Linear Integrated Circuit MCQ [Free PDF] - Objective Question Answer for Linear Integrated Circuit Quiz

51. 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.

52. 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)].

53. 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.

54. 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.

55. 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)].

56. 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).

57. 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.

58. Open loop configuration is not preferred in op-amps because

A. First break frequency is too large
B. First break frequency is very small
C. Second break frequency is too large
D. All of the mentioned

Answer: B

The bandwidth of the op-amp is simply the first break frequency. As the value of bandwidth is very small, the open-loop configuration is of little use or not preferred in the op-amp.

59. Calculate the value of the open-loop frequency response curve at any point beyond break frequency in 741C op-amp?

A. 1000000Hz
B. 1000Hz
D. 10000000Hz
D. 100000Hz

Answer: A

For a 741C op-amp the gain=200000 and the break frequency = 5Hz. Therefore, the unity-gain bandwidth or the product of the coordinates (gain and frequency) of any point beyond the break frequency

= 200000×5Hz =1MHz =1000000Hz.

60. When does a system said to be stable?

A. Output reaches a minimum value at finite time
B. Output reaches a maximum value at any time
C. Output reaches a fixed value at finite time
D. Output reaches a fixed value at any time

Answer: C

A circuit or a group of circuits connected together as a system is said to be stable if its output reaches a fixed value in a finite time.

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