Semiconductors Practice Questions PDF

ALLEN® Semiconductors 1 SEMICONDUCTORS 6. Zener breakdown occ...

ALLEN® Semiconductors 1 SEMICONDUCTORS 6. Zener breakdown occurs in a p-n junction 1. If an emitter current is changed by 4 mA, the having p and n both : collector current changes by 3.5 mA. The value (1) lightly doped and have wide depletion layer. of b will be : (2) heavily doped and have narrow depletion (1) 7 (2) 0.5 (3) 0.875 (4) 3.5 2. In connection with the circuit drawn below, the layer. value of current flowing through 2 kW resistor (3) lightly doped and have narrow depletion is ________ × 10–4 A. layer. 1 kW (4) heavily doped and have wide depletion 2 kW layer. 5V 7. A 5V battery is connected across the points X and Y. Assume D1 and D2 to be normal silicon 10V 3. A common transistor radio set requires diodes. Find the current supplied by the battery 12V (D.C.) for its operation. The D.C. source if the +ve terminal of the battery is connected to is constructed by using a transformer and a point X. rectifier circuit, which are operated at 220 V D1 10W (A.C.) on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are _______. D2 5W A 4. C B XY The logic circuit shown above is equivalent to : (1) ~ 0.5 A (2) ~ 1.5 A A (3) ~ 0.86 A (4) ~ 0.43 A (1) C 8. For extrinsic semiconductors; when doping B A level is increased: (2) C (1) Fermi-level of p-type semiconductor will go B A upward and Fermi-level of n-type (3) C B semiconductors will go downward. A (2) Fermi-level of p-type semiconductors will (4) C B go downward and Fermi-level of n-type 5. Given below are two statements : semiconductor will go upward. Statement I : PN junction diodes can be used to (3) Fermi-level of both p-type and n-type function as transistor, simply by connecting two semiconductros will go upward for T > TF node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors diodes, back to back, which acts as the base K and downward for T < TF K, where TF is terminal. Fermi temperature. Statement II : In the study of transistor, the (4) Fermi-level of p and n-type semiconductors amplification factor b indicates ratio of the will not be affected. collector current to the base current. In the light of the above statements, choose the correct answer from the options given below : (1) Statement I is false but Statement II is true (2) Both Statement I and Statement II are true (3) Both Statement I and Statement II are false (4) Statement I is true but Statement II is false E 2 Semiconductors ALLEN® 9. The truth table for the followng logic circuit 12. The circuit contains two diodes each with a is : forward resistance of 50 W and with infinite A reverse resistance. If the battery voltage is 6 V, the current through the 120 W resistance is_ mA. D1 130W Y B (1) A B Y (2) A B Y D2 100W 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1 1 1 120W (3) A B Y (4) A B Y 6V 0 0 1 0 0 0 13. Draw the output signal Y in the given 0 1 0 0 1 1 1 0 1 1 0 0 combination of gates :- 1 1 0 1 1 1 10. Match List I with List II. A List I List II 0 1 2 3 4 5 t(s) (a) Rectifier (i) Used either for stepping up or stepping down the B a.c. voltage 0 1 2 3 4 5 t(s) (b) Stabilizer (ii) Used to convert a.c. voltage into d.c. voltage A Y (c) Transformer (iii)Used to remove any ripple in the rectified B output voltage (d) Filter (iv) Used for constant (1) output voltage even 0 1 2 3 4 5 t(s) when the input voltage or load current change Choose the correct answer from the options given below : (2) (1) (a)–(ii), (b)–(iv), (c)–(i), (d)–(iii) 0 1 2 3 4 5 t(s) node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors (2) (a)–(iii), (b)–(iv), (c)–(i), (d)–(ii) (3) (a)–(ii), (b)–(i), (c)–(iv), (d)–(iii) (4) (a)–(ii), (b)–(i), (c)–(iii), (d)–(iv) (3) 11. LED is constructed from Ga-As-P 0 1 2 3 4 5 t(s) semiconducting material. The energy gap of this LED is 1.9 eV. Calculate the wavelength of light emitted and its colour. [h = 6.63 × 10–34 Js and c = 3 × 108 ms–1] (4) (1) 1046 nm and red colour 0 1 2 3 4 5 t(s) (2) 654 nm and orange colour (3) 1046 nm and blue colour (4) 654 nm and red colour E  ALLEN® Semiconductors 3 14. The zener diode has a Vz = 30 V. The current 20. An npn transistor operates as a common emitter passing through the diode for the following amplifier with a power gain of 106. The input circuit is......... mA. circuit resistance is 100W and the output load 4k resistance is 10 KW. The common emitter current gain 'b' will be ______. (Round off to 90 V 5k the Nearest Integer) 21. The correct relation between a (ratio of 15. The value of power dissipated across the zener collector current to emitter current) and b (ratio diode (Vz = 15 V) connected in the circuit as of collector current to base current) of a shown in the figure is x × 10–1 watt. RS = 35 W transistor is: a b (1) b = (2) a = 22V VZ = 15V RL = 90 W 1+ a 1- a 1 b (3) b = (4) a = 1- a 1+ b The value of x, to the nearest integer, is ______. 16. In the logic circuit shown in the figure, if input 22. The typical output characteristics curve for a A and B are 0 to 1 respectively, the output at Y transistor working in the common-emitter would be 'x'. The value of x is _______. configuration is shown in the figure. A IC(mA) Y 8 IB = 40µA B 17. The following logic gate is equivalent to : 6 IB = 30µA A 4 IB = 20µA Y B 2 IB = 10µA (1) NOR Gate (2) OR Gate 0 (3) AND Gate (4) NAND Gate VCE(V) 18. The output of the given combination gates The estimated current gain from the figure is represents : 23. For the circuit shown below, calculate the value A of Iz : Y node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors B (1) XOR Gate (2) NAND Gate (3) AND Gate (4) NOR Gate 19. Which one of the following will be the output of the given circuit ? A Y (1) 25 mA (2) 0.15 A B (3) 0.1 A (4) 0.05 A (1) NOR Gate (2) NAND Gate (3) AND Gate (4) XOR Gate E 4 Semiconductors ALLEN® 24. A zener diode having zener voltage 8 V and 27. Identify the logic operation carried out. power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of (1) OR (2) AND protective resistance Rp is...........W. (3) NOR (4) NAND 28. In a semiconductor, the number density of intrinsic charge carriers at 27°C is 1.5 × 1016 / m3. If the semiconductor is doped with impurity atom, the hole density increases to 4.5 × 1022 / m3. The electron density in the doped semiconductor is _______ × 109/m3. 29. A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a 25. For the forward biased diode characteristics resistor of 1000 W in the collector circuit is 0.6 shown in the figure, the dynamic resistance at ID = 3 mA will be_______W. V. If the current gain factor (b) is 24, then the base current is _______ mA. (Round off to the Nearest Integer) 30. Find the truth table for the function Y of A and B represented in the following figure. A B Y A B Y (1) (2) 0 0 0 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 26. In a given circuit diagram, a 5 V zener diode 1 1 0 1 1 1 node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors along with a series resistance is connected across a 50 V power supply. The minimum A B Y A B Y (3) (4) value of the resistance required, if the maximum 0 0 0 0 0 0 zener current is 90 mA will be _____ W. 0 1 0 