\nState the reason, why GaAs is most commonly used in making of a solar cell. (All India 2008)
\nAnswer:
\nGaAs is most commonly used in making of a solar cell because :
\n(i) It has high optical absorption (~ 104 cm-1<\/sup>) .
\n(ii) It has high electrical conductivity.<\/p>\n
Question 2.
\nWhy should a photodiode be operated at a reverse bias? (All India 2008)
\nAnswer:
\nAs fractional change in minority charge carriers is more than the fractional change in majority charge carriers, the variation in reverse saturation current is more prominent.<\/p>\n
Question 3.
\nGive the logic symbol of NOR gate. (All India 2009)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/12\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-204.png?resize=406%2C55&ssl=1\")
<\/p>\n
Question 4.
\nGive the logic symbol of NAND gate. (All India 2009)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-2.png?resize=250%2C57&ssl=1\")
<\/p>\n
Question 5.
\nGive the logic symbol of AND gate. (All India 2009)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-3.png?resize=227%2C67&ssl=1\")
<\/p>\n
Question 6.
\nIn a transistor, doping level in base is increased slightly. How will it affect
\n(i) collector current and
\n(ii) base current? (Delhi 2011)
\nAnswer:
\nIncreasing base doping level will decrease base resistance and hence increasing base current, which results in a decrease in collector current.<\/p>\n
Question 7.
\nWhat happens to the width of depletion layer of a p-n junction when it is
\n(i) forward biased,
\n(ii) reverse biased? (Delhi 2011)
\nAnswer:
\n(i) In forward biased, the width of depletion layer of a p-n junction decreases.
\n(ii) In reverse biased, the width of depletion layer of a p-n junction increases<\/p>\n
Question 8.
\nWhat is the difference between an H-type and a p-type intrinsic semiconductor? (Comptt. Delhi 2008)
\nAnswer:
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-200.png?resize=380%2C123&ssl=1\")
<\/p>\n
Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. Question 25. Question 26. Question 27. Question 28. Question 29. Question 30. Question 31. Question 32. Question 33. Question 34. Question 35. Question 36. Reverse biasing : If the positive terminal of a battery is connected to the 72-side and negative terminal to the p-side, then the p-n junction is said to be reverse biased. Question 37. Question 38. Question 39. Question 40. Question 41. Question 42. Question 43. When the photodiode is reverse biased then a certain current exits in the circuit even when no light is incident on the p-n junction of photodiode. This current is called dark current. A photodiode can turn its current ON and OFF in nanoseconds. Hence it can be used to detect the optical signals.<\/p>\n Question 44. Question 45. Question 46. Question 47. Question 48. Question 49. Question 50. Question 51. Question 52. Question 53. Question 54. Question 55. Question 56. Question 57. Question 58. Question 59. Question 60. Question 61. Question 62. Question 63. Question 64. Question 65. Question 66. Oscillator action : In an ideal n-p-n biased transistor, when input base emitters junction and output base collector junction are forward and reverse biased respectively, a high collector current IC<\/sub> flows through the circuit. If in circuit switch S is on, this current IC<\/sub> will start flowing in the emitter circuit through the inductive coupling between coils T1<\/sub> and T2<\/sub>, which provides the +ve feedback output to input and hence make IE<\/sub> maximum. In the absence of +ve feedback the IE thus decreases making the circuit back to its original state. This process continues and ocillations are produced. Question 67. Question 68. Question 69. During the negative half cycle of the input, the forward bias is decreased resulting in decrease in IE<\/sub> and hence IC<\/sub>. Thus VCC<\/sub> would increase making the collector more positive. Hence in a common-emitter amplifier, the output voltage is 180\u00b0 out of phase with the input voltage. Question 70. Question 71. Question 72. Question 73. Question 74. Question 75. Question 76. When the transistor is used in the cut off or saturation state, it acts as a switch.<\/p>\n As long as Vi<\/sub> is low and unable to forward bias the transistor, then V0<\/sub> is high. If Vi<\/sub> is high enough to drive the transistor into saturation, then V0<\/sub> is low. When the transistor is not conducting, it is said to be switched off and when it is driven into saturation, it is said to be switched on. This shows that a low input switches the transistor off and a high input switches it on. Question 77. Question 78. Question 79. Question 80. Question 81. (b) Working of photodiode : When the (c) Diagram of photodiode (d) Reason. It is easier to observe the change in the current, with change in light intensity, if a reverse bias is applied.<\/p>\n Question 82. Question 83. Question 84. Question 85. Question 86. Role of energy levels in conduction and valence bands : In the energy band diagram of n-type Si semiconductor, the donor energy level ED<\/sub> is slightly below the bottom EC<\/sub> of the conduction band and electrons from this level moves into conduction band with very small supply of energy. At room temperature, most of the donor atoms get ionised, but very few (~ 10-12<\/sup>) atoms of Si atom get ionised. So the conduction band will have most electrons coming from donor impurities, as shown in the figure.<\/p>\n For p-type semiconductor, the acceptance energy level EA<\/sub> is slightly above the top EV<\/sub> of the valence band. With very small supply of energy, an electron from the valence band can jump to the level EA<\/sub> and ionise the acceptor negatively. At room temperature, most of the acceptor atoms get ionised leaving holes in the valence band.<\/p>\n Question 87. When the transistor is used in the cut off or saturation state, it acts as a switch.<\/p>\n As long as Vi<\/sub> is low and unable to forward bias the transistor, then V0<\/sub> is high. If Vi<\/sub> is high enough to drive the transistor into saturation, then V0<\/sub> is low. When the transistor is not conducting, it is said to be switched off and when it is driven into saturation, it is said to be switched on. This shows that a low input switches the transistor off and a high input switches it on. (ii) Question 88. (ii) Question 89. (b) Rectifier. A rectifier is a circuit which converts an alternating current into direct current. Question 90. Question 91. Question 92. (b) The ‘NOR’ gate that gives a high output only when both the inputs are high, is an ‘AND’ gate. The required Question 93. Due to the diffusion of electrons and holes across the junction, a region of (immobile) positive charge is created on the n-side and a region of (immobile) negative charge is created on the p-side, near the junction; this is called depletion region.<\/p>\n Barrier potential is formed due to loss of electrons from n-region and gain of electrons by p-region. Its polarity is such that it opposes the movement of charge carriers across the junction.<\/p>\n Question 94. Question 95. This is because, in the breakdown region, Zener voltage remains constant even though the current through the Zener diode changes.<\/p>\n Question 96. Using the circuit arrangements shown in fig. (i) and fig (ii), we study the variation of current with applied voltage to obtain the V-I characteristics. Question 97. Question 98. Question 99. Question 100. (b) Common emitter (CE) transistor characteristics. The transistor is most widely used in the CE configuration. When a transistor is used in CE configuration, the input is between the base and emitter and the output is between the collector and emitter. Question 101. Question 102. (b) Photo diodes. Photo diode is a special type of photo-detector. Simplest photo-diode is a reverse biased as shown in Figure (i). Question 103. (ii) Drift: The drift of charge carriers occurs due to electric field. Due to built in potential barrier an electric field directed from n-region to p-region is developed across the junction. This field causes motion of electrons on p-side of the junction to n-side and motion of holes on n-side of junction to p-side. Thus a drift current starts. This current is opposite to the direction of diffusion current. Question 104. Question 105. (b) Working of photodiode : When the (c) Diagram of photodiode (d) Reason. It is easier to observe the change in the current, with change in light intensity, if a reverse bias is applied.<\/p>\n Question 106. From the circuit diagram, we find Question 107. Question 108. Light Emitting Diode (LED) : A light emitting diode is simply a forward biased p-n junction which emits spontaneous light radiation. When forward bias is applied, the electron and holes at the junction recombine and energy released is emitted in the form of light. V-I characteristics of LED are similar to that of Si junction diode but the threshold voltages are much higher and slightly different for each colour. No conduction or light emission occurs for reverse bias which, if it exceeds 5V, may damage the LED. Question 109. During the negative half cycle of the input, the forward bias is decreased resulting in decrease in IE<\/sub> and hence IC<\/sub>. Thus VCC<\/sub> would increase making the collector more positive. Hence in a common-emitter amplifier, the output voltage is 180\u00b0 out of phase with the input voltage. Question 110. This property of the Zener diode is used for regulating voltages so that they are constant. Semiconductor diode as a half wave Rectifier : The junction diode D, supplies rectified current to the band during one half of the alternating input voltage and is always in the same direction. During the first half cycles of the alternating input voltage, junction diodes D1<\/sub> will conduct each permitting current to flow during one half cycle whenever its p-terminal is positive with respect to the n-terminal. Question 111. (b) Device ‘X’ given here represents the full wave rectifier. (c) Logic gate is AND gate. Truth table of AND gate is Question 112. Oscillator action : In an ideal n-p-n biased transistor, when input base emitters junction and output base collector junction are forward and reverse biased respectively, a high collector current IC<\/sub> flows through the circuit. If in circuit switch S is on, this current IC<\/sub> will start flowing in the emitter circuit through the inductive coupling between coils T1<\/sub> and T2<\/sub>, which provides the +ve feedback output to input and hence make IE<\/sub> maximum. In the absence of +ve feedback the IE thus decreases making the circuit back to its original state. This process continues and ocillations are produced. (b) Conductors : Question 113. Using the circuit arrangements shown in fig. (i) and fig (ii), we study the variation of current with applied voltage to obtain the V-I characteristics. (b) Light emitting diode (LED) is a heavily doped p-n junction which under forward bias emits spontaneous radiations. Question 114. Oscillator action : In an ideal n-p-n biased transistor, when input base emitters junction and output base collector junction are forward and reverse biased respectively, a high collector current IC<\/sub> flows through the circuit. If in circuit switch S is on, this current IC<\/sub> will start flowing in the emitter circuit through the inductive coupling between coils T1<\/sub> and T2<\/sub>, which provides the +ve feedback output to input and hence make IE<\/sub> maximum. In the absence of +ve feedback the IE thus decreases making the circuit back to its original state. This process continues and ocillations are produced. Question 115.
