Question

What is extrinsic semiconductor? How many types of these are? Write their names. Explain the processes which occurred during the formation of the P-N junction. Determine the electric field produced at a P-N junction when width of depletion layer is $$1$$ micrometre and barrier potential is $$0.7volt$$.

Answer 1

Hint:In this question, we can define the extrinsic semiconductors as artificial semiconductors. We can solve the numerical problem by using the relation between the electric field, potential barrier and depletion layer.

Complete step by step solution: -
Extrinsic semiconductors- These types of semiconductors are those semiconductors which have been doped. This doping is done during the manufacture of the crystal. The element or chemical added or doped in the semiconductor is known as a doping agent.

There are two types of extrinsic semiconductors.
-N-type semiconductor- In this type of semiconductors, an electron donor is doped into an intrinsic semiconductor. In these semiconductors, the majority charge carriers are electrons.

-P-type semiconductor- In this type of semiconductor, an electron acceptor is doped into an intrinsic semiconductor. In these semiconductors, the majority charge carriers are holes.

P-N junction- A p-n junction, is a boundary between the p-type and n-type semiconductors. This is the combination semiconductor of p-type and n-type semiconductor. The p-side has an excess of holes while the n-type has the excess of the electrons.

When an impurity is added into a p-type semiconductor, then after doping some part of the p-type semiconductor converts into n-type. Now, this semiconductor contains both p-type and n-type regions. Now, two types of processes will follow in this semiconductor. These are diffusion and drift. In this semiconductor, there are uneven concentrations of holes and electrons in both regions. Now, for making these uneven concentrations to even concentration, the holes are diffused into excess electrons in n-type regions and the electrons are diffused into excess holes in p-type regions. This diffusion gives a rise to a current (diffusion current) which is due to mobility of the electrons and holes. An electric field is induced due to this diffusion current. Due to the mobility of electrons and holes a drift current starts in the opposite direction of the diffusion current. When diffusion current and drift current are equal then the p-n junction is formed.

Now, according to the question, we have the depletion layer which is equal to $1\mu m={10}^{-6}m$ and the barrier potential is equal to $0.7V$.
We know that the if the barrier potential is $dV$ and the width of the depletion layer is $dx$, then the electric field $E$ is given as-
 $E=-{\displaystyle \frac{dV}{dx}}$
Now, substituting the values, we get-
$$\begin{gathered} E=-{\displaystyle \frac{0.7}{{10}^{-6}}} \\ \Rightarrow E=-0.7\times {10}^{6}V/m \end{gathered}$$
Here the negative sign gives that when the potential barrier increases the width of the depletion layer decreases.
So, the electric field is $0.7\times {10}^{6}V/m$

Note: -In this question, we have to remember that in p-type semiconductors, the holes are charge carriers while in n-type semiconductors, the electrons are charge carriers. In the numerical problem, we have to keep in mind that the negative sign gives that if the width of the depletion layer increases then the potential barrier decreases.