Explore topic-wise InterviewSolutions in .

The correct option is (c) It is only useful for amplifying AC signals

For explanation: Operational amplifier is a HIGH voltage gain, DIRECT coupled amplifier which can be used to perform MATHEMATICAL operations on ANALOG signals. It can be used to amplify both DC and AC signals.

Correct answer is (a) A = Dual input and balanced OUTPUT differential amplifier, B = Dual input and UNBALANCED output differential amplifier, C = Level shifter, D = Power amplifier

The best I can explain: The correct BLOCK diagram is

The correct CHOICE is (a) 47.7 kHz

For EXPLANATION I would say: dVOUT(max)/dT =< SLEW rate

A.Vmω =< 1.5 V/μsec

10^3.5.10^-3.ω = < 1.5×10^6

ω=<0.3×10^6

Frequency f = 0.0477×10^6Hz.

Right OPTION is (a) 0.99% negative feedback

For explanation I would SAY: B2 = B1(1+βA) = 10(1+β10^6) = 100k

1 + β10^6 = 10k

β = 9.999×10^-3

In percentage, feedback β = 0.99% negative feedback.

The correct OPTION is (d) CMRR is zero

Easiest EXPLANATION: For an ideal op-amp, the open loop voltage gain is infinite. The output RESISTANCE is 0 and the INPUT resistance is infinite. Op-amp has zero input current, zero offset voltage, infinite bandwidth, infinite CMRR and infinite slew rate.

Right answer is (B) 0.48V

Best explanation: CMRR = 20 LOG(AD/ACM) = 20 log(4000×12/Vocm) = 100

Log(48000/VOCM) = 5

48000/VOCM = 10^5

VOCM = 0.48V.

Right OPTION is (b) 5000

For explanation: UGB = AOLfOL

fOL = 10M/2×10^6 = 5Hz

The AC GAIN |A| = \(\frac{AOL}{\sqrt{1+(f/5)^2}}\)

|A| = 2×10^6/400 = 5000.

Correct answer is (a) Unity GAIN bandwidth

Easiest explanation: The 3DB frequency is the cut-off frequency, where the gain is \(\frac{1}{\sqrt{2}}\) of the maximum. The gain is only constant for infinite frequency, that is, has infinite bandwidth if the op-amp is ideal, which PRACTICALLY doesn’t exist. The UGB is the point at which the gain of op-amp REACHES 1.

Correct option is (b) Decreases the slew RATE of op-amp

Easy EXPLANATION: COMPENSATION CAPACITOR in the internal structure of op-amp to improve its frequency response, increasing its stability. It ALSO decreases the slew rate of the op-amp.

Right choice is (d) A passive filter is USED at audio frequency and an active at radio frequency

Best explanation: A passive filter can consist of all R, L and C elements. An op-amp is used in an active filter, and it also provides amplification ALONG with filtering. There are no inductors used in active filters because they are bigger in size and BULKY. Active filters are used at audio frequency and passive filter at radio frequency.

Correct answer is (b) 55kΩ

For explanation I would SAY: The MAXIMUM GAIN is at LOW frequencies and at low frequencies, the capacitance reactance is infinite and thus the total feedback impedance ZF = RF = R2

Thus gain = -ZF/Z = -R2/R1

Amplitude = 5.5 = R2/R1 = R2/10k

R2 = 55kΩ.

Right CHOICE is (c) 1+R3/R2

The explanation is: The above CIRCUIT is a LOW pass filter of the non-inverting type, where the input is at the non-inverting END. The GAIN is A = 1+R3/R2 / 1+jωR1C1

Thus maximum gain is A = 1+R3/R2.

The CORRECT CHOICE is (d) 135°

To EXPLAIN: PHASE SHIFT = 180° – tan^-1f/fC

Phase shift = 180-45 = 135°.

The correct answer is (c) 60 dB/decade

To explain: Roll-off rate is the rate at which the GAIN of a FILTER decreases outside the pass-band. Roll of rate is increased as the ORDER INCREASES. For an nth order filter, roll-off rate is 20xn dB/decade.

The correct ANSWER is (a) Single LOW pass RC circuit, Voltage follower

Explanation: A first-order Butterworth LPF is an INTERCONNECTION of a single low pass RC circuit and a voltage follower.

The correct ANSWER is (C) 2.93 kΩ

Easiest explanation: MAXIMUM GAIN, A = 3 – α = 1.586

For the above circuit, the gain is 1 + R3/R2 = 1.586

R3/R2 = 0.586

R3 = 2.93 kΩ.

Correct choice is (B) 4 dB

The explanation: For a second-order Butterworth HPF, the maximum gain AMAX = 3 – α.

α is the damping factor = 1.414

AMAX = 1.586

In dB, the gain is 20 LOG(1.586) = 4dB.

The correct option is (c) Its phase shift is -2tan^-1RC, between 0 to -180°

Explanation: An all-pass FILTER passes all frequencies but provides a different phase shift to each frequency PRESENT in the CIRCUIT. It is used for phase equalization or delay equalization and is used in landline communication. Following is a circuit of an all-pass filter USING a single op-amp.

However, its phase shift is -2tan^-1(ωRC)

Correct CHOICE is (d) The LPF and HPF are connected in series

The explanation: When designing a band-stop filter, a HPF and LPF are connected in PARALLEL, and their output goes into the input of an ADDER to get the desired output. For correct output, the cut-off frequency of HPF should be much HIGHER than that of the LPF.

The CORRECT option is (b) 0.012 MV

Best explanation: The output is saturated at the input +VSat/AOL or –VSat/AOL.

For positive saturation, the maximum ALLOWABLE input is 12/10^6 = 0.012 mV.

The correct CHOICE is (b) NEGATIVE feedback is applied while INPUT is more than –VSat/AOL

Easiest explanation: If op-amp is used WITHOUT feedback or with POSITIVE feedback, the difference voltage VD will be large enough and hence op-amp can be considered to be in the saturation region. If op-amp is used with negative feedback then the input is smaller. Saturation of output occurs if the input is not between –VSat/AOL and +VSat/AOL.

Right answer is (c) The OUTPUT offset voltage does not depend on the supply voltage

The best explanation: Output offset voltage is the output voltage of the op-amp when the input at both TERMINALS is zero. It occurs due to the dissimilarities and MISMATCHES in the INTERNAL structure of the op-amp. The value of output offset voltage is in volts. It varies with changes in supply voltage.

Right option is (C) 0.36

Explanation: Duty cycle for a SIGNAL = Duration when output is VSat/Total period

Let input = VI = 10 sinθ

The output goes to +VSat whenever the input VI crosses AMPLITUDE of 4V.

Thus 4 = 10sinθ

Θ = sin^-1 0.4 = 0.411 rad = 23.57°

The duty cycle = [180 – θ – θ]/360 = 180-47.14/360 = 0.36.