\nThe figure shows the V-I characteristic of a semi conductor device. Identify this device. Explain briefly, using the necessary circuit diagram, how this device is used as a voltage regulator. (Comptt. Delhi 2011)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-4.png?resize=177%2C170&ssl=1\")
\nAnswer:
\n(i) The semiconductor diode used is a Zener diode.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-79.png?resize=350%2C211&ssl=1\")
\n(iii) Zener diode as a voltage regulator
\nPrinciple : When a zener diode is operated in the reverse breakdown region, the voltage across it remains practically constant (equal to the breakdown voltage Vz) for a large change in the reverse current. If the input voltage increases, the current through RS<\/sub> and zener diode also increases. This increases the voltage drop across RS<\/sub> without any change in the voltage across the zener diode. This is because in the breakdown region, zener voltage remains constant even though the current through the zener diode changes. Similarly, if the input voltage decreases, the voltage across RS<\/sub> decreases without any change in the voltage across the zener diode. Thus any increase\/decrease of the input voltage results in increase\/ decrease of the voltage drop across RS<\/sub> without any change in voltage across zener diode. Hence the zener diode acts as a voltage regulator.<\/p>\n
\nHow does the depletion region of a p-n junction diode get affected under reverse bias? (Comptt. Delhi 2011)
\nAnswer:
\nDepletion region widens under reverse bias.<\/p>\n
\nHow does the width of depletion region of a p-n junction diode change under forward bias?
\n(Comptt. Delhi 2011)
\nAnswer:
\nThe width of depletion region of a p-n junction<\/p>\n
\nThe graph shown in the figure represents a plot of current versus voltage for a given semi-conductor. Identify the region, if any, over which the semi-conductor has a negative resistance.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-5.png?resize=192%2C191&ssl=1\")
\nAnswer:
\nBetween the region B and C, the semiconductor has a negative resistance.<\/p>\n
\nWrite the truth table for a NAND gate as shown in the figure. (Comptt. All India 2013)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-6.png?resize=183%2C65&ssl=1\")
\nAnswer:
\nTruth table for NAND gate
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-201.png?resize=192%2C130&ssl=1\")
<\/p>\n
\nWhat is the function of a photodiode? (Comptt. All India 2013)
\nAnswer:
\nA photodiode is a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on diode. It is operated under reverse bias.<\/p>\n
\nWrite the truth table for a NOT gate connectedA as shown in the figure. (Comptt. All India 2013)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-15.png?resize=187%2C68&ssl=1\")
\nAnswer:
\nTruth Table
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-16.png?resize=387%2C154&ssl=1\")
<\/p>\n
\nWrite the truth table of a two point input NAND gate. (Comptt. All India 2013)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-17.png?resize=284%2C147&ssl=1\")
<\/p>\n
\nShow variation of resistivity of Si with temperature in a graph. (Delhi 2014)
\nAnswer:
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-18.png?resize=175%2C143&ssl=1\")
<\/p>\n
\nPlot a graph showing variation of current versus voltage for the material GaAs. (Delhi 2014)
\nAnswer:
\nA Graph showing variation of current versus voltage for GaAs
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-19.png?resize=235%2C185&ssl=1\")
<\/p>\n
\nDraw the logic symbol of NAND gate and give its Truth Table. (Comptt. All India 2015)
\nAnswer:
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-20.png?resize=372%2C175&ssl=1\")
<\/p>\n
\nIdentify the logic gate whose output equals 1 when both of its inputs are 0 each. (Comptt. Delhi 2015)
\nAnswer:
\nNAND gate or NOR gate.<\/p>\n
\nName the junction diode whose I-V characteristics are drawn below: (Delhi 2015)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-21.png?resize=230%2C144&ssl=1\")
\nAnswer:
\nSolar cell<\/p>\nSemiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions Short Answer Type<\/h4>\n
\nDistinguish between an intrinsic semiconductor and p-type semiconductor. Give reason, why, a p-type semiconductor crystal is electrically neutral although nh<\/sub> >> ne<\/sub>? (Delhi 2008)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-202.png?resize=369%2C327&ssl=1\")
\n(ii) In a p-type semiconductor, the trivalent impurity atom shares its three valence electrons with the three tetravalent host atoms while the fourth bond remains unbounded. The impurity atom as a whole is electrical neutral. Hence the p-type semiconductor is also neutral.<\/p>\n
\nThe given inputs A, B are fed to a 2-input NAND gate. Draw the output wave form of the gate.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-22.png?resize=225%2C145&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-23.png?resize=282%2C180&ssl=1\")
<\/p>\n
\nDraw the output wave form at X, using the given inputs A, B for the logic circuit shown below. Also identify the gate. (Delhi 2008)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-24.png?resize=319%2C197&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-25.png?resize=373%2C370&ssl=1\")
<\/p>\n
\nIf the output of a 2 input NOR gate is fed as both inputs A and B to another NOR gate, write down a truth table to find the final output, for all combinations of A, B. (Delhi 2008)
\nAnswer:
\nThe truth table is:
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-203.png?resize=408%2C134&ssl=1\")
<\/p>\n
\nThe following figure shows the input waveforms (A, B) and the output waveform (Y) of gate. Identify the gate, write its truth table and draw its logic symbol. (Delhi 2009)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-26.png?resize=271%2C164&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-27.png?resize=370%2C242&ssl=1\")
<\/p>\n
\nThe output of a 2-input AND gate is fed to a NOT gate. Give the name of the combination and its logic symbol. Write down its truth table. (Delhi 2009)
\nAnswer:
\nName : NAND gate.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-28.png?resize=373%2C195&ssl=1\")
<\/p>\n
\n(i) Sketch the output waveform from an AND gate for the inputs A and B shown in the figure.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-29.png?resize=248%2C134&ssl=1\")
\n(ii) If the output of the above AND gate is fed to a NOT gate, name the gate of the combination so formed. (Delhi 2009)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-30.png?resize=360%2C130&ssl=1\")
\n(ii) If this output of AND gate is fed to a NOT gate, the result will be a NAND gate.<\/p>\n
\nDraw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity? (Delhi 2010)
\nAnswer:
\nA measurement of the change in the reverse. saturation current on illumination can give the values of light intensity because photocurrent is pro-portional to incident light intensity.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-31.png?resize=208%2C143&ssl=1\")
<\/p>\n
\n(i) Identify the logic gates marked P and Q in the given logic circuit.
\n(ii) Write down the output at X for the inputs A = 0, B = 0 and A = 1, B = 1. (All India 2010)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-32.png?resize=215%2C92&ssl=1\")
\nAnswer:
\n(i) P is NAND gate and Q is OR gate.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-33.png?resize=316%2C85&ssl=1\")
<\/p>\n
\n(i) Identify the logic gates A marked P and Q B in the given logic circuit.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-34.png?resize=203%2C68&ssl=1\")
\n(ii) Write down the output at X for the inputs A = 0, B = 0 and A = 1, B = 1. (All India 2010)
\nAnswer:
\n(i) P is NOT gate
\nQ is OR gate
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-35.png?resize=308%2C107&ssl=1\")
<\/p>\n
\nDraw the output wave form at X, using the given inputs A and B for the logic circuit shown below. Also, identify the logic operation performed by this circuit. (Delhi 2011)
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-36.png?resize=409%2C170&ssl=1\")
\nAnswer:
\n<\/p>\n
\nName the semiconductor device that can be used to regulate an unregulated dc power supply. With the help of I-V characteristics of this device, explain its working principle. (Delhi 2011)
\nAnswer:
\nName : Zener diode is used to regulate an unregulated dc power supply.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-37.png?resize=235%2C167&ssl=1\")
\nWorking principle : When a zener diode is operated in the reverse break down region, the voltage across it remains practically constant (equal to the break down voltage V-I) for a large change in the reverse current.<\/p>\n
\nDraw the transfer characteristic curve of a base biased transistor in CE configuration. Explain clearly how the active region of the VD versus V, curve in a transistor is used as an amplifier. (Delhi 2011)
\nAnswer:
\nFor using the transistor as an amplifier we will use the active region of the V0 vs. V, curve. The slope of the linear part of the curve represents the rate of change of the output with input. It is negative, that is why as input voltage of the CE amplifier increases its output voltage decreases and the output is said to be out of phase with input.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-38.png?resize=252%2C212&ssl=1\")
<\/p>\n
\nDraw the output waveform at X, using the given inputs A and B for the logic circuit shown below. Also, identify the logic operation performed by this circuit.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-39.png?resize=396%2C168&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-40.png?resize=427%2C187&ssl=1\")
<\/p>\n
\nHow is forward biasing different from reverse biasing in a pn junction diode? (Delhi 2011)
\nAnswer:
\nForward biasing : If the positive terminal of a battery is connected to a p-side and the negative terminal to the 72-side, then the p-n junction is said to be forward biased. Here the applied voltage V opposes the barrier voltage VB. As a result of this<\/p>\n\n
\nThe applied voltage V and the barrier potential VB<\/sub> are in the same direction. As a result of this<\/p>\n\n
\nExplain how a depletion region is formed in a junction diode. (Delhi 2011)
\nAnswer:
\nAs soon as a p-n junction is formed, the majority charge carriers begin to diffuse from the regions of higher concentration to the regions of lower concentrations. Thus the electrons from the n-region diffuse into the p-region and where they combine with the holes and get neutralised. Similarly, the holes from the p-region diffuse into the n-region where they combine with the electrons and get neutralised. This process is called electron-hole recombination.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-84.png?resize=330%2C245&ssl=1\")
\nThe p-region near the junction is left with immobile -ve ions and n-region near the junction is left with +ve ions as shown in the figure. The small region in the vicinity of the junction which is depleted of free charge carriers and has only immobile ions is called the depletion layer. In the depletion region, a potential difference VB is created, called potential barrier as it creates an electric field which opposes the further diffusion of electrons and holes.
