250+ TOP MCQs on Integrator – 2 & Answers

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Linear Integrated Circuit Multiple Choice Questions on “Integrator – 2″.

1. Find the range of frequency between which the circuit act as integrator?
A. [1/(2πRFCF)]– (2πR1CF)
B. (2πRFCF) – [1/(2πR1CF)].
C. [1/(2πRFCF)]- [1/(2πR1CF)].
D. None of the mentioned

Answer: C

2. What will be the output voltage waveform for the circuit, R1×CF=1s and input is a step voltage. Assume that the op-amp is initially nulled.
A. Triangular function
B. Unit step function
C. Ramp function
D. Square function

Answer: C
Clarification: Input voltage Vin = 1.2v for 0≤t≤0.4ms. The output voltage at t=0.4ms is
Vo = (1/R×CF)×t∫0 Vindt+C =-(1/1) × 0.4∫01.2 dt
=> Vo =-[0.1∫01.2 dt + 0.2∫0.11.2 dt + 0.3∫0.21.2 dt + 0.4∫0.31.2 dt ] = -(1.2+1.2+1.2+1.2) = -4.8v

Therefore, the output voltage waveform is a ramp function.

3. Find R1 and RF in the lossy integrator so that the peak gain and the gain down from its peak is 40db to 6db. Assume ω=20,000 rad/s and capacitance = 0.47µF.
A. R1 = 10.6Ω, RF = 106Ω
B. R1 = 21.2Ω, RF = 212.6Ω
C. R1 = 42.4Ω, RF = 424Ω
D. R1 = 29.8Ω, RF = 298Ω

Answer: B
Clarification: The gain of the lossy integration is A=(RF/ R1)/√[1+(ωRFCF)]2
=> A(dB. = 20log{(RF/R1)/√[1+(ωRFCF)]2}
=> 40db = 20log×[(RF/R1)/√1
=> R1 = RF/20.
At ω=20000rad/s, the gain is down by 6db from its peak of 20db and thus is 14db. The gain at 14db => 14db= 20log ×{[ (RF/ RF/20)] / [√(1+(200000×0.47×10-6×RF)2]}
=> 20log[1+(9.4×10-3×RF)2] = 20-14
=> RF = √3/9.4×10-3 = 212.26Ω and R1 = 212.26Ω/10 = 21.2Ω.

4. Why a resistor is shunted across the feedback capacitor in the practical integrator?
A. To reduce operating frequency
B. To enhance low frequency gain
C. To enhance error voltage
D. To reduce error voltage

Answer: D
Clarification: The input current charging the feedback capacitor produces error voltage at the output of the integrator. Therefore, to reduce error voltages a resistor (RF) is connected across the feedback capacitor. Hence, RF minimizes the variation in the output voltage.

5. Find the application in which integrator is used?
A. All of the mentioned
B. Analog Computers
C. FM Detectors
D. AM detectors

Answer: B
Clarification: The integrator is most commonly used in analog computers mainly for signal wave shaping.

6. At what condition the input signal of the integrator is integrated properly
A. T = RFCF
B. T ≤ RFCF
C. T ≥ RFCF
D. T ≠ RFCF

Answer: C
Clarification: The input signal will be integrated properly, if the time period T of the signal is larger than or equal to RF×CF (Feedback resistor and capacitor).

7. Find the output waveform for an input of 5kHz.

Answer: A
Clarification: The output voltage of integrator, Vo = (1/R1×CF)×t∫0 Vindt+C = -[(1/10kΩ×10nF) 0.1ms∫01dt ]= -(104×0.1×10-3) = -1v.
The input is constant amplitude of 2v from 0 to 0.1ms and from 0.1ms to 0.2ms. The output for each of these half periods will be ramp. Thus, the expected output is triangular wave.

8. What happens if the input frequency is kept lower than the frequency at which the gain is zero?
A. Circuit act like a perfect integrator
B. Circuit act like an inverting amplifier
C. Circuit act like a voltage follower
D. Circuit act like a differentiator

Answer: B
Clarification: If the input frequency is lower that the lower frequency limit of the integrator (i.e. when gain = 0), there will be no integration results and the circuit act like a simple inverting amplifier.

9. Match the correct frequency range for integration. (Where f –> Input frequency and fa –> Lower frequency limit of integration)

1.f a i. Results in 50% accuracy in integration
2.f = fa ii. Results in 99% accuracy in integration
3.f = 10fa iii. No integration results

A. 1-iii, 2-i, 3-ii
B. 1-i, 2-ii, 3-iii
C. 1-ii, 2-iii, 1-i
D. 1-iii, 2-ii, 3-i

Answer: A
Clarification: The mentioned answer gives the exact ranges where the integration starts.