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## Linear Integrated MCQ Set 1

1. Choose the compensating network design for non-inverting amplitude

Answer: a [Reason:] If an op-amp is used as an non-inverting amplifier, the compensating network should be connected to the inverting input terminal of the op-amp.

2. Find the thevenin’s equivalent for resistance and voltage?

a) 1-iii, 2-ii, 3-1
b) 1-ii, 2-I, 3-iii
c) 1-I, 2-ii, 3-iii
d) 1-ii, 2-iii, 3-i

Answer: b [Reason:] The maximum thevenin equivalent resistance Rmax occurs when the wiper is at the center of the potentiometer and the maximum thevenin equivalent voltage Vmax is equal to +Vcc or –Vee, when wiper is uppermost or lowest in the potentiometer.

3. What is done to compensate the voltage, when V1 > V2?

a) Move the wiper towards +Vcc
b) Move the wiper towards –Vee
c) Keep the wiper at the center of potentiometer
d) None of the mentioned

Answer: a [Reason:] V1 > V2 implies that output offset voltage is positive. This means that V2 should be increased until it is equal to V1. The wiper can be moved towards +Vcc until output offset voltage is reduced to zero.

4. Calculate the maximum thevenin equivalent resistance, if a 10kΩ potentiometer is used?
a) 0.4kΩ
b) 5 kΩ
c) 2.5kΩ
d) 4kΩ

Answer: c [Reason:] Rmax= Ra/2 || Ra/2 = Ra/4. Given potentiometer, Ra = 10kΩ => Therefore, Rmax = 10kΩ/4 =2.5kΩ.

5. Find the input offset voltage for the circuit shown

a) Vio = (Rb*Vmax)/( Rmax+ Rb+ Rc)
b) Vio = Rmax/( Rmax+ Rb+ Rc)
c) Vio = (Rc*Vmax)/( Rmax+ Rb+ Rc)
d) Vio = Vmax/( Rmax+ Rb+ Rc)

Answer: c [Reason:] Compensating network using maximum thevenin’s equivalent for resistance and voltage circuit is shown. Since |V1-V2|- Vio, the maximum value of V2 can be equal to Vio.

6. Find the value of Ra and Rb from the circuit shown?

a) Ra =4.6kΩ ; Rb= 9kΩ
b) Ra =7.3kΩ ; Rb= 3.4kΩ
c) Ra =2.5kΩ ; Rb= 5.1kΩ
d) Ra =4kΩ ; Rb= 10kΩ

Answer: d [Reason:] We know that input offset voltage, Vio =(Rc*Vmax)/ Rb => Rb = Vmax*(Rc / Rb ) = (10v/10mv)*10Ω (∵ Vio specified on the datasheet is 10mv for LM307 op-amp). => Rb =10000 = 10kΩ. Since Rb > Rmax let us choose Rb = 10*Rmax. (Where Rmax = Ra/4). ∴ Rb = (10*Rb)/4 and Ra = Rb/2.5 = 10kΩ/2.5=4kΩ.

7. Why does an op-amp without feedback is not used in linear circuit application?
a) Due to high current gain
b) Due to high voltage gain
c) Due to high output signal
d) All of the mentioned

Answer: b [Reason:] In an op-amp without feedback, the voltage gain is extremely high (ideally infinite). Because of the high risk of distortion and clipping of the output signal, an op-amp in open loop configuration is not used in linear circuit application.

8. When the input voltage is reduced to zero in a closed loop configuration the circuit acts as
a) Inverting amplifier
b) Non-inverting amplifier
c) Inverting and non-inverting amplifier
d) None of the mentioned

Answer: c [Reason:] Since, the input signal voltage is reduced to zero, the internal resistance is negligibly small. The output offset voltage is expressed in terms of external resistance and the specified input offset voltage for a given op-amp. If the non-inverting input terminal is connected to ground, it acts as inverting op-amp and vice versa.

9. How the value of output offset voltage is reduced in closed loop op-amp?
a) By increasing gain
b) By reducing gain
c) By decreasing bandwidth
d) By reducing bandwidth

Answer: b [Reason:] The output offset voltage is a product of gain and specified input offset voltage for a given op-amp. Voo= Aoo*Vio. So, the value of output offset voltage can be reduced by reducing the gain value.

## Linear Integrated MCQ Set 2

1. Miniaturization of components with superior performance can be obtained in
a) Integrated circuits
b) Discrete circuits
c) Both integrated and discrete circuits
d) None of the mentioned

Answer: a [Reason:] All the components in integrated circuits are fabricated on the same chip, whereas in discrete circuits discrete components are used and occupies more area.

