Open-Loop Voltage Gain as Function of Frequency MCQ [Free PDF] - Objective Question Answer for Open-Loop Voltage Gain as Function of Frequency Quiz

11. What happens when the frequency increases?

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

Answer: A

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

A_OL(f) = A/ [√ 1+ (f/f_o)] A

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

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

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

Answer: C

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

13. Which of these statements is false?

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

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

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

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

Answer: C

When A_OL(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.

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

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

Answer: D

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

15. 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.

16. 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 A_OL 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 20logxA_OL 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 20logxA_OL in dB
2. f=f₁	Gain is 3dB down from the value of A_OL in dB
3. f>>f₁	Gain roll-off at the rate of 20 dB/decade

17. The specific frequency at which A_OL (dB. is called

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

Answer: D

A_OL (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.

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

A. A/[1+ j(f/f_o1)]x[1+ j(f/f_o2)]x[1+ j(f/f_o3)].

B. A/[1+ j(f/f_o)]³

C. A/A/[1+ j(f/f_o1)]+ [1+ j(f/f_o2)]+ [1+ j(f/f_o3)].

D. All of the mentioned

Answer: C

The gain equation for op-amp is A_OL(f) = A/[1+ j(f/f_o1)]x[1+ j(f/f_o2)]x[1+ j(f/f_o3)].

19. Find out the frequencies that are avoided in the frequency response plot?

A. None of the mentioned
B. Single break frequency
C. Upper break frequency
D. Lower break frequency

Answer: C

Op-amps have more than one break frequency because only a few capacitors are present. Often upper break frequencies are well above the unity-gain bandwidth. So, they are avoided in the frequency response plot.

20. In a phase response curve of MC1556, the phase shift is -162.5^o about 4MHz. If the approximate value of the first break frequency is 5.5Hz. Determine the approximate value of the second break frequency?

A. 945.89MHz
B. 945.89kHz
C. 945.89GHz
D. 945.89Hz

Answer: B

The phase angle equation is φ(f) = – tan^-1 (f/f_o1)- tan^-1(f/f_o2)

=> -162.5^o = – tan^-1(3MHz/5.5) – tan^-1(3MHz/f_o2)

=> tan^-1(3MHz/f_o2) = +162.5^o-89.99^o

=> 3MHz/f_o2= tan(72.50^o)

=> f_o2 = 3MHz/tan(72.50^o)=945.89kHz.

21. Consider a practical op-amp having two break frequencies due to a number of RC pole pairs. Determine the gain equation using the given specifications:
A ≅ 102.92dB; f_o1 =6Hz ; f_o2 =2.34MHz .

A. 140000/[1+ j(f/6)]x[1+ j(f/2.34)].
B. 140000/[1+ j(f/6)]x[1+ j(f/234)].
C. 140000/[1+ j(f/6)]x[1+ j(f/23400)].
D. 140000/[1+ j(f/6)]x[1+ j(f/2340000)].

Answer: D

Gain of op-amp at 0Hz, A = 10^102.92/20

=> A = 10^5.146 =139958.73 ≅ 140000.

The gain equation with two break frequency value is given as

A_OL(f) = A/[1+ j(f/f_o1)]x[1+ j(f/f_o2)]

=140000/[1+ j(f/6)]x[1+ j(f/2.34)].

22. What is the voltage transfer function of the op-amp in the s-domain.

A. S×A/(s+ω_o)
B. A/(s+ω_o)
C. (A×ω_o)/(s+ω_o)
D. None of the mentioned

Answer: C

For the voltage transfer function in s-domain is expressed as

A_OL(f) = A/[1+ j(f/f_o)]

=A/[1+j(ω/ω_o)]

= A×ω_o/(jω/jω_o)

= A×ω_o / (s+ω_o).

23. Calculate the unity-gain bandwidth of an op-amp. Given, A≅ 0.2×10⁶ & f_o =5Hz?

A. 10⁶
B. 10⁸
C. 10^-5
D. 10^-10

Answer: A

The break frequency, f_o = UGB/A

=f_o×A

= 0.2×10⁶ × 5 = 10⁶.

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