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.

  • \$\begingroup\$ For mathematical formatting please use \$ instead of a single $ symbol. \$\endgroup\$ Jan 28, 2018 at 5:56
  • \$\begingroup\$ Yes, it is due to recombination as far as I am aware. If you add many donor atoms, providing electrons, they will recombine with most of the holes, reducing the hole density. The converse is true if we add acceptor atoms. \$\endgroup\$
    – jramsay42
    Jan 28, 2018 at 7:23

2 Answers 2


In an intrinsic semiconductor at thermal equilibrium, carrier generation and recombination rate will be balanced so that net carrier density is constant. Also the concentrations n and p, of the electrons and holes are equal.

For any semiconductor,

$$ np = n_i^2 $$

Where ni is known as intrinsic career concentration of the semiconductor. This relation is called mass action law.

When an intrinsic semiconductor is doped with a dopor impurity like Phosphorus, electron concentration will increase, due to the surplus electron provided by each of the doper atom. Hole concentration remains the same. The net effect is -> \$ np> n_i^2 \$ . But mass action law has to be obeyed in the doped semiconductor too. So when it is doped, recombination rate will increase from its earlier rate to reduce the hole concentration. It drives the semiconductor back to the thermal equilibrium. i.e,\$ np = n_i^2 \$ .

So yes, Recombination Process is the sole reason for reducing the hole concentration while doping. Recombination Processes are of different types : band to band , auger recombination etc.

  • \$\begingroup\$ Hi, i was wondering if the mass action law would be valid in a current carrying pn junction.? Could you please help. \$\endgroup\$
    – Kashmiri
    Jul 16, 2020 at 13:19

Statistically speaking, yes; recombination is the reason. A greater number of electrons implies more recombination, thus reducing the lifetime of holes. And this can be understood as there being fewer holes present in the valence band, overall.

However, your second hypothesis is a more fundamental explanation. With the law of mass action, you are essentially starting out from the assumption that doping has no effect on the number of states, i.e. an unperturbed density of states. And with simple permutation & combination calculations (multinomial theorem in combinatorics) we arrive at law of mass action.


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