Few basic questions about p-n junction

ViolentCorpse

Hello everyone. I have a couple of questions about the depletion region in the p-n junction

1) Why don't the negative ions on p-side of the depletion region donate their respective extra electrons to the positive ions on the n-side of the region so they can both be happy (This sounds kind of weird, because those ions exist due to the very same process that I'm now asking to reverse :p ) ?

2) After the application of a forward voltage, the width of the depletion region is reduced due to electrons falling in the +ve ions on the n-side, and holes falling in the -ve ions on the p-side, right? But those electrons and holes didn't exist independently, they belonged to their parent atoms that they've now left a net charge on to become a part of these new atoms. So in effect, those ions haven't been neutralized, only displaced to the right in n-side and to the left in the p-side. But this is not what theory says, so what am I missing here?

Thank you for your time!

Simon Bridge

1. because it is energetically unfavorable due to the doping
2. a positive bias is usually accompanied by some more charge carriers from the wire used to connect the voltage - otherwise, yes, you'll just get a displacement of the charge carriers.

ViolentCorpse

But wouldn't that again leave the connecting wires with a net charge? It seems to me that one element or another would have to sacrifice its neutrality for filling the deficiency of the charged atoms (ions), unless carriers are being injected into the device from some external source and as far as I know, a battery merely "pressures" the carriers forward not inject them into the circuit.

Ratch

VC,

The p-type does not have "extra" electrons. The p-type impurity atoms accepted those electrons which diffused from the n-side, and became bound negative ions in order to fill out their outer valance orbital. Then they were able to fully bond covalently with their neighboring silicon or germanium atoms. The n-type had too many electrons, so those electrons filled the holes which diffused from the p-type material. Whether there are too many or too little holes or electrons is determined by whether full covalent bonds can be formed. Those electrons from the n-side and holes from the p-side moved into their neighbors regions due to diffusion. The n-type material wants to give away its extra electrons and the p-type material wants more electrons. Why would the process reverse? Does a pot of cold water become hot without an outside infusion of energy?

The voltage source supplies electrons to the n-type material, and supplies holes the p-type material by removing electrons, so I don't know why you see a problem. What does the theory say?

The battery pumps electrons and holes into the the semiconductor by moving electrons. Otherwise, there would not be a steady current.

Ratch

Simon Bridge

What he said :D
If there were no power supply then what you say would be correct: for something to gain charge, something else has to lose it.
However, the "bias" usually means the component is part of an electric circuit of some kind. That means a PSU somewhere.
With a PSU, there is a supply of charge to keep all the components happy.

ViolentCorpse

This makes perfect sense. I just had this misconception that it is unfavorable for atoms to gain a net charge, but you have cleared that up very well.

But then, if it's their covalent bonds that those atoms want to complete, why does part of the layer annihilate after a forward bias is applied, because that would disrupt their bonds again?


I'm so very sorry but I don't think I still quite get this point. Here's what I think the function of a power supply is:

a) It keeps the already existing electrons in the components flowing.
b) It does not generate/inject new electrons into the components.

So if the negative terminal of the supply "pressures" a free electron in one of the n-type atoms and that electron falls in one of the positive ions near the junction layer, that would leave the former atom with a net charge, which might be neutralized by another electron coming from behind it, but that would leave that atom of THAT electron with a net charge.

I think I'm going in circles here..

Thanks a lot for your responses, Simon Bridge and Ratch. I really appreciate it!

Ratch

VC,

Are you asking about the depletion region? Which exists because the mobile carrier concentration is below their thermal equilibrium values? Without forward voltage, bound ions in the p-type material repel the electrons trying to come over from the n-type material and vice versa. This sets up a barrier voltage and the depletion region. This barrier voltage lowers the mobile carrier transport. Diffusion stops when the barrier voltage equals the diffusion voltage. A forward voltage counteracts and lowers the barrier voltage, and allows a steady current to exist. This also causes the depletion region to narrow because fewer ions are involved in producing the lower barrier voltage. The current is supplied by the voltage source.

a) Without a PS, there would be no steady flow of electrons or holes.
b) The PS is in the series circuit, so any steady current in the semiconductor has to go through the PS also.

