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Linear Integrated MCQ Set 1

1. In which amplifier the output voltage is equal to the negative sum of all the inputs?
a) Averaging amplifier
b) Summing amplifier
c) Scaling amplifier
d) All of the mentioned

Answer: b [Reason:] In summing amplifier the output voltage is equal to the sum of all input. Since the total input is a sum of negative input, the amplifier is an inverting summing amplifier.

2. Determine the expression of output voltage for inverting summing amplifier consisting of four internal resistors? (Assume the value of internal resistors to be equal)
a) Vo = -(Rf/R )×(Va +Vb+Vc+Vd)
b) Vo = (RF/R)×(Va +Vb+Vc+Vd)
c) Vo = (R/ RF)×(Va +Vb+Vc+Vd)
d) None of the mentioned

Answer: a [Reason:] If the internal resistors of the circuit is same i.e Ra=Rb=Rc=Rd=R (since there are four internal resistor) Then, the output voltage for inverting amplifier is given as Vo= -(Rf/R)×(Va +Vb+Vc+Vd).

3. An inverting amplifier with gain 1 have different input voltage: 1.2v,3.2v and 4.2v. Find the output voltage?
a) 4.2v
b) 8.6v
c) -4.2v
d) -8.6v

Answer: d [Reason:] When the gain of the inverting summing amplifier gain is 1 then, the internal resistors and feedback resistors have the same value. So, the output is equal to the negative sum of all input voltages. VO= -(Va+Vb+Vc) =-(1.2+3.2+4.2)= -8.6v.

4. In which type of amplifier, the input voltage is amplified by a scaling factor
a) Summing amplifier
b) Averaging amplifier
c) Weighted amplifier
d) Differential amplifier

Answer: c [Reason:] The weighted amplifier is also called as scaling amplifier. Here each input voltage is amplified by a different factor i.e. Ra,Rb and Rc are different in values ( which are the input resistors at each input voltage).

5. An inverting scaling amplifier has three input voltages Va, Vb and Vc. Find it output voltage?
a) VO= – {[(RF/Ra)×Va] +[(RF/Rb)×Vb]+[(RF/Rc)×Vc]}
b) VO= – [(RF/Ra)+(RF/Rb)+(RF/Rc)]×[( Va +Vb+Vc)].
c) VO = – {[(Ra/RF)×Va] +[(Rb/RF)×Vb]+[(Rc/RF)×Vc]}
d) None of the mentioned

Answer: a [Reason:] Since three input voltages are given assume the input resistors to be Ra,Rb and Rc. In a scaling amplifier, the input voltages are amplified by a different factor => ∴ RF/Ra ≠ RF/Rb ≠ RF/Rc Therefore, output voltage Vo = -{[(RF/Ra) Va] +[(RF/Rb) Vb]+[(RF/Rc) Vc]}.

6. An amplifier in which the output voltage is equal to average of input voltage?
a) Summing amplifier
b) Weighting amplifier
c) Scaling amplifier
d) Averaging amplifier

Answer: d [Reason:] An averaging amplifier can be used as an averaging circuit, in which the output voltage is equal to the average of all the input voltages.

7. Find out the gain value by which each input of the averaging amplifier is amplified ?( Assume there are four inputs)
a) 0.5
b) 0.25
c) 1
d) 2

Answer: b [Reason:] In an averaging amplifier, the gain by which each input is amplified must be equal to lower number of input. => RF /R =1/n , where n=number of inputs ∴ RF /R=1/4 = 0.25 (Four inputs) So, each input in the averaging amplifier must be amplified by 0.25.

8. 3v, 5v and 7v are the three input voltage applied to the inverting input terminal of averaging amplifier. Determine the output voltage?
a) -5v
b) -10v
c) -15v
d) -20v

Answer: a [Reason:] The output voltage, Vo = -[(Va+Vb+Vc)/3] = -[(3+5+7)/3] =-5v.