\n(i) In forward biased, the width of depletion region is decreased.
\n(ii) In reverse biased, the width of depletion region is increased.<\/p>\n
\nWrite the truth table for the logic circuit shown below and identify the logic operation performed by this circuit.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-41.png?resize=253%2C88&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-42.png?resize=362%2C167&ssl=1\")
<\/p>\n
\nThe current in the forward bias is known to be more (~mA) than the current in the reverse bias (~\u00b5A). What is the reason, then, to operate the photodiode in reverse bias? (Delhi 2012)
\nAnswer:
\nThe fractional increase in majority carriers is much less than the fractional increase in minority carriers. Consequently, the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the majority carrier dominated forward bias current.<\/p>\n
\nDescribe briefly with the help of a circuit diagram, how the flow of current carriers in a p-n-p transistor is regulated with emitter-base junction forward biased and base-collector junction reverse biased. (All India 2012)
\nAnswer:
\nIn a p-n-p transitor, the heavily doped emitter which is p-type has a majority charge carrier of holes. These holes when move towards 21-type base get neutralized by e–<\/sup> in base. The majority carriers enter the base region in large numbers. As the base is thin and lightly doped, the majority carriers (holes) swamp the small number of electrons there and as the collector is reverse biased, these holes can easily cross the junction and enter the collector.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-44.png?resize=198%2C186&ssl=1\")
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-43.png?resize=380%2C180&ssl=1\")
<\/p>\n
\n(a) Why are Si and GaAs preferred materials for fabrication in solar cells?
\n(b) Draw V-I characteristic of solar cell and mention its significance.(Comptt. All India 2012)
\nAnswer:
\n(a) The important criteria for the fabrication of a material for solar cell fabrication are :
\n(i) Band gap of the order of 1.0 eV to 1.8 eV
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-45.png?resize=367%2C193&ssl=1\")
<\/p>\n
\nIn the given circuit diagram, a voltmeter ‘V’ is connected across a lamp ‘L’. How would
\n(i) the brightness of the lamp and
\n(ii) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’ is decreased? Justify your answer. (Delhi 2012)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-46.png?resize=174%2C144&ssl=1\")
\nAnswer:
\nWhen the value of R is decreased, forward biasing of emitter-base junction increases. As a result of this, the emitter current and hence the collector current increases. Therefore :
\n(i) The bulb glows more brightly.
\n(ii) The reading of voltmeter is increased.<\/p>\n
\nExplain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect the optical signals. (Delhi 2013)
\nAnswer:
\nWorking of a photo-diode: Its working is based on photo conduction from light. The conductivity of p-n junction photodiode increases with the increase in intensity of light falling in it.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-47.png?resize=168%2C125&ssl=1\")
\nWhen visible light of energy greater than forbidden energy gap (i.e. hv > Eg<\/sub>) is incident on a reverse biased p-n junction photodiode, additional electron-hole pairs are created in the depletion layer (or near the junction) due to the absorption of photons. The charge carriers will be separated by the junction field and made to flow across the junction, creating reverse current across the junction. The value of reverse saturation current increases with increase in the intensity of incident light. It is found that the reverse saturation current through the photodiode varies almost linearly with the light flux.<\/p>\n
\nMention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range? (Delhi 2013)
\nAnswer:
\nThe important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED) are :
\n(i) The Light Emitting efficiency is maximum.
\n(ii) The reverse breakdown voltage of LEDs are very low. Care should be taken that high reverse voltages do not appear across them.
\n(iii) The semiconductor used for fabrication of visible, LEDs must have a band gap of 1.8 eV (spectral range of visible light is from about 0.4 \u00b5m to 0.7 \u00b5m i.e. from about 3 eV to 1.8 eV).<\/p>\n
\nDraw typical output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine output resistance. (All India 2013)
\nAnswer:
\nTypical output characteristic curves :
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-48.png?resize=258%2C183&ssl=1\")
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-49.png?resize=377%2C146&ssl=1\")
\nThe reciprocal of the slope of the linear part of the output characteristic gives the value of output resistance (r0<\/sub>). The output resistance of the transistor is mainly controlled by the base-collector junction. The high magnitude of the output resistance (of the order of 100 K\u2126) is due to the reverse biased state of this diode. This also explains why the resistance at the initial part of the characteristic, when the transistor is in saturation, is very low.<\/p>\n
\nIn the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table.(All India 2013)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-50.png?resize=205%2C86&ssl=1\")
\nAnswer:
\nThe equivalent gate is OR.
\nTruth table :
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-51.png?resize=246%2C129&ssl=1\")
<\/p>\n
\nIn the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table.
\n(All India 2013)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-52.png?resize=208%2C106&ssl=1\")
\nAnswer:
\nAND Gate
\nTruth table:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-53.png?resize=249%2C133&ssl=1\")
<\/p>\n
\nIn the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table. (All India 2013)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-54.png?resize=232%2C94&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-55.png?resize=369%2C170&ssl=1\")
<\/p>\n
\nAssuming that the two diodes Dj and D2 used in the electric circuit shown in the figure are ideal, find out the value of the current flowing through 1\u2126 resistor. (Comptt. Delhi 2013)
\nAnswer:
\nSince the diodes used are ideal, the diode Dj in forward bias will conduct the current in forward direction, while diode D2<\/sub> in reverse bias will not allow any current to flow.
\nAs such, 2\u2126with D1<\/sub> and 1\u2126 are in series, the net resistance of the circuit will be
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-56.png?resize=151%2C114&ssl=1\")
\nHence the value of the current flowing through 1\u2126 resistor = 2A<\/p>\n
\nAssuming that the two diodes D1<\/sub> and D2<\/sub> used in the electric circuit shown in the figure are ideal, find out the value of the current flowing through 2.5 \u2126 resistor. (Comptt. Delhi 2013)
\nAnswer:
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-57.png?resize=150%2C116&ssl=1\")
\nValue of the current flowing through 2.5 \u2126 resistor = 2.5 A<\/p>\n
\nAssuming that the two diodes D1<\/sub> and D2<\/sub> used in the electric circuit shown in the figure are ideal, find out the value of the current flowing through 2 \u2126 resistor. (Comptt. Delhi 2013)
\nAnswer:
\nD1<\/sub> will conduct current while D2<\/sub> will not allow Hence R = 3\u2126 + 2\u2126 = 5\u2126 As such, 2\u2126 with D1<\/sub> and 2\u2126 are in series, the net resistance of the circuit will be
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-58.png?resize=158%2C115&ssl=1\")
\n\u2234 Value of the current flowing through 2\u2126 resistor = 0.4A<\/p>\n
\nWrite the truth table for the combination of the gates shown. Name the gates used. (Delhi 2013)
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-59.png?resize=157%2C95&ssl=1\")
\nAnswer:
\nR gate = OR
\nS gate = AND
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-60.png?resize=198%2C148&ssl=1\")
<\/p>\n
\nIdentify the logic gates marked ‘P’ and ‘Q’ in the A given circuit. Write the B truth table for the combination. (Delhi 2013)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-61.png?resize=167%2C87&ssl=1\")
\nAnswer:
\nP gate = NAND
\nQ gate = OR
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-62.png?resize=202%2C161&ssl=1\")
<\/p>\n
\nExplain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier. (All India 2013)
\nAnswer:
\nRectifier. A rectifier is a circuit which converts an alternating current into direct current.
\np-n diode as a half wave rectifier. A half wave rectifier consists of a single diode as shown in the circuit diagram. The secondary of the transformer gives the desired a.c. voltage across A and B.
\nIn the positive half cycle of a.c., the voltage at A is positive, the diode is forward biased and it conducts current.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-172.png?resize=372%2C210&ssl=1\")
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-173.png?resize=318%2C244&ssl=1\")
\nIn the negative half cycle of a.c., the voltage at A is negative, the diode is reversed biased and it does not conduct current.
