88 + Interview Questions in Application Of Diodes in Electronic Devices & Circuits Page 1 InterviewSolution

1.	A transistor series regulator has the following specifications: VIN=15V, VZ=8.3V, β=100, R=1.8KΩ, RL=2KΩ. What will be the Zener current in the regulator circuit?(a) 4.56mA(b) 3.26mA(c) 4.56mA(d) 3.68mAThe question was posed to me in an internship interview.The question is from Voltage Regulation Using Zener Diode topic in division Application of Diodes of Electronic Devices & Circuits
Answer» The CORRECT choice is (d) 3.68mA For EXPLANATION: We know, VO=VZ-VBE=8.3-0.7=7.6V. VCE=VIN-V0=15-7.6=7.4V.So, IR=(VIN-VZ)/R=(15-8.3)/1.8m=3.72mA. IL=VO/RL=7.6/2000=3.8mA. IB=IL/ β=3.8mA/100=0.038mA. FINALLY, IZ=IR-IB=3.72-0.038=3.682mA.

Discussion

2.	In a half wave rectifier, the sine wave input is 50sin50t. If the load resistance is of 1K, then average DC power output will be?(a) 3.99V(b) 2.5V(c) 5.97V(d) 6.77VThis question was posed to me during a job interview.Enquiry is from Half-Wave Rectifier in section Application of Diodes of Electronic Devices & Circuits
Answer» Correct OPTION is (B) 2.5V For EXPLANATION: The standard form of a sine WAVE is Vmsinωt. BY comparing the given information with this equation, VM =50. Power=Vm^2/RL=50*50/1000=2.5V.

Discussion

3.	The ripple factor (ϒ) of inductor filter is_________(a) ϒ = RZ3/√2ωL(b) ϒ = RZ/3√2ωL(c) ϒ = RZ3√2/ωL(d) ϒ = RZ3/√2ωLI had been asked this question during an interview.Origin of the question is Inductor Filters topic in division Application of Diodes of Electronic Devices & Circuits
Answer» Correct choice is (B) ϒ = RZ/3√2ωL For explanation: Ripple factor will DECREASE when L is INCREASED and RL. Inductor has a HIGHER dc resistance. It depends on PROPERTY of opposing the change of direction of current.

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4.	Determine the minimum value of load resistance that can be used in the circuit with (IZ)Min=3mA. The input voltage is 10V and the resistance R is 500Ω. The Zener diode has a VZ=6V 0and (IZ)MAX=90mA.(a) 1KΩ(b) 2.4KΩ(c) 1.2KΩ(d) 3.6KΩI got this question at a job interview.The doubt is from Voltage Regulation Using Zener Diode in portion Application of Diodes of Electronic Devices & Circuits
Answer» Right option is (c) 1.2KΩ Best explanation: The I=(VIN-VZ)/R=(10-6)/500=8mA. (IL)MAX=I-(IZ)MIN=8-3=5mA. (RL)MIN=VZ/(IL)MAX=6/5m=1.2KΩ.

Discussion

5.	The limiting value of the current resistor used in a Zener diode (when used as a regulator)(a) (R)min=[(Vin)max + VZ/R(b) (R)min=[(Vin)max-VZ]/R(c) (R)min=[(Vin)max-VZ]R(d) (R)min=[(Vin)max+ VZ]RI got this question in an interview for job.The origin of the question is Voltage Regulation Using Zener Diode topic in division Application of Diodes of Electronic Devices & Circuits
Answer» The correct answer is (b) (R)min=[(Vin)max-VZ]/R Explanation: When the input voltage is maximum, the LOAD current is minimum, the ZENER current should not INCREASE the maximum RATED value. THEREFORE there should be a minimum value of resistor.

Discussion

6.	The output waveform of CLC filter is superimposed by a waveform referred to as_________(a) Square wave(b) Triangular wave(c) Saw tooth wave(d) Sine waveI had been asked this question in a job interview.This intriguing question comes from CLC Filter in section Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (C) Saw tooth wave Explanation: Since the rectifier conducts current only in the forward direction, any energy DISCHARGED by the capacitor will FLOW into the LOAD. This result in a DC voltage upon which is superimposed a waveform referred to as a saw tooth wave.

