InterviewSolution

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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

51.	Heat dissipation from heat sink take place primarily by ___________(a) Conduction(b) Convection(c) Radiation(d) All of the mentionedI had been asked this question in an online interview.I would like to ask this question from Heat Sink for Power Transistor-2 in chapter Power Amplifier of Analog Circuits
Answer» Correct choice is (d) All of the mentioned The best explanation: Heat dissipation from heat sink take place primarily by convection since there is no actual CONTACT between heat sink and TRANSISTOR. Heat dissipation also takes place by Radiation but it is comparatively LOW.

52.	What is a heat spreader?(a) An older version of a heat sink(b) Another name for heat sink(c) Device to transfer heat from device to the sink(d) A connector for coolant to flowI have been asked this question by my school principal while I was bunking the class.I would like to ask this question from Heat Sink for Power Transistor-2 topic in portion Power Amplifier of Analog Circuits
Answer» The correct choice is (c) Device to transfer heat from device to the sink The BEST I can explain: A heat SPREADER CONDUCTS heat between a heat source and a heat sink. Most COMMONLY, a block of material with high THERMAL conductivity is used as a heat spreader.

53.	Heat sinks are provided with peripheral fins to ___________(a) Provide good appearance(b) Increase heat absorption(c) Increase surface area of heat dissipation(d) Provide material stabilityThis question was addressed to me in exam.I'd like to ask this question from Heat Sink for Power Transistor-2 topic in division Power Amplifier of Analog Circuits
Answer» RIGHT choice is (c) INCREASE SURFACE area of HEAT dissipation To elaborate: Heat sinks are provided with PERIPHERAL fins to increase surface area of heat conduction. Increase in dissipation also increases absorption but more correct option is “Increase surface area of heat dissipation”.

54.	Heat sinks are usually provided with black anodized finish to ___________(a) Enhance heat dissipation by radiation(b) Enhance heat dissipation by convection(c) Prevent electrical connection(d) Remove dust accumulationI have been asked this question in an online interview.My question is based upon Heat Sink for Power Transistor-2 in chapter Power Amplifier of Analog Circuits
Answer» RIGHT option is (a) Enhance heat DISSIPATION by radiation Best explanation: Heat sinks are usually provided with black anodized finish to enhance heat dissipation by radiation. ANODIZING THICKENS and toughens the naturally occurring protective oxide layer on the metal surface. It PROVIDES dielectric isolation between cooling components and the electronics in the device.

55.	Which substance is used to maintain electrical insulation between a heat sink and transistor?(a) plastic(b) fiber(c) ceramic(d) micaI got this question during an interview for a job.The question is from Heat Sink for Power Transistor-2 in section Power Amplifier of Analog Circuits
Answer» Right option is (d) mica For explanation I would SAY: Many transistors have metal cases which are CONNECTED to a LEAD and thus it BECOMES necessary to insulate the heat sink from the TRANSISTOR. Insulating kits of a mica sheet and a plastic sleeve are used.

56.	By dividing volumetric thermal resistance by required thermal resistance we will obtain ___________(a) Power dissipation of heat sink(b) Volume of heat sink(c) Maximum temperature of heat sink(d) Volume of surrounding airI have been asked this question by my college professor while I was bunking the class.My enquiry is from Heat Sink for Power Transistor-2 topic in portion Power Amplifier of Analog Circuits
Answer» RIGHT answer is (b) Volume of heat SINK Explanation: By dividing volumetric THERMAL RESISTANCE by required thermal resistance we will obtain the volume of heat sink.

57.	Cavities embedded in heat source is also known as ___________(a) Low sink(b) Sink hole(c) On-sink(d) Inverted sinkI have been asked this question at a job interview.I would like to ask this question from Heat Sink for Power Transistor-2 in division Power Amplifier of Analog Circuits
Answer» The CORRECT answer is (d) Inverted SINK Easiest explanation: Cavities EMBEDDED in heat source is ALSO KNOWN as an inverted sink.

58.	De-rating factor is usually expressed in ___________(a) W/°C(b) °C/W(c) W/s(d) kW/KI have been asked this question during an interview.The query is from Heat Sink for Power Transistor-2 in section Power Amplifier of Analog Circuits
Answer» Right answer is (a) W/°C Easy EXPLANATION: De-rating factors for DETERMINING the POWER dissipation rating is usually GIVEN at any temperature above 25°C. The de-rating factor is specified in W/°C.

