The Role of Doping and Electric Fields in P-N Junction Formation?

[Muhammad Usman]

I was reading the PN junction. During the formation of PN junction diffusion process takes place in which the free electrons cross the boundary of the P-N and joined with the holes in the P-type to form the negative ion and holes on the N side cause positive ions. So, the P-side has negative ions and N-side has positive that constitutes the depletion region
From doping we know there are other free electrons in the N-type and holes in the P-side. What I am not able to understand that why not the electron can join the N-side positive ions to make it neutral atom and why not on P side electrons from negative ions drift further away from the boundary to join with other holes. Why they always stay near the boundary (Depletion Region). Please tell me a simple answer as I am just a student.

Best Regards

 

[Drakkith]

What I am not able to understand that why not the electron can join the N-side positive ions to make it neutral atom and why not on P side electrons from negative ions drift further away from the boundary to join with other holes.

I believe the electric field in the depletion region prevents this. Electrons and holes can't move away from the depletion region once equilibrium is reached.
See the following picture. The bottom represents equilibrium.

 

[Muhammad Usman]

Hi,

I was talking about the mixing of bulk electrons (Carriers in the N-type region due to doping) in the newly formed positive ion. Because logically positive ion should attract negative electron and those ions by logic should be neutralized by why this phenomena is not happening ?

 

[Drakkith]

I was talking about the mixing of bulk electrons (Carriers in the N-type region due to doping) in the newly formed positive ion.

It does happen. But each time an electron on the N-side joins with a hole, it leaves a hole behind itself. The electric field across the depletion regions causes the trillions of electrons to feel a slight force away from the P-side, as they want to follow the electric field lines. This counteracts further diffusion of electrons from the N-side to the P-side and prevents further expansion of the depletion region into each side.

Because logically positive ion should attract negative electron and those ions by logic should be neutralized by why this phenomena is not happening ?

That's exactly what is happening. Electrons and holes are constantly recombining on both sides. But holes are left behind as the electrons move around, so there is always a new ion created when an electron and a hole recombine.

 

[Let'sthink]

You are imagining a static situation a positive immobile ion on the n-side and a free mobile hole in the p-region. But the situation is chaotic and equilibrium is dynamic. The majority charge carriers along-with minority ones are moving randomly in all possible directions. I a pure p region and also in a pure n region these moving charge carriers do meet and destroy each other creating neutral sights and positively charged sites which are mobile and electrons but there is symmetry so there is no directional steady electric field. This symmetry is broken when a p-n junction is formed. Because of preferential recombination occurring in the junction region n side becomes positive and p side becomes negative and an electric field is created in the direction N to P and the junction region is depleted of both hole and electron on the average and this equlibrium is dynamic and not static.

 

[sophiecentaur]

@Muhammad Usman There are three conditions that you need to discuss. The pn diode on its own, the diode forward biassed and the diode reverse biassed.
When un-biassed, a number of electrons in the N material find a lower energy position when they occupy holes in the P material. But this can only happen locally and the depletion region is limited in size.
If you reverse bias the diode the Positive potential applied to the n side will cause electrons to move towards it and fewer (negative) electrons will preferentially fill positive holes and the depletion region will get narrower.
If you forward bias the diode electrons will flow from the n region into the p region and the supply of electrons doesn't run out.

From doping we know there are other free electrons

I would not call them "free", so much as less tightly bound to their parent atom than they are attracted to a nearby hole in a p region. This migration of electrons from where 'they should be' to another atom is the way that Chemistry works and it explains how compounds are formed

 

[Muhammad Usman]

Hi Let'sthink,

Please help me to understand about this concept that why does not the electrons on the N-region (Free carriers) does not combine with the positive ions formed after P_N junction layer is formed.

Is it continuously joining and leaving like continues process of neutralization and new ion process takes place ?

 

[sophiecentaur]

Hi Let'sthink,Is it continuously joining and leaving like continues process of neutralization and new ion process takes place ?

Is your question, in fact, pretty simple?
Are you referring to the thermal effect? The energy involved in electron movement from hole to hole is very low so yes there will be a constant process of hole and electron movement around a junction.
You use the term "ion" as if it's a phenomenon just involving single atoms but there is a lattice of atoms - just as there is in a metal - and the electrons move around in the 'context' of the lattice and not atom-by-atom. The difference between a metal and a semiconductor is to do with the actual energy levels involved - which accounts for the different conductivity.

 

[Muhammad Usman]

Many thanks I got this concept but please help me to understand my Question :-

Please help me to understand about this concept that why does not the electrons on the N-region (Free carriers) does not combine with the positive ions formed after P_N junction layer is formed.

