p-n junction question


When various textbooks explain the operation of a p-n junction or a
diode, they usually do not talk about metal contacts. That is usually
a separate topic. But let's consider the whole thing together since
this the only way the diode can operate and let's assume that both
anode and cathode contacts are made from the same metal and therefore
have the same energy level of free electrons.

The energy of an electron flowing through the diode under a forward
bias condition would have to first be elevated from the metal energy
level to the energy level of the conduction band of the N-type
semiconductor, then it would drop when the electron crosses the
junction an falls into a hole in the P-type semiconductor (and
releases thermal or light energy) and then elevated again to the
original metal energy level.

The voltage drop on a diode will be equal to the band gap and the
energy spent to get an electron through the diode will be eaual to the
energy released during the recombination. This energy is needed to
elevate the energy of an electron before it plunges to a hole and,
again, to get it out of a hole back to the metal, but any textbook
would tell you that the voltage drop occurs on the junction itself to
overcome a barrier created there though carrier diffusion before an
external voltage was applied.

As far as I can see there is no barrier to overcome at the junction in
the forward direction - it's a fall, like in a waterfall. But you do
need to overcome a barrier between the metal terminal and the N-type
semiconductor of the cathode to get to that high energy level and
that's where an external energy is spent.

Am I missing something basic here? Does everyone else understand the
story about the depletion region and how a dynamic balance (diffusion
+ drift) existing before an external voltage is applied to a diode
somehow affects the operation of a diode ever after?

.

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