Tiêu đề 26. SEMI CONDUCTORS and ELECTRONIC DEVICES Medium.pdf ✅

1. At absolute zero, Si acts as a. (a) Metal (b) Semiconductor (c) Insulator (d) None of these 2. If the energy of a photon of sodium light (  = 589 nm) equals the band gap of semiconductor, the minimum energy required to create hole electron pair (a) 1.1eV (b) 2.1eV (c) 3.2eV (d) 1.5eV 3. The manifestation of band structure in solids is due to (a) Heisenberg uncertainty principle (b) Pauli’s exclusion principle (c) Bohr’s correspondence principle (d) Boltzmann law 4. The electrical conductivity of a semi conductor increases when electronmagnetic radiation of wave length shorter than 2480 nm is incident on it. The band gap (in eV) for semiconductor is (a) 0.9 (b) 0.7 (c) 0.5 (d) 1.1 5. Find the wavelength of light that may excite an electron in the valence band of diamond to the conduction band. The energy gap is 5.50eV (a) 226 nm (b) 312 nm (c) 432 nm (d) 550 nm 6. Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bandsseparated by energy band gap respectively equal to (E )c,(E )si g g and (E )Ge, g which of the following statements is true? (a) ( ) ( ) ( ) c g Ge g si Eg  E  E (b) ( ) ( ) ( ) Si Eg C  Eg Ge  Eg (c) ( ) ( ) ( ) Ge g Si Eg C  Eg  E (d) ( ) ( ) ( ) Ge g Si Eg C = Eg = E 7. The maximum wavelength of electromagnetic radiation which can create a hole electron pair in germanium (Given that forbidden energy gap in germanium is 0.72eV ) (a) 1.7 10 m −6  (b) 1.5 10 m −5  (c) 1.3 10 m −4  (d) 1.9 10 m −5  8. In good conductors of electicity the type of bonding that exist is (a) Van der walls (b) Covalent (c) Ionic (d) Metallic 9. An intrinsic semiconductor has a resistivity of 0.50 m at room temperature. Find the intrinsic carries concentration if the mobilities of electrons and holes are 2 1 1 0.39m V s − − and 2 1 1 0.11 m V s − − respectively (a) 18 3 1.2 10 m −  (b) 19 3 2.5 10 m −  (c) 20 3 1.9 10 m −  (d) 21 3 3.1 10 m −  10. In pure semiconductor, the number of conduction electrons is 18 610 per cubic metre. How many holes are there in a sample of size 1cm1cm1 mm? (a) 10 310 (b) 11 610 (c) 11 310 (d) 10 610 11. Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are 2 1 0.54 m V − 1 s − and 2 1 1 0.18m V s − − respectively. If the electron and hole densities are equal to 19 3 3.6 10 m −  the germanium conductivity is. (a) 1 4.14 Sm− (b) 1 2.12Sm − (c) 1 1.13 Sm− (d) 1 5.6 S m − 12. The probability of electorns to be found in the conduction band of an intrinsic semiconductor of finite temperature. (a) Increases exponentially with increasing band gap (b) Decreases exponentially with increasing band gap (c) Decreases with increasing temperature. (d) Is independent of the temperature and band gap 13. A block of pure silicon at 300 K has a length ofj 10 cm and an area of 2 1.0cm A battery of EMG 2 V is connected across it. The mobility of electons is 2 1 1 0.14 m V s − − and their number density is the electrons current is. (a) 6.72 10 A −4  (b) 6.72 10 A −5  (c) 6.72 10 A −6  (d) 6.72 10 A −7  14. Which of the following equations correctly represents the temperature variation of energy gap between the conduction and valence bands for Si? (a) E (T) 0.70 2.23 10 T eV 4 g − = −  (b) E (T) 0.70 2.23 10 T eV 4 g − = +  (c) E (T) 1.10 3.60 10 T eV 4 g − = −  (d) E (T) 1.10 3.60 10 TeV 4 g − = +  15. A pure Si crystal has 22 510 atoms 3 m − it is doped by 1 ppm concentration of pentavalent As. The number of holes is   p e i 2 n = n n (Take 16 3 ni 1.5 10 m − =  ) (a) 9 3 4.5 10 m −  (b) 6 3 4.5 10 m −  (c) 9 3 2.5 10 m −  (d) 6 3 2.5 10 m −  16. If a small amount of antimony is added to germanium crystal. (a) Its resistance is increased (b) It becomes a p – type semiconductor (c) There will be more free electrons than holes in the semiconductor (d) None of these 17. In n – type semiconductor when all donor states are filled, then the net charge density in the donor states becomes (a) 1 (b) > 1 (c) < 1, but not zero (d) Zero