9. The following circuit represents an inverting scaling amplifier. Compute the value of RoM and VO?

a) VO = -0.985v ; RoM = 111.11Ω
b) VO = -2.567v ; RoM = 447.89Ω
c) VO = -1.569v ; RoM = 212.33Ω
d) VO = -1.036v ; RoM = 320.56Ω

Answer: d [Reason:] VO = – {[(RF/Ra)×Va]+[(RF/Rb)×Vb] + [(RF/Rc)×Vc]} = – {[(10kΩ/1kΩ)×3.3mv] + [(10kΩ/1.25kΩ)×5mv] + [(10kΩ/820Ω)×7.9mv]} = -1.036v. RoM = [Ra||Rbc||RF] = [(Ra×Rb)/(Ra+ Rb)] || [(Rc×RF)/( Rc+ RF)] = [(1kΩ×1.25kΩ)/(1kΩ+1.25kΩ)] || [(820Ω×10kΩ)/(820Ω+10kΩ)] = 555.55||757.85 =[(555.55 ×757.85)/(555.55+757.85)] = 320.56Ω.

Linear Integrated MCQ Set 2

1. What is done in switching regulators to minimize its power dissipation during switching?
a) Uses external transistor
b) Uses 1mH choke
c) Uses external transistor and 1mH choke
d) None of the mentioned

Answer: c [Reason:] To minimize power dissipation during switching, the external transistor must be a switching power transistor and a 1mH choke smooth out the current pulses delivered to the load.

2. Fixed voltage regulators and adjustable regulators are often called as
a) Series dissipative regulators
b) Shunt dissipative regulators
c) Stray dissipative regulators
d) All the mentioned

Answer: a [Reason:] Series dissipative regulators simulate a variable resistance between the input voltage & the load and hence functions in a linear mode.

3. Linear series regulators are suited for application with
a) High current
b) Medium current
c) Low current
d) None of the mentioned

Answer: b [Reason:] In series dissipative regulator, conversion efficiency decreases as the input or output voltage differential increases and vice versa. So, linear series regulators are suited for medium current application with a small voltage differential.

4. A series switching regulators
a) Improves the efficiency of regulators
b) Improves the flexibility of switching
c) Enhance the response of regulators
d) All of the mentioned

Answer: a [Reason:] A series switching regulators is constructed such that, a series pass transistor is used as a switch rather than as a variable resistance in linear mode.

5. The switching regulators can operate in
a) Step up
b) Step down
c) Polarity inverting
d) All the mentioned

Answer: d [Reason:] The switching regulators can operate in any one of the three modes depending on the way in which the components are connected.

6. Find the diagrammatic representation of basic switching regulator?

Answer: a [Reason:] Basic switching regulators consist of four major components; 1) Voltage source 2) Switch 3) Pulse generator 4) Filters as mentioned in the diagram.

7. What are the conditions to be satisfied by a voltage source for using it in switching regulator.
1. It must supply the required output power and the losses associated with the switching regulator
2. It must be large enough to supply sufficient dynamic range line and load variations
3. It must be sufficient high to meet the minimum requirement of the regulator system to be designed
4. It may be required to store energy for a specified amount of time during power failure especially if the system is designed for a computer power supply.
a) 1 and 3
b) 1,2,3 and 4
c) 2,3 and 4
d) 1,3 and 4

Answer: b [Reason:] A voltage source must satisfy the entire given requirement to be used in switching regulator.

8. Which among the following act as a switch in switching regulator?
a) Rectifiers
b) Diode
c) Transistors
d) Relays

Answer: c [Reason:] A transistor is connected as power switch and is operated in the saturated mode.Thus, the pulse generator output alternatively turns the switch ON and OFF in switching regulator.

9. What should be the frequency range of pulse generator?
a) 250 kHz
b) 40 kHz
c) 120 kHz
d) 20 kHz

Answer: d [Reason:] The most effective frequency range for pulse generator for optimum efficiency and component size is 20kHz.

