CBSE 12th Standard Physics Subject Semiconductor Electronics Materials Devices And Simple Circuits Chapter Case Study Questions With Solution 2021
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CBSE 12th Standard Physics Subject Semiconductor Electronics Materials Devices And Simple Circuits Case Study Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Physics
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A silicon p-n junction diode is connected to a resistor R and a battery of voltage VB through milliammeter (mA) as shown in figure. The knee voltage for this junction diode is VN = 0.7 V. The p-n junction diode requires a minimum current of 1m. A to attain a value higher than the knee point on the J- V characteristics of this junction diode. Assuming that the voltage V across the junction is independent of the current above the knee point. A p-n junction is the basic building block of many semiconductor devices like diodes. Important process occurring during the formation of a p-n junction are diffusion and drift. In an n-type semiconductor concentration of electrons is more as compared to holes. In a p-type semiconductor concentration of holes is more as compared to electrons.
(i) If VB = 5 V, the maximum value of R so that the voltage V is above the knee point voltage is(a) 40 \(\Omega\) (b) 4.3 \(\Omega\) (c) 5.0 \(\Omega\) (d) 5.7 \(\Omega\) (ii) If VB = 5 V, the value of R in order to establish a current to 6 mA in the circuit is
(a) 833 \(\Omega\) (b) 717 \(\Omega\) (c) 950 \(\Omega\) (d) 733 \(\Omega\) (iii) If VB = 6 V, the power dissipated in the resistor R, when a current of 6 mA flows in the circuit is
(a) 30.2 mW (b) 30.8 mW (c) 31.2 mW (d) 31.8 mW (iv) When the diode is reverse biased with a voltage of 6 V and Vbi = 0.63 V. Calculate the total potential.
(a) 9.27 V (b) 6.63 V (c) 5.27 V (d) 0.63 V (v) Which of the below mentioned statement is false regarding a p-n junction diode?
(a) Diodes are uncontrolled devices. (b) Diodes are rectifying devices. (c) Diodes are unidirectional devices. (d) Diodes have three terminals -
The electron mobility characterises how quickly an electron can move through a metal of semiconductor when pulled by an electric field. There is an analogous quality for holes, called hole mobility. A block of pure silicon at 300 K has a length of 10 cm and an area of 1.0 cm2. A battery of emf 2 V is connected across it. The mobility of electron is 0.14 m2 y-1 s-1 and their number density is 1.5 x 1016 m-3. The mobility of holes is 0.05 m2 y-1 s-1.
(i) The electron current is(a) 6.72 x 10-4 A (b) 6.72 x 10-5 A (c) 6.72 x 10-6 A (d) 6.72 x 10-7 A (ii) The hole current is
(a) 2.0 x 10-7 A (b) 2.2 x 10-7 A (c) 2.4 X 10-7 A (d) 2.6 x 10-7 A (iii) The number density of donor atoms which are to be added up to pure silicon semiconductor to produce an n-type semiconductor of conductivity 6.4 \(\Omega\)-1 cm-1 is approximately (neglect the contribution of holes to conductivity)
(a) 3 x 1022 m-3 (b) 3 x 1023 m-3 (c) 3 x 1024 m-3 (d) 3 x 1021 m-3 (iv) When the given silicon semiconductor is doped with indium, the hole concentration increases to 4.5 x 1023 m-3. The electron concentration in doped silicon is
(a) 3 x 109 m-3 (b) 4 x 109 m-3 (c) 5 x 109 m-3 (d) 6 x 109 m-3 (v) Pick out the statement which is not correct.
(a) At a low temperature, the resistance of a semiconductor is very high. (b) Movement of holes is restricted to the valence band only (c) Width of the depletion region increases as the forward bias voltage increases in case of a p- n junction diode. (d) In a forward bias condition, the diode heavily conducts -
p-n junction is a single crystal of Ge or Si doped in such a manner that one half portion of it acts as p-type semiconductor and other half functions as n-type semiconductor. As soon as a p-n junction is formed, the holes from the p-region diffuse into the n-region and electron from n region diffuse in to p-region. This results in the development of VB across the junction which opposes the further diffusion of electrons and holes through the junction.
