CBSE 12th Standard Physics Subject Semiconductor Electronics Materials Devices And Simple Circuits Chapter Case Study Questions 2021
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CBSE 12th Standard Physics Subject Semiconductor Electronics Materials Devices And Simple Circuits Case Study Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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When the diode is forward biased, it is found that beyond forward voltage V = Vk, called knee voltage, the conductivity is very high. At this value of battery biasing for p-n junction,the potential barrier is overcome and the current increases rapidly with increase in forward voltage. When the diode is reverse biased, the reverse bias voltage produces a very small current about a few microamperes which almost remains constant with bias. This small current is reverse saturation current.
(i) In which of the following figures, the p-n diode is forward biased.
(ii) Based on the V-I characteristics of the diode, we can classify diode as
(a) bi-directional device (b) ohmic device (c) non-ohmic device (d) passive element (iii) The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is
(a) 100 (b) 106 (c) 10 (d) 10-6 (iv) In the case of forward biasing of a p-n junction diode, which one of the following figures correctly depicts the direction of conventional current (indicated by an arrow mark)?
(v) If an ideal junction diode is connected as shown, then the value of the current I is
(a) 0.013 A (b) 0.02 A (c) 0.01 A (d) 0.1 A -
Rectifier is a device which is used for converting alternating current or voltage into direct current or voltage. Its working is based on the fact that the resistance of p-n junction becomes low when forward biased and becomes high when reverse biased. A half-wave rectifier uses only a single diode while a full wave rectifier uses two diodes as shown in figures (a) and (b) .
(i) If the rms value of sinusoidal input to a full wave rectifier is \(\frac{V_{0}}{\sqrt{2}}\) then the rms value of the rectifier's output is\(\text { (a) } \frac{V_{0}}{\sqrt{2}}\) \(\text { (b) } \frac{V_{0}^{2}}{\sqrt{2}}\) \(\text { (c) } \frac{V_{0}^{2}}{2}\) \(\text { (d) } \sqrt{2} V_{0}^{2}\) (ii) In the-diagram, the input ac is actoss the terminals A and C. The output across Band D is
(a) same as the input (b) half wave rectified (c) zero (d) full wave rectified (iii) A bridge rectifier is shown in figure. Alternating input is given across A and C. If output is taken across BD, then it is
(a) zero (b) same as input (c) half wave rectified (d) full wave rectified (iv) A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit. The current (I) in the resistor(R) can be shown by
(v) With an ac input from 50 Hz power line, the ripple frequency is
(a) 50 Hz in the dc output of half wave as well as full wave rectifier (b) 100 Hz in the de output of half wave as well as full wave rectifier (c) 50 Hz in the de output of half wave and 100 Hz in dc output offull wave rectifier (d) 100 Hz in the dc output of half wave and 50 Hz in the de output of full wave rectifier -
From Bohr's atomic model, we know that the electrons have well defined energy levels in an isolated atom. But due to interatomic interactions in a crystal, the electrons of the outer shells are forced to have energies different from those in isolated atoms. Each energy level splits into a number of energy levels forming a continuous band.The gap between top of valence band and bottom of the conduction band in which no allowed energy levels for electrons can exist is called energy gap.
(i) In an insulator energy band gap is(a) Eg = 0 (b) Eg< 3eV (c) Eg > 3eV (d) None of the above (ii) In a semiconductor, separation between conduction and valence band is of the order of
(a) 0 eV (b) 1 eV (c) 10 eV (d) 50 eV (iii) Based on the band theory of conductors, insulators and semiconductors, the forbidden gap is smallest in
(a) conductors (b) insulators (c) semiconductors (d) All of these (iv) Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?
(a) The number of free electrons for conduction is significant only in Si and Ge but small in C. (b) The number of free conduction electrons is significant in C but small in Si and Ge. (c) The number of free conduction electrons is negligibly small in all the three. (d) The number offree electrons for conduction is significant in all the three. (v) Solids having highest energy level partially filled with electrons are
(a) semiconductor (b) conductor (c) insulator (d) none of these -
Light emitting diode is a photoelectric device which converts electrical energy into light energy. It is a heavily doped p-n junction diode which under forward biased emits spontaneous radiation. The general shape of the J- V characteristics of an LED is similar to that of a normal p-n junction diode, as shown. The barrier potentials are much higher and slightly different for each colour.
(i) The J- V characteristic of an LED is
(ii) The schematic symbol of light emitting diode is (LED)
(iii) An LED is constructed from a p-n junction based on a certain Ga-As-P semiconducting material whose energy gap is 1.9 eV. Identify the colour of the emitted light.
(a) Blue (b) Red (c) Violet (d) Green (iv) Which one of the following statement is not correct in the case of light emitting diodes?
(a) It is a heavily doped p-n junction. (b) It emits light only when it is forward biased. (c) It emits light only when it is reverse biased. (d) The energy of the light emitted is less than the energy gap of the semiconductor used (v) The energy of radiation emitted by LED is
(a) greater than the band gap of the semiconductor used (b) always less than the band gap of the semiconductor used (c) always equal to the band gap of the semiconductor used (d) equal to or less than the band gap of the semiconductor used. -
A photodiode is an optoelectronic device in which current carriers are generated by photons through photoexcitation i.e., photo conduction by light. It is a p-n junction fabricated from a photosensitive semiconductor and provided with a transparent window so as allow light to fall on its function. A photodiode can turn its current ON and OFF in nanoseconds. So, it can be used as a fastest photo-detector.