10. Filter used in switching regulator’s are also as called
a) DC – AC transformers
b) AC – DC transformers
c) DC transformer
d) AC transformer

Answer: c [Reason:] Filter converts the pulse waveform from the output of the switch into a dc voltage. Since this switching mechanism allows a conversion similar to transformers, the switching regulators is often referred to as a DC transformer.

11. Which of the following is considered to be the most important components of the switching regulator?
a) RC or RLC filter
b) RL or RLC filter
c) ORC or RL filter
d) RC, RLC or RL filter

Answer: b [Reason:] RL or RLC filter is the most important components of the switching regulator, because there are several areas that are affected by the choke of inductor including energy storage for the regulators output ripple, transient, response etc.

12. Which is the most commonly used low voltage switching regulators?
a) Powdered Permalloy toroids
b) Fermite EI, U and toroid cores
c) Silicon steel EI butt stacks
d) None of the mentioned

Answer: c [Reason:] The silicon steel EI butt stack exhibits high permeability high flux density and ease of construction and mounting therefore, it is most commonly used in low voltage switching regulators.

13. Find the value of Rsc, L and Co for a µA7840 switching regulator to provide +5 v at 3A, using the following specifications: toff= 24µs, ripple voltage = 400mA and ton=26µs.
a) Rsc = 55 mΩ , L = 25µH & Co = 750µF
b) Rsc = 550 mΩ , L = 25µH & Co = 75µF
c) Rsc = 650 mΩ , L = 25µH & Co = 65µF
d) Rsc = 720 mΩ , L = 25µH & Co = 250µF

Answer: a [Reason:] Peak current,Ipk= 400mA× 1.5 (since Ipk = 1.5 A for peak current) ∴ Rsc = 0.33ohm/Ipk = 0.33ohm/6 = 0.055ohm. => L= [(Vo +Vp) / I pk]×toff =[(5+1.25) /6 ]× 24×10-6 =25µH. => Co = [Ipk (Ton +Toff)]/[8×Vripple] ∵T = [ton + toff] = 26µs + 24µs = 50µs => Co = [ (6×50µs)]/(8×50mA) = 7.5×10-4 = 750µF.

14. Calculate the efficiency of the step down switching regulator given the input voltage Vin= 13.5v and output voltage =6v. Assume the saturating Voltage Vs=1.1v and the forward voltage drop Vd = 1.257v
a) η = 75%
b) η = 48.5%
c) η = 63.9%
d) η = 80.5%

Answer: d [Reason:] Efficiency of the step down switching regulator, η = {[(Vin-Vs+Vd)]/ [Vin]}×(Vo) / [(Vo+Vd)] = {[(13.5v-1.1v+1.257v)/13.5v]} ×[(6/(6 ×1.257)] => Efficiency of switching regulator, η = (1.012×0.7955)×100 = 0.8051×100 = 80.5%.

15. Match the characteristics for various switching regulators.

 Switching regulator Characteristics 1. Inverting (i) [ton / toff ] = [ Vo + Vd] / [Vin -Vs -Vd]; Rsc = ( 0.33/ Ipk) 2. step down (ii) [ton / toff ] = [ Vo + Vd -Vin] / [Vin-Vs] ; Rsc = ( 0.33/ Ipk) 3. step up (iii) [ton / toff ] = [ Modulus Vo +Vd] / (Vin-Vs); Rsc = ( 0.33/ Ipk

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

Answer: b [Reason:] Characteristics and design formula for step up, step down and converting mode of switching regulator.

Linear Integrated MCQ Set 3

1. Which factor affect the input offset voltage, bias current and input offset current in an op-amp
a) Change in temperature
b) Change in supply voltage
c) Change in time
d) All of the mentioned

Answer: d [Reason:] Any change in the mentioned parameters affect the values of input offset voltage, bias current and input offset current from remaining constant.

2. Thermal voltage drift is defined as
a) △Vio/△T
b) △VF/△T
c) △Iio/△T
d) △IB/△T

Answer: a [Reason:] The average rate of change of input offset voltage per unit change in temperature is called thermal voltage drift, i.e. △Vio/△T.

