250+ TOP MCQs on Open-Loop Voltage Gain as a Function of Frequency – 1 & Answers

Linear Integrated Circuit Multiple Choice Questions on “Open-Loop Voltage Gain as a Function of Frequency – 1”.

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

A. V_O ={ jX_C /[R_o+(iX_C)]} × AV_id
B. V_O = = AV_id / [1+ j2πfR_oC].
C. V_O = AV_id /(R_o+j2πfC.
D. V_O = V_id / [R_o-(j2πfR_oC.].
Answer: B
Clarification: The output voltage for an op-amp with single break frequency,
V_O = {(-jX_C) / [(R_o)-(jX_C)} ×AV_id
∵ -j=1/j & X_C =1/2πfC
=> V_O = {(1/j2ΠfC./[R_o + (1/ j2πfC.] } × AV_id = AV_id / [1+ j2πfR_oC].

2. Compute the break frequency of an op-amp, if the output resistance=10kΩ and capacitor connected to the output =0.1µF.
A. 159.2Hz
B. 6.28Hz
C. 318.4Hz
D. 1000Hz
Answer: A
Clarification: Break frequency of the op-amp is given as f_o = 1/(2πR_oC.= 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. A_OL(f) = V_O/V_in
B. A_OL(f) = V_O/V_id
C. A_OL(f) = V_O/V_f
D. All of the mentioned
Answer: B
Clarification: The open loop voltage gain as a function of frequency is defined as ratio of output voltage to the difference of input voltages.

4. Which among 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
Clarification: Break frequency f_o depends on the value of capacitors and on output resistance. Therefore, f_o is fixed for any op-amp.

5. Find the gain magnitude and phase angle of the op-amp using the specifications:
f= 50Hz; f_o=5Hz ; A=140000.
A. A_OL(f)= 22.92dB , Φ(f) = – 89.99^o
B. A_OL(f)= 66dB , Φ(f) = – 90^o
C. A_OL(f)= 26dB, Φ(f) = – 89.99^o
D. A_OL(f)= 20dB , Φ(f) = – 84.29^o
Answer: A
Clarification: The open loop gain magnitude |A_OL(f)|= 20log[A/√[1+ f/f_o)²] = 20logA-20 log[A/√ [1+(f/f_o)²] = 20log(140000)- 20log[√(1+(50,000/5)²)]
A_OL(f) dB= 102.922-80 = 22.92dB.
Phase angle, φ(f) = -tan^-1(f/f_o) = -tan^-1(50000/5) = -89.99^o.

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
Clarification: Open loop voltage gain, A_oL(f) = V_o/V_id
V_O = A_OL(f) × (V_in1-V_in2) = 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/f_o)²
B. 20log{A/√[1+(f/f_o)²}
C. A/[1+j(f/f_o)].
D. None of the mentioned
Answer: C
Clarification: The open loop gain of the op-amp A_OL(f) is a complex quantity and is expressed as A_OL(f) = A/[1+ j(f/f_o)] . The remaining equations are expressed in polar form.

8. Determine the difference between two A_OL(f) at 50Hz and 500Hz frequency? (Consider the op-amp to be 741C.
A. 40dB
B. 30dB
C. 20dB
D. 10dB
Answer: C
Clarification: A_OL(f) dB= 20log[√ [1+ (f/f_o)²]
At f= 50 Hz,
A_OL(f) dB = 20log(200000)- 20log(√(1+(50/5)²) = 106.02-20.04 ≅ 86dB
At f= 500Hz
A_OL(f) dB =20log(200000)-20log(√(1+(500/5)²) = 106.02-40 ≅ 66dB
Therefore, the difference between A_OL(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
Clarification: At 0Hz the phase shift between input and output voltage is zero.
At f=0Hz, φ(f) = – tan^-1 (f/f_o) = -tan^-1(0/5) = 0^o

10. At what frequency A_OL(f)=A?
A. 50Hz
B. 10Hz
C. 5Hz
D. 0Hz
Answer: D
Clarification: For any frequency less than break frequency (f_o =5Hz) the gain is approximately constant and is equal to A. For example, f_o =0Hz,
Then A_OL(f) dB= 20log(200000-20log[√1+(0/5)²)] = 106dB. Where A =20,000 ≅ 106dB.

11. What happen 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
Clarification: The open loop voltage gain as a function of frequency is given as A_OL(f) = A/ [√ 1+ (f/f_o)] at 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) continuous 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
Clarification: For any frequency above break frequency, the absolute value of phase shift increases towards 90^o with increase in frequency.

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