\nThus, we get output across RL<\/sub> during positive half cycles only. The output is unidirectional but varying<\/p>\n
\nDraw a circuit diagram of n-p-n transistor amplifier in CE configuration. Under what condition does the transistor act as an amplifier? (All India 2014)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-63.png?resize=382%2C253&ssl=1\")
\nCondition: The linear portion of the active region of the transistor is used as an amplifier.<\/p>\n
\nThe outputs of two NOT gates are fed to a NOR gate. Draw the logic circuit of the combination of gates. Give its truth table. Identify the gate represented by this combination. (Comptt. Delhi 2014)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-64.png?resize=344%2C247&ssl=1\")
<\/p>\n
\nName the gates \u2018P’ and ‘Q’ shown in the figure of logic circuit of logic circuit given below. Write the truth table for the combination of the gates and identify the equivalent gate. (Comptt. Delhi)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-65.png?resize=227%2C48&ssl=1\")
\nAnswer:
\nP gate : AND
\nQ gate : NOT
\nIdentification of gate : NAND
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-66.png?resize=284%2C140&ssl=1\")
<\/p>\n
\nName the gates \u2019F and ‘Q’ in the logic circuit shown in the figure. Write the truth table for the combination of the gates and identify the equivalent gate.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-67.png?resize=270%2C38&ssl=1\")
\nAnswer:
\nP gate : NOT
\nQ gate : AND
\nIdentification of equivalent gate : NAND
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-68.png?resize=269%2C141&ssl=1\")
<\/p>\n
\nThe input waveforms ‘A’ and ‘B’ and the output waveform ‘Y’ of a gate are shown. Name the gate it represents, write its truth table and draw the logic symbol of this gate. (Comptt. All India 2014)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-69.png?resize=281%2C173&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-70.png?resize=406%2C165&ssl=1\")
<\/p>\n
\n(a) Write the truth table for an OR gate and draw its logic symbol.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-71.png?resize=329%2C129&ssl=1\")
\n(b) The input waveforms A and B, shown here, are fed to an AND gate. Find the output waveform. (Comptt. All India 2014)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-72.png?resize=389%2C288&ssl=1\")
<\/p>\n
\n(i) Write the truth table for an AND gate and draw its logic symbol.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-73.png?resize=327%2C134&ssl=1\")
\n(ii) The input waveforms A and B, as shown, are fed to a NAND gate. Find the output waveform. (Comptt. All India 2014)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-74.png?resize=418%2C321&ssl=1\")
<\/p>\n
\nDistinguish between ‘intrinsic’ and ‘extrinsic’ semiconductors. (Delhi 2015)
\nAnswer:
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-75.png?resize=382%2C153&ssl=1\")
<\/p>\n
\nThe following data was obtained for a given transistor :
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-76.png?resize=222%2C80&ssl=1\")
\nFor this data, calculate the input resistance of the given transistor. (Comptt. Delhi 2015)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-77.png?resize=399%2C354&ssl=1\")
<\/p>\n
\nThe figure given below shows the V-I characteristic of a semiconductor diode.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-78.png?resize=297%2C254&ssl=1\")
\n(i) Identify the semiconductor diode used.
\n(ii) Draw the circuit diagram to obtain the given characteristic of this device.
\n(iii) Briefly explain how this diode can be used as a voltage regulator. (Delhi 2015)
\nAnswer:
\n(i) The semiconductor diode used is a Zener diode.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/12\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-207.png?resize=350%2C211&ssl=1\")
\n(iii) Zener diode as a voltage regulator
\nPrinciple : When a zener diode is operated in the reverse breakdown region, the voltage across it remains practically constant (equal to the breakdown voltage Vz) for a large change in the reverse current. If the input voltage increases, the current through RS<\/sub> and zener diode also increases. This increases the voltage drop across RS<\/sub> without any change in the voltage across the zener diode. This is because in the breakdown region, zener voltage remains constant even though the current through the zener diode changes. Similarly, if the input voltage decreases, the voltage across RS<\/sub> decreases without any change in the voltage across the zener diode. Thus any increase\/decrease of the input voltage results in increase\/ decrease of the voltage drop across RS<\/sub> without any change in voltage across zener diode. Hence the zener diode acts as a voltage regulator.<\/p>\n
\nDraw the labelled circuit diagram of a common-emitter transistor amplifier. Explain clearly how the input and output signals are in opposite phase. (All India 2008)
\nAnswer:
\nThe diagram shows the circuit diagram of a n-p-n trasistor as a CE amplifier. In this diagram it is evident that the base-emitter junction is forward biased whereas collector emitter junction is set to be reverse biased for an ideal operation as an amplifier. In absence of any input a.c. signal the p.d. between collector and emitter is given by
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-80.png?resize=374%2C217&ssl=1\")
\nIn the presence of an input a.c. signal, the forward biased voltage increases resulting in an increase in collector current IC<\/sub> during the positive half cycle, which further decreases the VC<\/sub> from equation (i) Whereas IE<\/sub> and IC<\/sub> both decrease during the negative half cycle as a result of reverse biasing of input section, the decrease in IC<\/sub> increases the VC<\/sub>. So the change in VC<\/sub> during the positive and negative half input cycle results in a 180\u00b0 phase difference between input and output.<\/p>\n
\nState briefly the underlying principle of a- transistor oscillator. Draw a circuit diagram showing how the feedback is accomplished by inductive coupling. Explain the oscillator action. (All India 2008)
\nAnswer:
\nPrinciple of transistor oscillator : “Sustained a.c. signals can he obtained from an amplifier circuit without any external input signal by giving a positive feedback to the input circuit through inductive coupling or RC\/LC network.”<\/p>\n
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-81.png?resize=362%2C253&ssl=1\")
<\/p>\n
\nThe inputs A and B are inverted by using two NOT gates and their outputs are fed to the NOR gate as shown:
\nAnalyse the action of the gates (1) and (2) and identify the logic gate of the complete circuit so obtained. Give its symbol and the truth table. (All India 2008)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-82.png?resize=222%2C116&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-83.png?resize=372%2C319&ssl=1\")
<\/p>\n
\nWith the help of a suitable diagram, explain the formation of depletion region in a p-n junction. How does its width change when the junction is
\n(i) forward biased, and
\n(ii) reverse biased? (All India 2008)
\nAnswer:
\nAs soon as a p-n junction is formed, the majority charge carriers begin to diffuse from the regions of higher concentration to the regions of lower concentrations. Thus the electrons from the n-region diffuse into the p-region and where they combine with the holes and get neutralised. Similarly, the holes from the p-region diffuse into the n-region where they combine with the electrons and get neutralised. This process is called electron-hole recombination.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/12\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-205.png?resize=330%2C245&ssl=1\")
\nThe p-region near the junction is left with immobile -ve ions and n-region near the junction is left with +ve ions as shown in the figure. The small region in the vicinity of the junction which is depleted of free charge carriers and has only immobile ions is called the depletion layer. In the depletion region, a potential difference VB is created, called potential barrier as it creates an electric field which opposes the further diffusion of electrons and holes.
\n(i) In forward biased, the width of depletion region is decreased.
\n(ii) In reverse biased, the width of depletion region is increased.<\/p>\n
\nGive a circuit diagram of a common emitter amplifier using an n-p-n transistor. Draw the input and output waveforms of the signal. Write the expression for its voltage gain. (All India 2009)
\nAnswer:
\n(i) (a) Common emitter configuration of n-p-n transistor
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-139.png?resize=351%2C615&ssl=1\")
\n(ii) Transistor as an amplifier (C.E. configuration) : The circuit diagram of a common emitter amplifier using n-p-n transistor is given below :
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-140.png?resize=384%2C152&ssl=1\")
\nThe input (base-emitter) circuit is forward biased and the output circuit (collector- emitter) is reverse biased.
\nWhen no a.c. signal is applied, the potential difference VCC<\/sub> between the collector and emitter is given by
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-141.png?resize=149%2C25&ssl=1\")
\nWhen an a.c. signal is fed to the input circuit, the forward bias increases during the positive half cycle of the input. This results in increase in IC<\/sub> and decreases in VCC<\/sub>. Thus during positive half cycle of the input, the collector becomes less positive.<\/p>\n
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-142.png?resize=307%2C46&ssl=1\")
<\/p>\n
\n(i) With the help of circuit diagrams, distinguish between forward biasing and reverse biasing of a p-n junction diode.
