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## Linear Integrated MCQ Set 1

1. How to obtain a desired amount of multiplication in frequency multiplier?
a) By decreasing the multiplication factor
b) By increasing the input frequency
c) By selecting proper divide by N-network
d) None of the mentioned

Answer: c [Reason:] The desired amount of multiplication can be obtained by properly selecting a divide by N-network. For example, to obtain the output frequency fout=5×fin, a divide by N = 5 network is needed.

2. Calculate the output frequency in a frequency multiplier if, fin = 200Hz is applied to a 7 divide by N-network.
a) 1.2kHz
b) 1.6kHz
c) 1.2kHz
d) 1.9kHz

Answer: c [Reason:] Since the VCO is actually running at a multiple of input frequency. fout=divide by N-network x fin=7x200Hz=1400Hz =>fout=1.4kHz.

3. For what kind of input signal, the frequency divider can be avoided frequency multiplier?
a) Triangular waveform
b) Square waveform
c) Saw tooth waveform
d) Sine waveform

Answer: a [Reason:] VCO can be directly locked to the nth harmonic of the input signal without connecting any frequency divider in between the input signal rich in harmonics like square wave.

4. What must the typical value of n for a frequency multiplication / division? (n->order of harmonics)
a) n ≤ 12
b) n > 11
c) n < 10
d) n = 7

Answer: d [Reason:] As the amplitude of the higher order harmonics becomes less, effective locking may not take place for high values of n. So, the typical value of n is less than 10 for frequency multiplication / division.

5. Determine the offset frequency of frequency translation, when the output and input frequency are given as 75kHz and 1000Hz.
a) 35 kHz
b) 20 kHz
c) 29 kHz
d) 14 kHz

Answer: b [Reason:] The output of the frequency translation fo= fs+f1 => f1 = fo– fs= 75kHz-55kHz =20kHz.

6. The frequency corresponding to logic 1 state in FSK is called
a) Space frequency
b) Mark frequency
c) Both mark and space frequency
d) None of the mentioned

Answer: b [Reason:] Frequency shift is usually accomplished by dividing a VCO with binary data signal. Therefore, the logic 1 state of the binary data signal corresponds to mark frequencies.

7. Find the frequency shift in FSK generator?
a) 230 Hz
b) 250 Hz
c) 180 Hz
d) 200 Hz

Answer: d [Reason:] Frequency shift is the difference between FSK signals of 1070 Hz and 1270 Hz frequency, which is 200 Hz.

8. Which filter is chosen to remove the carrier component in the frequency shift keying?
a) Three stage filter
b) Two stage filter
c) Single stage filter
d) All of the mentioned

Answer: a [Reason:] The high cut-off frequency of ladder filter is chosen to be approximately halfway between the maximum keying rate of 150Hz & twice the input frequency (≅ 2200Hz) which can be obtained using three stage filters.

## Linear Integrated MCQ Set 2

1. Output current in general purpose op-amp can be increased using
a) Power comparator
b) Power amplifier
c) Power resistor
d) Power booster

Answer: d [Reason:] A simple method of increasing output current of a general purpose op-amp is to connect a power booster in series with the op-amp.

2. Which type of power transistor is chosen for a discrete power booster?
a) Collector follower stage
b) Emitter follower stage
c) Base follower stage
d) None of the mentioned

Answer: b [Reason:] A discrete power booster is formed by an emitter follower stage using a power transistor. Because of unity gain, the emitter follower helps to retain the voltage gain characteristics of the general purpose op-amp.

3. What is the power dissipation of power transistor?
a) ≅ 0.5W
b) ≤ 0.5W
c) > 0.5W
d) ≠ 0.5W

Answer: c [Reason:] Transistors with power dissipation larger than 1/2 (0.5)W are called power transistors.

4. Burr-Brown 3553 power amplifier is suited for
a) Line driving applications
b) Power supply requirement
d) High frequency application

Answer: a [Reason:] Burr-Brown 3553 power amplifier is suited for line driving applications in which fast pulses or wideband signals are involved.

5. Find the IC that can be used for short circuit proof protection?
a) All of the mentioned
b) IC LM384
c) ICL8063
d) ICLM380

Answer: c [Reason:] The intersil ICL8063 monolithic power transistor driver and amplifier has a built-in safe area protection and short-circuit proof protection.