3. A completely compensated inverting amplifier is nulled at room temperature 25oC, determine the temperature at which the total output offset voltage will be zero?
a) 50oC
b) 25oC
c) 75oC
d) 125oC

Answer: b [Reason:] When amplifier is nulled at room temperature, the effect of input offset voltage and current is reduced to zero. Change in the total output offset voltage occurs only, if there is any change in the value of Vio and Iio. Therefore, the total output offset voltage will be zero at room temperature.

4. How the effect of voltage and current drift on the performance of an amplifier is determined?
a) △VooT/△T = {[1-RF/R1)]×(△Vio/△T)} + RF×(△Iio/△t)
b) △VooT/△T = {(-RF/R1)×(△Vio/△T)} + RF×(△Iio/△t)
c) △VooT/△T = {[1+(RF/R1)]×(△Vio/△T)} + RF×(△Iio/△t)
d) None of the mentioned

Answer: c [Reason:] As the amplifier is used in inverting configuration, the effect of voltage and current drift is given as, the average change in total output offset voltage per unit change in temperature. △VooT/△T = {[1+(RF/R1)]×(△Vio/△T)} + RF×(△Iio/△t).

5. The error voltage in a compensating inverting amplifier is obtained by
a) Multiplying △T to total output offset voltage
b) Multiplying △T to input offset voltage
c) Multiplying △T to input offset current
d) All of the mentioned

Answer: a [Reason:] The maximum possible change in the total output offset voltage △VooT results from a change in temperature △t. Therefore, error voltage is obtained by multiplying △T in the average total output offset voltage. Ev =( △VooT/△T)×△T = [1+(RF/R1)]×(△Vio/△T)×△T + RF×(△Iio/△T)×△T.

6. A 7.5kΩ internal resistor and a 12kΩ feedback resistor are connected to an inverting amplifier. Find the error voltage, if the output voltage is 3.99mv for an input of 1.33mv.
a) ±0.6v
b) ±0.6mv
c) ± 60mv
d) ±6mv

Answer: d [Reason:] The output voltage of inverting amplifier is Vo= -(RF/R1)×Vin±Ev => Ev= ± Vo+(RF/R1)×Vin = 3.99mv+(12kΩ/7.5kΩ)×1.33mv = ±6.118 ≅ ±6mv.

7. Consider the amplifier is nulled at 27oC. Calculate the output voltage , if the input voltage is 6.21mv dc at 50oC. Assume LM307 op-amp with specification: △Vio/△T=30µV/oC ; △Iio/△T = 300pA/oC; VS =±15v.

a) +0.53v or -0.68v
b) +0.52v or -0.78v
c) +0.54v or -0.90v
d) +0.51v or -0.86v

Answer: d [Reason:] Change in temperature △T = 50oC-27oC = 23oC. => Error voltage, Ev =[1+(RF/R1)]×(△Vio/△T)×△T + RF×(△Iio/△T)×△T = [1+(100kΩ/1kΩ)]×(30µv/1oC)× 23oC + 100kΩ×(300pA/1oC)× 23oC = 0.06969+ 6.9×10-9 => Ev= 0.0704 = 70.4mv. For an input voltage of 6.21mv dc, the output voltage, Vo=-(RF/R1)×Vin±Ev = -(100kΩ/1kΩ)×6.21mv±70.4mv = +0.69v or -0.55v.

8.

The error voltage for the above circuit is 0.93v. Compute the output voltage?
a) +15v to +17v
b) +17v or -15v
c) -17v or +15v
d) None of the mentioned

Answer: b [Reason:] The output voltage for the non-inverting amplifier is Vo=[1+(RF/R1+R2)]×Vin±Ev = [1+(50kΩ/3kΩ+10kΩ)]×3.3±0.93v = 15.99±0.93 => Vo = +16.92v or -15.06v ≅ +17v or -15v.

Linear Integrated MCQ Set 4

1. When does an integrated circuit exhibit greater degree of freedom and electrical performance?
a) In thin and thick film technology
b) In semiconductor technology
c) In semiconductor and films technology
d) In thick film technology only

Answer: c [Reason:] Combining films and semiconductor technology provide a better electrical performance than either technology can provide separately.