(i) In an unbiased p-n junction electrons diffuse from n-region to p-region because(a) holes in p-region attract them (b) electrons travel across thejunction due to potential difference (c) electron concentration in n-region is more as compared to that in p-region (d) only electrons move from n to p region and not the vice-versa (ii) Electron hole recombination in p-n junction may lead to emission of
(a) light (b) ultraviolet rays (c) sound (d) radioactive rays (iii) In an unbiased p-n junction
(a) potential at p is equal to that at n (b) potential at pis +ve and that at n is -ve (c) potential at p is more than that at n (d) potential at p is less than that at n. (iv) The potential of depletion layer is due to
(a) electrons (b) holes (c) ions (d) forbidden band (v) In the depletion layer of unbiased p-n junction
(a) it is devoid of charge carriers (b) has only electrons (c) has only holes (d) p-n junction has a weak electric field -
The potential barrier in the p-n junction diode is the barrier in which the charge recquires additional force for crossing the region. In other words, the barrier in which the charge carrier stopped by the obstructive force is known as the potential barrier.
When a p-type semiconductor is brought into a close contact with n-type semiconductor, we get a p-n junction with a barrier potential 0.4 V and width of depletion region is 4.0 x 10-7 m. This p-n junction is forward biased with a battery of voltage 3V and negligible internal resistance, in series with a resistor of resistance R, ideal millimeter and key K as shown in figure. When key is pressed, a current of 20 mA passes through the diode.
(i) The intersity of the electric field in the depletion region when p-n junction is unbiased is(a) 0.5 x 106 V m-1 (b) 1.0 x 106 V m-1 (c) 2.0 x 106 V m-1 (d) 1.5 x 106 V m-1 (ii) The resistance of resistor R is
(a) 150 \(\Omega\) (b) 300 \(\Omega\) (c) l30 \(\Omega\) (d) 180 \(\Omega\) (iii) In a p-n junction, the potential barrier is due to the charges on either side of the junction, these charges are
(a) majority carriers (b) minority carriers (c) both (a) and (b) (d) fixed donor and acceptor ions. (iv) If the voltage of the potential barrier is V0 A voltage V is applied to the input, at what moment will the barrier disappear?
(a) V (b) V=V0 (c) V>V0 (d) V< (v) If an electron with speed 4.0 x 105 m s-1 approaches the p-n junction from the n-side, the speed with which it will enter the p-side is
(a) 1.39 x 105 m S-1 (b) 2.78 x 105 m S-1 (c) 1.39 x 106 m S-1 (d) 2.78 x 106 m s-1 -
Solar cell is a p-n junction diode which converts solar energy into electric energy. It is basically a solar energy converter. The upper layer of solar cell is of p- type semiconductor and very thin so that the incident light photons may easily reach the p-n junction. On the top face of p-layer, the metal finger electrodes are prepared in order to have enough spacing between the fingers for the lights to reach the p-n junction through p-layer.
(i) The schematic symbol of solar cell is
(ii) The p-n jqnction which generates an emf when solar radiations fall an it, with no external bias applied,is a
(a) light emitting diode (b) photodiode (c) solar cell (d) None of these (iii) For satellites the source of energy is
(a) Solar cell (b) Fuel cell (c) Edison cell (d) None of these (iv) Which of the following material is used in solar cell?
(a) Barium (b) Silicon (c) Silver (d) Selenium (v) The efficiency of a solar cell may be in the range
(a) 2 to 5% (b) 10 to 15% (c) 30 to 40% (d) 70 to 80%
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CBSE 12th Standard Physics Subject Semiconductor Electronics Materials Devices And Simple Circuits Case Study Questions With Solution 2021 Answer Keys
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(i) (b) : Voltage drop across R.