\n(ii) Draw V-I characteristics of a p-n junction diode in
\n(a) forward bias,
\n(b) reverse bias. (All India 2009)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-85.png?resize=355%2C136&ssl=1\")
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-86.png?resize=379%2C224&ssl=1\")
<\/p>\n
\nExplain with the help of a circuit diagram how a zener diode works as a DC voltage regulator. Draw its I – V characteristics. (All India 2009)
\nAnswer:
\nZener diode is fabricated by heavily doping both p and n-sides. Due to this, depletion region formed is very thin (< 10-6<\/sup> n and the electric field of the junction is extremely high (~5 \u00d7 106<\/sup> V \/ m) even for a small reverse bias voltage of 5 volts. It is seen that when the applied reverse bias voltage (V) reaches the breakdown voltage (Vz) of the Zener diode, there is a large change in the current. After the breakdown voltage Vz, a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant even though current through the Zener diode varies over a wide range. This property of the Zener diode is used for regulating voltages so that they are constant. Semiconductor diode as a half wave Rectifier : The junction diode D, supplies rectified current to the band during one half of the alternating input voltage and is always in the same direction. During the first half cycles of the alternating input voltage, junction diodes D1<\/sub> will conduct each permitting current to flow during one half cycle whenever its p-terminal is positive with respect to the n-terminal.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-143.png?resize=382%2C230&ssl=1\")
\nThe resulting output current is a series of unidirectional pulses with alternate gaps.<\/p>\n
\nDraw a labelled diagram of a full wave rectifier circuit. State its working principle. Show the input-output waveforms. (All India 2009)
\nAnswer:
\np-n junction diode as full wave rectifier
\nA full wave rectifier consists of two diodes and special type of transformer known as centre tap transformer as shown in the circuit. The secondary of transformer gives the desired a.c. voltage across A and B.
\nDuring the positive half cycle of a.c. input, the diode D1<\/sub> is in forward bias and conducts current while D2<\/sub> is in reverse biased and does not conduct current. So we get an output voltage across the load resistor RL<\/sub>.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-87.png?resize=334%2C379&ssl=1\")
\nDuring the negative half cycle of a.c. input, the diode D1<\/sub> is in reverse biased and does not conduct current while diode D2<\/sub> in forward biased and conducts current. So we get an output voltage across the load resistor RL<\/sub>.
\nNOTE: This is a more efficient circuit for getting rectified voltage or current.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-88.png?resize=386%2C131&ssl=1\")
<\/p>\n
\nYou are given a circuit below. Write its truth table. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-89.png?resize=256%2C109&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-90.png?resize=431%2C211&ssl=1\")
<\/p>\n
\nYou are given a A circuit below. Write its truth table. Hence, identify the B logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to. (All India 2011)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-91.png?resize=209%2C77&ssl=1\")
\nAnswer:
\n<\/p>\n
\nYou are given a circuit below. Write its truth A table. Hence, identify the logic operation B carried out by this circuit. Draw the logic symbol of the gate it corresponds to. (All India 2011)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-92.png?resize=207%2C85&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-93.png?resize=379%2C310&ssl=1\")
<\/p>\n
\nDraw the transfer characteristic of a base-biased transistor in CE configuration. Mark the regions where the transistor can be used as a switch. Explain briefly its working. (Comptt. Delhi 2011)
\nAnswer:
\nTransistor as a switch. The circuit diagram of transistor as a switch is shown in Figure 1. Transfer characteristics. The graph between V0<\/sub> and Vi<\/sub> is called the transfer characteristics of the base-biased transistor, shown in Figure 2.<\/p>\n
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-94.png?resize=287%2C422&ssl=1\")
<\/p>\n
\nThe figure shows the V-I characteristics of a semiconductor device. Identify this device. Explain briefly, using the necessary circuit diagram, how this device is used as a voltage regulator. (Comptt. Delhi 2012)
\nAnswer:
\n(i) The semiconductor diode used is a Zener diode.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/12\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-206.png?resize=350%2C211&ssl=1\")
\n(iii) Zener diode as a voltage regulator
\nPrinciple : When a zener diode is operated in the reverse breakdown region, the voltage across it remains practically constant (equal to the breakdown voltage Vz) for a large change in the reverse current. If the input voltage increases, the current through RS<\/sub> and zener diode also increases. This increases the voltage drop across RS<\/sub> without any change in the voltage across the zener diode. This is because in the breakdown region, zener voltage remains constant even though the current through the zener diode changes. Similarly, if the input voltage decreases, the voltage across RS<\/sub> decreases without any change in the voltage across the zener diode. Thus any increase\/decrease of the input voltage results in increase\/ decrease of the voltage drop across RS<\/sub> without any change in voltage across zener diode. Hence the zener diode acts as a voltage regulator.<\/p>\n
\nOutput characteristics of an n-p-n transistor in CE configuration is shown in the figure.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-95.png?resize=277%2C181&ssl=1\")
\nDetermine
\n(i) dynamic output resistance
\n(ii) dc current gain and
\n(iii) ac current gain at an operating point
\nVCE<\/sub> = 10 V, IB<\/sub> = 30 \u00b5A (Delhi 2012)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-96.png?resize=364%2C345&ssl=1\")
<\/p>\n
\nDraw V-I characteristics of a p-n junction diode.
\nAnswer the following questions, giving reasons:
\n(i) Why is the current under reverse bias almost independent of the applied potential upto a critical voltage?
\n(ii) Why does the reverse current show a sudden increase at the critical voltage.
\nName any semiconductor device which operates under the reverse bias in the breakdown region.
\n(All India 2012)
\nAnswer:
\n(i) In reverse bias of p-n junction diode the small current is due to minority carrier and hence resistance is also very high. Increase in voltage leads to a very-very small increase in reverse bias currents so we conclude that in reverse bias reverse current is almost independent of applied potential upto a critical voltage because after this critical voltage, current increases suddenly.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-97.png?resize=317%2C285&ssl=1\")
\n(ii) In reverse bias, reverse current through junction diode is due to minority charge carriers. As reverse bias voltage is increased, electric field at junction becomes significant. When reverse bias voltage becomes equal to zener voltage, electric field strength across junction becomes high. Electric field across junction is sufficient to pull valence electrons from the atom on p- side and accelerate them towards n-side. The movement of these electrons across the function account for high current which is observed at breakdown reverse voltage. Zener diode and photo diode operate under reverse bias.<\/p>\n
\nWrite any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams. (All India 2012)
\nAnswer:
\nDistinguishing features between conductors, semiconductors and insulators :
\n(i) Insulator. In insulator, the valence band is completely filled. The conduction band is empty and forbidden energy gap is quite large. So no electron is able to go from valence band to conduction band even if electric field is applied. Hence electrical conduction is impossible. The solid\/ substance is an insulator.
\n(ii) Conductors (Metals). In metals, either the conduction band is partially filled or the conduction and valence band partly overlap each other. If small electric field is applied across the metal, the free electrons start moving in a direction opposite to the direction of electric field. Hence, metal behaves as a conductor.
\n(iii) Semiconductors. At absolute zero kelvin, the conduction band is empty and the valence band is filled. The material is insulator at low temperature. However the energy gap between valence band and conduction band is small. At room temperature, some valence electrons acquire thermal energy and jump to conduction band where they can conduct electricity. The holes left behind in valence band act as a positive charge carrier.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-98.png?resize=419%2C203&ssl=1\")
<\/p>\n
\nWith what considerations in view, a photodiode is fabricated? State its working with the help of a suitable diagram.
\nEven though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason? (Delhi 2014)
\nAnswer:
\n(a) Why is photodiode fabricated?<\/p>\n\n
\nphotodiode is illuminated with photons of energy (hv > Eg<\/sub>) greater than the energy gap<\/p>\n\n
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-99.png?resize=353%2C160&ssl=1\")
<\/p>\n
\nDraw a circuit diagram of a transistor amplifier in CE configuration.
\nDefine the terms :
\n(i) Input resistance and
\n(ii) Current amplification factor. How are these determined using typical input and output characteristics? (Delhi 2012)
\nAnswer:
\nCircuit diagram of Transistor Amplifier in CE configuration
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-100.png?resize=407%2C460&ssl=1\")
\nThe value of input resistance is determined from the slope of IB<\/sub> versus VBE<\/sub> plot at constant VCE<\/sub>.
\nThe value of current amplification factor is obtained from the slope of collector current IC<\/sub> versus VCE<\/sub> plot, using different values of IB<\/sub>.<\/p>\n
\nIdentify the gates P and Q shown in A – the figure. Write B” the truth table for the combination of the gates shown.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-108.png?resize=230%2C75&ssl=1\")
\nName the equivalent gate representing this circuit and write its logic symbol. (All India 2014)
\nAnswer:
\n(i) P acts as AND gate; Q as NOT gate.
\n(ii) Truth table for combination of gates P and Q
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-109.png?resize=342%2C220&ssl=1\")
<\/p>\n
\nDraw a circuit diagram of a C.E. transistor amplifier. Briefly explain its working and write the expression for
\n(i) current gain
\n(ii) voltage gain of the amplifier.
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-110.png?resize=383%2C497&ssl=1\")
\nDuring the positive half cycle of input signal, the forward bias of emitter-base junction increases.
\nDue to increased forward bias, emitter current (IE<\/sub>) increases and hence according to equation (i) collector current (IC<\/sub>) also increases. Therefore, the voltage drop across RL<\/sub> (i.e. IC<\/sub>RL<\/sub>) increases. According to equation (ii), the collector voltage or output voltage (V0<\/sub>) decreases. Thus collector is connected to the positive terminal of the battery (VCC<\/sub>)
\nso decrease in V0<\/sub> means that the collector voltage becomes 1 cm positive. In other words, amplified negative signal is obtained across the output.
