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I wanted to know the relationship between the band gap (Eg) of the p/n regions in a diode and the built in potential in equilibrium.

My intuition says that Eg = e*Vo. I did a small calculation for silicon at T=300K. I got the relation to be approximately correct.

Looking at the band gap diagram, the valence band of the p side seems to coincide with the conduction band of the n side in equilibrium.

Is my intuition correct? What is the actual relationship?

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No --- the built-in potential is equal to the difference of the Fermi levels in bulk semiconductor N and P, so for a given material it basically depends on the doping of the P and N zones. See the figure in In band diagram, why the Fermi energy (EF) is constant along the device?

Think about it --- if you were right, all the diodes of a given material would have the same built-in potential.

So the answer is that (using the simply common model of the diode):

$$ V_0 = \frac{k_BT}{q_e} log (\frac{N_AN_D}{n_i^2}) $$

with the usual meaning of the symbols. Notice though that in the term \$n_i^2\$ lie a dependency to the band-gap of the material, too.

(And be sure to not mistake the built-in potential with the threshold voltage, they are completely different beasts).

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  • \$\begingroup\$ These carrier densities are related to the band gap height. But the answer doesn’t elucidate it. \$\endgroup\$ – Incnis Mrsi Sep 3 '16 at 21:44
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from what I learnt, in a simple diode that has only a pair of pn junction, i do believe that based on equation, "the higher the Eg, the higher also the built-in potential"....below is the band diagram to give a little bit visualization

energy band diagram

energy band diagram with no bias voltage applied

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