By "pressuring", do you mean the PS supplies electrons to the n-type material? And removes electrons from the p-type material to make holes? In the n-type material, the positive ions do not want a free electron from the PS. They already have enough electrons to fully bond covalently, and they just gave away an electron so they are now positive ions. That free electron is going to want to go over to the p-type material and fill in a hole to make a negative ion.

Ratch

ViolentCorpse

The mechanism behind the underlined section is exactly what I'm inquiring about. How is the number of ions going to be reduced if those ions do not want a free electron/hole from the PS?

By '"pressuring" I mean it pumps the electrons from the n-type material into the p-type and back into the n-type in a loop. I don't think the PS needs to remove electrons from the p-type material to make holes, because a p-type material already has an abundance of holes. The only thing required is to move them

Ratch

VC,

It depends on which side and which charge carriers you are talking about . Let's take the p-side first. The PS is pumping electrons out of the p-side. It has to do that in order to sustain a current. This action tries to reduce the p-side negative ion count, and also tries to increase the p-side hole count in order to supply electrons to the PS. The p-side can only get electrons from the n-side. There is less diffusion of electrons from the n-side donor atoms because more free electrons are being being pumped into the negative side by the PS. These free electrons from the n-side and supplied by the PS can satisfy what the p-side needs with less diffusion taking place from the donor atoms than before when no PS voltage was present.

Less diffusion means a thinner depletion region.

The same thing happens in reverse for the n-side. The n-side needs more holes, and the PS can supply them from the p-side with less diffusion taking place from the p-side acceptor atoms.

Look at it this way. If an electron from a donor atom in the n-side goes over to the p-side, a positive ion is left behind in the n-side. If a free electron goes over to the p-side, a hole has effectively travelled over from the p-side and no positive ion is formed in the n-side.

As I said before, electrons are removed from the p-side to sustain a current.

Ratch

Simon Bridge

It does not need to - however, that is physically what it does. But it certainly does need to if a current is to be sustained.You could do that by just applying a static electric field across the component can't you? There is a difference between that and having a power supply in the circuit.

The action of a power supply is to remove electrons from one side and add them to the other. This happens no matter what the doping and applies to insulators and conductors as well as semi-conductors.

If there were no mechanism to get charge-carriers from one side to the other side, then, as you have intuited, a separated charge distribution would build up against the applied voltage and the current would eventually stop - as in a RC circuit. Since this does not happen, it follows that there is some mechanism for getting the charge-carriers from one side to the other.

ViolentCorpse

Oh, I think I'm getting it now. The PS reduces the p-side negative ions, and the electrons that made those ions are supplied by the negative terminal of the PS to the n-side positive ions, reducing the depletion region. Is this correct?

Ratch,

If an n-side free electron goes over to the p-side, is it correct to say that a hole has effectively crossed over to the n-side? I mean n-side most n-side atoms are donors, so even if they lose an electron, they'd still have their bonds complete. How can they accommodate a hole?

Thank you so much for your continuing help Ratch and Simon Bridge! I really appreciate it.

Ratch

VC,

Yes, by removing some of the electrons. Other electrons will come from the n-side to partially replenish the removed electrons. This causes a current.

Not quite. Without any PS connection, ions were formed at the junction barrier just by diffusion. The diffusion stopped when the back-voltage set up by the ions was in equilibrium with the diffusion voltage. The diffusion continued again when the PS lowered the back-voltage. This set up a new equilibrium with fewer ions. Fewer ions also means a reduced depletion region. This PS voltage also causes a current.

Yes, the n-side has lost a free electron and the p-side gained an electron. That is equivalent to a gain of a hole for the n-side and a loss of a hole for the p-side.

That is not true. Crystalline silicon has a density 5E23 atoms per cm. A typical concentration of dopant is 1E14 atoms per cm. So the dopant atoms are scattered very, very sparsely.

Losing an electron is equivalent to gaining a hole. So the question is really how can the n-side lose a electron? The answer is by diffusion and transport across the junction barrier into the p-type region.

Ratch

ViolentCorpse

Hm I see. I think I get it now.

Thank you so much Ratch and Simon Bridge!