Discussion

7.	What is the relation between time constant and load resistance?(a) They don’t depend on each other(b) They are directly proportional(c) They are inversely proportional(d) Cannot be predictedI got this question at a job interview.This is a very interesting question from CLC Filter topic in portion Application of Diodes of Electronic Devices & Circuits
Answer»

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8.	A single phase full wave rectifier makes use of pi section filter with 10µF capacitors and a choke of 10henry. The secondary voltage is 280V and the load current is 100mA. Determine the dc output voltage when f=50Hz.(a) 345V(b) 521V(c) 243V(d) 346VI have been asked this question in unit test.Query is from CLC Filter topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The CORRECT answer is (d) 346V The best I can EXPLAIN: GIVEN, VRMS=280V So, V¬m = 1.414280=396V. From theory of CAPACITOR filter, VDC = VM –IDC/4fC=396-0.1/ (4501010^-6)=346V.

Discussion

9.	The advantages of a pi-filter is_________(a) low output voltage(b) low PIV(c) low ripple factor(d) high voltage regulationThis question was addressed to me in exam.My question is based upon CLC Filter in section Application of Diodes of Electronic Devices & Circuits
Answer» Right option is (c) low ripple factor The explanation: Due to the use of two capacitors with an inductor, an improved filtering ACTION is provided. This leads to decrement in ripple factor. A low ripple factor signifies REGULATED and ripple free DC voltage.

Discussion

10.	The ripple to heavy loads by a capacitor is_______(a) high(b) depends on temperature(c) low(d) no ripple at allThis question was addressed to me in a national level competition.Question is taken from L-Section Filter topic in division Application of Diodes of Electronic Devices & Circuits
Answer» Right answer is (c) low To elaborate: Ripple FACTOR is INVERSELY proportional to LOAD RESISTANCE for a CAPACITOR filter. So, the ripples that are produced are low when the load is high.

Discussion

11.	The output voltage VDC for a rectifier with inductor filter is given by_________(a) (2Vm/π)-IDCR(b) (2Vm/π)+IDCR(c) (2Vmπ)-IDCR(d) (2Vmπ)+IDCRThis question was posed to me in an online quiz.This interesting question is from Inductor Filters topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Right option is (a) (2Vm/π)-IDCR Explanation: The INDUCTOR with HIGH RESISTANCE can CAUSE poor voltage REGULATION. The choke resistance, the resistance of half of transformer secondary is not negligible.

Discussion

12.	DC power output of bridge full wave rectifier is equal to (Im is the peak current and RL is the load resistance).(a) 2 Im^2RL(b) 4 Im^2RL(c) Im^2RL(d) Im^2 RL/2I had been asked this question in examination.The doubt is from Bridge Rectifier in section Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT answer is (b) 4 Im^2RL The EXPLANATION is: DC output power is the power output of the rectifier. We know VDC for a bridge rectifier is 2Vm and IDC for a bridge rectifier is 2Im. We also know VDC=IDC/RL. Hence output power is 4Im^2RL.

Discussion

13.	The rms value of ripple current for an L section filter is_______(a) IRMS=√2/3XLVDC(b) IRMS=√2/3XLVDC(c) IRMS=√2/3XL VDC(d) IRMS=√2/3XLVDCThis question was posed to me in an internship interview.The origin of the question is L-Section Filter in division Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (a) IRMS=√2/3XLVDC To explain I WOULD SAY: The ac current through L is determined primarily by XL=2ωL. It is directly proportional to VOLTAGE PRODUCED and indirectly proportional to the REACTANCE.

Discussion

14.	A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.(a) 368.98mW(b) 275.2mW(c) 145.76mW(d) 456.78mWThe question was posed to me during a job interview.My question comes from Full-wave Rectifier in division Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT option is (B) 275.2mW To explain I would SAY: The AC power input PIN=IRMS^2(RF+Rr). IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)1.414=13.5mA So, PIN=(13.510-3)2*(1500+10)=275.2mW.

