Hi Naseri,

Since the system only single junction (N-type semiconducto) not P-N junction system. Is it important to see EFp? The junction should be Schotky or Ohmic right? I am also curious why we need to see EFp in this juntion. Could you share any articles which discuss about this study? I am very appropriate if you could share it.

Thank you...

Some text books also fine if you have it and share to me.

Thank you.

Cheers~

Hi Randi,

It is an ohmic contact in this case. I think it is important to be able to draw the band diagram of the semiconductor. Efp is important to draw as clearly we are out of equilibrium. The usefulness of drawing the minority carrier band is subject to application, I guess... Solid State electronic devices by Streetman is ok, Also Fundamentals of semiconductor devices by B Anderson.

Thank you for the response Mrs Naseri,

I see... It should be interesting to study it then. Do you have any related reference to study this topic?

apart from the above two books, Sze is also good. Also look at Chenming Hu lecture notes from UCLA, in general https://people.eecs.berkeley.edu/~hu/Chenming-Hu_ch4.pdf

Thank you,

But I am looking for single junction study N-type only or P-type only which explain about both EFn and EFp in your case. Do you have it?

I have not come across anything that exactly deals with this case. But in any of those books look under non-equilibrium for a semiconductor under illumination.

I see... Okay thank you.

Hi Naseri,

I have tried to look and study about this kind of issue. I think if this N-type junction only, we just care about Efn more. Since the majority carrier is electron. The hole will not much affect on this condition, so we can ignore Efp line. This much I can get so far. If you have any better explantion, please share to me as well.

thank you.

Best!

Hi Randi,

I agree with you. That's why I have confidently drawn EFn it on the above picture. EFn is determining the general Fermi level as they are least perturbed by the low level injection. But minority carriers are very important in terms of characterising the semiconductor responses to the perturbation (such as even low injection) as EFp goes under a huge change, due to being so small to begin with. That's why I am curious to draw EFp, as well.

Hi

Have you tried to characterize the minority carrier? For example mobility or other physical behind it?

Best!

As long as the law of mass action holds (ne * nh = ni2) there is no need for separate Fermi levels. As far as I know, a stationary current flow doesn't change the carrier densities, so as long as the semiconductor remains in the dark, there is just a single Fermi level, being relevant both for majorities and minorities. However, as one starts considering illumination, one needs separate quasi-Fermi levels. So, what situation do you have in mind -- with or without illumination?

Dear Pourandokht,

In order to draw the quasi Fermi levels for electrons or holes you have to refer to their definition where n= Nc e^ -(Ec- Efn)/ kT where Efn is the quasi fermi level of the material. After disturbance n = n0 + delta n,

Likewise for the holes p= Nve^-( Efp- Ev)/kT, where Efp is the quasi Fermi level of holes. For a disturbed material p= p0+ delta p,

Normally because of quasi neutrality deltan = delta p.

When delta p = delta n =0 which is the equilibrium case n0p0= ni^2

At disturbance np= ni^2 e^ (Efn- Efp)/kT,

So,the splitting between Efn and Efp depends on the disturbance delta n and delta p. So if you know Efn you can determine Efp if np is known.

In your problem as you have ohmic contact and the material is not illuminated with light that can generate electron hole pairs then the carrier concentrations in the materials will not be disturbed and n=n0, p=n0,

So the quasi Fermi level of the holes will coincide the coincide with that of the quasi Fermi levels with electrons.

So, i believe that your question is fully answered.

Best wishes

Dear Jan-Martin and Abdelhalim,

Thank you for your replies. Are you suggesting that if the injection is by external voltage we are STILL at equilibrium, despite of having a net flow of current? I believe that although we are in steady state we have minority carrier injection (delta_p=delta_n) and hence out of eqilibrium and Efn and EFp MUST split.

EFp goes under a huge change, due to being so small to begin with, but Efn doesn't. So using np= ni^2 e^ (Efn- Efp)/kT, will give you back Efp, which doesn't make sense to me.

Dear Jan-Martin,

There is no illumination, here. But what is the difference? My understanding is that at non-equilibrium (when we have net current such as by applying a voltage or illumination) the mass action law doesn't hold. Hence we need to work with the quasi fermi levels.