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 31. Consider a situation in which reverse biased current of a particular P-N junction increases when it is exposed to a light of wavelength £ 621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly. (1) 2 eV (2) 4 eV (3) 1 eV (4) 0.5 eV E  ALLEN® Semiconductors 5 32. Identify the logic operation carried out by the 35. For the given circuit, the power across zener given circuit :- diode is............ mW. (1) OR (2) AND (3) NOR (4) NAND 36. For a transistor in CE mode to be used as an 33. Statement-I : By doping silicon semiconductor amplifier, it must be operated in : with pentavalent material, the electrons density (1) Both cut-off and Saturation increases. (2) Saturation region only Statement-II : The n-type semiconductor has (3) Cut-off region only net negative charge. (4) The active region only In the light of the above statements, choose the 37. A circuit is arranged as shown in figure. The output voltage V0 is equal to....... V. most appropriate answer from the options given below : (1) Statement-I is true but Statement-II is false. (2) Statement-I is false but Statement-II is true. (3) Both Statement-I and Statement-II are true. (4) Both Statement-I and Statement-II are false. 34. Four NOR gates are connected as shown in figure. The truth table for the given figure is : IC 38. For a transistor a and b are given as a = IE IC and b =. Then the correct relation between IB a and b will be : A B Y A B Y 1-b a (1) a = (2) b = 0 0 1 0 0 0 b 1- a b node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors (1) 0 1 0 (2) 0 1 1 (3) ab = 1 (4) a = 1 0 1 1 0 1 1-b 1 1 0 1 1 0 39. A zener diode of power rating 2W is to be used A B Y A B Y as a voltage regulator. If the zener diode has a breakdown of 10 V and it has to regulate 0 0 0 0 0 1 voltage fluctuated between 6 V and 14 V, the (3) 0 1 1 (4) 0 1 0 value of Rs for safe operation should be 1 0 0 1 0 0 _______ W. 1 1 1 1 1 1 E 6 Semiconductors ALLEN® 40. In the following logic circuit the sequence of 42. If VA and VB are the input voltages (either 5V the inputs A, B are (0, 0), (0,1), (1, 0) and (1, 1). or 0V) and Vo is the output voltage then the two The output Y for this sequence will be : gates represented in the following circuit (A) and (B) are:- (1) 1, 0, 1, 0 (2) 0, 1, 0, 1 (3) 1, 1, 1, 0 (4) 0, 0, 1, 1 41. Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one: (1) AND and OR Gate (2) OR and NOT Gate (3) NAND and NOR Gate (4) AND and NOT Gate 43. Statement–I : To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output (1) parallel to the load RL. Statement–II : To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RL. (2) In the light of the above statements, choose the most appropriate answer from the options given below : (1) Statement I is true but Statement II is false node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors (2) Statement I is false but Statement II is true (3) Both Statement I and Statement II are false (3) (4) Both Statement I and Statement II are true 44. In the given figure, each diode has a forward bias resistance of 30W and infinite resistance in reverse bias. The current I1 will be : (4) (1) 3.75 A (2) 2.35 A (3) 2 A (4) 2.73 A E  ALLEN® Semiconductors 7 SOLUTION 4. Official Ans. by NTA (4) Sol. Truth table of the given gate : 1. Official Ans. by NTA (1) A B C 0 0 0 Sol. Ie = IC + IB 0 1 1 Þ DIe = DIC + DIB 1 0 0 1 1 0 4mA = 3.5 mA + DIB Truth table of option (1) A B C Þ DIB = 0.5 mA 0 0 1 0 1 1 DI C 1 0 0 Þb= 1 1 1 DI B Truth table of option (2) 3.5 A B C b= 0 0 1 0.5 0 1 0 Þb=7 1 0 1 1 1 1 2. Official