18. A semiconductor has equal electron and hole concentration of 8 610 per 3 m on doping with certain impurity, electron concentration increases to 12 3 910 per m The new hole concentration is (a) 4 210 per 3 m (b) 2 210 per 2 m (c) 4 410 per 3 m (d) 2 410 per 3 m 19. In an n – type silicon, which of the following statements is true. (a) Electrons are majority carriesrs and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the copants. (d) Holes are majority carriers and trivalent atoms are the dopants. 20. The number density of electrons and holes in pure silicon at 27 C o are equal and its value is 16 3 2.0 10 m −  on doping with indium the hole density increases to 22 3 4.5 10 m −  the electrons density in doped silicon is. (a) 9 3 10 10 m −  (b) 9 3 8.86 10 m −  (c) 9 3 11 10 m −  (d) 9 3 16.78 10 m −  21. Which of the following statements is incorrect for the depletion region of a diode? (a) There the mobile charges exist. (b) Equal number of holes and electrons exist, making the region neutral. (c) Recombination of holes and electrons has taken place. (d) None of these 22. In an unbiased p – n junction, holes diffuse from the p – region to n – region because (a) Free electrons in the n region attract them (b) They move across the junction by the potential difference (c) Hole concentration in p – region is more as compared to n – region (d) All of these 23. A potential barrier of 0.3V exists across a p – n junction if the depletion region is 1 m wide, what is the intensity of electric field in this region? (a) 5 1 2 10 V m −  (b) 5 1 3 10 V m −  (c) 5 1 4 10 V m −  (d) 5 1 5 10 V m −  24. The dominant mechanism for motion of charge carriers in forward and reverse biased silicon p – n junction are (a) Drift in forward bias, diffusion in reverse bias (b) Diffusion in forward bias, drift in reverse bias (c) Diffusion in both forward and reverse bias (d) Drift in both forward and reverse bias 25. Potential barrier developed in a junction diode opposes the flow of (a) Minority carrier in both regions only (b) Majority carriers only (c) Electrons in p – region (d) Holes in p region 26. Region without free electrons and holes in a p – n junction is (a) N – region (b) P - region (c) Depletion region (d) None of these 27. The equivalent resistance of the circuit shown in figure between the points A and B if VA  VB is (a) 10 (b) 20  (c) 5  (d) 40  28. In the question number 27, the equivalent resistance between the points A and B if VA  VB is (a) 10 (b) 20  (c) 30  (d) 15  29. In the circuit shown if drift current for the diode is 20 A, the potential difference a cross the diode is (a) 2V (b) 4.5V (c) 4 V (d) 2.5V 30. Mark the incorrect statement. When a potential difference is applied across, the current passing through (a) An insulator at 0 K is zero (b) A semiconductor 0 K is zero (c) A metal at 0 K is zero (d) A p – n junction diode at 300 K is finite, if it is reverse biased. 31. Of the diodes shown in the following figures, which one is reverse biased? (a) (b) (c) (d)
32. The V – I characteristic of a silicon diode is shows in figure. The resistance of the diode at ID =15mA is (a) 5 (b) 10 (c) 2 (d) 20 33. The following table provides the set of values of V and I obtained for a given diode. Let the characteristics -be nearly linear, over this range, the forward and reverse bias resistance of the given diode respectively are V I Forward biasing 2.0V 60 mA 2.4V 80 m A Reverse biasing 0V 0A − 2V −0.25A (a)    6 10 ,8 10 (b)    5 20 ,4 10 (c)    6 20 , 8 10 (d) 10,10 34. Which of the junction diodes shown below are forward biased? (a) (b) (c) (d) 35. The circuit shown in the figure contains two diodes each with a forward resitance of 30 and with infinite backward resistance. If the battery is 3V, the current through the 50  resistance (in ampere) is (a) Zero (b) 0.01 (c) 0.02 (d) 0.03 36. When the voltage drop across a p – n junction diode is increased from 0.65 V to 0.70V, the change in the diode current is 5mA. The dynamic resistance of the diode is (a) 5 (b) 10 (c) 20 (d) 25 37. In a full wave junction diode rectifier the input ac has rms value of 20 V. the transformer used is a step up transformer having primary and secondary turn ration 1 : 2 The dc voltage in the rectified output is (a) 12 V (b) 24 V (c) 36 V (d) 42 V 38. With an ac input form 50 Hz power line, the ripple frequency is. (a) 50 Hz in the dc output of half wave as well as full wave rectifier (b) 100 Hz in the dc output of half wave as well as full wave rectifier. (c) 50 Hz in the dc output of half wave and 100 Hz in dc output of full wave rectifier (d) 100 Hz in the dc output of half wave and 50 Hz in the dc output of full wave rectifier 39. In a half wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be. (a) 25 Hz (b) 50 Hz (c) 70.7 Hz (d) 100 Hz 40. Which of the