\nSimilarly, during negative hay cycle, an amplified positive signal is obtained across the output.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-111.png?resize=398%2C295&ssl=1\")
<\/p>\n
\nDistinguish between \u00ab-type and p-type semi-conductors on the basis of energy band diagrams. Compare their conductivities at absolute zero temperature and at room temperature. (Comptt. Delhi 2014)
\nAnswer:
\nDistinction between n-type and p-type semiconductors on the basis of energy level diagram :
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-112.png?resize=417%2C247&ssl=1\")
\n(i) In n-type semi conductors an extra energy level (called donor energy level) is produced just below the bottom of the conduction band, while in the p-type semiconductor, this extra energy band (called acceptor energy level) is just above the top of the balanced band.
\n(ii) In n-type semiconductors, most of the electrons come from the donor impurity while in p-type semiconductor, the density of holes in the valence band is predominantly due to the impurity in the extrinsic semiconductors.
\n(iii) At absolute zero temperature conductivities of both types of semi-conductors will be zero.
\n(iv) For equal doping, an n-type semiconductor will have more conductivity than a p-type semiconductor, at room temperature.<\/p>\n
\nDraw the energy band diagram of
\n(i) n-type and
\n(ii) p-type semiconductor at temperature, T > OK. In the case n-type Si semiconductor, the donor level is slightly below the bottom of conduction band. whereas in p-type semiconductor, the aceceptor energy level is slightly above the top of the valence band. Explain, what role do these energy levels play in conduction and valence bands. (Comptt. All India 2014)
\nAnswer:
\nFor energy level diagrams of n-type and p-type semiconductors:
\nDistinction between n-type and p-type semiconductors on the basis of energy level diagram :
\n
\n(i) In n-type semi conductors an extra energy level (called donor energy level) is produced just below the bottom of the conduction band, while in the p-type semiconductor, this extra energy band (called acceptor energy level) is just above the top of the balanced band.
\n(ii) In n-type semiconductors, most of the electrons come from the donor impurity while in p-type semiconductor, the density of holes in the valence band is predominantly due to the impurity in the extrinsic semiconductors.
\n(iii) At absolute zero temperature conductivities of both types of semi-conductors will be zero.
\n(iv) For equal doping, an n-type semiconductor will have more conductivity than a p-type semiconductor, at room temperature.<\/p>\n
\nDraw a plot of transfer characteristic (V0<\/sub> vs Vi<\/sub> and show which portion of the characteristic is used in amplification and why?
\nDraw the circuit diagram of base bias transistor amplifier in CE configuration and briefly explain its working. (Comptt. All India 2014)
\nAnswer:
\n(i)
\nTransistor as a switch. The circuit diagram of transistor as a switch is shown in Figure 1. Transfer characteristics. The graph between V0<\/sub> and Vi<\/sub> is called the transfer characteristics of the base-biased transistor, shown in Figure 2.<\/p>\n
\n<\/p>\n
\nDuring the positive half cycle of input signal, the forward bias of emitter-base junction increases.
\nDue to increased forward bias, emitter current (IE<\/sub>) increases and hence according to equation (i) collector current (IC<\/sub>) also increases. Therefore, the voltage drop across RL<\/sub> (i.e. IC<\/sub>RL<\/sub>) increases. According to equation (ii), the collector voltage or output voltage (V0<\/sub>) decreases. Thus collector is connected to the positive terminal of the battery (VCC<\/sub>)
\nso decrease in V0<\/sub> means that the collector voltage becomes 1 cm positive. In other words, amplified negative signal is obtained across the output.
\nSimilarly, during negative hay cycle, an amplified positive signal is obtained across the output.
\n<\/p>\n
\n(i) Write the functions of three segments of a transistor.
\n(ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained. (Delhi 2014)
\nAnswer:
\n(i)
\n(a) All the three segments of a transistor have different thickness and their doping levels are also different. A brief description of the three segments of a transistor is given below :<\/p>\n\n
\nCommon emitter (CE) transistor characteristics. The transistor is most widely used in the CE configuration. When a transistor is used in CE configuration, the input is between the base and emitter and the output is between the collector and emitter.
\nThe input and the output characteristics of an n-p-n transistor in CE configuration can be studied by using the circuit as shown in Figure 1.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-165.png?resize=309%2C189&ssl=1\")
\n(i) Input characteristics. The variation of the base current IB with the base emitter voltage VBE<\/sub> is called the input characteristic keeping VCE<\/sub> fixed. A curve is plotted between the base current IB<\/sub>
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-166.png?resize=362%2C367&ssl=1\")
\n(ii) Output characteristics. The variation of the collector current IC<\/sub> with the collector emitter voltage VCE<\/sub>, keeping the base current IB constant is called output characteristics.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-167.png?resize=302%2C201&ssl=1\")
\nThe plot of IC<\/sub> versus VCE<\/sub> for different fixed values of IB<\/sub> gives one output characteristic. The different output characteristics for different values of IB<\/sub> is shown in Figure 3.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-168.png?resize=173%2C60&ssl=1\")
<\/p>\n
\n(i) Explain with the help of a diagram the formation of depletion region and barrier potential in a pn junction.
\n(ii) Draw the circuit diagram of a half wave rectifier and explain its working. (All India 2016)
\nAnswer:
\n(a) (i) Depletion layer. The layer containing unneutralized acceptor and donor ion across a p-n junction is called depletion layer. It is called depletion layer because it is depleted of mobile charge carriers.
\n(ii) Barrier potential. The electric field between the acceptor and donor ions is called the barrier. The difference of potential from one side of the barrier to the other side is called barrier potential.
\n(i) The increase of doping concentration will reduce width of depletion layer in semi conductor.
\n(ii) depletion layer widens under reverse bias and vice versa.<\/p>\n
\np-n diode as a half wave rectifier. A half wave rectifier consists of a single diode as shown in the circuit diagram. The secondary of the transformer gives the desired a.c. voltage across A and B.
\nIn the positive half cycle of a.c., the voltage at A is positive, the diode is forward biased and it conducts current.
\n
\n
\nIn the negative half cycle of a.c., the voltage at A is negative, the diode is reversed biased and it does not conduct current.
\nThus, we get output across RL<\/sub> during positive half cycles only. The output is unidirectional but varying.<\/p>\n
\nFor a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2k\u2126 is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 k\u2126. (All India 2016)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-113.png?resize=349%2C172&ssl=1\")
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-114.png?resize=275%2C92&ssl=1\")
<\/p>\n
\nGive reasons for the following :
\n(i) High reverse voltage do not appear across a LED.
\n(ii) Sunlight is not always required for the working of a solar cell.
\n(ill) The electric field, of the junction of a Zener diode, is very high even for a small reverse bias voltage of about 5V. (Comptt. Delhi 2016)
\nAnswer:
\n(i) It is because reverse breakdown voltage of LED is very low, i.e., nearly 5V.
\n(ii) Solar cell can work with any light whose photon energy is more than the band gap energy.
\n(iii) The heavy doping of p and n sides of pn junction makes the depletion region very thin, hence for a small reverse bias voltage, electric field is very high.<\/p>\n
\nIt is required to design a (two-input) logic gate, using an appropriate number, of :
\n(a) NAND gates that gives a ‘low’ output only when both the inputs are ‘low’.
\n(b) NOR gates that gives a ‘high’ output only when both the inputs are ‘high’.
\nDraw the logic circuits for these two cases and write the truth table, corresponding to each of the two designs. (Comptt. All India 2017)
\nAnswer:
\n(a) The ‘NAND’ gate that gives a ‘low’ output only when both its inputs are low, is an ‘OR’ gate
\nThe required design and the truth table are as follow :
\nTruth Table
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-115.png?resize=280%2C210&ssl=1\")
<\/p>\n
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-116.png?resize=206%2C143&ssl=1\")
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-117.png?resize=324%2C153&ssl=1\")
<\/p>\n
\nWrite the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. (Delhi 2016)
\nAnswer:
\nDiffusion and Drift are the two processes which take place in the formation of p-n junction.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-118.png?resize=287%2C136&ssl=1\")
<\/p>\n
\nFor a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 k\u2126 is 2V. Given the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 k\u2126 (Delhi 2017)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-119.png?resize=363%2C263&ssl=1\")
<\/p>\n
\nA zener diode is fabricated by heavily doping both p- and n-sides of the junction. Explain, why? Briefly explain the use of zener diode as a dc voltage regulator with the help of a circuit diagram. \u2018 (Delhi 2017)
\nAnswer:
\nZener Diode : By heavily doping both p and n sides of the junction, depletion region formed is very thin, i.e. < 10-6<\/sup> m. Hence, electric field, across the junction is very high (~5 \u00d7 106<\/sup> V\/m) even for a small reverse bias voltage. This can lead to a ‘breakdown’ during reverse biasing.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-120.png?resize=314%2C149&ssl=1\")
\nIf the input voltage increases\/decreases, current through resistor RS<\/sub>, and Zener diode, also increases\/decreases. This increases\/decreases the voltage drop across Rs without any change in voltage across the Zener diode.<\/p>\n
\nExplain briefly with the help of necessary diagrams, the forward and the reverse biasing of a p-n junction diode. Also draw their characteristic curves in the two cases. (Delhi 2017)
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-150.png?resize=366%2C137&ssl=1\")
\nThe battery is connected to the silicon diode through a potentiometer (or rheostat), so that the applied voltage can be changed for different values of voltages, the corresponding values of current are noted.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-151.png?resize=231%2C191&ssl=1\")
<\/p>\n
\nFrom the V-I characteristics of a junction diode, it is clear that it allows the current to pass only when it is forward biased. So when an alternatively voltage is applied across the diode, current flows only during that part of the cycle when it is forward biased.<\/p>\n
\n(a) In the given diagram, is the junction diode forward biased or reverse biased?
