250+ TOP MCQs on AC Amplifiers with Single Supply Voltage & Answers

Linear Integrated Circuit Multiple Choice Questions on “AC Amplifiers with single supply voltage”.

1. How an AC amplifier can be powered by a single supply voltage, 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
Clarification: 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. Given below are the circuits connected to the non-inverting input terminal of AC amplifier. Choose the circuit which produces positive output swing equal the negative?

Answer: B
Clarification: The circuit is generally a positive dc level (+Vcc/2), so the positive output swing will be equal to negative and this is accomplished by selecting R₂ = R₃.

3. What is the output waveform at the point V_O1 in the given circuit? (Take R₁=1kΩ and R_F=5kΩ)


Answer: C
Clarification: The input is applied at the inverting terminal so, the AC output will be riding on a DC level of (+Vcc/2) volts and will be inverted.
V_o = -(R_F/R₁)×V_in = -(5kΩ/1kΩ)×6V_p = 30V_p sinewave.

4. The input waveform of an AC non-inverting amplifier with single supply is given below. Find the output waveform?


Answer: B
Clarification: The coupling capacitor at the output blocks the AC output signal riding on a DC level of (+Vcc/2) volts and the resultant waveform will be AC. In non-inverting amplifier, the output will be of the same phase as that of input signal.

5. 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
Clarification: The value of power supply for 741 op-amp=±15v. Therefore, the ideal maximum output voltage swing for an AC amplifier with single power supply = +Vcc = +15v.

6. 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
Clarification: The input resistance of the amplifier is R_IF =(R₂ ||R₃) || [ri×(1+AB.] –> equ 1
As [ri×(1+AB.] >> R₂
=> Therefore, equ 1 becomes
R_IF ≅ R₂ || R₃ = 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.

7. 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
Clarification: 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 scaling amplifier.

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