Title 26. SEMI CONDUCTORS and ELECTRONIC DEVICES Med Ans.pdf ✅

1. (c): At absolute zero, Si acts as an insulator due to the absence of free electrons in the conduction band. 2. (b): using,  = = hc E Eg eV eV j 2.1 589 10 1.6 10 6.6 10 3 10 589 10 6.6 10 3 10 9 19 34 8 9 34 8 =       =     = − − − − − 3. (b): According to the Pauli’s exclusion principle, the electronic configuration of number of subshells existing in a shell and number of electrons entering each sub shell is found. Hence, on the basis of the Pauli’s exclusion principle, manifestation of band structure in solids can be explained. 4. (c): eV nm hc eVnm Eg 0.5 2480 1240 = =  = 5. (a): Energy gap, g g E hc hc E =  = Here, energy gap = 5.50eV Take hc=1240eV nm nm eV eV nm 226 5.5 1240  = = 6. (c): Eg c Eg si Eg Ge ( ) ( )  ( ) 7. (a): here, E eV j g 19 0.72 0.72 1.6 10− = =   If  is the maximum wavelength of electromagnetic radiation which can create a hole- electron pair in germanium, then  = hc Eg Or m E hc g 6 19 34 8 1.7 10 0.72 1.6 10 6.62 10 3 10 − − − =       = = 8. (d): Metallic 9. (b): here, =0.50m 2 1 1 2 1 1 0.11 0.39 − − − −  =  = m V s m V s h e The resistivity of intrinsic semiconductor is ( ) 1 ni e ni h =e  +   Where i n is the intrinsic carrier concentration ( ) 1 e h i e n   +   = Substituting the given values, we get 19 3 19 2.5 10 (0.5)(1.6 10 )(0.39 0.11) 1 − − =   + ni = m 10. (b): here, 18 3 6 10 − ne =  m Volume of the sample 7 3 1cm 1cm 1mm 10 m − =   = Number of holes in the sample= Number of electrons in the sample 11 18 7 6 10 6 10 10  =   =  − ne V 11. (a): As ( ) ne e nn n =e  +  ( ) i e n =en  + 1 19 19 4.147 1.6 10 3.6 10 (0.54 0.18) − − = =    + Sm 12. (b): Decrease exponentially with increasing band gap. 13. (d): As 1 20 0.1 2 − = = = Vm m V l V E 2 4 2 A 1.0cm 1.0 10 m − = =  1 0.14 20 2.8 − v = E=  = ms e e The electron current is A I n Aev e e e 7 16 4 19 6.72 10 (1.5 10 ) (1.0 10 ) (1.6 10 ) 2.8 − − − =  =       = 14. (c): the energy gap Eg depends on the temperature For silicon, Eg T TeV 4 ( ) 1.10 3.60 10− = −  For germanium Eg T TeV 4 ( ) 0.70 2.23 10− = −  15. (a): The electron and holes concentration in a semiconductor in thermal equilibrium is 2 e h i n n =n Or e i h n n n 2 = Here, 16 3 22 3 1.5 10 , 5 10 − − ni =  m ne =  m 9 3 22 32 22 16 2 4.5 10 5 10 2.25 10 (5 10 ) (1.5 10 ) − =    =   nh = m
16. (c): Adding fifth group element to germanium makes it an n- type semiconductor. Antimony is a fifth group element and so germanium becomes n-type semi conductor 17. (b): if all the donor states in n-type semiconductor are filled, the number of electrons in donor states will increase, due to it, the charge density in donor states will become more than one. 18. (c): As 2 e h i n n =n Here, 8 3 ni =610 perm and 12 3 ne =910 perm 4 3 12 8 2 4 10 9 10 (6 10 ) perm n n n e i h =     = = 19. (c): Holes are minority carriers and pentavalent atoms are the dopants. 20. (b): using, 2 e i n =n Here, 16 3 2 10 − ni =  m 22 3 4.5 10 − =  m 9 3 22 2 16 2 8.89 10 4.5 10 (2 10 ) − =    =   = m n n i 21. (a): There the mobile charges exist 22. (c): In an unbiased p-n junction, the diffusion of charge carries across the junction takes place from higher concentration to lower concentration, thus option (c) is correct 23. (b): Electric field 5 1 6 3 10 1 10 0.3 − − =   = = Vm d V E 24. (b): in p-n junction, the diffusion of majority carries takes place when junction is forward biased and drifting of minority carries takes place across the junction, when it is reverse biased. 25. (b): Potential barrier developed in a junction diode opposes the majority carriers only. 26. (c): The depletion region created at the junction is devoid of free charge carriers. 27. (b): When , VA VB the diode gets reverse biased and offers infinite resistance. No current flows through the upper branch R=20 28. (a): when VA  VB the diode gets forward biased and offers no resistance =  +   = 10 20 20 20 20 R 29. (c): since the diode is reversed biased, only drift current exists in circuit which is 20A Potential drop across 15 resistor V V A 300 0.0003 15 20 =  = =   Potential difference across the diode =4 − 0.0003= 3.99 = 4V 30. (c): a metal at 0K is zero 31. (c): p-n junction is reverse biased when p side is at a lower potential than n side. It is for the circuit (c). 32. (b): In diagram, at I = 20mA,V =0.8V and At I =10mA,V =0.7V = =    = 10 10 0.1 mA V I V Brfb 33. (c): for forward biasing, V =2.4 − 2.0=0.4V I =80− 60 = 20mA =   =    = − 20 20 10 0.4 3 I V rfb For reverse biasing V = −2 − 0 = −2V =   −  − =  =− − = −  − 6 6 8 10 0.25 10 2 0.25 0 0.25 rb r I A 34. (a): The p-n junction diode is forward biased when p is at high potential with respect to n. Hence option (a) is correct. 35. (c): In the circuit the upper diode D1 is reverse biased and the lower diode D2 is forward biased. Thus there will be no current across upper diode junction. The effective circuit will be as shown in figure. Total resistance of circuit R=50+ 70+ 30=150 Current in circuit, A V R V I 0.02 150 3 =  = = 36. (b): Dynamic resistance is
