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I am learning about intrinsic semiconductors, and have just come across the intrinsic carrier concentration equation -

$n_{i}=BT^{3/2}e^{\cfrac{-E_{g}}{2kT}}$

Which is supposed to give the number of conduction band carriers per cubic centimeter. Is this value an average? Surely the number of carriers in the conduction band cannot be the same everywhere in each $cm^3$ within the semiconductor at any given time?

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The formula you mention for $n_i$ is empirically supported (i.e., it is an empirical fit to experimental data for a solid material) and is indeed an average in a cubic centimeter of material. $E_g$ is the band gap energy for the material of interest and B is an empirically determined constant. Thus, normally, with access to a table of such constants and band gap energies, you can determine the intrinsic carrier concentration for a given temperature of interest.

$n_i$ plays a significant role in the design of semiconductors devices because of the relationship, $n_i^2 = np$ (law of mass action) which holds in doped and undoped material. This is important for, among other reasons, doping profiles and for understanding how a material acts a different temepratures. However, given that most semiconductor components are much much smaller than a $cm^3$ in volume (state of the art transistors have a linewidth of $14-22 nm$ and the junction depth is a few hundred angstrom), you can see that the of scale for $n_i$ is rather meaningless to an individual component and an average would be necessary anyway.

One last point, the reason why the equilibrium relation, $n_i^2 = np$, holds true is because in a crystal structure, the generation of hole-electron pairs and recombination is constantly occurring in the crystal because of thermal agitation in the material. Take the equation for $n_i$ and let the temperature go to absolute zero, you find that the intrinsic carrier concentration also becomes 0. The meaning of the temperature increasing is an increase in kinetic energy/vibration of atoms or equivalently an average increase in the internal energy in the crystal. This is because temperature is a function of the internal energy (U), $T = (\frac{\partial{S}}{\partial{U}})^{-1} = (\frac{Nk_b}{U})^{-1}$. So, there are more opportunities for bonds in crystal to break and reform.

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  • $\begingroup$ So, if for some reason I had a semiconductor that was 4 $cm^3$, then the average number of carriers per cubic centimeter at any given time would be given by that equation? Also, in reality, you mean that its really just used as a measure of how certain temperatures affect the carrier density of an intrinsic semiconductor of an arbitrary size? $\endgroup$
    – Ben Granger
    Commented Jun 4, 2016 at 4:42
  • $\begingroup$ @BenGranger, I updated my answer to address your comment about the temperature. You are right that the above equation would give you the number of carriers per cubic centimeter at a given temperature for an intrinsic material. But remember that the equation only holds within a certain range because you could theoretically increase the temperature until the material transforms into a gas or plasma and then the experimentally data the B constant is based on and $E_g$ are no longer valid. $\endgroup$
    – ansebbian0
    Commented Jun 4, 2016 at 5:28
  • $\begingroup$ Awesome, so the only reason its used is to give us a quantitative value/idea describing the carrier density in an intrinsic semiconductor of arbitrary size? e.g. if we have two $14nm$ transistors, the one with a higher $n_i$ will have more charge carriers than the other at some temperature $T$, even though they are both much smaller than $1cm^3$? $\endgroup$
    – Ben Granger
    Commented Jun 4, 2016 at 5:35
  • $\begingroup$ Actually, the key point here is that it helps you to understand how many charge carriers are available to you in a given material at a given temperature for an arbitrary volume (so you can make comparisons). This is important because to make useful semiconductor devices which work at a given $T$, you need to have enough charge carriers in the material to allow a current to flow at a reasonable threshold voltage, $V_{th}$. So, you use $n_i^2=np$ to determine how much dopant (boron, phosphorus, etc) you need to implant. Again, though, this is just one use of this formula. $\endgroup$
    – ansebbian0
    Commented Jun 4, 2016 at 5:56
  • $\begingroup$ Ah, ok, so you end up manipulating the $cm^3$ dimension to fit the dimensions of your device? $\endgroup$
    – Ben Granger
    Commented Jun 4, 2016 at 5:57

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