Let's have a microscopic look at the situation: What happens in the bulk of the semiconductor, and what happens at the contacts?

Let's look at the bulk first: What is the difference in the behaviour of the carriers with and without the current flow? Current flow means the application of an electric field (according to Ohm's law: j = sigma * E). Without applied electric field, all quasi-free carriers move at a speed according to thermal equilibrium. With the field applied, the carriers gain the drift velocity (according to vD = mu * E), which is much, much smaller than the equilibrium speed. So, a current flow leads to very little deviation from equilibrium in the bulk.

What happens at the contacts? Obviously, for a semiconductor in the dark, the contacts are the only place where injection of whatever kind of charge carrier may occur. Without current flow, there is equilibrium between the n-type semiconductor and the metal of the wire. As a consequence, the Fermi level is the same in both materials. This Fermi level typically is energetically just a little below the conduction band, but well above the valence band. Now, let's have some current flowing, considering both carrier types, majorities and minorities, separately:

An electron comes from the metal, on an energy level being on average identical to the Fermi level, so by entering the semiconductor it needs to "jump up" into the conduction band. For this jump the electron indeed needs additional energy, which is taken from the heat of the surrounding -- in other words, we encounter the Peltier effect; here as cooling. (The opposite happens when the electron "jumps out" from the semiconductor into the metal at the other end; there, Peltier heating occurs.) This is what usually happens.

To inject a hole into the semiconductor is equivalent to extract an electron from the valence band and bring it up to the Fermi level. However, since the energetical difference from the valence band to the Fermi level of the metal is large, there is only very little probability for this process to happen. This is the reason why, although not "fully forbidden", in an n-type semiconductor typically no additional holes are injected when a current is applied. However, if the doping level is low and the band gap of the semiconductor is small, then the probability for hole injection increases. (Unfortunately I have no idea whether for practial situations this really plays a role.)

If your contacts are ohmic, how would you inject excess holes in your n-type material? This is a key point to locate the quasi Feri levels?

In your device with ohmic contacts and with no external electron hole pair generation then the electron concentration n=n0, and p=p0 , that at their thermal equilibrium values.

I would like that you follow the link for more information about the disturbance of the interior of the semiconductor materialsBook Electronic Devices

Best wishes

Dear All,

Very nice information and discussion.

Thank you

Regards

Dear Abdelhalim and Jan-Martin,

Thank you very much for your detailed replies. I am still trying to work this out...

I would appreciate if I could know your responses in terms of just a simple yes or no, so I am more clear in following your above explanations.

1) "If the injection is by external voltage, are we STILL at equilibrium, despite of having a net flow of current?" yes /no

My understanding was that if we have current we are automatically out of equilibrium.

2)In the following phrase are you referring to applying a voltage in the dark? Yes/no? "In your device with ohmic contacts and with no external electron hole pair generation then the electron concentration n=n0, and p=p0 , that at their thermal equilibrium values."

Regards

Dear Pourandokht,

your starting question about a possible difference in quasi-Fermi levels is equivalent to the question about the presence of an appreciable amount of minority carriers. Therefore, what does "equilibrium" mean in your first question to be answered "yes/no"?

Of course, current flow is a non-equilibrium phenomenon, but please have a closer look: What "driving force" causes the current flow? And in which way does this "driving force" cause a deviation from equilibrium: what is changed, what is slightly perturbed, and what remains unchanged under its infleunce?

(The second yes/no question I leave to Abdelhalim.)

Dear Jan-Martin,

If you are saying (that I agree with) outside equilibrium we have current, then it follows that mass-action law is no longer true. Assuming that we DO have a current (steady or otherwise) when APPLYING A VOLTAGE IN THE DARK (i.e. voltage is the driving force) then the Fermi level SHOULD split according to your above explanation. But then it seems that at the same time this conclusion is being contradicted. What am I missing here?

Thank you,

This is exactly the pseudo-reasoning I wanted you not to follow! The point is that we are dealing with "one-way conclusions": If from A follows B, it is not necessarily the case that when B is present, A was the reason for it -- because there might be a reason C that also implies B (details later).