Introduction to Electronics Devices & Circuits

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This topic includes multiple choice questions along with simulation results

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MET’s Institute of Engineering Bhujbal Knowledge CityNashik Department of Electrical Engineering Subject : Introduction to Electronic DevicesAnd 1 Subject : Introduction to Electronic DevicesAnd Circuits Prepared by Mr. Kishor Tukaram Ugale

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Objectives 1 Understand and apply basic and semiconductor principles to the device to observe itsperformance 2 Comply and verify parameters after exciting devices by any stated method. 2 3 Simulate electronics circuits using computer simulation software to obtain desired results.

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1. Calculate the power dissipation of a diode having I D 40 mA. a 28 mW b 28 W Prof.K.T.Ugale METs IOE BKC Nashik 3 c 280 mW d Undefined

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Ans. a Discussion : V D I D 0.7 0.040W 0.028 W 28 mW Power Dissipationof a diode 28 mW 4

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- biased 2.Each diode in a center-tapped full-wave rectifier is a d conducts for of the input cycle. a forward 90º b reverse 180º 5 b reverse 180º c forward 180º d reverse 90º

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Ans. c Circuit Schematic: Input Output Waveforms : T T 6

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3. The output frequency of a full-wave rectifier is of input frequency. a One half b Equal to b Equal to c Twice d One quarter 7 Prof.K.T.Ugale METs IOE BKC Nashik

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Ans. c Circuit Schematic: Input Output Waveforms : T T 8

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4. Determine the peak value of the current through the load resistor. I a 2.325 mA b 5 mA c 1.25 mA d 0 mA 9

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Ans. a Discussion : V i Voltage across diode + 2K+2KI ---------------------------------- I peak value of current V i 0.7 + 4KI ---------------------- For Si diode V D 0.7V I V i – 0.7 / 4K 10 Peak value of current flowing through load resistor 2.325 mA

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5. Determine the peak for both half cycles of the output waveform. V a 16V -4 b 16 V 4 V c –16 V 4 V d –16 V –4 V 11

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6. Determine the reading on the meter when V CC 20 V R C 5 k and I C 2 mA. a 10V b 0.2V c 0.7V d none of above 12

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Ans. d Discussion : Apply KVL at Collector-Emitter loop we have +V CC – I C R C +V CE 0 V CC I C R C – V CE V CC I C R C – V CE After substituting the values 20 5k 2m – V CE V CE 10V – 20V 13 V CE -10V

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7. Refer to this figure. If the value of Determine the value of A v . .6 a 49 b 5 c 100 d 595 14

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Ans. b Discussion : A v Rc/r e where r e 26mV/Ie where Ie6mA r 26/64.34ohms. r e 26/64.34ohms. 15 A v 25/4.34 A v V oltage Gain 5

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8. Refer to this figure. Determine the minimum value of I B that will produce saturation. a 0.25 mA b 5.325 μA c 1.065μA d 10.425 μA 16

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Ans : d Discussion : Apply kirchoffs voltage law at the Base to Emitter loop V CC I C R C + V CE 10 4.710 3 Ic + 0.2 10 4.710 Ic + 0.2 on solving Ic 2.085 mA. For CE configuration we know Ic βI b where β 200 So I b 2.085/200. I b 10.425 µA 1 7

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9. For given circuit In this circuit β DC 100 and V IN 8 V. The value of R B that will produce saturation is: a 92 kΩ b 9.1 kΩ c 100 kΩ d 150kΩ

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Ans : a Discussion : The condition for saturation need to be consider that is V CE 0.2 V Then I 20- 0.2/2.5 I csat 0.2/2.5 7.92 mA. I B 7.92/100 0. 079 mA. Now R B 8 - 0.7 / 0.079 Rb 92 k approx

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10. Refer to this figure I Csat is: 0.05 mA a 0.05 mA b 2.085 mA c 1.065 mA d 7.04 mA

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Ans. b Discussion : Apply KVL in Collector to Emitter loop 10 4.710 3 I c + 0.2 I c 2.085mA

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11. Refer to this figure. The value of V BC is: a 9.2 V b 9.9 V c -9.2 V d -9.9 V

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Ans : c V BC V B - V C O R V BC V BE -V CE----------------- ------ Now applying KVL between base-emitter loop -5+5000I B +0.7 0 or I B 8.610 -4 A ----------------------- ------ We know I C β I B or Ic 258.610 -4 0.0215 A ------------------------ 1 2 A We know I C β I B or Ic 258.610 0.0215 A ------------------------ -------- Again applying KVL between collector-emitter loop -20 + 470Ic + V CE 0 V CE 9.895V After substituting the values of V BE and V CE in equation 1we have V BC 0.7 - 9.895 V BC -9.2 V . 2 2 3