I V rrd   = Here, V = 0.7 − 0.65V =0.05V I mA A 3. 5 5 10−  = =  =    = − 10 5 10 0.05 d 3 r 37. (c): here, input Vrms =20V Peak value of input voltage Vo = 2Vrms = 2  20 = 28.28V Since the transformer is a step up transformer having transformer ratio 1: 2 the maximum value of output voltage of the transformer applied to the diode will be Vo 2 Vo 2 28.28V ' =  =   dc voltage V Vo 36 22/ 7 2 2 2 28.28 ' =   =  = 38. (c): 50 Hz is the dc output of half wave and 100 Hz in dc output of full wave rectifier 39. (b): as the output voltage obtained in a half wave rectifier circuit has a single variation in one cycle of ac voltage, hence the fundamental frequency in the ripple of output voltage would be =50 Hz. 40. (d) 41. (a): The maximum permissible current is mA V p I Z Z 40 9.1 364 10 3 max =  = = − 42. (d): during regulation action of a Zener diode, the current through the series resistance changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant. 43. (d): Here, Eg =2eV Wavelength of radiation corresponding to this energy is nm eV eV nm E hc g 620 2 1240 = = = Frequency Hz m c ms 14 9 8 1 5 10 620 10 3 10 =    =   = − − 44. (d): The voltage drop across R2 is VR =VZ = 2 10V The current through R2 is A V R V I R R 2 2 0.667 10 1500 10 2 2 − =   = = The voltage drop across R1 is () V V V VR V VR 15 10 5 15 1 2 = − = = − The current through R1 is A A mA V R V I R R 10 10 10 10 500 5 2 3 1 1 1 = =  =  = = − − The current through the zener diode is mA mA I I I Z R R (10 6.67) 3.33 1 2 = − = = − 45. (d): the detection occurs only when the energy of incident photon greater than or equal to the energy band gap eV hc =2.5  nm Å eV eV eV hc 496 5000 2.5 1240 2.5  = = =  46. (b): Energy of incident photon,  = hc E 2.06eV 6 10 1.6 10 6.6 10 3 10 7 19 34 8 =       = − − − The incident radiation can be detected by a photodiode if energy of incident photon is greater than the band gap. As , D2 =2eV D2 will detect these radiations . 47. (b): 10  48. (c): the voltage gain of a common emitter amplifier B C i o i o v R R V V or R R A  =−  =− Output voltage V R R V V B C i o 2.5 1 50 5 0.01 =−   =−   =− 49. (c): current gain, VCE B C I I           = And voltage gain, in out V R R A = Here, 1000 1000 200 5 103 =   AV =
50. (b): using , mA R V I C CE C 10 1 2 10 2 3 3 = =  = = − B C I I = A A I I C B = =   = − 20 50 10 3 51. (a): current gain, 21 20 840 800 Voltage gain Powergain = = = Now, 20 1 (20/ 21) (20/ 21) 1 = − = −   = As, B C I I  = IC =IB =201.2 = 24mA (Given IB =12mA ) 52. (b): here, C E E I of I I 100 80 =80% = Or mA I I C E 12.5 0.8 10 0.8 = = = IB =IE − IC =12.5−10=2.5mA 53. (b): here, VCE =1.5V 3 3 10 ; 50 3 RC = k=   = A R V I C CE C 3 3 0.5 10 3 10 1.5 − =    = = A I I C B 3 3 0.01 10 50 0.50 10 − − =   =  = A mA I I I E C B 0.49 10 0.49 (0.50 0.01) 10 3 3 =  = = − = −  − − 54. (b): here, 100 100 10 , 120 6 =  =   = − B A A ac I Using, 6 100 10 120 −    =    = C B C ac I I I Changes in collector current IC 120 100 10 12 10 12mA 6 3  =   =  = − − 55. (d): As 1 1 1 1 , 1 +  = +    = −   = so and 56. (a): Voltage gain i o V R R = A =  Also, current gain 49 1 0.98 0.98 1 = − = −   =          = i V R A 3 500 10 (49) Power gain = Current gain × voltage gain Power gain 49 500 10 6.0625 10 49 3 6           =  Ri Or Ri =198 57. (a): As, t n e I E E  = And E C E C I t n e t n e I =   =  = 100 (98/100) 98 Current transfer ratio, 0.98 100 98 = = = IlE IC Current amplification factor, 0.02 49 0.98 49 1 0.98 0.98 1 = =   =  − = −   = 58. (c): in a bipolar junction transistor, emitter is heavily doped, base is lightly doped and collector is moderately doped 59. (c) 60. (b): Voltage gain in 10 Av dB = 20log 20 2 40dB 20log10 100 =  = = 61. (a): current gain, B C I I  = For common emitter configuration E C I =I IC = 7mA mA I A I C B 0.1 69 7 10 3 =  =   = − 62. (b): Voltage gain = current gain×resistance gain = current gain 180 1 6  =30 = i C R R 63. (b) here, IB A A 6 100 100 10−  =  =  IC mA A 3 10 10 10−  = =   Ac current gain 100 100 10 10 10 6 3 =   =    = − − B C ac I I 64. (b): Voltage gain, I C B C I C V R R I I R R A =   =