 

[Let'sthink]

Many thanks I got this concept but please help me to understand my Question :-

Please help me to understand about this concept that why does not the electrons on the N-region (Free carriers) does not combine with the positive ions formed after P_N junction layer is formed.

Ok Mr Usman I try to explain the real answer to your question. For clarity let us be specific. Consider germanium pure crystal at room temperature. At any time some bonds are broken so that there is mobile positive ion vacancy, which we call hole and the corresponding electron is the property of the whole crystal. Even the hole is also considered the property of whole crystal because it also can reach anywhere although it is bit sluggish. So we have only positive ion vacancies (holes), electrons and larger number of neutral germanium atoms intact. If by chance electron jumps into a hole then both vanish and create the neutral site. Rate of this occurrence may be low but it is not zero. We do not bother about it much because for every such event a parallel event may also occur where a neutral site gives rise to an electron-hole pair. Now if I ask you to explain a p -type material with let us say Indium as impurity. I want you to describe it to me. To help you at this stage I refer to the first figure which you yourself have uploaded and tell you that the positive ions which you are showing are the immobile phosphorous ions in the N region and the negative ions in the region are the immobile indium ions in the p region. I shall await your reply. This gives the answer to you question.

 

[Drakkith]

Please help me to understand about this concept that why does not the electrons on the N-region (Free carriers) does not combine with the positive ions formed after P_N junction layer is formed.

They do, just like I already explained. Can you be more specific with your question or explain exactly what you don't understand?

 

[Let'sthink]

Please help me to understand about this concept that why does not the electrons on the N-region (Free carriers) does not combine with the positive ions formed after P_N junction layer is formed.

In extrinsic semiconductors there are two types of ions. Positive Germanium or Sillicon ions which are mobile and are called holes and other immobile positive and negative impurity ions. Think about the their distribution and then you will get the answer to your specific question.

There are two ways to understand semiconductors one is band picture and another is bond picture. Although they are equivalent bond picture is more convenient to understand the type of question you have asked. Just being a positive ion means it can attract an electron to itself but can it keep it with itself you need to look into. Is there a structural vacancy you need to inquire!

 

[sophiecentaur]

Just being a positive ion means it can attract an electron to itself

But the action of semiconductors is more to do with the structure as a whole and not the individual atoms / ions. An isolated ion will eventually recombine with or lose and electron to become stable. When you put donor atoms in an intrinsic semiconductor, the 'extra' electron in the lattice is more shared with all the other intrinsic atoms because the 'bonding' is with all the local atoms. The whole lattice is held together due to a co-valent system of bonds between all the atoms; it's easy to ignore this in the diagrams that are drawn.

 

[Muhammad Usman]

@Letsthink, Many thanks for helping me to understand the topic. What I analyzed is in P-type material the impurity is added and there is a mobile hole (Or in other words mobile electron that changes its position) Since here the number of holes are more therefore the charge moves more freely in the crystal.

Please correct my understanding if you find some flaw in it. Thanks

Immobile ions are the one which are impurity ions (Lets say negative atoms with five valence electrons). These five valence electrons when loose electron then they become positive ion due to the concentration of electrons and protons in an atom but on the other hand their octet is completed (8 Electrons Rule) therefore no further electron could come in and in case if they come in the positive ion attraction force is less than energy level of electron so it becomes completely mobile and dynamically electrons move in and move out but since the concentration of electrons in millions which are involve in this process therefore no such visible impact is observed (However the impact could be observed if some day we have high frequency electron Microscope). Now in case of forward biased these ions are neutralized and once neutralized the electric field they have created with the peer has also become less effective and the deplition region gets thinner and thinner and more electrons are able to pass through.

 

[Muhammad Usman]

Now in case of reverse biased I have one more confusion. In case of reversed biased the opposite polarity of battery are added with P-type and N-type, In this case one side is electron injected (Negative Terminal of the Battery) and the other side is hole injection (Or in other words electrons Removal). Since more electrons are removed from N-region (Went back to the Ground or battery terminal) there are more positive ions on (N-Side) and more negative ions on P-Side (Lets forget the P-Type for now). Since more electrons are inject out from the N-region due to reverse biased, Then how does the depletion region shrink (And returns back to its original position) after some time when the biasing is off. Where does the electrons come from, to neutralize those ions which are formed from the reverse biased.

I have really apologize as I am still in the basic but I want to Learn it from the deep. Thanks to all who responded to my Query, Really appreciated your efforts who spread their knowledge to World.