2. Why Integrated Circuits are considered to be economical?
a) Simple manufacturing process
b) Provide trouble free service
c) Due to batch production
d) Easy to use and reuse

Answer: c [Reason:] Integrated Circuits are considered to be economical because, it reduces the cost due to batch production. A standard 10cm diameter wafer can be divided into 8000 rectangular chip (approx). So, if 10 wafers are processed in one batch, then 80,000 ICs are produced simultaneously.

3. Which of the following circuit is not made of digital ICs?
a) Shift registers
b) Multipliers
c) Demultiplexers
d) Counters

4. Controlling the characteristics of operating region is not taken care in digital ICs because
a) It involves only two voltages
b) It is used to form transistor logic
c) Digital IC requires only power supply, input and output
d) All of the mentioned

Answer: a [Reason:] Digital circuits are primarily concerned with only two level of voltages (or currents): ‘high’ and ‘low’. Therefore, accurate control of operating region characteristic is not required in digital circuits, unlike in linear circuits.

5. Linear Integrated Circuit can also be called as
a) Monolithic circuit
b) General purpose circuit
c) Special purpose
d) Analog circuit

Answer: d [Reason:] The output electrical signal and the input signal applied in linear circuit vary either in proportion or in physical quantities they represent. Since, the electrical signals are analogous to the physical quantities, linear circuit are also referred to as analog circuit.

6. How are components array available in IC form?
a) Array of silicon wafer
b) Group of isolated transistor, diodes and resistor
c) Array of individual stages of differential and cascade amplifier
d) Group of gates logic circuits arranged in array format

Answer: b [Reason:] Majority of linear ICs are available as operational amplifier. So, these components array consists of group of isolated transistor, diodes and resistors.

7. Which type of integrated circuit is suitable for small quantity custom circuits?
a) Monolithic IC
b) Linear IC
c) Hybrid IC
d) Digital IC

Answer: c [Reason:] In hybrid circuits, separate components parts are attached to a ceramic substrate and interconnected by means of either metallization pattern or wire bond. This technology is adaptable to small quantity custom circuits only.

8. What is the isolation technique used in monolithic IC?
a) PN-junction
b) Dielectric junction
d) All of the mentioned

Answer: d [Reason:] Electrical isolation between the components in monolithic ICs can be achieved by any one of the three isolation techniques dielectric, beam-lead or PN-junction.

9. Monolithic Integrated Circuits exhibits
a) Increase in power consumption
b) Non-matched devices
c) Thermal stability
d) All of the mentioned

Answer: c [Reason:] Monolithic ICs exhibits good thermal stability because all the components are integrated on the same chip very close to each other.

10. How are thin film hybrid ICs obtained?
a) Evaporation method
b) Spraying method
c) Silk screening method

Answer: a [Reason:] In thin film hybrid IC the resistors, capacitors and interconnections are separately formed on the substrate and a suitable material is evaporated on the substrate to make connection.

11. Which of the following belongs to special purpose op-amp?
a) LM741
b) LM380
c) IC351
d) None of the mentioned

Answer: b [Reason:] LM380 is a special purpose op-amp, used only for audio power application.

12. How many op-amps will be present in a typical op-amp package?
a) Three
b) Four
c) Five
d) Nine

Answer: b [Reason:] Op-amp package IC consists of single, dual (two) or Four (quad) op-amps.

## Linear Integrated MCQ Set 3

1) What is the best choice of IC package used for experimental purpose?
a) DIP package
b) Metal can package
c) Flat pack
d) Transistor pack

Answer: a [Reason:] DIP package are used as it is easy to mount. The mounting does not require bending or soldering of the leads.

2. What is the general information specified in ordering an IC?
a) Temperature range
b) Device type
c) Package type
d) All of the mentioned

Answer: d [Reason:] Generally, in ordering an IC, all the three informations must be specified.

3. Find the ordering information for µA741TC.
a) Sprague 741 DIP with Industrial temperature range
b) Intersil 741 DIP with commercial temperature range
c) Fairchilds 741 DIP with commercial temperature range
d) Texas instrument 741 metal can with Industrial temperature range

Answer: c [Reason:] Here “µA” represents the identifying initials used by Fairchild, T represents Mini DIP package and C represents Commercial temperature range.

4. How a Motorola IC with plastic DIP and commercial temperature range is ordered?
a) ICLxxxP -> 0o to 75oc
b) CAxxE -> -55o to +125oc
c) LMxxxxA -> -40o to+85oc
d) MCxxxP -> 0o to 70oc

Answer: d [Reason:] The ordering format for a typical Motorola IC is, MCxxxx –> Device type P –> Package type(Plastic DIP) 0o to 70oc –> Temperature range (Commercial).