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-121.png?resize=274%2C64&ssl=1\")
\n(b) Draw the circuit diagram of a full wave rectifier and state how it work. (All India 2017)
\nAnswer:
\n(a) The junction diode is reverse biased in the given circuit diagram.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-122.png?resize=363%2C207&ssl=1\")
\nWorking : The diode Dj is forward- biased during one half cycle and current flows through the resistor, but diode D2<\/sub> is reverse-biased and no current flows through it. During the other half cycles, current through the resistor flows in the same direction.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-123.png?resize=308%2C333&ssl=1\")
<\/p>\n
\n(a) Write the functions of the three segments of a transistor.
\n(b) The figure shows the input waveforms A and B for ‘AND’ gate. Draw the output waveform and write the truth table for this logic gate. (All India 2017)
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-124.png?resize=273%2C172&ssl=1\")
\nAnswer:
\n(a) All the three segments of a transistor have different thickness and their doping levels are also different. A brief description of the three segments of a transistor is given below :<\/p>\n\n
![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-125.png?resize=322%2C355&ssl=1\")
<\/li>\n<\/ul>\n
\n(a) In the given diagram, which bulb out of B1<\/sub> and B2<\/sub> will glow and why ?
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-126.png?resize=233%2C113&ssl=1\")
\n(b) Draw the circuit diagram of a full wave rectifier and state how it works.
\n(c) Explain briefly the three processes due to which generation of emf takes place in a solar cell. (All India 2017)
\nAnswer:
\n(a) Bulb B1<\/sub> will glow, because Diode D1<\/sub> is forward biased.
\n(b) Diagram of Solar Cell :
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-127.png?resize=242%2C157&ssl=1\")
\n(c) Three processes in a solar cell for generation of emf:
\nGeneration : Incident light generates electron-hole pairs.
\nSeparation : Electric field of the depletion layer separates the electrons and holes. Collection : Electrons and holes are collected at the n and p side contacts.<\/p>\n
\n(a) Draw the circuit diagram for studying the characteristics of a transistor in common emitter configuration. Explain briefly and show how input and output characteristics are drawn.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-128.png?resize=302%2C195&ssl=1\")
\n(b) The figure shows input waveforms A and B to a logic gate. Draw the output waveform for an OR gate. Write the truth table for this logic gate and draw its logic symbol. (All India 2017)
\nAnswer:
\n(a) The base is made very thin so as to control current flowing between emitter and collector. The base is lightly doped to make a thin depletion layer between emitter and collector.<\/p>\n
\nThe input and the output characteristics of an n-p-n transistor in CE configuration can be studied by using the circuit as shown in Figure 1.
\n
\n(i) Input characteristics. The variation of the base current IB with the base emitter voltage VBE<\/sub> is called the input characteristic keeping VCE<\/sub> fixed. A curve is plotted between the base current IB<\/sub>
\n
\n(ii) Output characteristics. The variation of the collector current IC<\/sub> with the collector emitter voltage VCE<\/sub>, keeping the base current IB constant is called output characteristics.
\n
\nThe plot of IC<\/sub> versus VCE<\/sub> for different fixed values of IB<\/sub> gives one output characteristic. The different output characteristics for different values of IB<\/sub> is shown in Figure 3.
\n<\/p>\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-129.png?resize=318%2C458&ssl=1\")
<\/p>\n
\n(a) Draw the circuit diagram of an n-p-n transistor amplifier in common emitter configuration.
\n(b) Derive an expression for voltage gain of the amplifier and hence show that the output voltage is in opposite phase with the input voltage. (All India 2017)
\nAnswer:
\n
\nDuring the positive half cycle of input signal, the forward bias of emitter-base junction increases.
\nDue to increased forward bias, emitter current (IE<\/sub>) increases and hence according to equation (i) collector current (IC<\/sub>) also increases. Therefore, the voltage drop across RL<\/sub> (i.e. IC<\/sub>RL<\/sub>) increases. According to equation (ii), the collector voltage or output voltage (V0<\/sub>) decreases. Thus collector is connected to the positive terminal of the battery (VCC<\/sub>)
\nso decrease in V0<\/sub> means that the collector voltage becomes 1 cm positive. In other words, amplified negative signal is obtained across the output.
\nSimilarly, during negative hay cycle, an amplified positive signal is obtained across the output.
\n<\/p>\n
\n(a) In the given following diagram ‘S’ is a semiconductor. Would you increase or decrease the value of R to keep the reading of the ammeter A constant when S is heated? Give reason for your answer.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-130.png?resize=200%2C132&ssl=1\")
\n(b) The figure shows input waveforms A and B to a logic gate. Draw the output waveform for an OR gate. Write the truth table for this logic gate and draw its logic symbol. (All India 2017)
\nAnswer:
\n(a) The value of ‘R’ would be increased since the resistance of ‘S’, a semi conductor decreases on heating.<\/p>\n
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-176.png?resize=383%2C164&ssl=1\")
\nWhen a p-n diode is illuminated with light photons having energy \/xv > and intensities Iv I2, I3 etc. the electron and hole pairs generating in the depletion layer will be separated by the junction field and made to flow across the junction.
\nGraph showing variation in reverse bias currents for different intensities are shown in Figure (ii).<\/p>\n
\nExplain the two processes involved in the formulation of a p-n junction diode. Hence define the term ‘barrier potential’. (Comptt. Delhi 2017)
\nAnswer:
\n(a) Two important processes that occur during the formation of a p-n junction are
\n(i) diffusion and
\n(ii) drift.
\n(i) Diffusion: In n-type semiconductor, the concentration of electrons is much greater as compared to concentration of holes; while in p-type semiconductor, the concentration of holes is much greater than the concentration of electrons. When a p-n junction is formed, then due to concentration gradient, the holes diffuse from p side to n side (p \u279d n) and electrons diffuse from n side to p-side (n \u279d p). This motion of charge carriers gives rise to diffusion current across the junction.
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-159.png?resize=302%2C181&ssl=1\")
<\/p>\n
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-160.png?resize=312%2C186&ssl=1\")
<\/p>\n
\nUsing the wave forms of the input A and B, draw the output waveform of the given logic circuit. Identify the logic gate obtained. Write also the truth table. (Comptt. Delhi 2017)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-131.png?resize=235%2C200&ssl=1\")
\nAnswer:
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-132.png?resize=304%2C290&ssl=1\")
<\/p>\n
\nState the reason, why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode, has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of wavelength of 400 nm incident on it. (Comptt. All India 2017)
\nAnswer:
\n(a) Why is photodiode fabricated?<\/p>\n\n
\nphotodiode is illuminated with photons of energy (hv > Eg<\/sub>) greater than the energy gap<\/p>\n\n
\n<\/p>\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-133.png?resize=273%2C204&ssl=1\")
<\/p>\n
\nDraw the circuit diagram of a common emitter transistor amplifier. Write the expression for its voltage gain. Explain, how the input and output signals differ in phase by 180\u00b0. (Comptt. All India 2017)
\nAnswer:<\/p>\n
\nDuring the positive half cycle of input signal, the forward bias of emitter-base junction increases.
\nDue to increased forward bias, emitter current (IE<\/sub>) increases and hence according to equation (i) collector current (IC<\/sub>) also increases. Therefore, the voltage drop across RL<\/sub> (i.e. IC<\/sub>RL<\/sub>) increases. According to equation (ii), the collector voltage or output voltage (V0<\/sub>) decreases. Thus collector is connected to the positive terminal of the battery (VCC<\/sub>)
\nso decrease in V0<\/sub> means that the collector voltage becomes 1 cm positive. In other words, amplified negative signal is obtained across the output.
\nSimilarly, during negative hay cycle, an amplified positive signal is obtained across the output.
\n<\/p>\n
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-135.png?resize=249%2C70&ssl=1\")
\nHence, change in output is negative when the input signal is positive.
\nThis shows that input and output signals differ in phase by 180\u00b0.<\/p>\n
\nDraw the circuit diagram of a full wave rectifier. Explain its working principle. Draw the input and output waveforms. (Comptt. All India 2017)
\nAnswer:
\nWorking of a full wave rectifier :
\n1. A full wave rectifier uses two diodes and gives the rectified output voltage corresponding to both the positive and negative half-cycle of alternating current.
\n2. The p-side of the two diodes are connected to the ends of the secondary of the transformer and, the n-sides of the diodes are connected together.
\n3. Output is taken from between the common- point of the two diodes and secondary of the transformer. Hence, the secondary of the transformer is provided with center tapping and is also called the centre-tap transformer.
\n4. Let, the input voltage to A with respect to the centre be positive and, at the same instant, voltage at B being out-of-phase will be negative. Therefore, diode D1<\/sub> is forward biased and starts conducting whereas, D2<\/sub> being reverse biased does not conduct.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-136.png?resize=288%2C427&ssl=1\")
\n5. Thus, we get an output current and an output voltage across the load resistance RL in the first positive half-cycle.