Ans. by NTA (25) Truth table of option (3) A B C Sol. Current through 2kW resistance 0 0 1 0 1 0 5 I= = 2.5 × 10–3 A 1 0 0 2 ´103 1 1 0 I = 25 × 10–4 A Truth table of option (4) A B C Ans. 25 0 0 0 0 1 1 3. Official Ans. by NTA (440) 1 0 0 1 1 0 node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors N P VP Sol. = Since option (1) has same truth table, hence NS VS answer is option (4) is correct. Alternative solution : NP 220 = Given Boolean expression can be written as 24 12 A+B= C NP = 220 ´ 24 \ C = A. B = A.B 12 Hence option (4) is correct 5. Official Ans. by NTA (1) NP = 440 Sol. Back to back diode will not the make a transistor Ans. 440 turns i b= c ib E 8 Semiconductors ALLEN® 6. Official Ans. by NTA (2) 12. Official Ans. by NTA (20) Sol. Zener diode is heavily doped and have narrow D1 R1 i depletion layer. Option (2) is correct. i D2 R2 7. Official Ans. by NTA (4) Sol. D1 10W D2 5W Sol. i In this circuit D1 will be forward bias and D2 will be revers bias. X Y 5V \ There will be no current through D2 and R2 Here only D1 will work and we know Apply KVL in circuit we get for silicon diode, potential drop on D1 will be + 6 – 50i – 130i – 120i = 0 0.7V 6 6 i= A= ´ 1000mA 5 - 0.7 300 300 I= = 0.43 A Þ 20 mA 10 8. Official Ans. by NTA (2) 13. Official Ans. by NTA (4) Sol. (2) conceptual Sol. According to gates 9. Official Ans. by NTA (2) by Demorgan's law Sol. y = ( AB + AB) A+B= A.B y = AB. AB By observation. y = ( A + B ). ( A + B) 14. Official Ans. by NTA (9) y = A. A + AB + A.B + BB 60V 4 kW i 30V i1 30V y = AB + AB 90V –– i – i1 A B Y = AB + AB Sol. 90 V 5 kW 30V 0 0 1 0 1 0 1 0 0 0V 0V 0V 1 1 1 60 i= A 10. Official Ans. by NTA (1) 4000 node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors Sol. (a) Rectifier ® AC to DC 30 (b) Stabilizer ® used for constant output i1 = A 5000 voltage even when input voltage or current change. 60 30 9 i – i1 = - = A (c) Transformer ® Step - up or step - down ac 4000 5000 1000 voltage. current from zener diode (d) Filter ® used to iz = i – i1 = 9mA remove any ripple in the rectified output voltage. 11. Official Ans. by NTA (4) hc 6.6 ´ 10 -34 ´ 3 ´ 10 8 Sol. l = = -19 = 6.54 ´ 10 -7 E 1.9 ´ 1.6 ´ 10 = 654 nm Red color E  ALLEN® Semiconductors 9 15. Official Ans. by NTA (5) 21. Official Ans. by NTA (4) I RS = 35 W I1 I I Sol. a = C , b = C I2 IE IB Sol. 22V VZ = 15V RL = 90 W IE = IB + IC IC 1 a= = Voltage across RS = 22 – 15 = 7V IB + IC I B +1 7 1 IC Current through RS = I = = A 35 5 1 b a= a= 15 1 1 1+ b Current through 90W = I2 = = A +1 90 6 b ; 1 1 1 22. Official Ans. by NTA (200) Current through zener = – = A 5 6 30 DI C 2 ´ 10 -3 Sol. b= = Power through zener diode DI B 10 ´ 10 -6 P = VI 1 1 b = ´ 103 P = 15 × = 0.5 watt 5 30 b = 2 × 102 P = 5 × 10–1 watt b = 200 16. Official Ans. by NTA (0) 0 0 23. Official Ans. by NTA (1) A 0 50 Sol. 1 1 0 Y Sol. I= = 50mA 1 1000 B 0 1 1 0 17. Official Ans. by NTA (1) Sol. Truth table for the given logic gate : A B Y 0 0 1 0 1 1 1 0 1 50 I= = 25mA node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors 1 1 0 2000 The truth table is similar to that of a NOR gate. IZ = I1000 - I 2000 = 50 – 25 = 25 mA 18. Official Ans. by NTA (2) 24. Official Ans. by NTA (192 ) Sol. By De Morgan's theorem, we have Sol. P = Vi 1 A A A·B A·B = NAND 0.5 = 8i; i = A Y 16 B B E = 20 = 8 + i RP RP = 12 × 16 = 192W 19. Official Ans. by NTA (4) 25. Official Ans. by NTA (25) Sol. (4) Conceptual dV 1 1 20. Official Ans. by NTA (100) Sol. Rd = = = di di 5 - 1 ´ 10 -3 R dv 0.75 - 0.65 Sol. 106 = b2 ´ 0 Ri 100 = 25W 104 4 106 = b2 ´ 102 b2 = 10 4 Þ b = 100 E 10 Semiconductors ALLEN® 