following circuits provides full wave rectification of an ac input? (a) (b) (c) (d) 41. A zener diode is specified as having a breakdown voltage of 9.1V with a maximum power dissipation of 364 mW. What is the maximum current the diodecan handle? (a) 40 mA (b) 60 Ma (c) 50 mA (d) 45 mA 42. What happens during regulation action of a zener diode? (a) The current through the series resistance ( ) RS changes. (b) The resistance offered by the zener changes (c) The zener resistance is constant. (d) Both (a) and (b) 43. A p – n photodiode is made of a material with band gap of 2 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly (Take hc = 1240 eV nm) (a) 1 10 Hz 14  (b) 20 10 Hz 14 
(c) 10 10 Hz 14  (d) 5 10 Hz 14  44. In the circuit given, the current through the zener diode is. (a) 10 mA (b) 6.67 Ma (c) 5 mA (d) 3.33 mA 45. A p – n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. The signal wavelength is (a) 6000 o A (b) 6000 nm (c) 4000 nm (d) 5000 o A 46. Three photo diodes D , 1 D2 and D3 are made of semiconductors having bandgap of 2.5 eV, 2 eV and 3 eV, respectively. Which one will be able to detect light of wavelength 6000 o A ? (a) D1 (b) D2 (c) D3 (d) D1 and D2 both 47. A transistor has a current amplification factor (current gain) of 50. In a common emitter amplifier circuit, the collector resistance is chosen as 5 and the input resistance is 1 The output voltage if input voltage is 0.01V is (a) - 2 V (b) – 5 V (c) − 2.5V (d) – 1 V 48. In p – n –p transistor circuit, the collector current is 10 m A. if 90%oftheholes reach the collector, the emitter and base currents respectively are. (a) 10 mA, 1 mA (b) 22 mA, 11 mA (c) 11 mA, 1 mA (d) 20 mA, 10 mA 49. The input resistance of transistor is 1000  on charging its base current by 10 A, the collector current increases by 2 mA if a load resistance is 5 k is used in the circuit, the voltage gain of the amplifier is (a) 100 (b) 500 (c) 1000 (d) 1500 50. A transistor connected in common emitter mode,the voltage drop across the collector is 2V and  is 50, the base current if RC is 2k is (a) 40A (b) 20A (c) 30A (d) 15A 51. The power gain for common base amplifier is 800 and the voltage amplification factor is 840. The collector current when base current is 1.2mA is (a) 24 mA (b) 12 Ma (c) 6 m A (d) 3 mA 52. In an n – p – n circuit transistor, the collector current is 10 mA. If 80% electron emitted reach the collector, then (a) The emitter current will be 7.5mA (b) The emitter current will be 12.5mA (c) The base current will be 3.5mA (d) The base current will be 1.5mA 53. The potential difference across the collector of a transistor, used in common emitter mode is 1.5V, with the collector resistance of 3 k the emitter current is  = 50 (a) 0.70mA (b) 0.49mA (c) 1.1mA (d) 1.9mA 54. The ac current gain of transistor is 120. What is the change in the collector current in the transistor whose base current changes by 100 A ? (a) 6 mA (b) 12 mA (c) 3 mA (d) 24 mA 55. The current amplification factor - of a common base transistor and the current amplification factor  of a common emitter transistor are not related by (a) +   = 1 (b) −   = 1 (c) 1 1 1 =  −  (d) +    = 1 56. The input resistance of common emitter transistor amplifier, if the output resistance is 500 k  the current gain  = 0.98 and the power gain is 6 6.062510 is (a) 198 (b) 300 (c) 100 (d) 400  57. In an n – p – n transistor 10 10 electron enter the emitter in 6 10− s. if 2% of the electrons are lost in the base, the ratio of current transfer ratio and the current amplification factor is. (a) 0.02 (b) 7 (c) 33 (d) 4.9 58. The heavily and lightly doped regions of a bipolar junction transistor are respectively. (a) Base and emitter (b) Base and collector (c) Emitter and base (d) Collector and emitter 59. Two amplifiers are connected one after the other in series (Cascade). The first amplifier has a voltage gain 10 and the second has a voltage gain of 20. If the input signal is 0.01 V. the out ac signal will be (a) 4V (b) 1 V (c) 2V (d) 6V 60. An amplifier has a voltage gain of 100the voltage gain in dB is (a) 20 dB (b) 40 dB (c) 30 dB (d) 50 dB 61. The current gain for a common emitter amplifier is 69 if the emitter current is 7 mA, the base current is (a) 0.1 mA (b) 1 mA (c) 0.2mA (d) 2 mA 62. A transistor has a current gain of 30. If the collector resistance is 6 k input resistance is 1 k its voltage gain is