Discussion

15.	In a bridge full wave rectifier, the input sine wave is 40sin100t. The average output voltage is_________(a) 22.73V(b) 16.93V(c) 25.47V(d) 33.23VThis question was posed to me by my school teacher while I was bunking the class.Question is taken from Bridge Rectifier topic in division Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT OPTION is (c) 25.47V The BEST I can explain: The equation of SINE wave is in the form Emsinωt. Therefore, Em=40. HENCE output voltage is 2Em=80V.

Discussion

16.	The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by _________(a) Vm/√2RL(b) Vm/(RF+RL)π(c) 2Vm/√π(d) Vm/RLI had been asked this question by my college director while I was bunking the class.The doubt is from Half-Wave Rectifier topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT ANSWER is (B) VM/(RF+RL)π The explanation is: For a half wave RECTIFIER, the IDC=IAVG=Im/π I= Vmsinωt/(RF+RL)=Imsinωt Im =Vm/ RF+RL So, IDC=Im/π=Vm/(RF+RL).

Discussion

17.	When is a regulator used?(a) when there are small variations in load current and input voltage(b) when there are large variations in load current and input voltage(c) when there are no variations in load current and input voltage(d) when there are small variations in load current and large variations in input voltageThe question was asked in an online quiz.I'd like to ask this question from Voltage Regulation Using Zener Diode in portion Application of Diodes of Electronic Devices & Circuits
Answer» Correct CHOICE is (a) when there are small variations in load current and input VOLTAGE For EXPLANATION: The regulator has following limitations: 1.It has low efficiency for heavy load currents 2. The output voltage changes slightly DUE to ZENER impedance. Hence, it is used when there are small variations in load current and input voltage.

Discussion

18.	A full wave rectifier with a load resistance of 5KΩ uses an inductor filter of 15henry. The peak value of applied voltage is 250V and the frequency is 50 cycles per second. Calculate the ripple factor (ϒ).(a) 0.1(b) 0.6(c) 0.5(d) 0.4The question was posed to me in unit test.I'm obligated to ask this question of Inductor Filters in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Correct OPTION is (d) 0.4 The explanation: ϒ=IAC/IDC, IAC=2√2Vm/3π(RL^2+4ω^2L^2)^1/2 By PUTTING the VALUES, IAC=4.24Ma. VDC=2Vm/π, IDC=VDC/RL=2Vm/RL π IDC=2250/(3.1415*10^3)=10.6mA. ϒ=4.24/10.6=0.4.

Discussion

19.	The voltage in case of a full wave rectifier in a CLC filter is_________(a) Vϒ = IDC/2fC(b) Vϒ = IDC fC(c) Vϒ = IDC/fC(d) Vϒ = 2IDCfCThis question was addressed to me in class test.The doubt is from CLC Filter topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Right choice is (a) Vϒ = IDC/2fC For explanation I would SAY: T he filter CIRCUIT is a COMBINATION of capacitors and inductors. The RMS value DEPENDS on the peak value of charging and discharging magnitude, VPEAK.

Discussion

20.	Ripple factor of a half wave rectifier is_________(Im is the peak current and RL is load resistance)(a) 1.414(b) 1.21(c) 1.4(d) 0.48The question was posed to me by my college director while I was bunking the class.I'm obligated to ask this question of Half-Wave Rectifier in section Application of Diodes of Electronic Devices & Circuits
Answer» The correct choice is (b) 1.21 To explain I would say: The ripple factor of a RECTIFIER is the measure of DISTURBANCES PRODUCED in the output. It’s the EFFECTIVENESS of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 1.21.

Discussion

21.	If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be_________(a) 150Hz(b) 200Hz(c) 100Hz(d) 300HzI got this question in unit test.The question is from Half-Wave Rectifier in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Correct choice is (c) 100Hz Easiest EXPLANATION: The RIPPLE FREQUENCY of the output and input is same. This is because, one half cycle of input is passed and other half cycle is SEIZED. So, effectively the frequency is the same.

Discussion

22.	What isthe output as a function of the input voltage (for positive values) for the given figure. Assume it’s an ideal op-amp with zero forward drop (Di=0)(a) 0(b) -Vi(c) Vi(d) 2ViI have been asked this question in a national level competition.This interesting question is from Half-Wave Rectifier topic in section Application of Diodes of Electronic Devices & Circuits
Answer» The correct answer is (C) Vi Easiest explanation: When the INPUT of the inverted mode op-amp is POSITIVE, the output is negative. The diode is reverse BIASED. The input appears at the output.

