Op-Amp Open-Loop Voltage Gain MCQ [Free PDF] – Objective Question Answer for Op-Amp Open-Loop Voltage Gain Quiz

1. 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].


2. 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.


3. 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.


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


5. 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.


6. 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.


7. Express the open-loop gain of the op-amp in complex form?

A. A/√ [1+(f/fo)2
B. 20log{A/√[1+(f/fo)2}
C. A/[1+j(f/fo)].
D. None of the mentioned

Answer: C

The open-loop gain of the op-amp AOL(f) is a complex quantity and is expressed as

AOL(f) = A/[1+ j(f/fo)] .

The remaining equations are expressed in polar form.


8. Determine the difference between two AOL(f) at 50Hz and 500Hz frequencies? (Consider the op-amp to be 741C.

A. 40dB
B. 30dB
C. 20dB
D. 10dB

Answer: C

AOL(f) dB= 20log[√ [1+ (f/fo)2]

At f= 50 Hz,

AOL(f) dB = 20log(200000)- 20log(√(1+(50/5)2) = 106.02-20.04 ≅ 86dB

At f= 500Hz

AOL(f) dB =20log(200000)-20log(√(1+(500/5)2) = 106.02-40 ≅ 66dB

Therefore, the difference between AOL(f)dB = 86-66 = 20dB.


9. At what frequency, the phase shift between input &output voltage will be zero?

A. -40Hz
B. 0Hz
C. -22Hz
D. 20Hz

Answer: B

At 0Hz the phase shift between input and output voltage is zero.

At f=0Hz

φ(f) = – tan-1 (f/fo)

= -tan-1(0/5) = 0o


10. At what frequency AOL(f)=A?

A. 50Hz
B. 10Hz
C. 5Hz
D. 0Hz

Answer: D

For any frequency less than break frequency (fo =5Hz) the gain is approximately constant and is equal to A.

For example, fo =0Hz,

Then AOL(f) dB= 20log(200000-20log[√1+(0/5)2)]

= 106dB.

Where A =20,000 ≅ 106dB.

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