Why diode with larger band gap has larger onset potential?

I don't understand why voltage increases with band gap. I understand why from the equations, but not conceptually. From my understanding, the onset voltage is what is needed to overcome the contact potential that forms when the p-type and n-type samples are joined together in a p-n junction. But the contact potential doesn't seem to be affected by the band gap, only how easy it is for electron carriers to move from the p side to n side, and for holes to move the other way. So I don't see why band gap should influence onset voltage, and I'd like some insight/clarification for what I'm missing.

I think what you might be looking for is that the semiconductor material in it has a band gap energy. Different semiconductors have higher or lower band gap energies which effect their conductivity.

osilmag said:

I think what you might be looking for is that the semiconductor material in it has a band gap energy. Different semiconductors have higher or lower band gap energies which effect their conductivity.

Hi, thanks for your response! I'm still a bit confused, though. I might be misunderstanding something; I thought current flows through the diode in forward bias when the electrons on the n-side flow to the p-side (and vice versa for holes). Since these are carriers introduced through doping, they don't need to jump the band gap energy to start flowing. They only need to move from their localized level into the conduction band, then jump the contact potential. That's my understanding, and I'm unsure why it's incorrect.

You're welcome. Is the onset voltage the voltage required to form the conduction band (i.e. close the gap)? It's been awhile since I've studied semiconductors specifically.

prox777 said:

TL;DR Summary: I'm confused why onset voltage increases with band gap, conceptually.

I understand why from the equations, but not conceptually

prox777 said:

Since these are carriers introduced through doping, they don't need to jump the band gap energy to start flowing.

It's been a long time for me but iirc, it is the dopants that cause the band gap. There is no band gap across an arbitrary division in the intrinsic semiconductor.

Potential Difference is the same as the energy needed to cross the band gap.The presence of the dopants means that there is not much energy (volts) required for the current to flow through the bulk of the doped semiconductor (low resistance) so the voltage drop is all across the junction.

The equations that you feel that you understand, are derived using the presence of the dopant atoms. So what do you find confusing?

If the potential needed to move charge carriers across the junction were zero, particles would continue to just fall into holes all day long each releasing one band gap worth of energy in its fall. This source of free power is halted because the charge flow forms a charge barrier at the gap which is just high enough to halt current flow. The height of the barrier that does this is just the band gap.

Paul Colby said:

If the potential needed to move charge carriers across the junction were zero, particles would continue to just fall into holes all day long each releasing one band gap worth of energy in its fall. This source of free power is halted because the charge flow forms a charge barrier at the gap which is just high enough to halt current flow. The height of the barrier that does this is just the band gap.

This process only occurs once, of course, and that is when the junction is formed. Work is done when the band gap is formed - just as in chemical reactions.
Pushing charge across a forward biased junction also involves doing work per unit charge (the band gap). This, by definition, is the Voltage. Each coulomb passed takes about 0.7J of energy. There’s the 0.7V drop, almost whatever current is passing, so you can't assign an (Ohms style) resistance.

prox777 said:

Since these are carriers introduced through doping, they don't need to jump the band gap energy to start flowing.

I just spotted this: Could this be your problem? When current flows, new electrons enter from one side and need to be 'raised' to the other potential. This requires WORK; it's not free.