Discussion

23.	A Zener regulator has to handle supply voltage variation from 19.5V to 22.5V. Find the magnitude of regulating resistance, if the load resistance is 6KΩ. The Zener diode has the following specifications: breakdown voltage =18V, (IZ)Min=2µA, maximum power dissipation=60mW and Zener resistance =20Ω.(a) 0 < R < 500Ω(b) 77.8 < R < 500Ω(c) 77.8 < R < 100Ω(d) 18 < R < 500ΩThis question was addressed to me in an internship interview.This is a very interesting question from Voltage Regulation Using Zener Diode in portion Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (b) 77.8 < R < 500Ω Easy EXPLANATION: (PZ)MAX/rZ=(IZ)MAX2 . So, (IZ)MAX =60m/20=54.8µA. IL=VO/RL=18/6000=3mA. RMAX=(VMin-VZ)/[( IZ)Min+( IL)MAX]=(19.5-18)/(2µ+3m)=500Ω. RMIN=(VMAX-VZ)/[( IZ)MAX+( IL)Min]=(22.5-18)/(54.8m+3m)=77.8Ω.

Discussion

24.	The percentage voltage regulation (VL) is given by_________(a) (VNL-VL)/VNL100(b) (VNL+VL)/VNL100(c) (VNL-VL)/VL100(d) (VNL+VL)/VL100This question was posed to me in an online interview.The doubt is from Voltage Regulation Using Zener Diode topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (a) (VNL-VL)/VNL*100 The best explanation: The CHANGE in the output voltage from no load to full load condition is called as voltage REGULATION, where VNL is the voltage at no load condition. It is used to maintain a NEARLY CONSTANT output voltage. If the regulation is high, the output voltage is stable.

Discussion

25.	Calculate LC for a full wave rectifier which provides 10V dc at 100mA with a maximum ripple of 2%. Input ac frequency is 50Hz.(a) 4010^-6(b) 1010^-6(c) 3010^-6(d) 9010^-6I got this question in an interview for job.I'm obligated to ask this question of L-Section Filter topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The correct CHOICE is (a) 4010^-6 To EXPLAIN: LC=1/6√2ω^2ϒ ω=2πf=314 By putting the values, LC=1/6√2(314)^2 0.02=4010^-6.

Discussion

26.	A full wave rectifier uses a capacitor filter with 500µF capacitor and provides a load current of 200mA at 8% ripple. Calculate the dc voltage.(a) 15.56V(b) 20.43V(c) 11.98V(d) 14.43VI have been asked this question in an international level competition.The origin of the question is Capacitor Filters topic in division Application of Diodes of Electronic Devices & Circuits
Answer» The correct answer is (d) 14.43V The BEST I can EXPLAIN: The RIPPLE factor ϒ=IL/ 4√3 fCVDC VDC=20010^-3/ 4√3 505008 =14.43.

Discussion

27.	Efficiency of bridge full wave rectifieris_________(a) 81.2%(b) 50%(c) 40.6%(d) 45.33%I had been asked this question in an interview for job.My question is from Bridge Rectifier in division Application of Diodes of Electronic Devices & Circuits
Answer» Right answer is (a) 81.2% Easy explanation: It’s obtained by taking ratio of DC power OUTPUT to maximum AC power delivered to LOAD. Efficiency of a rectifier is the effectiveness of rectifier to convert AC to DC. It’s usually expressed inn percentage. For BRIDGE full WAVE rectifier, it’s 81.2%.

Discussion

28.	Number of diodes used in a full wave bridge rectifier is_________(a) 1(b) 2(c) 3(d) 4I have been asked this question in class test.The query is from Bridge Rectifier in division Application of Diodes of Electronic Devices & Circuits
Answer» RIGHT option is (d) 4 To elaborate: The model of a bridge RECTIFIER is same as Wein Bridge. It NEEDS 4 resistors. Bridge rectifier needs 4 DIODES while CENTRE tap configuration requires only one.