59.	Which of these is not true for a class B amplifier?(a) It has zero DC bias(b) They have an efficiency less than that of class A amplifiers(c) The quiescent power dissipation is zero(d) The conduction angle is only 180°I had been asked this question in an interview for internship.Query is from Push-Pull Class B Amplifier topic in section Power Amplifier of Analog Circuits
Answer» Right OPTION is (b) They have an efficiency LESS than that of class A amplifiers To explain I would SAY: The class B AMPLIFIER has zero DC bias as the transistors are biased at cut-off only. Each transistor conducts when the input is greater than the base-emitter voltage. The conduction angle is only 180° for this amplifier. They have higher efficiency than class A amplifiers.

60.	What is the output of a class B amplifier for sinusoidal input?(a) Sinusoidal amplifier(b) Half-sinusoidal(c) Sinusoidal with higher frequency(d) Square waveThis question was posed to me in unit test.This intriguing question comes from Push-Pull Class B Amplifier in section Power Amplifier of Analog Circuits
Answer» CORRECT option is (b) HALF-sinusoidal The best explanation: If Q-point is in cut-off, then IC varies only in the positive direction, for saturation, it varies in the negative direction. So the output of CLASS B AMPLIFIER is half sinusoidal. There is no effect in the SHAPE or the frequency of the wave.

61.	How do we obtain sinusoidal output out of a class B amplifier?(a) By using non-sinusoidal inputs(b) By utilizing two transistors(c) By biasing it in the active region(d) By adding a capacitor to the outputI had been asked this question in final exam.Asked question is from Push-Pull Class B Amplifier topic in chapter Power Amplifier of Analog Circuits
Answer» RIGHT CHOICE is (b) By UTILIZING two transistors For explanation: To obtain sinusoidal output from a class B amplifier, two transistors must be used. Such a circuit is a class B push-pull amplifier, used in unturned power amplifiers and audio FREQUENCY power amplifiers.

62.	In a class B amplifier, it is found that DC power is 25W, find the ac power.(a) 10 W(b) 62.5 W(c) 25 W(d) 50 WThe question was asked during an online interview.This interesting question is from Push-Pull Class B Amplifier in division Power Amplifier of Analog Circuits
Answer» Right choice is (b) 62.5 W Explanation: For a CLASS B AMPLIFIER, FIGURE of merit = 0.4 = dc POWER/ac power Thus AC power = DC power/0.4 = 25/0.4 = 62.5 W.

63.	Why does no DC current flow in the primary winding of the output transformer of class B push-pull amplifier?(a) Because DC currents from both transistors flow in opposite directions(b) Because the net impedance is very high to allow flow of current(c) The winding only allows AC current to flow(d) Current only flows in secondary winding due to the presence of load at that sideI had been asked this question in an online quiz.The origin of the question is Push-Pull Class B Amplifier topic in section Power Amplifier of Analog Circuits
Answer» CORRECT option is (a) Because DC currents from both transistors flow in opposite directions The BEST I can explain: The net DC current in the PRIMARY winding of the output transformer is zero because DC collector currents of both transistors being used flow in opposite directions and HENCE transformer saturation doesn’t occur.

64.	When is the maximum efficiency of class B amplifier achieved?(a) When VMAX = VCC(b) When two transistors are used(c) When VMIN = 0(d) Efficiency is always constantThe question was posed to me during a job interview.My question is from Push-Pull Class B Amplifier in chapter Power Amplifier of Analog Circuits
Answer» RIGHT option is (C) When VMIN = 0 Explanation: Efficiency % = [1-VMIN/VCC] x 78.5 Maximum efficiency OCCURS when VMIN=0 and efficiency is 78.5%.

65.	What is the disadvantage of a class B push-pull amplifier?(a) The efficiency reduces(b) The figure of merit increases(c) The cross-over distortion occurs(d) The Q-power dissipation is very largeThis question was addressed to me in an international level competition.I'm obligated to ask this question of Push-Pull Class B Amplifier topic in portion Power Amplifier of Analog Circuits
Answer» CORRECT answer is (c) The cross-over DISTORTION occurs To elaborate: A class B amplifier helps increase efficiency, and the figure of merit reduces. The q power dissipation reduces and cross over distortion INCREASES. Due to two transistors, when ONE transistor turns off the other does not begin conduction immediately, HENCE output current is zero for a short interval.

66.	Which of these is incorrect for complementary symmetry push-pull amplifiers?(a) During positive cycle NPN transistor conducts(b) It is easier to fabricate on IC(c) Size of the transformer required reduces(d) Efficiency and figure of merit are same as transformer coupled push-pull amplifierThis question was addressed to me in an interview for internship.Enquiry is from Push-Pull Class B Amplifier topic in portion Power Amplifier of Analog Circuits
Answer» The correct CHOICE is (c) Size of the TRANSFORMER required reduces The EXPLANATION is: The complementary symmetry push-pull amplifier uses one NPN and one PNP transistor to conduct in positive and negative cycles respectively. It does not affect efficiency or figure of merit, but since no transformer is being used, it is easier to fabricate on ICs.