5. What does the 1-2-3 numbering system used in National Semiconductor IC denotes
a) Validity in years
b) Temperature range
c) Package type
d) Ordering information

Answer: c [Reason:] In National linear ICs, a 1-2-3 numbering system is used to represent the temperature range.

6. How does a industrial temperature range device in National Semiconductor IC is represented?
a) LM305
b) LM101
c) LM201
d) All of the mentioned

Answer: c [Reason:] In LM201, the number 2 denotes an industrial temperature range device.

7. Use device identification method to find the IC of Fairchild chip manufactured in the year 1980.

Answer: b [Reason:] In the chip, 80 represent the manufactured year

8. Dual-In-Line pack is considered to be suitable for mounting because,
a) Easy to handle
b) Fits mounting hardware
c) Inexpensive
d) All of the mentioned

Answer: c [Reason:] DIP pack is easy to handle, fit standard mounting hardware and is inexpensive when moulded on plastic.

9. What is the use of notch and dot in DIP ICs?
a) Determine the pin configuration
b) Designed to represent device type
c) Represent property of IC
d) Find the pin number

Answer: d [Reason:] A notch and dot as viewed form top view is used to find the pin terminal. The terminals are numbered counter clockwise.

10. How an eight pin Dual-In-Line Package is shortly named
a) 8p DIP
b) Maxi DIP
c) Mini DIP
d) ES DIP

Answer: c [Reason:] An eight pin Dual-In-Line Package is called as Mini DIP as it is used for devices with minimum number of inputs and outputs.

## Linear Integrated MCQ Set 4

1. The circuit in which the output voltage waveform is the integral of the input voltage waveform is called
a) Integrator
b) Differentiator
c) Phase shift oscillator
d) Square wave generator

Answer: a [Reason:] Integrator circuit produces the output voltage waveform as the integral of the input voltage waveform.

2. Find the output voltage of the integrator
a) Vo = (1/R×CFt0 Vindt+C
b) Vo = (R/CFt0 Vindt+C
c) Vo = (CF/R)×t0 Vindt+C
d) Vo = (R×CFt0 Vindt+C

Answer: a [Reason:] The output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RCF. Vo = (1/R×CFt0 Vindt+C Where C-> Integration constant and CF-> Feedback capacitor.

3. Why an integrator cannot be made using low pass RC circuit?
a) It require large value of R and small value of C
b) It require large value of C and small value of R
c) It require large value of R and C
d) It require small value of R and C

Answer: c [Reason:] A simple low pass RC circuit can work as an integrator when time constant is very large, which require large value of R and C. Due to practical limitations , the R and C cannot be made infinitely large.

4. How a perfect integration is achieved in op-amp?
a) Infinite gain
b) Low input impedance
c) Low output impedance
d) High CMRR

Answer: a [Reason:] In an op-amp integrator the effective input capacitance becomes CF×(1-Av). Where Av is the gain of op-amp. The gain is infinite for ideal op-amp. So, effective time constant of the op-amp integrator becomes very large which results in perfect integration.

5. The op-amp operating in open loop result in output of the amplifier to saturate at a voltage
a) Close to op-amp positive power supply
b) Close to op-amp negative power supply
c) Close to op-amp positive or negative power supply
d) None of the mentioned

Answer: c [Reason:] In practice, the output of op-amp never becomes infinite rather the output of the op-amp saturate at a voltage close to op-amp positive or negative power supply depending on the polarity of the input dc signal.

6. The frequency at which gain is 0db for integrator is
a) f=1/(2πRFCF)
b) f=1/(2πR1CF)
c) f=1/(2πR1R1)
d) f=(1/2π)×(RF/R1)

Answer: b [Reason:] The frequency at which the gain of the integrator becomes zero is f=1/(2πR1CF).

7. Why practical integrator is called as lossy integrator?
a) Dissipation power
b) Provide stabilization
c) Changes input
d) None of the mentioned

Answer: d [Reason:] To avoid saturation problems, the feedback capacitor is shunted by a feedback resistance(RF). The parallel combination of RF and CF behave like a practical capacitor which dissipates power. For this reason, practical integrator is called as a lossy integrator.

8. Determine the lower frequency limit of integration for the circuit given below.

a) 43.43kHz
b) 4.82kHz
c) 429.9kHz
d) 4.6MHz

Answer: b [Reason:] The lower frequency limit of integration, f= 1/(2πRFCF) = 1/(2π×1kΩ×33nF) = 4.82kHz.

## Linear Integrated MCQ Set 5

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 [Reason:] 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.

a) Triangular function
b) Unit step function
c) Ramp function
d) Square function

Answer: c [Reason:] 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 + 0.20.11.2 dt + 0.30.21.2 dt + 0.40.31.2 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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

Answer: b [Reason:] 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