Discussion

29.	Which of the following instruments cannot be applied for ac measurements?(a) Hot wire(b) PMMC(c) Electrostatic(d) Induction typeI have been asked this question in an interview.Query is from The Rectifier Voltmeter in division Application of Diodes of Electronic Devices & Circuits
Answer» The correct option is (b) PMMC The best explanation: The MOVING coil instrument can only be used on D.C SUPPLY as the reversal of current PRODUCES a reversal of torque on the coil. It’s very delicate and sometimes USES AC circuit with a rectifier. It’s costly as compared to moving coil iron instruments.

Discussion

30.	What causes to decrease the sudden rise in the current for a rectifier?(a) the electrical energy(b) The ripple factor(c) The magnetic energy(d) Infinite resistanceThis question was addressed to me in class test.My question comes from Inductor Filters in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The CORRECT option is (C) The MAGNETIC energy For explanation: When the output current of a rectifier increases above a certain value, magnetic energy is stored in the inductor. This energy tends to decrease the sudden rise in the current. This also HELPS to prevent the current to fall down too MUCH.

Discussion

31.	An L section filter with L=2henry and C= 49µF is used in the output of a full wave single phase rectifier that is fed from a 40-0-40 V peak transformer. The load current is 0.2A. Calculate the output dc voltage.(a) 20.76V(b) 24.46V(c) 34.78V(d) 12.67VThis question was posed to me in an interview.Asked question is from L-Section Filter topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» Correct choice is (b) 24.46V To explain I WOULD say: GIVEN, VL= 40V. VDC=2/πVL=2/π40=25.46V.

Discussion

32.	In electronic voltmeter, the range of input voltages can be extended by using _______(a) Functional switch(b) Input attenuator(c) Rectifier(d) Balanced bridge dc amplifierI got this question in an online quiz.I would like to ask this question from The Rectifier Voltmeter in section Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (B) Input attenuator To explain I would say: The function of the attenuator is that it HELPS to select a particular range of VOLTAGE VALUES. The rectifier is essential in a voltmeter for the conversion of AC voltage into DC voltage.

Discussion

33.	If peak voltage on a bridge full wave rectifier circuit is 5V and diode cut in voltage os 0.7, then the peak inverse voltage on diode will be_________(a) 4.3V(b) 9.3V(c) 8.6V(d) 3.6VI had been asked this question during an online interview.This is a very interesting question from Bridge Rectifier in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Correct choice is (d) 3.6V To explain I would say: PIV is the maximum reverse BIAS voltage that can be appeared across a DIODE in thecircuit. If PIV RATING of diode is less than this value breakdown of diode may OCCUR.. Therefore, PIV rating of diode should be greater than PIV in the circuit, For bridge RECTIFIER PIV is Vm-VD = 5-1.4=3.6.

Discussion

34.	When the regulation by a Zener diode is with a varying input voltage, what happens to the voltage drop across the resistance?(a) Decreases(b) Has no effect on voltage(c) Increases(d) The variations depend on temperatureThe question was asked in exam.The question is from Voltage Regulation Using Zener Diode topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» Right answer is (C) INCREASES To explain I would say: When the INPUT voltage varies, the input current ALSO varies. This makes more current to flow in the diode. This increase in the current should balance a change in the load current. Hence the voltage DROP increases across the resistor.

Discussion

35.	At f=50Hz, the ripple factor of CLC filter is_________(a) ϒ=5700RL / (LC1C2)(b) ϒ=5700/ (LC1C2RL)(c) ϒ=5700LC1/ (C2RL)(d) ϒ=5700C1C2/ (LRL)I got this question in an internship interview.Origin of the question is CLC Filter topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» The correct option is (b) ϒ=5700/ (LC1C2RL) To EXPLAIN: The ripple FACTOR of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple VOLTAGE. The ratio of ripple voltage to DC output voltage is ripple factor which is 5700 / (LC1C2RL) at 50HZ.

Discussion

36.	Which of the following are true about capacitor filter?(a) It is also called as capacitor output filter(b) It is electrolytic(c) It is connected in parallel to load(d) It helps in storing the magnetic energyI had been asked this question in examination.This intriguing question originated from Capacitor Filters in division Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT choice is (B) It is electrolytic The EXPLANATION: The rectifier MAY be full wave or half wave. The capacitors are usually electrolytic even THOUGH they are largein size.

