CBSE Class 12 Physics Chapter-14 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 14 – Semiconductor Electronic: Material, Devices And Simple Circuits 2 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter 14 – Semiconductor Electronic: Material, Devices And Simple Circuits
CBSE Class 12 Physics Important Questions Chapter 14 – Semiconductor Electronic: Material, Devices And Simple Circuits
2 Marks Questions
1.Draw a pn junction with reverse bias? Which biasing will make the resistance of a
p-n-junction high?
Ans.
Reverse biasing will make the resistance high as it will not allow the current to pass.
2.Write the truth table for the following combination of gates?
Ans.
| A | B | Y’ | Y |
| 0 1 1 | 0 1 1 | 0 1 1 1 | 0 1 1 |
3.Draw the voltage current characteristics of a zener diode?
Ans.
4.For a extrinsic semiconductor, indicate on the energy band diagram the donor and
acceptor levels?
Ans. N-type Extrinsic Semiconductor P-type Extrinsic Semiconductor
5.What do you mean by depletion region and potential barrier in junction diode?
Ans.A layer around the junction between p and n-sections of a junction diode where charge carriers electrons and holes are less in number is called depletion region. The potential difference created across the junction due to the diffusion of charge carriers across the junction is called potential barrier.
6.A transistor has a current gain of 30. If the collector resistance is 6k
, input resistance is 1k, calculate its voltage gain?
Ans.Given 
Voltage gain = current gain 
Rgain
Voltage gain = 306 = 180
7. What are the advantages and disadvantages of semiconductor devices over vacuum tubes?
Ans.Advantages – Semiconductor devices are very small in size as compared to the vacuum tubes. It requires low voltage for their operation
Disadvantage – Due to the rise in temperature and by applying high voltage it can be damaged.
8.The base of a transistor is lightly doped. Explain why?
Ans.In a transistor, the majority carries form emitter region moves towards the collector region through base. If base is made thick and highly doped, majority carriers will combine with the other carriers within the base and only few is collected by the collector which leads to small output collector current. Thus in order to have large output collector current, base is made thin and lightly doped.
9. Determine the currents through resistance R of the circuits (i) and (ii) when similar diodes 
are connected as shown in the figure.
Ans.In figure (i) are forward biased
In figure (ii) 
is forward biased but 
is reverse biased due to which offers infinite resistance
10.What do you mean by hole in a circuit? Write its two characteristics?
Ans.A vacancy created in a covalent bond in a semiconductor due to the release of electron is known as hole in a semiconductor.
Characteristics of hole
(i) Hole is equivalent to a positive electronic charge.
(ii) Mobility of hole is less than that of an electron
11. Diode used in the figure has a constant voltages drop at 0.5V at all currents and
a maximum power rating of 100mW. What should be the value of the resistance R, connected in series for maximum current?
Ans.
I = 0.2A
For the given circuit
IR = + 0.5 – 1.5 = 0
I R = 0.5
IR=0.5-1.5
IR-1=0
0.2 x R=1
12. Zener diode 
has saturation current of 20A and reverse breakdown voltage of 100V where as the corresponding value of 
are 40
A and 40. Find the current through the circuit?
Ans.Here is forward biased where as reverse biased hence behaves as a conductor and reverse saturation current will flow from
Thus 
Now 50V will appear across so
13. In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Ans.The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.
14. Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Ans.The correct statement is (d).
In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.
15. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to 
Which of the following statements is true?
Ans.The correct statement is (c).
Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least.
The energy band gap of these elements are related as: 
16. In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Ans.The correct statement is (c).
The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.
17. When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.
Ans.The correct statement is (c).
When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.
18. For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Ans.The correct statement is (b), (c).
For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased.
19. For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
Ans.The correct statement is (c).
The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.
20. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
Ans.Input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
Output frequency = 
= 100 Hz