26. Official Ans. by NTA (500) 29. Official Ans. by NTA (25) Sol. I Sol. b = C = 24 ; R = 1000 IB C DV = 0.6 0.6 IC = 1000 IC = 6 × 10–4 I 6 ´ 10 -4 IB = C = = 25 mA b 24 30. Official Ans. by NTA (2) Voltage across RL = 5V 5 Þ i2 = Sol. RL Also voltage across R = 50 – 5 = 45 volt v 45 By v = iR Þ R = = Y = A·B + B i i i + i2 A B Y 45 R= 0 0 1 5 90mA + RL 0 1 0 Current in zener diode is maximum when RL 1 0 1 ® ¥ (i2 ® 0 and ii = i) 1 1 1 45 So R = = 500W 31. Official Ans. by NTA (1) 90mA 27. Official Ans. by NTA (2) hc Sol. Band gap = l0 Sol. l0 ; threshold wavelength 1242 ev - nm Band gap = = 2eV 621nm 32. Official Ans. by NTA (3) A B X Y Z 1 1 0 0 0 28. Official Ans. by NTA (5) 1 0 0 1 0 Sol. node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors Sol. nenh = ni2 n 2 (1.5 ´ 1016 )2 1.5 ´ 1.5 ´ 1032 0 1 1 0 0 ne = i = = 0 0 1 1 1 nh 4.5 ´ 10 22 4.5 ´ 1022 5 ×109 /m3 Option (3) 33. Official Ans. by NTA (1) Sol. Pentavalent activities have excess free e– So e– density increases but overall semiconductor is neutral. Option (1) E  ALLEN® Semiconductors 11 34. Official Ans. by NTA (4) 37. Official Ans. by NTA (5) Sol. As diodes D1 and D2 are in forward bias, so they Sol. acted as neligible resistances Þ Input voltage become zero y = (A + A + B) + (B + A + B) Þ Input current is zero y = (A + A + B).(B + A + B) Þ Output current is zero Þ V0 = 5 volt A B y 38. Official Ans. by NTA (2) 0 0 1 I I Sol. a = C , b = C ; IE = IC + IB 0 1 0 IE IB 1 0 0 IC I /I b a= = C B= + 1 1 1 IC + I B I C b +1 +1 IB Ans.4 1 1 1 1– a a 35. Official Ans. by NTA (120) 1+ = = b= b a; b a ; 1– a 39. Official Ans. by NTA (20) Sol. Sol. When unregulated voltage is 14 V voltage across zener diode must be 10 V So potential difference across resistor DVRs = 4V and Pzener = 2W VI = 2 2 10V 14V I= = 0.2 A i= = 2mA I= = 14mA 10 5kW 1kW DVRs = I Rs node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors \ Iz = 12mA 40 4 × 0.2 Rs Þ Rs = = 20W 2 \ P = IzVz = 120 mW 40. Official Ans. by NTA (3) Ans. 120 Sol. 36. Official Ans. by NTA (4) Sol. Active region of the CE transistor is linear region and is best suited for its use as an amplifier Y = ( A × B) × ( A + B) Y )( 0,0) = 1 Y )( 0,1) = 1 Y )(1,0) = 1 Y )(1,1) = 0 Option (3) is correct E 12 Semiconductors ALLEN® 41. Official Ans. by NTA (3) 44. Official Ans. by NTA (3) Official Ans. by ALLEN (1) Sol. When Vi > 3 volt, VR > 0 Because diode will be in forward biased state When Vi £ 3volt ; VR = 0 Because diode will be in reverse biased state. 42. Official Ans. by NTA (2) Sol. Sol. VA = 5V Þ A=1 VA = 0 V Þ A=0 VB = 5 V Þ B=1 VB = 0 V Þ B=0 If A = B = 0, there is no potential anywhere here V0 = 0 As per diagram, If A = 1, B = 0, Diode D1 is forward biased, Diode D1 & D2 are in forward bias i.e. R = 30W here V0 = 5V whereas diode D3 is in reverse bias i.e. R = If A = 0, B = 1, Diode D2 is forward biased infinite hence V0 = 5V If A = 1, B = 1, Both diodes are forward biased Þ Equivalent circuit will be hence V0 = 5V Applying KVL starting from point A Truth table for Ist A B Output 0 0 0 0 1 1 1 0 1 1 1 1 \ Given circuit is OR gate For IInd circuit VB = 5V, A = 1 VB = 0V, A = 0 When A = 0, E–B junction is unbiased there is no current through it \ V0 = 1 When A = 1, E–B junction is forward biased V0 = 0 æI ö æI ö - ç 1 ÷ ´ 30 - ç 1 ÷ ´ 130 - I1 ´ 20 + 200 = 0 node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Physics\English\ Semiconductors \ Hence this circuit is not gate. è2ø è2ø 43. Official Ans. by NTA (4) Þ – 100 I1 + 200 = 0 Sol. To convert pulsating dc into steady dc both of mentioned method are correct. I1 = 2 Option (3) E