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this question is partially (but not totally) discussed here:

Why do diodes have a voltage drop?

Does the power drop on a diode become entirely heat?

The conclusion is that, in a forward-biased diode with steady DC current:

  1. the voltage drop represents the necessary energy (per unit charge) to push charge carries (electrons and holes) from one semiconductor to the other.

  2. the power drop on a forward-biased diode is entirely heat.

Let's focus more in detail on both statements. Let's consider the following band diagram in case of forward biasing:

enter image description here

I'd say that:

  • when the diode is forward biased, holes are pushed from P to N and electrons from N to P, correct?

  • both electrons and holes' levels change along the junction, since they correspond respectively to the conduction and valence bands. Their jump is the same since both levels are represented by parallel curves. Is the voltage drop of the diode equal to this jump, to twice this jump or other?

  • what do the charge carriers injected in each semiconductor do after they have passed the discontinuity? Do they recombine and produce heat? Do they pass undisturbed and continue flowing as electric current? In the first stack question it's written that there is a loss of carriers. How can the KCL law be true if some carriers disappear at the discontinuity?

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when the diode is forward biased, holes are pushed from P to N and electrons from N to P, correct?

Yes. [However, "forward biased", really only means that the P side is more positive than the N side. When the forward bias is very small, "leakage" current may predominate. Leakage current is caused by minority carriers. Although majority carriers are (usually) more common (hence "majority") they have a harder time crossing a PN junction than minority carriers. So, when the bias is very low, the few minority carriers that easily cross the junction may form a larger current than the many carriers that have a hard time crossing the junction.]

what do the charge carriers injected in each semiconductor do after they have passed the discontinuity?

They become minority carriers. There are multiple fates for such minority carriers. They may recombine with majority carriers, they may diffuse back across the junction, they may survive and reach the metal-semiconductor junction.

Do they recombine and produce heat?

Whatever their fate, they will eventually interact with the crystal lattice, with other carriers, with metal, etc, and their trajectory will become randomized. That is they will produce heat. However, the electric field will cause those carriers or new carriers to accelerate, thus maintaining the current.

How can the KCL law be true if some carriers disappear at the discontinuity?

When an electron and a hole recombine (at the discontinuity or elsewhere) there is electron current in one direction and conventional current in the other. So algebraicly, current is conserved.

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  • \$\begingroup\$ Thank you. All clear except for the last point. Isn't in general the hole current contribution added to that of electron (conventional) current? \$\endgroup\$
    – Kinka-Byo
    Apr 13, 2021 at 14:34
  • \$\begingroup\$ the holes traveling in one direction create a conventional current. The elections traveling in the opposite direction create an equal conventional current. They "annihilate" each other at some point, but the (conventional) current flowing into the anode is equal to the (conventional) current flow out of the cathode. \$\endgroup\$ Apr 13, 2021 at 14:42
  • \$\begingroup\$ Keep in mind that a "hole" is not a physical entity, like an electron (ignoring quantum effect). A hole is the lack of an electron in the lattice. If an electron moves from A to B in the lattice, the hole moves from B to A. \$\endgroup\$
    – SteveSh
    Apr 13, 2021 at 15:08
  • \$\begingroup\$ @SteveSh Nope, not true. Holes are more real than you describe. With the holes as "just" missing electrons model they wouldnt be distinguishable from electrons with the Hall effect. In fact electrons and holes within the same band move in the same direction under applied fields, not opposite directions, which is quite counterintuative, like a lot of things involving semiconductors. \$\endgroup\$
    – Matt
    Apr 13, 2021 at 19:22
  • \$\begingroup\$ @Matt- Guess I'm going to have to go back and review my semiconductor physics <wink>. \$\endgroup\$
    – SteveSh
    Apr 13, 2021 at 19:57

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