Electronic Devices - Semiconductors - Discussion

In case of forward biased pn junction the energy band digram of p type move downward and n type move upward so the gap between both intrinsic fermi level decreased.ie. fi(n)+fi(p)-Vf=Vbi. Thus Vbi decreased.

What happens to the barrier potential when dopping increases?

When diode reverse bias act as open circuit and no free of the electron. So no current in diode.

What happens when the diode is reverse bias?

What is meant by barrier potential?

It does not depend on "band gap" in diode. It depends on the "barrier". Barrier potential depends on amount of depletion region width formed. This depends upon the no. Free charges available for recombination at the junction and no immobile ions formed. So if you have more free charges during the formation of pn junction.

More recombination takes place. So we have more immobile ions at junction. Smaller gap is formed with higher density. As gap is lesser knee voltage decreases. Many have spoken about band gap of "conduction and valence bands".

Mathematically Vk = kT/e ln(NaNd/Ni^2).

With increase in temp 'T' increases but increases in Ni is high. So that Vk decreases.

The built-in voltage decreases with increase in temperature due to exponential increase of intrinsic carrier concentration with temperature.

In fb if tempr is increased the distance b/w vb & cb is decreased due to thermal conductivity.

According to PVT graph, it happens.

As the temperature increases the energy gap between valence band and conduction band decreases and hence barrier potential may also decrease.