Frequency Response of Non-Compensating Op-Amp MCQ [Free PDF] – Objective Question Answer for Frequency Response of Non-Compensating Op-Amp Quiz

11. 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.

 

12. 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.

 

13. 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.

 

14. 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.

 

15. 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.

 

16. 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.

 

17. 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

 

18. 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.

 

19. What happens when the frequency increases?

A. AOL(f) continues to drop
B. A increases
C. fo –> 0Hz
D. None of the mentioned

Answer: A

The open-loop voltage gain as a function of frequency is given as

AOL(f) = A/ [√ 1+ (f/fo)] A

t a frequency above fo, the denominator value increases, causing the gain, AOL(f) to decrease. Thus, as frequency increases, the gain AOL(f) continues to drop.

 

20. What will be the absolute value of phase shift, if the frequency keeps increasing?

A. Increase towards 45o
B. Decrease towards 45o
C. Increase towards 90o
D. Decrease towards 90o

Answer: C

For any frequency above break frequency, the absolute value of phase shift increases towards 90o with an increase in frequency.

 

21. Which of these statements is false?

A. The open-loop gain AOL(f) dB is approximately constant from 0Hz to fo

B. When input signal frequency and f is equal to break frequency fo, the gain frequency is called -3dB frequency

C. The open-loop gain AOL(f) dB is approximately constant upto break frequency fo, but thereafter it increases 20dB each time there is a tenfold increase in frequency.

D. At unity gain crossover frequency, the open-loop gain AOL(f) dB is zero

Answer: C

When AOL(f) is approximately constant up to break frequency, there will be a 20dB decrease each time there is a tenfold increase in frequency. Therefore it may be considered that the gain roll of at the rate of 20dB/decade.

 

22. What is the maximum phase shift that can occur in an op-amp with a single capacitor?

A. 180o
B. 60o
C. 270o
D. 90o

Answer: D

At corner frequency, the phase angle is -45o (lagging) and at the infinite frequency, the phase angle is -90o. Therefore, a maximum of 90 o phase change can occur in an op-amp with a single capacitor.

 

23. How can the gain roll-off be represented in dB/octave?

A. 12 dB/octave
B. 6 dB/octave
C. 10 dB/octave
D. 8 dB/octave

Answer: B

Octave represents a two-fold increase in frequency. Therefore, 20 gain roll-off at the rate of 20 dB/decade is equivalent to 6 dB/octave.

 

24. Select the correct magnitude and phase for the frequency range.

List-I List-II
1. f1 i. Gain is 3dB down from the value of AOL in dB
2. f=f1 ii. Gain roll-off at the rate of 20 dB/decade
3. f>>f1 iii. Magnitude of the gain is 20logxAOL in dB

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

Answer: A

Properties of magnitude and phase angle characteristics equations.

List-I List-II
1. f1 The magnitude of the gain is 20logxAOL in dB
2. f=f1 Gain is 3dB down from the value of AOL in dB
3. f>>f1 Gain roll-off at the rate of 20 dB/decade

 

25. The specific frequency at which AOL (dB. is called

A. Gain bandwidth product
B. Closed-loop bandwidth
C. Small signal bandwidth
D. All of the mentioned

Answer: D

AOL (dB. is zero at some specific value of input signal frequency called unity-gain bandwidth. The mentioned terms are the equivalent terms for unity-gain bandwidth.

 

26. Find out the expression for open loop gain magnitude with three break frequency?

A. A/[1+ j(f/fo1)]x[1+ j(f/fo2)]x[1+ j(f/fo3)].

B. A/[1+ j(f/fo)]3

C. A/A/[1+ j(f/fo1)]+ [1+ j(f/fo2)]+ [1+ j(f/fo3)].

D. All of the mentioned

Answer: C

The gain equation for op-amp is AOL(f) = A/[1+ j(f/fo1)]x[1+ j(f/fo2)]x[1+ j(f/fo3)].

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