL(f) = VO/Vin
B. AOL(f) = VO/Vid
C. AOL(f) = VO/Vf
D. All of the mentioned

Answer: B

The open-loop voltage gain as a function of frequency is defined as the ratio of output voltage to the difference of input voltages.

AOL(f) = VO/Vid

 

38. Which of the following factor remain fixed for an op-amp?

A. Open loop voltage gain
B. Gain of the op-amp
C. Operating frequency
D. Break frequency of the op-amp

Answer: D

Break frequency of the op-amp fo depends on the value of capacitors and on output resistance. Therefore, fo is fixed for an op-amp.

 

39. Find the gain magnitude and phase angle of the op-amp using the specifications:

f= 50Hz; fo=5Hz ; A=140000.

A. AOL(f)= 22.92dB , Φ(f) = – 89.99o
B. AOL(f)= 66dB , Φ(f) = – 90o
C. AOL(f)= 26dB, Φ(f) = – 89.99o
D. AOL(f)= 20dB , Φ(f) = – 84.29o

Answer: A

The open loop gain magnitude

|AOL(f)|= 20log[A/√[1+ f/fo)2]

= 20logA-20 log[A/√ [1+(f/fo)2]

= 20log(140000)- 20log[√(1+(50,000/5)2)]

AOL(f) dB= 102.922-80 = 22.92dB.

Phase angle, φ(f) = -tan-1(f/fo) = -tan-1(50000/5) = -89.99o.

 

40. Consider an op-amp where the inverting input voltage =3.7mv, non-inverting input voltage=6.25mv and open-loop voltage gain =142dB. Find the output voltage.

A. 0.21v
B. 0.45v
C. 0.78v
D. 0.36v

Answer: D

Open loop voltage gain, AoL(f) = Vo/Vid

VO = AOL(f) × (Vin1-Vin2)

= 142 dB×(6.25-3.7)

= 142×2.55 = 0.36v.

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