2. Give the thickness range of the film used in thin film technology
a) 0.5-2.5 mils
b) 0.02-8 mils
c) 10-20 mils
d) 0.05-0.0 7mils

Answer: b [Reason:] Thin films have thickness varying from 50 Å to 20,000 Å. W.k.t, 1 Å=0.4 µmil, =>50 Å=50 × 0.4µmil=0.02 mmil, =>20,000 Å=20, 000*0.4µmil=8 mmil, =>therefore, the thickness range from 0.02-8 mmil.

3. Which technology is used to get cheap resistors and capacitors?
a) Thick film technology
b) Thin film technology
c) Thin and thick film technology
d) None of the mentioned

Answer: b [Reason:] Thick film technology produces cheap and rugged components, whereas thin film technology provides greater precision in manufacturing but is quite expensive. The processing equipment for thick film circuit is relatively inexpensive and is easy to use.

4. How is the process of film deposition carried out in cathode sputtering?
a) Slower than evaporation method
b) Faster than evaporation method
c) Similar to same as evaporation method
d) All of the mentioned

Answer: a [Reason:] Cathode sputtering and vacuum evaporation uses identical system. However, the process of film deposition in cathode sputtering is slower than evaporation method. Since depositing a micron-thick film takes minutes to hours, compared to seconds to minutes for evaporation.

5. How a uniform film with good crystal structure is attained in cathode sputtering process?
a) By hitting high energy particle directly on the substrate
b) Allowing Less time for the particles to deposit on the substrate
c) High energy particle diffuse through low pressure gas and deposits on the substrate
d) Heavy inert gas is used for film deposition on the substrate

Answer: c [Reason:] The process of cathode sputtering is performed at a low pressure (about 10-12 torr). So, when the high energy particle landing on the substrate actually results in a very uniform film and adhesion.

6. Which process is used to deposit metals on glass, ceramic and plastic?
a) Silk plating technique
b) Gas plating technique
c) Electroless plating technique
d) Electroplating technique

Answer: c [Reason:] In electroless plating, a metal ion in solution is reduced to the free metal and deposited as a metallic coating without the use of a coating without the use of an electric current. Thus, this process is used in plating on glass, ceramic and plastic.

7. Electroplating technique is suitable for
a) Making conduction films ceramic
b) Coating with considerable thickness
c) Coating without use of electric current
d) Making conduction films of gold or copper

Answer: d [Reason:] Electroplating is a process of coating an object with one or more layers of different metal. When dc is passed through the electrolytic solution, the positive metal ions migrate from anode (metal) and deposit on the cathode (substrate).

8. Which of the following process is involve in thick film technology
a) Screen printing
b) Ceramic firing
c) Silk screening
d) All of the mentioned

Answer: c [Reason:] Silk screening is one of the processes of thin film technology.

9. An ancient process used till today for production of circuit films is,
a) Silk Screening technique
b) Surface Mount Technology
c) Ceramic Printing technique
d) Screen Printing technique

Answer: d [Reason:] The process of screen printing pattern is an ancient one. The Egyptian used this technique thousands of years ago to decrease potter and wall of building.

10. What is the advantage of using Surface Mount Technology?
a) All of the mentioned
b) Low power consumption
c) Reduces heat dissipation in components

Answer: d [Reason:] Surface Mount Technology utilizes micro-miniature leaded or leadless components called Surface Mount Device (SMD) which are directly soldered to the specified areas on the surface without hole. Also, the compact size of SMDs reduces the area in PCB and increases the packing density.

Linear Integrated MCQ Set 5

1. How a triangular wave generator is derived from square wave generator?
a) Connect oscillator at the output
b) Connect Voltage follower at the output
c) Connect differential at the output
d) Connect integrator at the output

Answer: d [Reason:] The output waveform of the integrator is triangular, if its input is square wave. Therefore, a triangular wave generator can be obtained by connecting an integrator at the output of the square wave generator.