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12. For agiven circuit which input is more suitable to drive the relay a sine b square c triangular d all of the above

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Ans. b

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13. To analyze the common-emitter amplifier what must be done to determine the dc equivalent circuit a leave circuitunchanged b Replacing coupling bypass capacitors with open b Replacing coupling bypass capacitors with open c Replacing coupling bypass capacitors with shorts d Replace VCCwith ground Ans : b

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14. Refer to this figure. The output signal from the first stage of this amplifier is 0 V. The trouble could be caused by a an open C4 b an openC2 c an openbase-emitterof Q1 d A shorterd C4

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Ans. c

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15. Determine the lower cutoff frequency of this network a 15.8 Hz b 46.13 Hz c 238.73 Hz d 1575.8Hz

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Ans. c Discussion : f 1/2ΠR sig + R G C G 1/2Π 10k + 1M100p f 238.73Hz

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16. Determine the break frequency for this circuit a 15.915 Hz b 159.15 Hz c 31.85 Hz d 318.5 Hz

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Ans. b Discussion : f 1/2ΠRC 1/2Π 10k 0.1µ f 159.15 Hz

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17. Refer to this figure. The upper cutoff frequency of this amplifier is 22 kHz. The output at that frequency is 6.71 V p-p. What is the output voltage at 220 kHz a 9.49 V p-p b 6.71 V p-p c 0.671 V p-p d 0.0671 V p-p 3 3

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Ans. c Discussion : As the frequency has increased 10 times the cutoff frequency so output voltage will reduce up to the same fraction. Simple math question: fc 22KHz x 10 220KHz. So Vout 6.71/10 0.671 V .

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18. Refer to this figure. If the value of R 1 decreases the voltage gain will and the input impedance will . a increase increase b increase decrease c decrease decrease d decrease. increase

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Ans. b

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19. Refer to the given figure. The input impedance of this circuit is a 500kΏ b 50kΏ c 10kΏ d 5k Ώ

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Ans. b Discussion : As Impedance ri+jxi. ri50k and xi0. So by above equation. ri50k and xi0. So by above equation. So input impedance 50kΏ

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20. Calculate the overall voltage gain of the circuit if R1 100 and Rf 1 k . a -1 b -10 c 1 1 d 9

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Ans. c Discussion : As the input is applied to non-inverting terminal of op-amp sothis amplifier is called as non-inverting amplifier According to output voltage equation of non-inverting amplifier V o 1+R f /R 1 V i Voltage gain Vo/Vi 1+R f /R 1 1+1K/100 V oltage gain 1 1

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21. An emitter-follower amplifier has an input impedance of 107 k . The input signal is 12 mV. The approximate output voltageis a 8.92 V b 1 1 2mV b 1 1 2mV c 16mV d 8.5 V

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22. Calculate the output voltage if R 1 R 2 R 3 100 R f 1 k and V 1 V 2 V 3 50 mV. a -1.5V b 1.5V c 0.5V d none of above

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Ans. a Discussion : By using formula V o - V 1 /R 1 +V 2 /R 2 +V 3 /R 3 R f V o -1.5V V o -1.5V

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23. Calculate the output voltage for this circuit when V 1 2.5 V and V 2 2.25 V. a –5.25V b 2.5V c 2.25V d 5.25V 4 4 Prof.K.T.Ugale METs IOE BKC Nashik

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Ans. d Discussion : Vo 1+2R 1 /R G R f /R 1 V 2 -V 1 1+25k/5005k/5K2.5-2.25 1+2010.25 1+2010.25 210.25 5.25 V V o 5.25V

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24. You have a light-dimmer circuit using an SCR. In testing the circuit you find that I G 0 mA and the light is still on. You conclude that the trouble might be one of the following: a The SCR isopen Theswitch isfaulty b Theswitch isfaulty c The gate circuit is shorted d This is normalnothing is wrong

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Static V-I Characterictics of Thyristor 4 7

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Ans. D Discussion : : If I g 0 i.e .gate current then SCR is ‘ON’ condition i.e SCR anode current crosses the latching current once the SCR is in latching stateafter removal of gate pulse or pulses SCR is still is in‘ON’ state and the light is still on. 4 8 Prof.K.T.Ugale METs IOE BKC Nashik

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