So,here's the equation of Mass-Action Law:-

$ np =n_i^2 $

Here, $ n= $ Electron Charge Density ; $ p= $ Hole Charge Density
And, $ n_i^2= $ Intrinsic Charge Carrier Concentration

The equation is true for both intrinsic and extrinsic semiconductors and tells 2 important facts about extrinsic semiconductors:-

For n-type semiconductor, $ n>n_i $ so $ p<n_i $
For p- type semiconductor, $ p>n_i $ so $ n<n_i $

My book says that

"If a pure semiconductor is doped with n-type impurities, the number of electrons in the conduction band increases above a level and the number of holes decreases below a level. Similarly, the addition of p-type impurities to a pure semiconductor increases the number of holes in the valence band above a level and decreases the number of electrons in the conduction band below a level."

The question is:-

Suppose we have an intrinsic semiconductor. It is known that, in a intrinsic semiconductor, $ n=p=n_i $
It is doped and turned into a n-type semicondctor.

My book says the number of holes in valence band/hole density, $p$ decreases. My question is that does the hole density (minority charge carriers) in n-type semiconductor decreases only due to recombination process or is there any other reason other than the recombination process?

Why has the question arisen in my mind?

See, in n-type semiconductor, holes can be generated by using thermal excitation only.

Now, if in the intrinsic semiconductor(sc), the hole density is $ p $ (which can be produced only by thermal agitation) and then that intrinsic sc is changed to n-type sc, the hole density, $ p $ will not change but the electron density $ n $ will increase by large amount($ n>>>p $) due to the presence of pentavalent impurity atoms and due to thermal agitation.

So the only way left with which the hole density, $ p $ can decrease is the process of recombination.

A way to Describe the Decrease in Hole Density other than Recombination Process in n-type semiconductor(According to Me):-

Suppose a crystal of intrinsic sc having 10 atoms(not possible but assume). 4 out of 10 atoms gets thermally excited and thus, 4 electron-hole pairs are created($n=4;p=4$).
It is turned into a n-type semiconductor by doping it with 2 pentavalent impurity atoms. The 2 impurity atoms replaces the 2 host atoms so to maintain the total number of atoms, i.e., 10.

One of the 2 pentavalent impurity atoms replaces that host atom which was earlier generating an electron-hole pair. So,now,

Electron-Hole pairs generated due to thermal excitation=3
Electrons generated due to pentavalent impurity atoms=2

Thus, $ n=3+2=5 ; p=3 $
So, hole density decreases from 4 to 3

So, while doping, if pentavalent impurity atoms replaces host atoms which were earlier creating electron-hole pairs, then there will be decrease in the value of hole density in n-type sc.
And that's how, the hole density can be reduced other than recombination process.

But the whole hypothesis is just based on the probability of replacement of thermally excited host atoms with donor atoms.

The hypothesis given most probably is wrong. But that's how I am visualising it.

But my main question is why the hole density in n-type semiconductor decreases and electron density in p-type semiconductor decreases? Is the recombination the main cause or something else?

Please, bear with me. I am studying introductory electronics.


The reason why your mental image of a semiconductor doesn't really help here, is because it's actually impossible to distinguish whether a free electron in the conduction band comes from a donor dopant or from a thermally generated electron-hole pair: they are simply indistinguishable.

Instead, you have to think in terms of statistics. Electron (n) and hole (p) concentrations are just the average number of carriers per unit volume present at a certain temperature. But from a microscopic point of view they are the result of a ceaseless occurrence of creations and recombinations. At thermal equilibrium, the creation rate must be equal to the recombination rate and the average electron and hole concentrations are then forced to stay constant.

Now, let's imagine to alter the number of electrons by means of doping as in the thought experiment you were proposing. This will bring the system momentarily out of equilibrium. The creation rate, which depends on the temperature, doesn't change, but the recombination rate, which depends on the electron and hole concentrations, is now increased. This higher rate of recombination will consume some of the carriers and will reduce both electron and hole concentrations, until a new equilibrium is found when their product is again back to $np=n_i$.

Although the equilibrium is restored by a decrease of both electron and hole concentrations, because they usually differ by many orders of magnitude, the relative variation of the majority carrier concentration is negligible, and only the minority carrier concentration appears to significantly decrease.


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