 

[sophiecentaur]

Where does the electrons come from,

Is there a reason why there would be a build up of electrons? For every extra electron in one (n) place there will be a missing electron in the other (p) place. But we are dealing with the statistics of large numbers in a massive structure and the Energy Bands, not with individual Ions.

 

[Muhammad Usman]

Actually the main question here is where would those electrons come from to neutralize the ions and how the depletion region will shrink back to its normal position when the reverse biasing is removed. Thanks

 

[Muhammad Usman]

Let me re-phrase it, because the electrons will move back to the battery terminal and ionization have been increased and more positive ions will be formed on the N-side of the region as the electrons are removed. When the battery biasing is removed it means that the electrons concentration is lessen due to connection with the positive terminal of battery and attracting the electrons from the N-Region. Now after the reverse biasing is removed the depletion region phenomena is returned back to its normal position it means the atoms which are ionized required electrons to be neutralized, OK we can use the phenomena and think like the electrons will have random behavior but doesn't this will reduce the concentration of electrons in N-region overall and making the diode reversed biased 3 or more times will make it run out of electrons.

 

[sophiecentaur]

Why would the electrons move to the battery terminal? It may not even be there but they can't flow into the terminal because there will be a potential stopping them. The nearest place they can go would be into the holes on the other side of the junction, which is also waiting to be filled. Any scenario you want to introduce into this will be symmetrical with equal numbers of + and 1 charges. The lowest energy (equilibrium) condition will be with the normal depletion region in the state as it was before all this started.
I think you are missing something fundamental here because you are getting preoccupied with the details rather than the overall principles. I can't think what it is that makes you think there's something wrong here.

 

[Let'sthink]

Mr Usman you have almost got the answer to your question. Forget about biasing and concentrate on the depletion region and majority of ions in P and N region which causes the potential difference which finally brings the dynamic equilibrium. When excess electrons from N region cross over and neutralize the excess holes in P region, making the region depleted of mobile carrier and do not come back what do they leave on the N side they leave positively charged Ph ions in the vicinity of the junction. Now these immobile ions electron octet is complete so they can no doubt attract an electron but cannot keep it. That is what you wanted to ask. Now try to describe the state of affairs in te P region about which you have not asked a question.

 

[Muhammad Usman]

Mr Usman you have almost got the answer to your question. Forget about biasing and concentrate on the depletion region and majority of ions in P and N region which causes the potential difference which finally brings the dynamic equilibrium. When excess electrons from N region cross over and neutralize the excess holes in P region, making the region depleted of mobile carrier and do not come back what do they leave on the N side they leave positively charged Ph ions in the vicinity of the junction. Now these immobile ions electron octet is complete so they can no doubt attract an electron but cannot keep it. That is what you wanted to ask. Now try to describe the state of affairs in te P region about which you have not asked a question.

--> In the P-type there are access number of holes (The reason why that I didnt ask about P-type because its another name of electron movement although the
movement is not considered to be negative charge as does the electron but its a positive charge and its easy to know about the Negative electrons and assume the similar case for the P-type) so when holes combine with the electrons they are also become neutral but overall has the negative charge. This negative charge attracts the hole (Due to Electron replusion) but since holes is not an actual physical particle so it attracts electrons from nearest negative ions. So if the electrons doesn't move it means that the force of attraction for breaking the octet rule to get the electron into the hole is not enough to make the things work that cause them to be static ion near the deplition region. I guess this is some different concept than the case of negative ions which I observed as dynamic equilibrum.
--> Or in other case electrons move but attraction cause other electron to come in and that's a dynamic equilibrum just like I mention in the above.

 

[Muhammad Usman]

Why would the electrons move to the battery terminal? It may not even be there but they can't flow into the terminal because there will be a potential stopping them. The nearest place they can go would be into the holes on the other side of the junction, which is also waiting to be filled. Any scenario you want to introduce into this will be symmetrical with equal numbers of + and 1 charges. The lowest energy (equilibrium) condition will be with the normal depletion region in the state as it was before all this started.
I think you are missing something fundamental here because you are getting preoccupied with the details rather than the overall principles. I can't think what it is that makes you think there's something wrong here.

Lets take a picture below as a reference for the reverse biasing

the Free electrons are now attracted towards positive terminal of the battery and holes towards the negative terminal of the battery (Actually this process is injection of electrons in the diode). The number of free electrons are technically reduced on N-Side and so does the holes in P-Side. Now what I don't understand how does the deplition region returns back to its normal position when it now have access positive ions (Less electrons) on the N-Side (Lets focus on N-side only first)

 

[Lord Jestocost]

Now what I don't understand how does the deplition region returns back to its normal position when it now have access positive ions (Less electrons) on the N-Side (Lets focus on N-side only first)

When you disconnect the battery, charges are internally redistributed between the positively and negatively charged depletion regions until the original equilibrium state (constant Fermi level throughout the pn-junction) is established again.