\(V_{R}=V_{B}-V_{N}=5-0.7=4.3 \mathrm{~V}\)
Here, Imin = 1 x 10-3 A
\(R_{\max }=\frac{V_{R}}{I_{\min }}=\frac{4.3}{1 \times 10^{-3}}=4.3 \times 10^{3} \Omega=4.3 \mathrm{k} \Omega\)
(ii) (b) : \(I=6 \mathrm{~mA}=6 \times 10^{-3} \mathrm{~A}\)
\(\begin{array}{l} V_{R}=V_{B}-V_{N}=5-0.7=4.3 \mathrm{~V} \\ R=\frac{V_{R}}{I}=\frac{4.3}{6 \times 10^{-3}}=717 \Omega \end{array}\)
(iii)(d) : Here, VB = 6 V; VN = 0.7V,
\(V_{R}=6-0.7=5.3 \mathrm{~V}\)
Power dissipated in R = l x VR
= (6 x 10-3) x 5.3 = 31.8 x 10-3 W
= 31.8 mW
(iv) (b) : Vt = Vbi + VR = 0.63 + 6 = 6.63 V
(v) (d) : Diode is two terminal device, anode and cathode are the two terminals. -
(i) (d): \(E=\frac{V}{l}=\frac{2}{0.1}=20 \mathrm{~V} / \mathrm{m} ;\)
\(A=1.0 \mathrm{~cm}^{2}=1.0 \times 10^{-4} \mathrm{~m}^{2}\)
\(\begin{array}{l} v_{e}=\mu_{e} E=0.14 \times 20=2.8 \mathrm{~m} \mathrm{~s}^{-1} \\ I_{e}=n_{e} A e v_{e} \end{array}\)
\(\begin{array}{l} =\left(1.5 \times 10^{16}\right) \times\left(1.0 \times 10^{-4}\right) \times\left(1.6 \times 10^{-19}\right) \times 2.8 \\ =6.72 \times 10^{-7} \mathrm{~A} \end{array}\)
(ii) (c): In a pure semiconductor,
\(n_{e}=n_{h}=1.5 \times 10^{16} \mathrm{~m}^{-3}\)
\(\begin{array}{l} v_{h}=\mu_{h} \times E=0.05 \times 20=1.0 \mathrm{~ms}^{-1} \\ I_{h}=n_{h} \mathrm{Aev}_{h} \end{array}\)
\(\begin{array}{l} =\left(1.5 \times 10^{16}\right) \times\left(1.0 \times 10^{-4}\right) \times\left(1.6 \times 10^{-19}\right) \times 1.0 \\ =2.4 \times 10^{-7} \mathrm{~A} \end{array}\)
(iii) (a): \(\sigma=e n_{e} m_{e}\)
\(\text { or } n_{e}=\frac{\sigma}{e \mu_{e}}=\frac{6.4 \times 10^{2}}{\left(1.6 \times 10^{-19}\right) \times 0.14}\)
\(=3.14 \times 10^{22} \approx 3 \times 10^{22} \mathrm{~m}^{-3}\)
(iv) (c): \(n_{e}=\frac{n_{i}^{2}}{n_{h}}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}=5 \times 10^{9} \mathrm{~m}^{-3}\)
(v) (c): In case of a p-n junction diode, width of the depletion region decreases as the forward bias voltage increases. -
(i) (c) : Electron concentration in n-region is more as compared to that in p-region. So electrons diffuse from n-side to p-side.
(ii) (a) : When an electron and a hole recombine, the energy is released in the form of light.
(iii) (a) : In an unbiased p-n junction, potential at p is equal-to that at n.
(iv) (c) : The potential of depletion layer is due to ions.
(v) (a) : In the depletion layer of unbiased p-n, junction has no charge carriers. -
(i) (b) : \(E=\frac{V_{B}}{d}=\frac{0.4}{4.0 \times 10^{-7}}=1.0 \times 10^{6} \mathrm{Vm}^{-1}\)
(ii) (c) : Potential difference across = R = 3 - 0.4 = 2.6 V
\(\text { Resistance } R=\frac{\text { Potential difference }}{\text { Current }}\)
\(=\frac{2.6}{20 \times 10^{-3}}=130 \Omega\)
(iii) (d)
(iv) (b) : When the voltage will be the same that of the potential barrier disappears resulting in flow of current.
(v) (a) : \(\frac{1}{2} m v_{1}^{2}=e V_{B}+\frac{1}{2} m v_{2}^{2}\)
\(\Rightarrow \frac{1}{2} \times\left(9.1 \times 10^{-31}\right) \times\left(4 \times 10^{5}\right)^{2}\)
\(=1.6 \times 10^{-19} \times(0.4)+\frac{1}{2} \times 9.1 \times 10^{-31} \times v_{2}^{2}\)
On solving, we get
\(v_{2}=1.39 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}\) -
(i) (a)
(ii) (c)
(iii) (a) : Solar cells are the source of energy for satellites.
(iv) (b) : Silicon is used in solar cell.
(v) (b) : 10 to 15%.
