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From what I understand is that, electrons from the N region diffuse into the holes in the P region. In doing so, they form a layer of positive and negative ions and thus an electric field.

When a voltage is applied and the diode is in forward bias, do the free electrons combine with the layer of positive ions and then through to the p region?

If these free electrons are initially bound to an atom, when it leaves it will also produce a positive ion...

What effect does this have on the drift current?

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  • \$\begingroup\$ Not an exact answer, but some of what you are looking for can be found in the first image here. \$\endgroup\$
    – jonk
    Jan 5 '19 at 1:41
  • \$\begingroup\$ In Reverse bias, i dont understand why the holes would move towards the negative terminal of the battery? \$\endgroup\$
    – AskJheeze
    Jan 5 '19 at 3:41
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From what I understand is that, electrons from the N region diffuse into the holes in the P region. In doing so, they form a layer of positive and negative ions and thus an electric field.

This is essentially correct. However, I would phrase it as "leaves behind a layer of positive and negative ions" rather than "forms". Here is a quick recap which will be useful for answering the other parts of your question:

There are two types of current in a diode, diffusion current and drift current. Diffusion current arises when there are different concentrations of carriers. If you imagine bringing a piece of N type silicon into contact with a piece of P type silicon, you will have formed a junction with a whole lot of electrons on one side and a whole lot of holes on the other side. As soon as this junction is formed they will diffuse into the other region. If these carriers had the same charge, or no charge, they would diffuse until they were of equal concentration everywhere.

This brings us to the second type of current, drift current. Drift current arises when charge carriers are in an electric field. At equilibrium a block of semiconductor is charge neutral. There are a number of electrons and holes and their charge is exactly canceled out by ionized donor or acceptor (dopant) ions in the semiconductor. When the electrons and holes initially diffuse into the other side, they leave behind ionized dopants that are not cancelled out by a corresponding electron or hole. This creates an electric field, which causes a drift current.

For a given depletion region width the drift current exactly cancels out the diffusion current and the diode is stable, and no net current flows.

When a voltage is applied and the diode is in forward bias, do the free electrons combine with the layer of positive ions and then through to the p region?

Free electrons and free holes are added to the n and p sides, respectively. This causes the depletion region to shrink and the energy barrier to lower. This lowered energy barrier allows an increase in the diffusion current and leads to a net current through the diode. The density of electrons in the conduction band is an exponential function, which leads to the exponential diode current as a function of voltage.

The depletion region is never completely gone though. You could theoretically apply enough voltage to make this happen, but the current through the diode would be so high that any real diode would burn out before this happened.

If these free electrons are initially bound to an atom, when it leaves it will also produce a positive ion...

They are not. They are injected from the terminals.

What effect does this have on the drift current?

None. The drift current remains constant.

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  • \$\begingroup\$ In Reverse bias, i dont understand why the holes would move towards the negative terminal of the battery? \$\endgroup\$
    – AskJheeze
    Jan 5 '19 at 3:41
  • \$\begingroup\$ @AskJheeze What? Where did this battery come from? I don't think either of us previously mentioned reverse bias at all. Could you clarify your confusion? \$\endgroup\$
    – Matt
    Jan 5 '19 at 3:44
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When you forward bias a PN junction, the external field that you are applying opposes the internal field of the depletion region. You're foprcing more electrons into the N side of the junction and more holes into the P side. Eventually, at around 650 mV, the depletion region is completely filled in with charge carriers and large currents can flow.

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  • \$\begingroup\$ Then where does the exponential I = exp(V) come from? \$\endgroup\$ Jan 5 '19 at 3:21
  • \$\begingroup\$ In Reverse bias, i dont understand why the holes would move towards the negative terminal of the battery? \$\endgroup\$
    – AskJheeze
    Jan 5 '19 at 3:41

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