Discussion

37.	Major part of the filtering is done by the first capacitor in a CLC filter because _________(a) The capacitor offers a very low reactance to the ripple frequency(b) The capacitor offers a very high reactance to the ripple frequency(c) The inductor offers a very low reactance to the ripple frequency(d) The inductor offers a very high reactance to the ripple frequencyI have been asked this question by my college professor while I was bunking the class.Query is from CLC Filter in portion Application of Diodes of Electronic Devices & Circuits
Answer» Correct option is (a) The CAPACITOR offers a very low reactance to the ripple FREQUENCY The best I can EXPLAIN: The CLC filters are used when high voltage and low ripple frequency is needed than L SECTION filters. The capacitor in a CLC filter offers very low reactance to the ripple frequency. So, maximum of the filtering is done by the first capacitor across the L section part.

Discussion

38.	A shunt capacitor of value 500µF fed rectifier circuit. The dc voltage is 14.43V. The dc current flowing is 200mA. It is operating at a frequency of 50Hz. What will be the value of peak rectified voltage?(a) 18.67V(b) 16.43V(c) 15.98V(d) 11.43VThis question was posed to me in an international level competition.The doubt is from Capacitor Filters topic in portion Application of Diodes of Electronic Devices & Circuits
Answer» The correct ANSWER is (B) 16.43V The BEST I can explain: We know, Vm=Vdc+Idc/4fC =14.43+ {200/ (200*500)} 10^3 =14.43+2=16.43V.

Discussion

39.	A dc voltage of 380V with a peak ripple voltage not exceeding 7V is required to supply a 500Ω load. Find out the inductance required.(a) 10.8henry(b) 30.7henry(c) 28.8henry(d) 15.4henryThe question was posed to me in examination.This is a very interesting question from Inductor Filters topic in division Application of Diodes of Electronic Devices & Circuits
Answer» The CORRECT CHOICE is (c) 28.8henry Explanation: Given the ripple voltage is 7V. So, 7=1.414VRMS ϒ=VRMS/VDC=4.95/380=0.0130. ϒ=1/3√2(RL/Lω) So, L=28.8henry.

Discussion

40.	Ripple factor of bridge full wave rectifier is?(a) 1.414(b) 1.212(c) 0.482(d) 1.321I have been asked this question during an interview.This key question is from Bridge Rectifier in chapter Application of Diodes of Electronic Devices & Circuits
Answer»

Discussion

41.	In a choke l section filter_______(a) the inductor and capacitor are connected across the load(b) the inductor is connected in series and capacitor is connected across the load(c) the inductor is connected across and capacitor is connected in series to the load(d) the inductor and capacitor are connected in seriesThis question was addressed to me during an interview.Question is from L-Section Filter topic in division Application of Diodes of Electronic Devices & Circuits
Answer» CORRECT answer is (b) the inductor is connected in series and capacitor is connected ACROSS the load For explanation: The choke filter is sometimes also called as L section filter because the inductor and capacitor are connected in the shape ’L’ or INVERTED MANNER. But several SECTIONS of thischoke filter are employed to smooth the output curve and make it free from ripples.

Discussion

42.	The rectifier current is a short duration pulses which cause the diode to act as a_________(a) Voltage regulator(b) Mixer(c) Switch(d) OscillatorThe question was posed to me in semester exam.This interesting question is from Capacitor Filters topic in division Application of Diodes of Electronic Devices & Circuits
Answer» Correct choice is (C) Switch The best I can explain: The diode permits CHARGE to flow in capacitor when the transformer VOLTAGE EXCEEDS the capacitor voltage. It DISCONNECTS the power source when the transformer voltage falls below that of a capacitor.

Discussion

43.	The inductor filter should be used when RL is consistently small because_________(a) Effective filtering takes place when load current is high(b) Effective filtering takes place when load current is low(c) Current lags behind voltage(d) Current leads voltageThis question was posed to me in an interview for job.My question is taken from Inductor Filters topic in division Application of Diodes of Electronic Devices & Circuits
Answer» The correct answer is (a) Effective filtering takes place when load CURRENT is high Easy EXPLANATION: When RLis infinite, the ripple FACTOR is 0.471. This value is CLOSE to that of a rectifier. So, the resistance should be small.

