Tiêu đề 26. SEMI CONDUCTORS and ELECTRONIC DEVICES Hard Ans.pdf ✅

1. (a) Since diode in upper branch is forward biased and in lower branch is reversed biased So current through circuit d R r V i + = ; here d r = diode resistance in forward biasing = 0 So A R V i 0.2 10 2 = = = . 2. (b) The diode in lower branch is forward biased and diode in upper branch is reverse biased  i A 50 5 20 30 5 = + = 3. (a) Diode is reverse biased. Therefore no current will flow through the circuit. 4. (b) The current through circuit A V P i 0.2 0.5 100 10 3 =  = = −  voltage drop across resistance = 1.5 – 0.5 = 1 V  = = 5  0.2 1 R 5. (d) In common emitter configuration current gain 6 3 1 25 10 10 50 1 +   − = + − = − oe L fe i h R h A = – 48.78. 6. (c) Voltage gain Input voltage Output voltage =  Vout = Vin  Voltage gain  Vout = Vin  Current gain  Resistance gain = Vin    BE L R R = 1 . 1 10 10 100 3   = V − 7. (a) 82 8.3 8.2 8.2 = − =           = Vce b c fe i i h 8. (d) A A R V i i i i c  b  500 10 500  1000 0.01 50 6 = =  =  =  = − 9. (a) i c =  i e = 0.98  2 = 196 mA  ib = i e − i c = 2 − 1.96 = 0.04 mA 10. (a) Balancing length is for emf of cell which does not depend on internal resistance. 11. (b) R = g I V – Rg R1 = g I 3 – Rg = 2980 = 2.98k (R1 + R2) = g I 15 – Rg = 14980 = 14.98k (R1 + R2 + R3) = g I 150 – Rg = 149980  = 149.98 k  R1 = 2.98 k R2 = 14.98 – 2.98 = 12 k R3 = 149.98 – 14.98 = 135 k 12. (b) E2 + E1 = x(825) E2 – E1 = x (225) 13. (a) balancing length =  E, r E, r E, 2r Enet = E  balancing length =  14. (a) r = (RB) 2 1 2         −    15. (c) Q' P' = S' R'  2 1 = Q 2  Q = 4  16. (d) r =         –1 2 1   R =       –1 150 165 × 5 = 0.5  17. (c) VAB = E A E L B Potential gradient  = L VAB = L E  Potential difference across length  is V = K = L E ×  = 3 6 × 0.5 = 1 volt 18. (c)
S R =   100 –  3 2 =   100 –  = 40 cm 19. (a) Potential drop across 3m = 6Volt  Potential drop across 2 1 m = 1 volt 20. (d) 0.6 0.4 R R 2 1 =  3 2 R R 2 1 = 0.4 0.6 R R 10 2 1 = +  2 3 3/2R R 10 1 1 = + R1 = 8  21. (c) G = R S RS −  500 = 5000 S 5000S − S = 455  22. (b)  = L 0     = 30 100 E   = 0.3 E 23. (a) 1 = 560 cm, 2 = 412 cm, R = 10  r = R         −1 2 1   , r =  =       −1 3.6 412 560 10 24. (a) i = .01A 100 100 2 = + , i = k k = i/ = .01/20 = 5 × 10–4 A/ div. 25. (b) Q P = S R  80 20 = R 55  R = 220  26. (c) E =         + +  w h o w R r R R L E ×  ( r = 0, Rh = 2 )  1.08 =         +  R 2 R 1m 2 w w × 0.90 m 2 0.90 1.08  = R 2 R w w + Hence : 0.6 = R 2 R w w + 0.6 Rw + 1.2 = Rw  1.2 = 0.4 Rw  3 = Rw  Resistance per unit length is 3 /meter 27. (c) V = e – ir  r = i e – V 28. (a) It is given that presence of switch in the circuit is meaningless. Since R and G now become in series, so, IR = IG 29. (b) = (Rwire) ×  I I = (R R ) wire + external  = (20 10) 3 + = 30 3 = 0.1A  = 10 200.1 = 0.2 V/m 30. (d) I = Req. V = 2 1 3 4 + + = 6 4 = 0.67A Resistance of 50 cm wire = 2 3 = 1.5   = 1.5 × 0.67 = 1V 31. (a) Zener diode is in parallel to load resistance and is connected in reverse bias. 32. (d) Phosphorus is pentavalent impurity. Its doping will not effect the concentration of holes. So number of holes will be equal to same as in intrinsic semiconductor. So nh= 1.41 × 1016 m–3 33. (c) Atomic radius for fcc crystal is r = 2 2 a = Å 2 2 3.6 = 1.27 Å 34. (c)  = eE hc = Å 1.6 10 57 10 6.62 10 3 10 10 –19 –3 –34 8 10        = 217100 Å 35. (c) The diode will be forward biased in one half cycle and will conduct where as it will be reverse biased in negative half cycle and will not conduct. I Input t
