Why n*p always equal to ni square? (semiconductor)

In summary, the equation np=ni^2 holds true for both intrinsic and doped semiconductors due to the thermal recombination and generation equilibrium being affected by the excess of mobile electrons introduced through doping. This leads to a decrease in hole generation and explains why p becomes smaller in doped semiconductors.
  • #1
paulzhen
33
0
Why n*p always equal to ni square?? (semiconductor)

Hi,

For you guys who studied semiconductor physics must be familiar with the equation:

np=ni2

I can understand why this is true for the intrinsic case (the broken bonds would always provide electron and hole in pairs )

But why is this still true for dopped semiconductors? Take Si for example, n=p=ni=1010 in intrinsic case (that's we all know). However, if dope 1015 Nd into the material, then n~1015 and p~105. The highlighted part is my confusion! Why p become smaller? Where do the holes go?

Thanks for helping!
 
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  • #2


Dopant brings in extra electrons that quenches the holes?
 
  • #3


rollingstein said:
Dopant brings in extra electrons that quenches the holes?

Don't think so. I have also thought of that but if this is true there must be some relevant theories to explain the "dopping and quenching relationship", or, do the scientists consider this is a neglectable question?
 
  • #4


paulzhen,

Don't think so. I have also thought of that but if this is true there must be some relevant theories to explain the "dopping and quenching relationship", or, do the scientists consider this is a neglectable question?

You are correct in being suspicious of that explanation. See the attachment for a mathematical explanation.

Ratch
 

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  • #5


Ratch said:
paulzhen,
You are correct in being suspicious of that explanation. See the attachment for a mathematical explanation.

Ratch

Thanks a lot Ratch. Actually I saw a similar maths explanation before, you both use the fact of neutrality to form the equation.
But could you tell me where do the holes go? Are they being eliminate by extra electrons? This is also the only physical explanation i can imagine. But since Nd>>p, the explanation of eliminating seems ... just not right, cause the eliminating efficiency will be too small to be (felt) right.
Please enlighten me
 
  • #6


paulzhen,

But could you tell me where do the holes go? Are they being eliminate by extra electrons? This is also the only physical explanation i can imagine. But since Nd>>p, the explanation of eliminating seems ... just not right, cause the eliminating efficiency will be too small to be (felt) right.
Please enlighten me

In intrinsic semiconductor material, the electron and hole concentration is caused by thermal recombination and generation (R-G). At room temperature, the balance for silicon is Nd = Na =10^10. If more mobile electrons are added to the semiconductor by doping, then the R-G equilibrium changes, and does not need to generate as many electrons thermally to support the equation np=ni^2. Since thermally generated charge carriers come in pairs, then the hole generation is also reduced. So the holes don't go anywhere, it is just that fewer of them are generated by the R-G thermal process, which has slowed down due to the excess of electrons. That is what those equations mean, and it the basis for their derivation. Ratch
 
  • #7


Ratch said:
paulzhen,



In intrinsic semiconductor material, the electron and hole concentration is caused by thermal recombination and generation (R-G). At room temperature, the balance for silicon is Nd = Na =10^10. If more mobile electrons are added to the semiconductor by doping, then the R-G equilibrium changes, and does not need to generate as many electrons thermally to support the equation np=ni^2. Since thermally generated charge carriers come in pairs, then the hole generation is also reduced. So the holes don't go anywhere, it is just that fewer of them are generated by the R-G thermal process, which has slowed down due to the excess of electrons. That is what those equations mean, and it the basis for their derivation. Ratch

I think I got what you mean, thanks!
 

1. Why is n*p always equal to ni square in semiconductors?

The product of electron concentration (n) and hole concentration (p) in a semiconductor is always equal to the square of the intrinsic carrier concentration (ni) due to the law of mass action. This law states that at thermal equilibrium, the product of the majority carrier concentration and minority carrier concentration is constant. In an intrinsic semiconductor, the majority and minority carrier concentrations are both equal to ni, hence the product n*p is equal to ni square.

2. How is the value of ni determined in semiconductors?

The value of ni (intrinsic carrier concentration) is determined by the material's bandgap energy, temperature, and effective density of states in the conduction and valence bands. It can also be calculated using the expression ni = sqrt(Nc*Nv*e^(-Eg/2kT)), where Nc and Nv are the effective densities of states in the conduction and valence bands, Eg is the bandgap energy, k is the Boltzmann constant, and T is the temperature in Kelvin.

3. Can the value of n*p be greater than ni square in semiconductors?

No, the product of n and p in a semiconductor can never be greater than ni square at thermal equilibrium. This is because any excess carriers generated in the material will quickly recombine until the product n*p returns to its equilibrium value of ni square.

4. Why is n*p equal to ni square not always observed in semiconductors?

The condition n*p = ni square is valid only at thermal equilibrium, where the Fermi level of the material is constant. In real-world applications, semiconductors are often subject to external influences such as electrical bias or light irradiation, which can change the Fermi level and therefore affect the value of n*p.

5. How does the value of n*p affect the conductivity of a semiconductor?

The product of n and p in a semiconductor is directly related to its electrical conductivity. At thermal equilibrium, where n*p = ni square, the conductivity is at its lowest point due to the balance between electron and hole concentrations. If n*p is increased, either by changing the temperature or introducing impurities, the conductivity of the material will also increase due to the higher carrier concentrations.

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