I was watching this video.

enter image description here

I understand that the dopant atoms introduce huge numbers of free electrons that essentially overshadow any additional electrons that break free due to higher temperatures.

What I don’t understand is why the hole concentration increases. Surely, since we have so many free electrons, any additional holes that are created due to higher temperatures should get instantly filled by the large number of electrons. Hence, shouldn’t P remain constant?


4 Answers 4


point is: nothing is ever instant. Nothing: So, while the number of free charge carriers of poth polarities increases, and so do the recombinations, the number of available ones at every point in time increases just as well.


When temperature increases, the energy gap between conduction band and valence band decreases. Effectively, it will be easier for electron-hole pairs (both to break). There are more free electrons and they might combine with the holes. But in parallel, there will be electron-hole pairs breaking and forming free electrons and holes.

This is dynamic; pairs are being formed and get broken. But to understand, we take the net effect, i.e., 'the number of pairs generated - pairs broken'. So, when more pairs are broken than generated, we say free electrons and holes are produced.

  • \$\begingroup\$ I agree that electron-hole pair generation is easier at high temperatures. But bandgap variation is not the major reason for this. \$\endgroup\$
    – nidhin
    Aug 12, 2021 at 17:23
  • \$\begingroup\$ I really appreciate your review , but then isnt it the increase in thermal energy causing the reduction in energy gap? And in your answer you have explained it w.r.t thermal energy and I did using energy gap? And I felt that was obvious and tried to explain it as simple as possible. And down voting doesnt make sense since its not wrong but its your wish so yeah!!! \$\endgroup\$ Aug 12, 2021 at 20:04
  • \$\begingroup\$ The temperature-dependent intrinsic carrier concentration is \$n_i(T)=K(T)^{3/2}e^{-E_g/(kT)}\$. Where \$K\$ is a material-dependent constant and \$E_g\$ is the bandgap. The point is, even if the energy gap remains the same, the carrier concentration will increase with temperature. \$\endgroup\$
    – nidhin
    Aug 13, 2021 at 19:53

The energy required to break electron free out of atom decreases with temperature. Thus at higher temperatures it's harder for an atom to bound to a free electron.


First, we should understand how the carrier concentration varies with temperature in an intrinsic (pure) semiconductor.

Intrinsic semiconductor

At zero Kelvin (absolute zero), there are no free carriers in the semiconductor. In the 'bond model', this means that all the electrons are bound to the atom, as there is no energy available to break the bond. In the band-diagram, this is equivalent to the condition where all the electrons are residing in the valence band, no free electrons are in the conduction band.

As the temperature increases, the thermal energy increases, a few electrons attain enough energy to jump from the conduction band to the valence band (or to break free from the atom). As the temperature further increases, the thermal energy increases and the free carriers also increase (both electrons and holes, equally). This increase in intrinsic carrier concentration (\$n_i\$) is exponential in nature. (see image below). This increase in electron and hole concentrations happen even in doped semiconductors. (This is what is shown in the question).

enter image description here

Doped semiconductor

Now consider the case when the semiconductor is doped with donor atoms. The dopant atoms contribute additional electrons to the semiconductor to make it n-type. Note that the doping concentration (\$N_D\$) will be so much higher than the intrinsic carrier concentration of the semiconductor, so that the intrinsic concentration is negligible (almost zero when plotted together) at room temperature. The electron concentration will thus be equal to the dopant concentration. \$n=N_D+n_i\approx N_D\$.

The dopant concentration is fixed. But the electron and hole concentration increases with temperature. Above some high temperature value, the value of hole concentration becomes significant and the increase will be visible as shown in the question. Note that both electron and hole concentrations are increasing and at very high temperatures, when the intrinsic carrier concentration becomes very much higher than that of the doping concentration, the semiconductor becomes intrinsic (it's no longer n-type). \$n=N_D+n_i\approx n_i\$, and \$p=n_i\$.

NB: This is one of the reasons why semiconductors have an upper limit on operating temperature.

  • \$\begingroup\$ Oops, apologies @Jonathan_the_seagull, I've since realised my mistake on your edit. Kelvin is not in units of degrees, as you rightly corrected it. \$\endgroup\$
    – TonyM
    Sep 17, 2021 at 10:11

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