# Mass-Action Law of Semiconductors

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?

Note:-
Please, bear with me. I am studying introductory electronics.

• For mathematical formatting please use \$instead of a single$ symbol. Jan 28, 2018 at 5:56
• 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. Jan 28, 2018 at 7:23

In an intrinsic semiconductor at thermal equilibrium, the carrier generation and recombination rates 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 $$\n_i\$$ is known as the intrinsic carrier concentration of the semiconductor. This relation is called the mass action law.

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

So yes, the recombination process is the sole reason for reducing the hole concentration from doping. Recombination processes are of different types: band-to-band, Auger recombination, etc.

• Hi, i was wondering if the mass action law would be valid in a current carrying pn junction.? Could you please help. 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.

The question is five years old; I believe the original poster has decided this problem for themself in the course of their studies, and this answer is for visitors coming from a search engine.

The statement the hole density in n-type semiconductor decreases due to recombination process implies that there are two concurrent processes in the semiconductor bulk, namely, recombination and generation of electron-hole pairs, similar to mutually reversible reactions in chemistry. By analogy, in dynamic equilibrium, therefore, the concentrations of "reactants" (electrons and holes) obey the law of mass action.

This chemistry analogy, being probably suitable for "visualizing", however, has no bearing on textbook derivation of the concentration formulas as well as the equation for the product of the free electron concentration and the free hole concentration $$\np = n_i^2\$$ ("Mass-Action Law of Semiconductors"). The concentration formulas do not depend on the mobilities nor on the cross sections of electron and hole quasiparticles. The product np is an approximation using Maxwell Boltzmann statistics and depends only on the densities of states at band edges, the band gap energy and the temperature. The concentrations depend also on the Fermi level.

The density of states is calculated from the Pauli principle of exclusion ($$\p_n(E), p_p(E)\$$ -- electron/hole momentum): $$D_{conductance} = 4π(1/{\hbar}^2)·m_n·p_n(E) = 4π(1/{\hbar}^2)·m_n\sqrt{2m_n(E-E_c)} \\ D_{valence} = 4π(1/{\hbar}^2)·m_p·p_p(E) = 4π(1/{\hbar}^2)·m_p\sqrt{2m_p(E_v-E)}$$ Each n, p state contributes $$\\exp(-(E-E_{Fermi})/kT)\$$, $$\\exp(-(E_{Fermi}-E)/kT)\$$ respectively. Integrating over all the energies we arrive at the concentration formulas $$n = \int_{E_c}^∞{D_c\exp(-(E-E_{Fermi})/kT)dE} = N_c\exp(-(E_c-E_{Fermi})/kT) \\ p = \int_{-∞}^{E_v}{D_v\exp(-(E_{Fermi}-E)/kT)dE} = N_v\exp(-(E_{Fermi}-E_v)/kT)$$

The derivation never mentions recombination.

Suppose a crystal of intrinsic sc having 10 atoms (not possible but assume): one need not assume impossible, all the more superfluous. The silicon unit cell has 8 atoms, and with only 10 atoms the cluster, even if exists, has no attributes essential for ensuing consideration: it would have no band structure (conduction, valence bands), only molecular orbitals, and a quasiparticle concept is irrelevant here -- so, no holes. Also, problematic is doping of this small cluster without an essential re-build of its molecular-orbital structure.

The smallest semiconductor-like Si clusters available today are silicon quantum dots. The smallest quantum dots even have chemical formulas, like $$\\text{Si}_{134} \text{B}_{24} \text{P}_{15} \text{H}_{52}\$$ -- a silicon quantum dot doped with boron and phosphorus. The hydrogen atoms terminate surface dangling bonds. Having only a finite, relatively small, count of atoms, it certainly has no (continuous) bands, but tight bunches of energy levels split due to atom interaction approximate bands: HOMO (highest occupied molecular orbital) plays the role of the valence band, LUMO (lowest unoccupied molecular orbital) plays the role of the conduction band.

At this small size (<2 nm), QD "operates" in the 'strong confinement regime', and quantum effects dominate. With small bandgap, the Fermi-Dirac statistics cannot be approximated with Maxwell-Boltzmann, and $$\np = n_i^2\$$ does not hold.

Validation of "mass action law for semiconductors" aside, the electron-hole recombination is a hot research topic in quantum electronics and photonics. It is a multifaceted process happening through multiple mechanisms: radiative recombination, trap-assisted recombination, exciton recombination.

Answering the main question of the post: the carrier densities in extrinsic semiconductor change w.r.t. the carrier densities in corresponding intrinsic semiconductor. The dopant atoms change the distribution of energy levels and state density. Not only the carriers from dopants become added, increasing the majority carrier concentration, but the minority carrier concentration also changes (decreases, and under certain conditions $$\np = n_i^2\$$ holds).

In a non-ideal semiconductor bulk (that is, practically in all cases) there are two concurrent processes running, generation and recombination. Dependent on mechanism, recombination can be fast or slow; in equilibrium, generation adjusts so that the concentrations remain constant.

In this context, Is the recombination the main cause or something else? is sort of the 'chicken or the egg' dilemma.