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NCERT Class 12 Physics Textbook Chapter 14 With Answer PDF Free Download
Chapter 14: Semiconductor Electronics
Chapter 14: Semiconductor Electronics- Material, Devices And Simple Circuits
14.1 Introduction
Devices in which a controlled flow of electrons can be obtained are the basic building blocks of all electronic circuits.
Before the discovery of the transistor in 1948, such devices were mostly vacuumed tubes (also called valves) like the vacuum diode which has two electrodes, viz., anode (often called a plate) and cathode; triode which has three electrodes – cathode, plate, and grid; tetrode and pentode (respectively with 4 and 5 electrodes).
In a vacuum tube, the electrons are supplied by a heated cathode and the controlled flow of these electrons in a vacuum is obtained by varying the voltage between its different electrodes.
A vacuum is required in the inter-electrode space; otherwise, the moving electrons may lose their energy on collision with the air molecules in their path.
In these devices, the electrons can flow only from the cathode to the anode (i.e., only in one
direction). Therefore, such devices are generally referred to as valves.
These vacuum tube devices are bulky, consume high power, operate generally at high voltages (~100 V), and have limited life and low reliability.
The seed of the development of modern solid-state semiconductor electronics goes back to the 1930s when it was realized that some solid-state semiconductors and their junctions offer the possibility of controlling the number and the direction of flow of charge carriers through them.
Simple excitations like light, heat, or small applied voltage can change the number of mobile charges in a semiconductor.
Note that the supply and flow of charge carriers in the semiconductor devices are within the solid itself, while in the earlier vacuum tubes/valves, the mobile electrons were obtained from a heated cathode and they were made to flow in an evacuated space or vacuum.
No external heating or large evacuated space is required by the semiconductor devices.
They are small in size, consume low power operates at low voltages and has a long life and high reliability.
Even the Cathode Ray Tubes (CRT) used in television and computer monitors which work on the principle of vacuum tubes are being replaced by Liquid Crystal Display (LCD) monitors with supporting solid-state electronics.
Much before the full implications of the semiconductor devices were formally understood, a naturally occurring crystal of galena (lead sulfide, PbS) with a metal point contact attached to it was used as a detector of radio waves.
On the basis of energy bands
According to the Bohr atomic model, in an isolated atom the energy of any of its electrons is decided by the orbit in which it revolves.
But when the atoms come together to form a solid they are close to each other. So the outer orbits of electrons from neighboring atoms would come very close or could even overlap.
This would make the nature of electron motion in a solid very different from that in an isolated atom.
Inside the crystal, each electron has a unique position and no two electrons see exactly the same pattern of surrounding charges.
Because of this, each electron will have a different energy level. These different energy levels with continuous energy variation form what are called energy bands.
The energy band which includes the energy levels of the valence electrons is called the valence band.
The energy band above the valence band is called the conduction band. With no external energy, all
the valence electrons will reside in the valence band.
If the lowest level in the conduction band happens to be lower than the highest level of the valence band, the electrons from the valence band can easily move into the conduction band.
Normally the conduction band is empty. But when it overlaps on the valence band electrons can move freely into it. This is the case with metallic conductors.
If there is some gap between the conduction band and the valence band, electrons in the valence band all remain bound and no free electrons are available in the conduction band.
This makes the material an insulator. But some of the electrons from the valence band may gain
external energy to cross the gap between the conduction band and the valence band.
Then these electrons will move into the conduction band. At the same time, they will create vacant energy levels in the valence band where other valence electrons can move.
Thus the process creates the possibility of conduction due to electrons in the conduction band as well as vacancies in the valence band.
| Author | NCERT |
| Language | English |
| No. of Pages | 46 |
| PDF Size | 2.8 MB |
| Category | Physics |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 12 Physics Chapter 14 Semiconductor Electronics
Q 6. 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:
For a half-wave rectifier, the output frequency is equal to the input frequency, in this case, the input frequency of the half-wave rectifier is 50 Hz.
On the other hand, the output frequency for a full-wave rectifier is twice the input frequency.
Therefore, the output frequency is 2 × 50 = 100 Hz.
Q 7. A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Ans:
No, the photodiode cannot detect the wavelength of 6000 nm because of the following reason:
The energy bandgap of the given photodiode, Eg = 2.8 eV
The wavelength is given by λ = 6000 nm = 6000 × 10−9 m
We can find the energy of the signal from the following relation:
E = hc/λ
In the equation, h is Planck’s constant = 6.626 × 10−34 J, and c is the speed of light = 3 × 108 m/s.
Substituting the values in the equation, we get
E = (6.626 x 10-34 x 3 x 108) / 6000 x 10-9 = 3.313 x 10-20 J
But, 1.6 × 10 −19 J = 1 eV
Therefore, E = 3.313 × 10−20 J = 3.313 x 10-20 / 1.6 x 10-19 = 0.207 eV
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
Q 8. The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that nI = 1.5 × 1016m–3. Is the material n-type or p-type?
Ans:
The following values are given in the question:
Number of silicon atoms, N = 5 × 10 28 atoms/m3
Number of arsenic atoms, nAS =5×1022atoms/m3
Number of indium atoms, nIn=5×1022atoms/m3
ni=1.5×1016electrons/m3
ne=5×1022−1.5×1016=4.99×1022
Let us consider the number of holes to be nh
In the thermal equilibrium, nenh = ni2
Calculating, we get
nh=4.51×109
Here, ne>nh, therefore the material is an n-type semiconductor.
NCERT Class 12 Physics Textbook Chapter 14 With Answer PDF Free Download