Discussion

44.	The inductor filter gives a smooth output because_________(a) It offers infinite resistance to ac components(b) It offers infinite resistance to dc components(c) Pulsating dc signal is allowed(d) The ac signal is amplifiedThis question was addressed to me in semester exam.Asked question is from Inductor Filters topic in section Application of Diodes of Electronic Devices & Circuits
Answer» Correct answer is (a) It OFFERS infinite resistance to ac components The BEST I can explain: The inductor does not ALLOW the ac components to PASS through the filter. The main PURPOSE of using an inductor filter is to avoid the ripples. By using this property, the inductor offers an infinite resistance to ac components and gives a smooth output.

Discussion

45.	If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be?(a) 5V(b) 4.9V(c) 4.3V(d) 6.7VI had been asked this question in an interview for internship.I need to ask this question from Half-Wave Rectifier in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The correct option is (c) 4.3V For explanation: PIV is the MAXIMUM reverse bias voltage that can be appeared across a DIODE in the given CIRCUIT, If the PIV rating is less than this VALUE of breakdown of diode will OCCUR. For a rectifier, PIV=Vm-Vd=5-0.7=4.3V.

Discussion

46.	DC average current of a bridge full wave rectifier (where Im is the maximum peak current of input).(a) 2Im(b) Im(c) Im/2(d) 1.414ImI had been asked this question during an internship interview.Asked question is from Bridge Rectifier in section Application of Diodes of Electronic Devices & Circuits
Answer» The correct CHOICE is (b) Im Easiest explanation: Average DC current of half wave rectifier is Im. Since output of half wave rectifier contains only one half of the INPUT. The average value is the half of the AREA of one half CYCLE of sine wave with peak Im. This is EQUAL to Im.

Discussion

47.	Efficiency of a centre tapped full wave rectifier is _________(a) 50%(b) 46%(c) 70%(d) 81.2%This question was posed to me during an interview.The above asked question is from Full-wave Rectifier topic in section Application of Diodes of Electronic Devices & Circuits
Answer» The correct choice is (d) 81.2% For explanation: Efficiency of a rectifier is the effectiveness to convert AC to DC. It’s obtained by taking ratio of DC power output to MAXIMUM AC power delivered to load. It’s usually expressed in percentage. For CENTRE tapped full WAVE rectifier, it’s 81.2%.

Discussion

48.	What makes the load in a choke filter to bypass harmonic components?(a) capacitor(b) inductor(c) resistor(d) diodesI got this question during an online exam.My query is from L-Section Filter topic in chapter Application of Diodes of Electronic Devices & Circuits
Answer» The correct option is (a) CAPACITOR For explanation I would say: When the capacitor is shunted across the LOAD, it bypasses the harmonic components. This is because it offers LOW reactance to ac ripple component. In another hand the inductor offers high impedance to harmonic TERMS.

Discussion

49.	The rms ripple voltage (Vrms) of a shunt filter is_________(a) IDC/2√3(b) IDC2√3(c) IDC/√3(d) IDC√3I had been asked this question in examination.The question is from Capacitor Filters topic in division Application of Diodes of Electronic Devices & Circuits
Answer» Correct option is (a) IDC/2√3 To explain I WOULD say: The ripple waveform will be TRIANGULAR in nature. The rms value of this wave is independent of slopes or lengths of straight LINES. It depends only on the peak value.

Discussion

50.	_________ diodes are preferred in AC rectifier voltmeter arrangements.(a) Silicon(b) Germanium(c) Gallium(d) ArsenideThis question was addressed to me in final exam.The doubt is from The Rectifier Voltmeter in portion Application of Diodes of Electronic Devices & Circuits
Answer» The correct answer is (a) Silicon The BEST I can explain: Silicon diodes are preferred because of their low REVERSE current and HIGH forward current ratings. A PMMC movement is ALSO used by the rectifier type instruments along with a rectifier arrangement.