6. Which among the following is ideal for consumer applications?
a) NE5018
b) LM380
c) MC1408
d) SE5018

Answer: b [Reason:] LM380 is a power audio amplifier designed to deliver a minimum of 2.5w to an 8Ω load and hence is ideal for consumer applications.

7. Bridge power audio amplifier can deliver power upto
a) Twice as much as output of single LM380 amplifier
b) Thrice as much as output of single LM380 amplifier
c) Four times as much as output of single LM380 amplifier
d) Half of the output of single LM380 amplifier

Answer: c [Reason:] Two LM380s are used in the bridge configuration.Therefore, it produce a maximum output voltage swing which is equal to twice that of a single LM380 amplifier.

8.

What precautionary measure should be taken if the circuit is used in a RF-sensitive environment?
a) Connect a parallel combination of RC at the output
b) Connect a series combination of Resistors at the output
c) Connect parallel capacitor at the output
d) Connect a series combination of RC at the output

Answer: d [Reason:] When LM380 audio power amplifier is used in an RF-sensitive environment, an RC combination should be added at the output terminal (Pin 8) to eliminate 5 to 10-MHz oscillation.

9. Which amplifier provides twice output swing as that of LM380 amplifier?
a) Hybrid power amplifier
b) Bridge power audio amplifier
c) Monolithic power audio amplifier
d) Dual power amplifier

Answer: b [Reason:] In bridge power audio amplifier, two LM380s are used in bride configuration. Therefore, the maximum output voltage swing will be twice that of a single LM380 amplifier.

10. A LM380 power amplifier is used in a intercom system with amplifier gain = 50 and the transformer turns ratio is given as 35.Find the overall gain of the circuit.
a) 1880
b) 1750
c) 1370
d) 1580

Answer: b [Reason:] Given, N1/N2=35; Gain of amplifier = 50.Therefore, the overall gain of the circuit = 50×35 =1750.

11. Determine the work done by the intercom system depending on the position of the switch

a) Remote speaker act as microphone
b) Master speaker act as microphone
c) Remote and master speaker act as microphone
d) None of the mentioned

Answer: a [Reason:] When the switch of the intercom system is in the listen mode, the remote speaker acts as the microphone.

## Linear Integrated MCQ Set 3

1. How can a first order low pass filter can be converted into second order low pass filter
c) By adding RC || LC network
d) None of the mentioned

2. Consider the following specifications and calculate the high cut-off frequency for the circuit given?

a) 95Hz
b) 48Hz
c) 14Hz
d) 33Hz

Answer: d [Reason:] The high cut-off frequency, fH= 1/[2 π√(R2 ×R3× C2× C3)] = 1/[2π√(33kΩ×15kΩ×0.47µF×0.1µF)]= 1/[2π× 4.82×(10-3)]= 33Hz.

3. Find the gain and phase angle of the second order low pass filter?
Where pass band gain of the filter is 5, frequency and the high cut-off frequency of the filter are 3000Hz and 1kHz.
a) None of the mentioned
b) Gain magnitude = -1.03dB , φ =63.32o
c) Gain magnitude = -5.19dB , φ =71.56o
d) Gain magnitude = -4.94dB , φ =90o

Answer: c [Reason:] The gain of the second order low pass filter, [VO /Vin] =AF/ √ [1+(f/fh)2] =5/ √[1+(3000/1000)4] =5/9.055 =0.55. => [VO /Vin] = 20log(0.55) =-.519dB. Phase angle of second order low pass filter is given as φ= tan-1(f/fH) => φ =71.56o.

4. Design a second order low pass butterworth filter at a high cut-off frequency of 2.2kHz. Given RF=20kΩ and capacitor 0.047µF.

Answer: c [Reason:] Given fH=21.2kHz, C2=C3 =0.047µF R3 = 1/(2πfHC3 = 1/(2π×2.2kHz×0.047µF) =1/5.9032×10-4 =1.69kΩ. =>R3 =R2=1.67kΩ Since, RF=0.586R1 => R1=RF/0.586 =20kΩ/0.586 R1 = 34.13kΩ.

5. A second order low pass filter is given an input frequency of 30kHz and produce a output having phase angle of 79o. Determine the pass band gain of the filter?
a) 11 dB
b) 89.11 dB
c) 46.78 dB
d) None of the mentioned

Answer: c [Reason:] Phase angle of the filter, φ = tan-1(f/fH) => fh =f×tan(φ) =30kHz × tan(79)= 154.34kHz. Therefore, the pass band gain AF = fH/0.707 = 154.34kHz/0.707 AF= 218.3 =20log(218.3)= 46.78dB.

