# Confused on relation between bandgaps and pn-junction operation

I am an EE not a physicist but I wanted to understand semiconductor operations a little better.

I have looked at a couple books on semiconductor physics and I still don't understand the relation between bandgaps in semiconductors and the operations of a PN junction. I am hoping someone here could provide an explanation linking the two together and why a diode only conducts when forward biased.

• This is a very broad question, usually taking several chapters in a semiconductor physics book. Do you have a more finite area of concern that could be addressed? Mar 2, 2017 at 21:55
• Ok, in other words why is drift current low? When a junction is reverse-biased, why don't electrons just travel across the P region? How does bandgap relate to this issue? Mar 3, 2017 at 1:46
• In the depletion layer there are no free carriers (which is why you can sustain the built-in voltage across it). (Well, you do have carrier generation in the depletion layer, which leads to the observed reverse current). The bandgap (relative to temperature) influences the carrier generation rate, and hence the reverse current. The higher the bandage, the lower the carrier generation rate which is why wide band gap semiconductors are desirable for voltage stand off. Mar 3, 2017 at 1:52
• Ok thank you, I guess the next question is - why exactly can't electrons move freely in P doped material though? Why is it only holes can move? Thank you. Mar 3, 2017 at 3:00
• Electrons can move in p-type material. There just aren't many of them since they are the minority carriers there. Just like there are holes in n-type material - again, just not that many. Remember, $np = n_{i}^{2}$ so the majority carrier density will be orders of magnitude higher than the minority carrier density in a doped region. Mar 3, 2017 at 3:04

The band gap energy is related to the maximum built-in voltage corresponding to the maximum band bending in an abrupt pn-junction: $$V_{bi}≲ \frac{E_g}{q}$$ This can be easily seen by drawing a band diagram in equilibrium. In the n-type region, the Fermi level is close to the conduction band edge and in the p-type region the Fermi level it is close to the valence band edge. The higher the doping, the closer the Fermi level comes to the respective band edge. The (ideal) saturation current of a diode decreases exponentially with the band gap energy $$J_s∝n_i^2∝\exp{(-\frac{E_g}{kT})}$$ For the dominant generation current in a Si diode it decreases as $$J_s∝n_i∝\exp{(-\frac{E_g}{2kT})}$$ thus the band gap energy determines the magnitude of the pn-junction current. Under forward bias the current increases exponentially because the applied voltage decreases the built-in potential barrier $V_{bi}$.