I output t 36. (a)  = e c I I = c b c I I I + = 0.985 Ic = 0.985 (Ic + Ib) Ic = 0.985 Ic + 0.985 Ib 0.985 Ib = 0.015 Ic = 0.015 × 2 mA Ib = 0.985 0.0152 = 0.03 mA Ib  0.03 mA 37. (b) D1 → F.B. , D2 → R.B I 6V 100 50 50 150 + –  = net net R V I = 300 6 = 0.02A 38. (b) (Vr.m.s.) H.W.R. = 2 V0 = 2 200 = 100 volt 39. (a) Diode in F.B. 10 10 10 a b + – 10 5 I  30V a • b • • + – 30V •  = 15 30 = 2A , Vab = IRab = 2 × 5 = 10V 40. (a) From chemistry of semiconductors 41. (a) Sote that zener diode is in parallel to load resistance and is connected in reverse bias. 42. (d) VC = V0 = 2 Vrms= 2 × 220 volt 43. (a)  = E C i i iC =  i = 0.96×7.2 mA 44. (c) (a) Z = (P + Q) 0 1 0 0 0 0 P Q Z (b) Z = (P.Q) 0 0 1 P Q Z (c) Z = (Q.(P + Q)) 1 1 1 1 0 1 0 1 1 0 0 0 P Q Z (d) Z = (P.Q) 1 0 0 0 1 0 0 0 0 P Q Z 45. (b) Ideal junction in forward bias behaves like a conducting wire. 46. (b) Gate is AND gate. Because ‘1’ and ‘0’ gives ‘0’. 47. (b) R P A A = R R 2 A  A =  2 48. (a) ie = iB = 80 × 250 μA 49. (a)  – B junction is forward bias and C – B junction is reversed bias. 50. (a) p-side at higher potential and n-side at lower potential. 51. (a) Upper diode is in forward bias, So, i = V/R = 2V/20 = 0.1 A 52. (a)  = E hc = 1.41 12400 Å = 10877 Å 53. (c)
Y = (A + B)  1 1 1 1 0 0 0 1 0 0 0 0 A B Y = AND 54. (c) ic = ie = 0.98 × 5 mA = 4.9 mA 55. (d) Using  = −  1 we get  = 0.1 0.9 = 9 56. (d) Using  = b c I I We get  Ic = Ib ×  = 250 × 80 A 57. (a) Use E =  12400 Ev 58. (b) Majority charge carrier in p-type is hole. 59. (c) Depletion layer consists of immobile ions. 60. (b) 61. (b) In forward bias, current is in milliampere. 62. (a) Diode is in reverse bias. 63. (c) Zener diode works only in reverse bias. 64. (c) Resistively of a semiconductor decrease with increase in temperature. 65. (d) Rectifier is used to convert AC into DC. 66. (a) Emitter is highly doped than collector and base is least doped. 67. (b) Ic = Ib = 0.8 × 6 mA = 4.8 mA 68. (c)  = 50 RL = 4k , , Ri = 500 Vce = IcRL , Vbe = IbRi  = b c I I = 50 AV = be ce V V = 500 50 4 1000 R R I I i L b e    = = 400 69. (d) (i) Without using voltmeter V4k = 2 volt (ii) With voltmeter RAB = 2k  VAB = 3 1 × 4 = 3 4 volt  % error = 2 2 − 4 / 3 ×100 = 6 200 = 3 100 =33.3% 70. (a) P.D of ext. ckt =  × balance length E/2 =         +  R r R L E w w ×  or E/2 = 600 E × 16r 15r ×    = 30 60016 = 320 cm 71. (b) x R1 = 60 40 = 3 2 .... (1), x R2 = 50 50 = 1 ..... (2) x R1 + R2 =   100 – .... (3) Add eq (1) & (2) x R1 + R2 = 3 2 + 1 = 3 5 ... (4) Placing eq (4) in (3) 5/ 3 =   100 –  500–5 = 3  8=500   = 62.5 cms 72. (d) S = A g g g I – I I R  S = 100 –1 199 = 1 73. (d) E1 = Q × 1 ..... (1)