2. The increase in the frequency of triangular wave generator.
a) Ramp the amplitude of triangular wave
b) Increase the amplitude of triangular wave
c) Decrease the amplitude of triangular wave
d) None of the mentioned

Answer: a [Reason:] As the resistor value increase or decrease, the frequency of triangular wave will decrease or increase, respectively. Therefore, the amplitude of the triangular wave decreases with an increase in it frequency and vice verse.

3. Which among the following op-amp is chosen for generating triangular wave of relatively higher frequency?
a) LM741 op-amp
b) LM301 op-amp
c) LM1458 op-amp
d) LM3530 op-amp

Answer: b [Reason:] The frequency of the triangular wave generator is limited by the slew rate of the op-amp. LM301 op-amp has a high slew rate.

4. What is the peak to peak (PP) output amplitude of the triangular wave?
a) VO(pp) = + VRamp + (- VRamp)
b) VO(pp) = – VRamp + (+ VRamp)
c) VO(pp) = + VRamp – (- VRamp)
d) VO(pp) = – VRamp – (+ VRamp)

Answer: c [Reason:] The peak to peak output waveform, VO(pp) = + VRamp-(-VRamp) Where, – VRamp –> Negative going ramp ; + VRamp–> positive going ramp.

5. Determine the output triangular waveform for the circuit.

Answer: b [Reason:] The voltage at which A1 switch from +Vsat to -Vsat => -Vramp =(-R2 / R3) × (+Vsat) = (-10kΩ/40kΩ) ×15v =-3.75v Similarly, the voltage at which A1 switch from -Vsat to +Vsat => +Vramp = (-R2 / R3) × (-Vsat) = 10kΩ/40kΩ ×15v =3.75v ∴ Time period, T = (4R1C1R2) / R3 = (4×10kΩ×0.05µF×10kΩ) /40kΩ = 0.5 ms.

6. Find the capacitor value for a the output frequency, fo = 2kHz & VO(pp) = 7v , in a triangular wave generator. The op-amp is 1458/741 and supply voltage = ±15v. (Take internal resistor=10kΩ)
a) 0.03nF
b) 30nF
c) 0.3nF
d) 3nF

Answer: d [Reason:] Given, Vsat =15v ∴ VO(pp) = (2R2/R3) × Vsat => R2 =(VO(pp) ×R3) / (Vsat×2) = [7/(2×15)]×R3 = 0.233R3 ∵ Internal resistor, R2 = R1= 10kΩ => R3 = 0.233×10kΩ = 2.33kΩ. So, the output frequency fO = R3 / ( 4×R1 ×C1× R2) => 2khz = 2.33khz/ (4×10kΩ ×10kΩ×C1) => C1 = 2.33kΩ / (8×10-11) = 2.9 ×10-9 ≅3nF.

7. Triangular wave form has
a) Rise time < fall time
b) Rise time = fall time
c) Rise time ≥ fall time
d) None of the mentioned

Answer: b [Reason:] The triangular wave form has rise time of the triangular wave always equal to its fall time, that is, the same amount of time is required for the triangular wave to swing from -VRamp to +VRamp as from +VRamp to -VRamp.

8. Output of an integrator producing waveforms of unequal rise and fall time are called
a) Triangular waveform
b) Sawtooth waveform
c) Pulsating waveform
d) Spiked waveform

Answer: b [Reason:] Sawtooth waveform has unequal rise and fall times. It may rise positively many times faster than it falls negatively or vice versa.

9. Find out the sawtooth wave generator from the following circuits.

Answer: c [Reason:] The triangular wave generator can be converted into a sawtooth wave generator by inserting a variable dc voltage into the non-inverting terminal of the integrator.

10. Consider the integrator used for generating sawtooth wave form. Match the list I with the list II depending on the movement of wiper.

 List-I List-II Rise time =fall time (Triangular wave) Longer fall time and short rise time (Sawtooth wave) Longer rise time and short fall time (Sawtooth wave)

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