Why Does N-Type Doping Result in Higher Hole than Electron Concentration?

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In the discussion, a user presents a problem involving an n-type doped semiconductor with a bandgap of 0.1 eV and specific densities of states. They initially calculate the electron and hole concentrations but find an unexpected result where hole concentration exceeds electron concentration. Other participants clarify that the user likely mixed up the doping concentrations, suggesting that under the given conditions, the electron concentration should indeed be greater than the hole concentration. The conversation emphasizes the importance of correctly applying the equations related to semiconductor doping. Ultimately, the user acknowledges the error in their calculations.
heehar
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Homework Statement


Suppose an unknown semiconductor material has bandgap energy of .1 eV. Density of states for conduction and valence bands are both = 10^18cm^-3.
If the material is doped with shallow donors (n-type) Nd=10^17cm^-3, what is electron and hole concentrations?


Homework Equations


ni=sqrt(Nc*Nv)*exp(-Eg/2kT)
n+Na=p+Nd
p=ni^2/n

The Attempt at a Solution


so I set Na = 0. and solve the quadratic equation but I end up with a hole concentration greater than an electron concentration...why is that the case? isn't the fact that its doped n-type mean the n concentration is supposed to be greater?
 
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You did the math wrong. With those conditions, I get n>p, with n~2*p. Did you mix up Na and Nd?
 
you are right haha thanks
 

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