Discussion

Explore topic-wise InterviewSolutions in .

The ripple factor (ϒ) of inductor filter is_________(a) ϒ = RZ3/√2ωL(b) ϒ = RZ/3√2ωL(c) ϒ = RZ3√2/ωL(d) ϒ = RZ3/√2ωLI had been asked this question during an interview.Origin of the question is Inductor Filters topic in division Application of Diodes of Electronic Devices & Circuits

The advantages of a pi-filter is_________(a) low output voltage(b) low PIV(c) low ripple factor(d) high voltage regulationThis question was addressed to me in exam.My question is based upon CLC Filter in section Application of Diodes of Electronic Devices & Circuits

The ripple to heavy loads by a capacitor is_______(a) high(b) depends on temperature(c) low(d) no ripple at allThis question was addressed to me in a national level competition.Question is taken from L-Section Filter topic in division Application of Diodes of Electronic Devices & Circuits

The voltage in case of a full wave rectifier in a CLC filter is_________(a) Vϒ = IDC/2fC(b) Vϒ = IDC fC(c) Vϒ = IDC/fC(d) Vϒ = 2IDCfCThis question was addressed to me in class test.The doubt is from CLC Filter topic in chapter Application of Diodes of Electronic Devices & Circuits

If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be_________(a) 150Hz(b) 200Hz(c) 100Hz(d) 300HzI got this question in unit test.The question is from Half-Wave Rectifier in chapter Application of Diodes of Electronic Devices & Circuits

Efficiency of bridge full wave rectifieris_________(a) 81.2%(b) 50%(c) 40.6%(d) 45.33%I had been asked this question in an interview for job.My question is from Bridge Rectifier in division Application of Diodes of Electronic Devices & Circuits

Number of diodes used in a full wave bridge rectifier is_________(a) 1(b) 2(c) 3(d) 4I have been asked this question in class test.The query is from Bridge Rectifier in division Application of Diodes of Electronic Devices & Circuits

Which of the following instruments cannot be applied for ac measurements?(a) Hot wire(b) PMMC(c) Electrostatic(d) Induction typeI have been asked this question in an interview.Query is from The Rectifier Voltmeter in division Application of Diodes of Electronic Devices & Circuits

At f=50Hz, the ripple factor of CLC filter is_________(a) ϒ=5700RL / (LC1C2)(b) ϒ=5700/ (LC1C2RL)(c) ϒ=5700LC1/ (C2RL)(d) ϒ=5700C1C2/ (LRL)I got this question in an internship interview.Origin of the question is CLC Filter topic in portion Application of Diodes of Electronic Devices & Circuits

Ripple factor of bridge full wave rectifier is?(a) 1.414(b) 1.212(c) 0.482(d) 1.321I have been asked this question during an interview.This key question is from Bridge Rectifier in chapter Application of Diodes of Electronic Devices & Circuits

DC average current of a bridge full wave rectifier (where Im is the maximum peak current of input).(a) 2Im(b) Im(c) Im/2(d) 1.414ImI had been asked this question during an internship interview.Asked question is from Bridge Rectifier in section Application of Diodes of Electronic Devices & Circuits

Efficiency of a centre tapped full wave rectifier is _________(a) 50%(b) 46%(c) 70%(d) 81.2%This question was posed to me during an interview.The above asked question is from Full-wave Rectifier topic in section Application of Diodes of Electronic Devices & Circuits

What makes the load in a choke filter to bypass harmonic components?(a) capacitor(b) inductor(c) resistor(d) diodesI got this question during an online exam.My query is from L-Section Filter topic in chapter Application of Diodes of Electronic Devices & Circuits

The rms ripple voltage (Vrms) of a shunt filter is_________(a) IDC/2√3(b) IDC2√3(c) IDC/√3(d) IDC√3I had been asked this question in examination.The question is from Capacitor Filters topic in division Application of Diodes of Electronic Devices & Circuits

_________ diodes are preferred in AC rectifier voltmeter arrangements.(a) Silicon(b) Germanium(c) Gallium(d) ArsenideThis question was addressed to me in final exam.The doubt is from The Rectifier Voltmeter in portion Application of Diodes of Electronic Devices & Circuits