6. The pass band voltage gain of a second order low pass butterworth filter is
a) 1.586
b) 8.32
c) 0.586
d) 0.707

Answer: a [Reason:] Second order low pass filter has a pass band voltage gain equal to 1.586 because of equal resistor and capacitor values. This gain is necessary to guarantee butterworth response.

7. Arrange the series of step involved in designing a filter for first order low pass filter
Step 1: Select a value of C less than or equal to 1µF
Step 2: Choose a value of high cut-off frequency fH
Step 3: Select a value of R1C and RF depending on the desired pass band gain
Step 4: Calculate the value of R
a) Steps- 2->4->3->1
b) Steps- 4->1->3->2
c) Steps- 2->1->4->3
d) Steps- 1->3->4->2

Answer: b [Reason:] The mentioned option is the sequence of steps followed for designing a low pass filter.

8. Frequency scaling is done using
a) Standard capacitor
b) Varying capacitor
c) Standard resistance
d) None of the mentioned

Answer: a [Reason:] In frequency scaling standard capacitors are chosen, because for non standard value of resistor, a potentiometer is used.

## Linear Integrated MCQ Set 4

1. Slew rate is defined as the rate of change of
a) Output voltage with respect to time
b) Input voltage with respect to time
c) Both output input voltage with respect to time
d) None of the mentioned

Answer: a [Reason:] Slew rate is the maximum rate of change of output voltage caused by a step input voltage with respect to time.

2. How the slew rate is represented?
a) 1V/ms
b) 1V/s
c) 1V/µs
d) 1mv/S

Answer: c [Reason:] Slew rate are usually specified in V/µs. For example, 1 V/µs means that the output rises or falls no faster than 1V every microsecond.

3. The natural semiconductor LH0063C has a slew rate of
a) 1400V/µs
b) 6000V/µs
c) 500V/µs
d) None of the mentioned

Answer: b [Reason:] National semiconductor LH0063C has a slew rate of 6000V/µs. Generally practical op-amp is available with slew rate from 0.1V/µs to well above 1000V/µs.

4. Rise time is specified for
a) Large signal
b) Medium signal
c) Small signal
d) All of the mentioned

Answer: c [Reason:] Rise time is specified for small signal, usually when the peak output voltage is less than one volt.

5. Op-amps with wide bandwidth will have
a) Increase in output
b) Higher slew rate
c) Low response time
d) None of the mentioned

Answer: b [Reason:] An op-amp slew rate is related to its frequency response. Usually op-amps with wider bandwidth have higher (better) slew rate.

6. Which factor is responsible for causing slew rate?
a) Internal capacitor
b) External resistor
c) None of the mentioned
d) Both internal and external capacitor

Answer: d [Reason:] Capacitors require a finite amount of time to charge and discharge. Thus, a capacitor inside or outside the op-amp causes slew rate.

7. Find the value of capacitor, if the rate of change of voltage across the capacitor is 0.78V/µs and current= 12µA.
a) 5µF
b) 2µF
c) 10µF
d) 15µF

Answer: d [Reason:] Rate at which the voltage across the capacitor increases is given as dVc/dt= I/C => C= I/ (dVc/ dt) = 12µA/0.78V/µs = 15.38 ≅15µF.

8. Find the slew rate of op-amp from the output waveform given below?

a) 3.4V/µs
b) 10V/µs
c) 20.66V/µs
d) 16V/µs

Answer: c [Reason:] Slew rate is defined as the maximum rate of change of the output SR = 3.1-(-3.1v)/ (0.6/2)µs =6.3/0.3 = 20.66Vµs.

9. For the circuit shown below, calculate the rate of change of output signal

a) Vm ωcosωt
b) Vm cosωt
c) Vm cosωt/ωt
d) Vm cosωt/ω

Answer: a [Reason:] The given circuit is a voltage follower circuit. So, input voltage = output voltage => Vo= Vmsinωt =>∴ the rate of change of the output voltage = dVo /dt= d(Vmsinωt)/ dt = Vm ωcosωt.

10. At what condition, the output of op-amp will be free of distortion?
a) Slew rate > 2πfVm /106 V/µs
b) Slew rate > 2πfVm /106 V/µs
c) Slew rate ≥ 2πfVm /106 V/µs
d) Slew rate = 2πfVm /106 V/µs

Answer: b [Reason:] As long as the value of right hand side equation is less than the slew rate of op-amp, the output wave form will always be undistorted.

