1. How an AC amplifier can be powered by a single supply voltage and produces voltage swing?

A. By inserting a voltage divider at the inverting input
B. By inserting a voltage divider at the non-inverting input
C. By inserting a voltage divider at the output
D. By inserting a voltage divider at the feedback circuit

Answer: B

A positive dc level is intentionally inserted using a voltage-divider network at the non-inverting input terminal so that output can swing both positively as well as negatively.

2. Find the maximum output voltage swing of an AC inverting amplifier using op-amp 741C?

A. +15V_{pp}
B. ±15V_{pp}
C. ±13V_{pp}
D. +13V_{pp}

Answer: A

The value of power supply for 741 op-amp=±15v. Therefore, the ideal maximum output voltage swing for an AC amplifier with a single power supply = +Vcc = +15v.

3. Determine the lower cut-off frequency of the circuit.

A. 21.3Hz
B. 12.15Hz
C. 1.35Hz
D. None of the mentioned

Answer: D

The input resistance of the amplifier is R_{IF} =(R_{2} ||R_{3}) || [ri×(1+AB.] –> equ 1

As [ri×(1+AB.] >> R_{2}

=> Therefore, equ 1 becomes

R_{IF} ≅ R_{2} || R_{3} = 100kΩ || 100kΩ

= (100×100)/(100+100) = 50kΩ.

=> R_{in} = R_{o} = 150Ω.

∴ Lower cut off frequency

f_{L}= 1/[2πC_{i}×(R_{IF}+R_{o})]

= 1/[2π×0.47µF×(50kΩ+150Ω)] = 6.75Hz.

4. In differential op-amp configuration a subtractor is called as

A. Summing amplifier
B. Difference amplifier
C. Scaling amplifier
D. All of the mentioned

Answer: C

In a subtractor, input signals can be scaled to the desired values by selecting appropriate values for the external resistors. Therefore, this circuit is referred to as a scaling amplifier

5. How the peaking response is obtained?

A. Using a series LC network with op-amp
B. Using a series RC network with op-amp
C. Using a parallel LC network with op-amp
D. Using a parallel RC network with op-amp

Answer: C

The peaking response is the frequency response that peaks at a certain frequency. This can be obtained by using a parallel LC network with the op-amp.

6. The expression for the resonant frequency of the op-amp

A. f_{p} = 1/[2π×√(LC.].
B. f_{p} = (2π×√L)/C
C. f_{p} = 2π×√(LC.
D. f_{p} = 2π/√(LC.

Answer: A

The resonant frequency is also called peak frequency, which is determined by the combination of L and C.

f_{p} = 1/(2π√LC).

7. From the circuit given below find the gain of the amplifier

A. 1.432
B. 9.342
C. 5.768
D. 7.407

Answer: D

Frequency,

f_{p}= 1/[2π×√(LC.]

=1/[2π√(0.1µF×8mH)]

=1/1.776×10_{-4}= 5.63kHz.

=> X_{L} = 2πf_{p}L

= 2π×5.63kHz×8mH = 282.85.

The figure of merit of coil

Q_{coil}= X_{L}/R_{1}= 282.85/100Ω = 2.8285.

∴ R_{p} = (Q_{coil})^{2} ×R_{1}

= (2.8285^{2})×100Ω= 800Ω.

The gain of the amplifier at resonance is maximum and given by

A_{F} =-(R_{F}||R_{p})/R_{1}

= -(10kΩ||800)/100Ω

=-740.740/100 = -7.407.

8. The parallel resistance of the tank circuit and for the circuit is given below. Find the gain of the amplifier?

A. -778
B. -7.78
C. -72.8
D. None of the mentioned

Answer: B

The gain of the amplifier at resonance is the maximum and given by,

A_{F} =-(R_{F}||R_{p})/ R_{1}

=-[(R_{p}×R_{F})/ (R_{F}+R_{p})] /R_{1}

= -[ (10kΩ × 35kΩ)/ (10kΩ+35kΩ)] /1kΩ

=> A_{F} =- 7.78kΩ/1kΩ= -7.78.

9. The band width of the peaking amplifier is expressed as

A. BW = (f_{p}× X_{L})/ (R_{F}+R_{p})
B. BW =[ f_{p}×(R_{F}+R_{p})× X_{L} ] / (R_{F}×R_{p})
C. BW =[ f_{p}×(R_{F}+R_{p})] / (R_{F}×R_{p})
D. BW = [f_{p}×(R_{F}+R_{p}) ]/ X_{L}

Answer: B

The bandwidth of the peaking amplifier,

BW = f_{p}/Q_{p},

where Q_{p} – figure of merit of the parallel resonant circuit