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Revision Notes for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits
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Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Notes Class 12 Physics
understand the chapter and score better marks in the examinations. 
11. (a) OR operation, Y= A+B
(b) AND operation Y=A.B
(c) NOT operation Y= A
12. Combination of gates
(a) NAND gate is combination of AND and NOT gates.
(b) NOR gate is combination of NOT and OR gates.
(c) XOR gate is combination of two NOT gates, two AND gates and one OR gate.
Level -01
(Numerical direct formula Based)
Q.1 : What is relation between voltage gain and trans conductor of a trimester amplifier?
Ans :- Voltage gain = Trans – Conductance X Output resistance.
Q. 2 : A transistor is being used as a common emitter amplifier. What is the value of phase difference, if any ,between the collector-emitter voltage and input signal?
Ans.: 1800 or π radian
Q.3. Write is the phase relationship between the output and input voltage in the common faze transmitter amplifier?
Ans: Output voltage is in phase with the input signal voltage.
Q.4. Write the relation between current gains ∞ or β.
Ans:
Q.5. Calculate the Current gain β of a transistor, if the current gain ∞ = 0.98
Ans:
Q. 6 For a Transmitter the value of β is 100, what is the value of ∞ .
Ans:
Q.7. When the voltage drop across a p.n. Junction is increased from 0.65 v to 0.70, the charge in the diode current is 5 ma . What is the dynamic resistance of the diode ?
Dynamic resistance of junction diode is
Ans:
Q.8. p – n – p transistor circuit, the collector is 10 ma , If 90 % of the reach the Collector, find emitter and base currents.
Ans: Here, I E = 10 m A
As 90 % of the holes reach the collector, so the collector current ,
I c = 90 % of I E = 90/100 IE
I E = 100/900 Ic = 100/90 x 10 = 11 m A.
Base Current, I B = I E – I c = 11-10 = 1 mA.
Q.9. A photodiode is fabricated froma semi conductor with band gap of 2.8 e V . Can it detect a wave of 6000 nm? Justify.
Ans : Energy Corresponding to Wave length 6000 nm is
The photon energy ( E = 0.2 ev ) of given waveleanth is much less then band gap ( Eg. ) , hance it caneot detevt the given wavelength.
Q.10. The number of silicon atoms per m3 is 5 x 1022 atom per 33 of Anesenice and 5 x 1020 per m3 atoms of Indian. Calculate the number of electrons and holes . Given that Ni = 1.5 X 1016 per m3 . In the material N-type on P-Type?
Ans : Arnesic is n-type impurty and indium is P-type impurity Number of electron, ne = n0 – nA = 5 x 1022 – 5 x 1020= 4.95 x 1022m-3
Q.1. When the voltage drop across a p-n junction diode is incrase from 0.65 v to 0.70 v , the change in the diode current is 5mA. What is the dynamic resistance of the diode?
Q.2. Diode used in figure has a constant voltage drop at 0.5 V at all current and a maximum power rating of 100mw. What should be the value of resistance R, coneected in series for maximum current.
Q.9. On the figure shown, find out the current passing through RL and Zener diode :
Please click the link below to download CBSE Class 12 Phyiscs - Electronic Devices Formulae
| CBSE Class 12 Physics Alternating Current Notes Set A |
| CBSE Class 12 Physics Alternating Current Notes Set B |
| CBSE Class 12 Physics Wave Optics Notes Set A |
| CBSE Class 12 Physics Wave Optics Notes Set B |
CBSE Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Notes
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Notes for Physics CBSE Class 12 Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits
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Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Notes for Physics CBSE Class 12
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Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits CBSE Class 12 Physics Notes
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Notes for CBSE Physics Class 12 Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits
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