## Linear Integrated MCQ Set 5

1. How are the square wave output generated in op-amp?
a) Op-amp is forced to operate in the positive saturation region
b) Op-amp is forced to operate in the negative saturation region
c) Op-amp is forced to operate between positive and negative saturation region
d) None of the mentioned

Answer: c [Reason:] Square wave outputs are generated where the op-amp is forced to operate in saturated region, that is, the output of the op-amp is forced to swing repetitively between positive saturation, +Vsat and negative saturation, -Vsat.

2. The following circuit represents a square wave generator. Determine its output voltage

a) -13 v
b) +13 v
c) ± 13 v
d) None of the mentioned

Answer: a [Reason:] The differential output voltage Vid = Vin1 – Vin2= 3-7v = -4v. The output of the op-amp in this circuit depends on polarity of differential voltage V0= -Vsat ≅ -Vee = -13 v.

3. Determine the expression for time period of a square wave generator
a) T= 2RC ln×[( R1+ R2) / ( R2)].
b) T= 2RC ln×[( 2R1+ R2) / ( R2)].
c) T= 2RC ln×[( R1+ 2R2) / ( R2)].
d) T= 2RC ln×[( R1+ R2) / (2 R2)].

Answer: b [Reason:] The time period of the output waveform for a square wave generator is T= 2RC ln×[(2R1+ R2)/( R2)].

4. Determine capacitor voltage waveform for the circuit

Answer: c [Reason:] When the op-amp output voltage is at negative saturation, V1 = [(R1) / (R1+ R2 )] × (-Vsat) = [10kΩ / ( 10 kΩ +11.6 kΩ)] × (-15v) = -7v. Similarly, when the op-amp’s output voltage is at positive saturation, V1 = [(R1) / (R1+ R2 )] × (+Vsat) = [10kΩ/ ( 10 kΩ +11.6 kΩ)] × (+15v) = +7v The time period of the output waveform,T= 2RC ln ×[( 2R1+ R2) / ( R2)] = 2× 10kΩ × 0.05 µF× ln (2×10kΩ + 11.6kΩ) / 11.6kΩ] = 1×10-3 × ln2.724 = 1ms. The voltage across the capacitor will be a triangular wave form.

5. What will be the frequency of output waveform of a square wave generator if R2 = 1.16 R1?
a) fo = (1/2RC)
b) fo = (ln/2RC)
c) fo = (ln /2 ×√RC)
d) fo = (ln/√(2 RC))

Answer: b [Reason:] When R2= 1.16 R1, then fo = 1/2RC× ln[ (2R1+ R2) / R2] = 1/2RC ×ln [(2R1 + 1.161R1 )/ (1.161R1)] = 1/( 2RC×ln2.700)= 1/2RC.

6. What could be the possible output waveform for a free running multivibrator whose op-amp has a supply voltage of ±5v operating at 5khz?

Answer: c [Reason:] In a free running multivibrator, the output is forced to swing repetitively between positive and negative saturation to produce square wave output. Therefore, +Vsat ≅ +Vcc =+5v and -Vsat ≅ -Vcc =-5v. => Frequency= 5khz , f =1/t = 0.2ms.

7. Determine the output frequency for the circuit given below

a) 28.77 Hz
b) 31.97 Hz
c) 35.52 Hz
d) 39.47 Hz

Answer: d [Reason:] The output frequency fo = 1/2RC×ln [ (2R1+ R2)/ R2] = 1 / {(2×33kΩ ×0.33µF)×ln[(2×33kΩ +30kΩ)/30kΩ]} = 1/ (0.02175×ln 32) = 39.47 Hz.

8. The value of series resistance in the square wave generator should be 100kΩ or higher in order to
a) Prevent excessive differential current flow
b) Increase resistivity of the circuit
c) Reduce output offset voltage
d) All of the mentioned

Answer: a [Reason:] In practice, each inverting and non-inverting terminal needs a series resistance to prevent excessive differential current flow because the inputs of the op-amp are subjected to large differential voltages.

9. Why zener diode is used at the output terminal of square wave generator?
a) To reduce both output and capacitor voltage swing
b) To reduce output voltage swing
c) To reduce input voltage swing
d) To reduce capacitor voltage swing