Reverse current

Question

I'm getting started with electronics and couldn't help but wonder about this. When the PN junction diode is reverse biased, P connected to the negative terminal and N connected to the positive terminal, shouldn't there be a large reverse current as the electrons (which are essentially the equivalent of minority charge carriers in the P region) get injected into the P region through the conductor? The depletion region can't really hinder the movement of electrons from P to N region as the field itself is from N region to the P region. Considering the very high concentration of free electrons in a conductor, shouldn't there be a huge reverse current?

Please do correct me if any of my assumptions are wrong and it would be really helpful if someone can thoroughly explain the actual concept behind this.

Answer 1

The metal making an "ohmic contact" to the p-region of the pn-diode (and analogously the metal contacting the n-region) is in (quasi-)equilibrium with the semiconductor and thus has the same (quasi-) Fermi level. Therefore, except for a very thin transition layer, the electron concentration in the semiconductor p-region is not influenced by the metal contact and remains at the very low (minority carrier) density determined by the p-doping. Depending on the semiconductor, the (usually) very low reverse current of the diode is mainly due to the diffusion of minority carriers at the depletion zone edges, or the electron-hole generation in the reverse biased depletion zone.