This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses 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
Explanation: 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?
Explanation: 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 VO1 in the given circuit? (Take R1=1kΩ and RF=5kΩ)
Explanation: 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.
Vo = -(RF/R1)×Vin = -(5kΩ/1kΩ)×6Vp = 30Vp sinewave.
4. The input waveform of an AC non-inverting amplifier with single supply is given below. Find the output waveform?
Explanation: 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?
Explanation: 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.
d) None of the mentioned
Explanation: The input resistance of the amplifier is RIF =(R2 ||R3) || [ri×(1+AB)] –> equ 1
As [ri×(1+AB)] >> R2
=> Therefore, equ 1 becomes
RIF ≅ R2 || R3 = 100kΩ || 100kΩ = (100×100)/(100+100) = 50kΩ.
=> Rin = Ro = 150Ω.
∴ Lower cut off frequency, fL= 1/[2πCi×(RIF+Ro)] = 1/[2π×0.47µF×(50kΩ+150Ω)] = 6.75Hz.
Sanfoundry Global Education & Learning Series – Linear Integrated Circuit.