(i) A p-n photo diode is fabricated from a semiconductor with a band gap of 2.5 ev' It can detect a signal of wavelength(a) 4000 nm (b) 6000 nm (c) 4000 \(\dot A\) (d) 6000 \(\dot A\) (ii) Three photo diodes D1 D2 and D3 are made of semiconductors having band gap of 2.5 eV, 2 eV and 3 eV, respectively. Which one will be able to detect light of wavelength 6000 \(\dot A\) ?
(a) D1 (b) D2 (c) D3 (d) D1, and D2 both (iii) Photodiode is a device
(a) which is always operated in reverse bias (b) which of always operated in forward bias (c) in which photo current is independent of intensity of incident radiation (d) which may be operated in forward or reverse bias. (iv) To detect light of wavelength 500 nrn, the photodiode must be fabricated from a semiconductor of minimum bandwidth of
(a) 1.24 eV (b) 0.62 eV (c) 2.48eV (d) 3.2 eV (v) Photo diode can be used as a photo detector to detect
(a) optical signals (b) electrical signals (c) both (a) and (b) (d) none of these
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CBSE 12th Standard Physics Subject Semiconductor Electronics Materials Devices And Simple Circuits Case Study Questions 2021 Answer Keys
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(i) (c) :The p-n diode is forward biased when p-side is at a higher potential than n-side.
(ii) (c)
(iii) (d) : Forward bias resistance,
\(R_{1}=\frac{\Delta V}{\Delta I}=\frac{0.8-0.7}{(20-10) \times 10^{-3}}=\frac{0.1}{10 \times 10^{-3}}=10\)
\(\text { Reverse bias resistance, } R_{2}=\frac{10}{1 \times 10^{-6}}=10^{7}\)
Then, the ratio of forward to reverse bias resistance
\(\frac{R_{1}}{R_{2}}=\frac{10}{10^{7}}=10^{-6}\)
(iv) (d) : In p-region the direction of conventional current is same as flow of holes.
In n-region the direction of conventional current is opposite to the flow of electrons.
(v) (c) : In the given circuit the junction diode is forward biased and offers zero resistance
\(\therefore \text { The current, } I=\frac{3 \mathrm{~V}-1 \mathrm{~V}}{200 \Omega}=\frac{2 \mathrm{~V}}{200 \Omega}=0.01 \mathrm{~A}\) -
(i) (a) :The rms value of the output voltage at the load resistance \(V_{\mathrm{rms}}=\frac{V_{0}}{\sqrt{2}}\)
(ii) (d)
(iii) (a)
(iv) (c): The given circuit works as a half wave rectifier. In this circuit, we will get current through R when p-n junction is forward biased and no current when p-n junction is reverse biased. Thus the current (I) through resistor (R) will be shown in option (c).
(v) (c) -
(i) (c) :In insulator, energy band gap is > 3 eV
(ii) (b) : In conductor, separation between conduction and valence bands is zero and in insulator, it is greater than 1 eV. Hence in semiconductor the separation between conduction and valence band is 1 eV.
(iii) (a): According to band theory the forbidden gap in conductors Eg = 0, in insulators Eg > 3 eV and in semiconductors Eg < 3 eV.
(iv) (a): The four valence electrons of C, Si and Ge lie respectively in the second, third and fourth orbit.Hence energy required to take out an electron from these atoms (i.e. ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, the number of free electrons for conduction in Ge and Si are significant but negligibly small for C.
(v) (b) -
(i) (b) : The l- V characteristics of an LED is similar to that of a Si junction diode. But the threshold voltages are much higher and slightly different for each colour.
(ii) (b)
(iii) (b) : \(\text { As } E_{g}=\frac{h c}{\lambda} \quad \therefore \lambda=\frac{h c}{E_{g}}\)
Here, Eg = 1.9 eV, he = 1240 eV nm
\(\therefore \quad \lambda=\frac{1240 \mathrm{eVnm}}{1.9 \mathrm{eV}}=652.6 \mathrm{nm}\)
Hence, the emitted light is of red colour.
(iv) (c): A light emitting diode is a heavily doped p-n junction diode which emits light only when it is forward biased.
(v) (d) -
(i) (c) : \(\lambda_{\max }=\frac{h c}{E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2.5 \times 1.6 \times 10^{-19}}\)
\(=5000 \dot A\)
\(\therefore \quad \lambda=4000 \dot A<\lambda_{\max }\)
(ii) (b) : Energy of incident photon \(E=\frac{h c}{\lambda}\)
\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7} \times 1.6 \times 10^{-19}}=2.06 \mathrm{eV}\)
The incident radiation can be detected by a photo diode if energy of incident photon is greater than the band gap.
As D2 = 2 eV, therefore D2 will detect these radiations.
(iii) (a): Photodiode is a device which is always operated in reverse bias.
(iv) (c) : Let Eg be the required bandwidth. Then
\(E_{g}=\frac{h c}{\lambda}\)
Here, he = 1240 eV nm, \(\lambda\) = 500 nm
\(\therefore \quad E_{g}=\frac{1240 \mathrm{eV} \mathrm{nm}}{500 \mathrm{nm}}=2.48 \mathrm{eV}\)
(v) (a): A photo diode is a device which is used to detect optical signals.
