Linear Integrated Circuit Questions and Answers – Integrator – 2


This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

Answer: c
Find the range of frequency from the circuit acting as integrator
For a practical integrator, the gain limiting frequency is fa= 1/(2πRF CF). After fa, the gain decreases at a rate 20db/decade and reaches 0db. The frequency at which gain is 0db is fb= 1/(2π RF×CF). So, the circuit act as integrator between fa and fb.

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.
Find the output voltage waveform from the circuit
a) Triangular function
b) Unit step function
c) Ramp function
d) Square function
View Answer

Answer: c
Explanation: Input voltage Vin = 1.2v for 0≤t≤0.4ms. The output voltage at t=0.4ms is
Vo = (1/R×CFt0 Vindt+C =-(1/1) × 0.401.2 dt
=> Vo =-[0.101.2 dt + dt + dt + dt ] = -(1.2+1.2+1.2+1.2) = -4.8v
Ramp function output voltage waveform
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
View Answer

Answer: b
Explanation: 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
View Answer

Answer: d
Explanation: 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.
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5. Find the application in which integrator is used?
a) All of the mentioned
b) Analog Computers
c) FM Detectors
d) AM detectors
View Answer

Answer: b
Explanation: 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
View Answer

Answer: c
Explanation: 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.
Find the output voltage of integrator from the given diagram
Find the constant amplitude from the given diagram
Find the half periods ramp from the given diagram
d) None of the mentioned
View Answer

Answer: a
Explanation: The output voltage of integrator, Vo = (1/R1×CFt0 Vindt+C = -[(1/10kΩ×10nF) 0.1ms01dt ]= -(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
View Answer

Answer: b
Explanation: 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 << fa 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
View Answer

Answer: a
Explanation: The mentioned answer gives the exact ranges where the integration starts.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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