 

[Drakkith]

This negative charge attracts the hole (Due to Electron replusion) but since holes is not an actual physical particle so it attracts electrons from nearest negative ions. So if the electrons doesn't move it means that the force of attraction for breaking the octet rule to get the electron into the hole is not enough to make the things work that cause them to be static ion near the deplition region.

I'm sorry but I can't quite understand what you're trying to say here.

Now what I don't understand how does the deplition region returns back to its normal position when it now have access positive ions (Less electrons) on the N-Side (Lets focus on N-side only first)

Once the external voltage is removed electrons in the P-side simply flow back to the N-side until equilibrium is reached.

 

[sophiecentaur]

Now what I don't understand how does the deplition region returns back to its normal position

Basically it's not a stable situation when suddenly disconnected. Why do you want to focus on just one side? The whole point is what happens in the total structure of the junction. When you remove any biasing PD the lowest energy condition is when the junction has returned to its original state. The disconnected diode has zero net charge and no polarisation of charge. Excess negative charges end up where there was a lack of negative charges.
You seem to be determined to address this problem on your own terms and it is getting you nowhere. Why not look at it as the other contributors to this thread are doing?

 

[Let'sthink]

so when holes combine with the electrons they are also become neutral but overall has the negative charge.

This is incorrect. when free electron combines with a hole both vanish and the site is neutral.

 

[Let'sthink]

Now if I ask you to explain a p -type material with let us say Indium as impurity. I want you to describe it to me. To help you at this stage I refer to the first figure which you yourself have uploaded and tell you that the positive ions which you are showing are the immobile phosphorous ions in the N region and the negative ions in the region are the immobile indium ions in the p region. I shall await your reply. This gives the answer to you question.

Here I had given you my reply for description of N region. Do you agree with that? You describe to me the P region in a similar way then I can make you understand what I am trying to say in response to your question.

 

[Muhammad Usman]

This is incorrect. when free electron combines with a hole both vanish and the site is neutral.

There are holes in the p-region so when the electrons are added in the P-type holes, their octet is completed but overall negative charge is there as the concentration of protons and electrons are not equal.

 

[Muhammad Usman]

A germanium has indium as an impurity in the atom, Germanium has 4 outermost electrons and when the impurity is added in the atom. Indium has three electrons in its valence shell. so it means that when the covalent bonding is formed there are 7 electrons in the molecule inspite of eight electrons and left a vaccency. This vaccency is attracted to the electrons. If the electron is fell into the vaccency it will make the octet complete but it will also make the molecule itself negative (I don't know whether it is right to use the word molecule because two atoms are making the covalent bonding) as the negative ion. This electron that fell into the bonding will cause the ion to be negative. The nearbvy hole will attract this negative ion in return cause the electron to attract towards itself. But each time the electron attracts it also left back the vaccency. What i get from the original description is that the equilibrium is dynamic. Which means negative ions are continusely made and changed to the neutral atoms but since bonding is immobile and electrons enter through the border therefore the depletion region is formed here.

 

[Let'sthink]

You are almost right for for a P-region. In a P region although immobile negative ions are there the overall region is neutral as there are equal number of mobile positively charged holes. But when this P region is beside n region the excess electrons their come on the P side and neutralize holes and thus the unbalanced immobile Indium ions help in creating electric field along with the unbalanced positively charged Phosphorous ions. So in brief preferred electron hole recombination creates depletion region and immobile impurity ions on both side create the electric field in equilibrium. Now you ask your original questio nad you get the correct answer. On this you can superimpose the applied bias and see what happens to the current flow depletion region and the electric field present in the depletion region.

 

[Let'sthink]

In a P region although immobile negative ions are there the overall region is neutral as there are equal number of mobile positively charged holes. But when this P region is beside The N region the excess electrons present there come on the P side and neutralize holes and thus the unbalanced immobile Indium ions help in creating electric field along with the unbalanced positively charged Phosphorous ions. So in brief preferred electron hole recombination creates depletion region and immobile impurity ions on both side create the electric field in equilibrium. Now you ask your original question and you get the correct answer. On this you can superimpose the applied bias and see what happens to the current flow, depletion region and the electric field present in the depletion region.

I have corrected some typographical mistake in the copied portion above.