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Barrier potential and width in a pn step junction, L-VI.

Today we will discuss about the depletion region in greater detail than before. We will derive a quantitative relation among barrier potential and its width which are created in the depletion region, as discussed before. We will also derive an expression for the electric field that is established due to this potential.

Lets first recall how electric field and electric potential are related to each other. Electric field is the negative potential gradient. Mathematically, in 1 dimension this is expressed by the following formula.

read more Barrier potential and width in a pn step junction, L-VI.

valence band and conduction band after diffusion of electrons takes place from n side to p side and creation of depletion layer.

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Energy levels in semiconductors. L-V.

Today we will discuss about energy levels and energy bands in semiconductors as well as breakdown in reverse bias.

This is because there is an attractive interaction between the nucleus and the electrons. Electrons must gain energy to go into higher energy levels. This is achieved by sources of heat, light or applied potentials. When the electrons fall from the higher energy levels they release their extra energy again in the form of heat, light or other radiation.

The energy of an electron is proportional to the size of the orbit it is found in. Thus specifying the radius of an orbit, of the electron, is equivalent to specifying its energy level. So electrons in the smallest or innermost orbit are in the first energy level. Electrons in the next higher orbit. i.e. second orbit belong to the second energy level, which is higher. Thee next orbit corresponds to the next energy level which is still higher, and so on for higher orbits and energy level.

read more Energy levels in semiconductors. L-V.

Forward and reverse bias of a pn junction diode.

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PN junction diodes (L-IV)

Today we will discuss about what are PN junction diodes and various conditions they can be subjected to, viz. the forward and reverse bias. We will discuss what is a depletion layer and whats a built-in potential barrier.

In our previous lecture we saw what are extrinsic semiconductors. We discussed that they are of two types, viz. p-type and n-type. By themselves the p-type and n-type semiconductors are not so useful.
But when crystals are doped so that one-half of the same is p-type and the other-half is n-type they serve very important purposes. They are now called PN-junction diode. The border or interface between the p-type part and n-type part is known as PN-junction. The PN junction finds application in almost all sorts of electronics through diodes, transistors and integrated circuits.

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Semiconductors and charge carriers: the silicon atom's electronic configuration. There are 14 electrons and 14 protons in the copper atom which makes it electrically neutral. Together with 10 electrons in the first 23 shells ( K, L ) and the 14 protons in the nucleus the copper atom's core has a net charge of + 4 e. The electron in the outermost M shell has 4 electrons, known as the valence electrons. Photo Credit:

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Conductivity and mobility in semiconductors, L-III

We will discuss in this lecture about drift velocity of electrons and holes in semiconductors which leads to the conductivity and mobility of free charge carriers in the same.

To smooth-sail through this lecture you might wanna first brush up the concepts discussed in the last two lectures: lecture I and lecture II.

Lets begin with Ohm’s law: I=V/R. — eqn 1

Here I is the electric current, V is the applied potential difference and R is the resistance of the material considered. The resistance R depends upon the length l and area of cross-section A of the given material. Let us cast Ohm’s law into a form which is independent of l and A.

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A cylindrical wire carrying current I. Two situations arise. Case A: uniform surface current on the outside surface. Case B: a volume current density that depends upon radial distance from axis of wire. Whats the magnetic field of such a configuration?

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Problem 5.13 Application of Ampere’s Law.

Yesterday we saw an interesting application of the Ampere’s Law (– in magnetostatics and sometimes called Ampere’s circuital law also) for the infinite uniform surface current. Today we will see yet another display of the elegance and efficacy of this law in the following problem. This problem is inherited from Griffith’s text on Electrodynamics (3rd edition)

I have tried to be a bit more explanatory than the basic solution available (in instruction manual, if you have a copy). Thats the whole idea of this labor I have taken up. I also strongly suggest anyone who want to sharpen his saber to try the problem on his/her own effort before looking into the solution. That way one can prepare oneself for the pitfalls of one’s own understanding before taking up help and damaging the opportunity of developing of a better sense of solving such problems.

A steady current I flows down a cylindrical wire of radius a. What would be the magnetic field outside the wire and inside of it? We need to find the same in two different scenarios given.

Here are the two different scenarios.

A. Its a surface current density on the outside surface and its uniform across the surface.

B. Its a volume current density and its distributed in the volume of theĀ  wire, but this time its not uniform. In-fact the volume current density J is directly proportional to s; the distance from the axis of the wire where we are referring the value of J.

read more Problem 5.13 Application of Ampere’s Law.