Si has an intrinsic concentration of 1.08·1010. What happens when it is doped with a concentration less than the intrinsic concentration?
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\$\begingroup\$ I'm guessing, much the same sort of things that happen at pressures less than perfect vacuum or temperatures below 0K. \$\endgroup\$– user16324Commented Sep 28, 2019 at 14:06
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1 Answer
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TL;DR: band structure basically stays the same since your dopant doesn't increase the electron/hole density enough.
Let's take an example: let's say \$n_i = 10^{-10}\$ and you have initially an intrinsic semiconductor. Then \$n_e = n_i = 10^{-10}\$.
Let's say you dope N-type with density \$N_d = 10^{-11}\$. Assuming full ionization of the dopant, you get \$n_e = n_i + N_d\$ (approximately, I believe there's an equation for this).
This means that you have \$n_e = 1.1 \times 10^{-10} \$ and \$n_h = n_i^2 / n_e = 0.909 \times 10^{-10} \$.
Basically the concentrations don't change enough to matter. This is why when you dope you dope much much greater than \$n_i\$ since if you don't dope enough nothing really happens.
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\$\begingroup\$ Did you write \$10^{-10}\$ instead of \$10^{10}\$? \$\endgroup\$– nidhinCommented Oct 6, 2019 at 8:42
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