\n6. During the course of the negative half-cycle, that is, when voltage at A becomes negative and voltage at B becomes positive, we will have D1<\/sub> as reverse biased and D2<\/sub> forward biased.
\n7. In the negative part of the cycle, only diode D2<\/sub> will conduct giving an output current and output voltage across RL<\/sub>.
\n8. For both positive and negative half cycle we will get the output voltage. This rectified output voltage has the shape of half sinusoids.<\/p>\n
\nDraw the V-I characteristic of an LED. State two advantages of LED lamps over conventional incandescent lamps. Write the factor which controls
\n(a) wavelength of light emitted,
\n(b) intensity of light emitted by an LED. (Comptt. All India 2017)
\nAnswer:
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-137.png?resize=307%2C154&ssl=1\")
<\/p>\n
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-138.png?resize=310%2C288&ssl=1\")
\nAdvantages of LED over conventional lamps :
\n(i) Low operational voltage.
\n(ii) Less power consumption.
\n(iii) Long life.
\n(iv) Ruggedness
\nControlling factors :
\n(a) Energy band gap controls the wavelength of light emitted.
\n(b) Forward current controls the intensity of emitted light.<\/p>\nSemiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions Long Answer Type<\/h4>\n
\n(i) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor in its common emitter configuration. Draw the typical input and output characteristics.
\n(ii) Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common emitter amplifier. (Delhi 2017)
\nAnswer.
\n(i) (a) Common emitter configuration of n-p-n transistor
\n
\n(ii) Transistor as an amplifier (C.E. configuration) : The circuit diagram of a common emitter amplifier using n-p-n transistor is given below :
\n
\nThe input (base-emitter) circuit is forward biased and the output circuit (collector- emitter) is reverse biased.
\nWhen no a.c. signal is applied, the potential difference VCC<\/sub> between the collector and emitter is given by
\n
\nWhen an a.c. signal is fed to the input circuit, the forward bias increases during the positive half cycle of the input. This results in increase in IC<\/sub> and decreases in VCC<\/sub>. Thus during positive half cycle of the input, the collector becomes less positive.<\/p>\n
\n<\/p>\n
\nHow is a zener diode fabricated so as to make it a special purpose diode? Draw I-V characteristics of zener diode and explain the significance of breakdown voltage.
\nExplain briefly, with the help of a circuit diagram, how a p-n junction diode works as a half wave rectifier.
\nAnswer:
\nZener diode is fabricated by heavily doping both p and n-sides. Due to this, depletion region formed is very thin (< 10-6<\/sup> n and the electric field of the junction is extremely high (~5 \u00d7 106<\/sup> V \/ m) even for a small reverse bias voltage of 5 volts. It is seen that when the applied reverse bias voltage (V) reaches the breakdown voltage (Vz) of the Zener diode, there is a large change in the current. After the breakdown voltage Vz, a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant even though current through the Zener diode varies over a wide range.<\/p>\n
\n
\nThe resulting output current is a series of unidirectional pulses with alternate gaps.<\/p>\n
\n(a) Explain the formation of depletion layer . and potential barrier in a p-n junction.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-144.png?resize=273%2C54&ssl=1\")
\n(b) In the figure given below the input waveform is converted into the output waveform by a device ‘X’. Name the device and draw its circuit diagram.
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-145.png?resize=218%2C52&ssl=1\")
\n(c) Identify the logic gate represented by the circuit as shown and write its truth table. (Delhi 2017)
\nAnswer:
\n(a)
\nAs soon as a p-n junction is formed, the majority charge carriers begin to diffuse from the regions of higher concentration to the regions of lower concentrations. Thus the electrons from the n-region diffuse into the p-region and where they combine with the holes and get neutralised. Similarly, the holes from the p-region diffuse into the n-region where they combine with the electrons and get neutralised. This process is called electron-hole recombination.
\n
\nThe p-region near the junction is left with immobile -ve ions and n-region near the junction is left with +ve ions as shown in the figure. The small region in the vicinity of the junction which is depleted of free charge carriers and has only immobile ions is called the depletion layer. In the depletion region, a potential difference VB is created, called potential barrier as it creates an electric field which opposes the further diffusion of electrons and holes.
\n(i) In forward biased, the width of depletion region is decreased.
\n(ii) In reverse biased, the width of depletion region is increased.<\/p>\n
\nWorking of full wave rectifier : AC input to be rectified is applied to the primary (P) of a step up transformer. Two ends of the secondary of the transformer are connected to P end of two junction diodes. It is centre-trapped at M which is connected to an end through the load resistance RL<\/sub>. Two crystals
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-146.png?resize=279%2C168&ssl=1\")
\nare formed biased and reverse biased alternately. During half cycle of A.C. input, current flows through one crystal diode and during the next half cycle the current flows through the other crystal diode. However across the load RL<\/sub>, current always flows in the same direction. Thus a continuous pulsating D.C. output voltage is obtained across the load resistance RL. This rectified signal is made smooth with the help of the fitter circuit.<\/p>\n
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-147.png?resize=218%2C151&ssl=1\")
<\/p>\n
\n(a) With the help of the circuit diagram explain the working principle of a transistor amplifier as an oscillator.
\n(b) Distinguish between a conductor, a semiconductor and an insulator on the basis of energy band diagrams. (Delhi 2017)
\nAnswer:
\n(a)
\nPrinciple of transistor oscillator : “Sustained a.c. signals can he obtained from an amplifier circuit without any external input signal by giving a positive feedback to the input circuit through inductive coupling or RC\/LC network.”<\/p>\n
\n<\/p>\n
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-148.png?resize=316%2C221&ssl=1\")
\n![\"Important](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-149.png?resize=359%2C262&ssl=1\")
<\/p>\n
\n(a) Draw the circuit diagrams of a p-n junction diode in
\n(i) forward bias,
\n(ii) reverse bias. How are these circuits used to study the V – I characteristics of a silicon diode? Draw the typical V – I characteristics.
\n(b) What is a light emitting diode (LED)?
\nMention two important advantages of LEDs over conventional lamps. (All India 2017)
\nAnswer:
\n
\nThe battery is connected to the silicon diode through a potentiometer (or rheostat), so that the applied voltage can be changed for different values of voltages, the corresponding values of current are noted.
\n<\/p>\n
\nFrom the V-I characteristics of a junction diode, it is clear that it allows the current to pass only when it is forward biased. So when an alternatively voltage is applied across the diode, current flows only during that part of the cycle when it is forward biased.<\/p>\n
\nTwo important advantages of LEDs:
\n(i) Low operational voltage and less power.
\n(ii) Fast on-off switching capacity.<\/p>\n
\n(a) Draw the circuit arrangement for studying the input and output characteristics of an n- p-n transistor in CE configuration. With the help of these characteristics define
\n(i) input resistance,
\n(ii) current amplification factor.
\n(b) Describe briefly with the help of a circuit diagram how an n-p-n transistor is used to produce self-sustained oscillations. (All India 2017)
\nAnswer:
\n(a) Circuit for studying the common emitter characteristics of an n-p-n transistor :
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-152.png?resize=349%2C146&ssl=1\")
\n(i) Input resistance : The input resistance r, of the transistor in CE configuration is defined as the ratio of the small change in base-emitter voltage to the corresponding small change in the base current, when the collector-emitter voltage is kept fixed.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-153.png?resize=180%2C62&ssl=1\")
\n(ii) Current amplification factor (\u03b2) : It is defined as the ratio of the change in collector current to the small change in base current at constant collector emitter voltage (VCE<\/sub>) when the transistor is in the active state.
\n![\"Important](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-154.png?resize=185%2C66&ssl=1\")
\n(b)
\nPrinciple of transistor oscillator : “Sustained a.c. signals can he obtained from an amplifier circuit without any external input signal by giving a positive feedback to the input circuit through inductive coupling or RC\/LC network.”<\/p>\n
\n<\/p>\n
\nDraw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, Av, of the amplifier is given by \\(\\mathbf{A}_{\\mathbf{v}}=-\\frac{\\boldsymbol{\\beta}_{a \\mathbf{c}} \\mathbf{c}_{\\mathbf{L}}}{r_{i}}\\), where \u03b2ac<\/sub> is the current gain, RL<\/sub> is the load resistance and ri<\/sub> is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain? (Delhi 2012)
\nAnswer:
\nn-p-n transistor as a common emitter amplifier : Working: According to Kirchoff’s law, emitter current It<\/sub>, base current (IB<\/sub>) and collector current (IC<\/sub>) are related as It<\/sub> = IB<\/sub> + IC<\/sub> …(i)
\nWhen current (IC <\/sub>) flows through the load resistance (RL<\/sub>),
\nOutput or collector voltage (V0<\/sub>)
\n![\"Important](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2020\/10\/Important-Questions-for-Class-12-Physics-Chapter-14-Semiconductor-Electronics-Materials-Devices-and-Simple-Circuits-Class-12-Important-Questions-155.png?resize=399%2C318&ssl=1\")
\nDuring the positive half cycle of input signal, the forward bias of emitter-base junction increases. Due to increased forward bias, emitter current (IE<\/sub>) increases and hence according to equation (i) collector current (IC<\/sub>) also increases. Therefore, the voltage drop across RL<\/sub> (i.e. IC<\/sub>R