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Lets say the diode is in reverse bias. Then the holes in the $p$-type will go away from the depletion layer, and the electrons from the $n$-type will go away from the depletion layer. As I understand in the $p$-type, the depletion layer we will have negative ions, and in the $n$-type, the depletion layer we will have positive ions.

My quesion:
Why can't the extra electron in the negative ion in the depletion layer in the $p$-type go over to a positive ion in the $n$-type, and hence we get a current?

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  • $\begingroup$ The negative(positive) ion in the depletion layer are made of the lack of hole(electron), not the extra of electron(hole). Remember, the neutral part of p-type(n-type) is actually where silicon atoms in the lattice is substituted by atoms from III(V) which are neutral atoms. $\endgroup$
    – Codelearner777
    Commented Mar 1, 2021 at 19:11

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The electron of negative ion in p-type semiconductor is in valence band which means it constitutes covalent bond. So it can't move to other atom freely. It must go up to conduction band of p-type to become "free electron" first. And this takes energy which is not available "on average". So there is no current from p-type to n-type macroscopically.

enter image description here

This is the situation. Reverse bias, p-type has no hole and negatively charged, and n-type has no "free electron" and positively charged.

The purple arrow (2) is the what you are asking. And this is not happening because the electron is in covalent bond. (it can actually happen in quantum effect, but the depletion area have to be thin enough and this is not the general situation what you are wondering about)

For electron to move, it have to take route of up arrow (1) first. Overcoming this band gap is like the current of intrinsic semiconductor(pure, no doping) which is not able in low temperature.

The electrical potential is even favorable to the current(p to n) but it's stuck at the first step.

In analogy, it's like a chemical reaction whose reactant is higher energy than product, but the activation energy is too high for the reaction to happen.

PS.

On the other hand, the hole can move, though it is in valence band.

For charge to move, it should have a room to move which means it's not octet. And the hole is the room. But the electron in your question have no more room in valence band and it have to go up to conduction band.

If you see periodic table, you will notice the metals are in middle, which means it's hard to make covalent bond to get octet, so there are free electron(conduction band).

(Non metal can easily have octet because they have to get or remove "few electrons")

Edit for comment by oliver

enter image description here

As you see in this graph, the conductivity of intrinsic semi-conductor is very low in room temperature. The x-axis start from 700K which is much higher than room temperature.

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    $\begingroup$ +1, but the OP might ask what happens at normal temperatures T>>0 when the energy to jump the bandgap is available, if not "on average" then at least statistically in the Fermi distribution. The answer is that there is always a small reverse leakage current due to this effect and it increases exponentially with temperature. One further aside: jumping the bandgap in reverse bias operation due to an external photon is even the desired working principle of a photodiode. $\endgroup$
    – oliver
    Commented Aug 25, 2021 at 14:11
  • $\begingroup$ Because my knowledge is high school level, I don't know what exactly the Fermi distribution is and how it's exponential increase. But I learned that in reverse bias, there is no current. Considering what I learned and your comment, is it true only when the temperature is low that there is no current in reverse bias? Or, there is no current also in room temp and my answer is wrong? $\endgroup$
    – Moohyun Roh
    Commented Aug 25, 2021 at 14:45
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    $\begingroup$ There is nothing wrong with your answer. The Fermi distribution describes how Fermions (i.e. electrons) "evaporate" across the bandgap for increasing temperature. Indeed, only at T=0K the reverse bias current would be exactly zero according to the usual semiconductor model. At room temperature, the current is small but nonzero. At yet much higher temperatures, the reverse bias current can become as big as to compete with the forward current. $\endgroup$
    – oliver
    Commented Aug 29, 2021 at 14:04
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When a PN junction is formed, because of the excess electrons in the N-types conduction band and the excess holes in the P-type valence band, they form a depletion region as a result of diffusion currents and recombination mainly.

The result of this depletion region is an Electric field at the junction which opposes P to N current; this is because in the N-type we have positive charges, and in the P-type we have negative charges. This is clear in the band diagram:

enter image description here

When a reverse bias voltage is applied, it only adds to the 'built in voltage' you can see in the picture, and the Electric field opposing current gets even stronger, which is why it acts as an insulator.

However in real diodes there is most often a breakdown, referred to as Avalanche Breakdown, or Zener Effect (these two are technically different but normally always happen at the same time), which is when the the strong enough electric fields enable passage of minority carriers across the depletion region of a semiconductor, leading to numerous free charges. This generation of carriers rapidly increases the reverse current and gives rise to the high slope conductance.

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  • $\begingroup$ Sorry, but I don't see you answering what I asked about. $\endgroup$
    – user394334
    Commented Jan 23, 2021 at 10:23
  • $\begingroup$ The missing bit, that could be added to the explanation, is that an electron needs energy $E = e\Delta V$ in order to "climb" the voltage barrier. If the only available energy is thermal, and $\Delta V$ is sufficiently big, no electrons will have enough energy to make it across. $\endgroup$
    – Paul T.
    Commented Aug 25, 2021 at 13:50
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Why can't the extra electron in the negative ion in the depletion layer in the p-type go over to a positive ion in the n-type, and hence we get a current?

Those charges are ionized dopants, either donor or acceptor atoms. They became ionized when they added their corresponding electron or hole to the semiconductor. These ions are fixed in their locations in the semiconductor and have no way to move around and contribute to current as you describe.

If you wanted to, for example, take an electron from an un-ionized donor and move it to an ionized donor, you would have to first find an un-ionized donor and then give it enough energy to free its electron. There aren't very many un-ionized dopants and there isn't energy available to do this. Therefore, this doesn't happen very often and there won't be any significant current because of this.

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