How do diodes (pn junction) work?

[DaTario]

Hi all,

I would like to know how pn junction work?

thanks

DaTario

 

[mjsd]

PN junction: putting a P-type material next to N-type material to form the PN junction

P-type is where you have more "holes"; N-type is where you have more electrons in the material. Initially, when you put them together to form a junction, holes near the junction tends to "move" across to the N-region, while the electrons in the N-region drift across to the p-region to "fill" some holes. This current will quickly stop as the potential barrier is built up by the migrated charges. So in steady state no current flows.

Then now when you put a potential different across the terminals you have two cases:

1. +ve end to P-type, -ve end to N-type: The electric field from the external potential different can easily overcome the small internal field (in the so-called depletion region, created by the initial drifting of charges): usually anything bigger than 0.6V would be enough. The external field then attracts more e- to flow from n-region to p-region and more holes from p-region to n-region and you have a forward biased situation. the diode is ON.

2. +ve end to N-type, -ve end to P-type: in this case the external field pushes e- back to the n-region while more holes into the p-region, as a result you get no current flow. Only the small number of thermally released minority carriers (holes in the n-type region and e- in the p-type region) will be able to cross the junction and form a very small current, but for all practical purposes, this can be ignored

of course if the reverse biased potential is large enough you get avalanche break down and current flow in the opposite direction. In many cases, except for Zener diodes, you most likely will destroy the diode.

 

[DaTario]

Ok, let me call your explanation the basic explanation. But now I would like to understand some points behind this.

If all atoms are electrically neutral, why electrons from the donor type go to the acceptor side?

Why this migration stops? Does it have to do with temperature in some sense?

I don´t understand the energy level diagram. Neither before the migration nor after, when the potential forms a smooth curve between the junction.

Do holes produce current? How so?

Thank you in advance for the attention,

best regards

DaTario

 

[marlon]

DaTario,

You are asking about the fundamentals of pn-junctions. To start, you need to understand the concept of a semiconductor. Do you know what that is ? Also, do you know what p type and n type semiconductors are ?

Find out the answers to those questions and you have already done a great part in understanding the basic ingredients that make up such a junction.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html#c1

marlon

 

[DaTario]

One thing I would like to understand about p-n junctions, and I have not seen the explanation in any book is the reason why, in the energy diagram showing the junction, the valence band in the As doped (donor) semiconductor lies bellow the valence band of the Ga doped (acceptor) semiconductor.
Since the doping with As tends to lower the average atractiveness of the net (to the electrons) it would be reasonable to have its valence band situated above the Ga doped Si cristal.

Best wishes

DaTario

 

[t-money]

DaTario
that is a good question, I would also expect the scenario you mentioned. The only thing that I can think of is that the donor is not at a state that is neutral relative to the acceptor, regardless of the energy band since stability is based on valence shell and how close it is to being full, but this is just a speculation.

 

[LydiaAC]

If all atoms are electrically neutral, why electrons from the donor type go to the acceptor side?
DaTario

Atoms are neutral. If you have a semiconductor at absolute zero, all the electrons are attached to their atoms. Electrons cannot move, so there is no conductivity.

At room temperature, some electrons have separated from their atoms and are running free. You must remember that the vertical axis in the band diagram is energy, so you must interpret any free electron as belonging to the conduction band. These electrons leave unoccupied states in the valence band. Roughly speaking, these are the "holes". Actually, the concept of hole is somewhat complicated.

Now, some "atoms" are no longer "atoms" but positive ions, because they lack an electron. You can also identify this positive ions with holes, but it is also an oversimplification. The number of positive ions is exactly the same as the number of electrons, so the net charge is zero. Thus, the semiconductor material is, globally, neutral. However, locally, there could be zones with positive or negative charge.

This is what happens in an "intrinsic semiconductor", for example, pure silicon. In an intrinsic semiconductor there is exactly the same number of electrons as of holes, and this number depends on temperature.

In an "extrinsic semiconductor", you put impurity atoms in the crystal. Silicon has four electrons in its outer layer, so if you put a boron atom in the lattice, there will be only three electrons to bond this atom with their silicon neighbors. A silicon neighbor will have an incomplete bond, and this unoccupied state can be filled with an electron from other silicon atom. This positive silicon ion is also a hole. The electron which this atom lost is with the negative boron ion. The number of this kind of holes must be the same as the number of boron ions. Now you can have more holes than electrons, and the material is P-type.

On the contrary, if you put phosphorus in the lattice, there will be four electrons for four bonds, and an extra electron. This electron can also abandon the phosphorus atom, leaving a positive phosphorus ion. The number of this kind of electrons must be exactly the same as the number of phosphorus ions. You can have more electrons than holes, and the material is N-type.

Why this migration stops? Does it have to do with temperature in some sense?

DaTario

When you put together a P-type and an N-type material, by methods of microelectronic fabrication. The system is globally neutral but, as I said before, there is no reason for it to be locally neutral. The P-type material has a high concentration of holes and the N-type has a low concentration. So, by Ficks first law, a diffusion current will arise, which take off holes from P-side and send them to N-side. The P-side is no longer neutral, it has a negative charge. The same happens in the N-side which acquires a positive charge. A similar process causes a diffusion current of electrons from N-side to P-side. Both processes are not reversal of each other but reinforcing: both works towards building a negative charge in P-side and a positive charge in N-side.

Actually, this only happens in the neighborhood of the "junction" (where P and N materials are in contact). The material far from the junction remains neutral. The charges produce an electric field pointing from N-side to P-side. This electric field pushes electrons back to N-side and holes back to P-side. This drift process does work against the diffusion process. In some point both processes reach an equilibrium and you will end with a fixed electric field pointing from N to P and no more net transference of electrons or holes from one material to the other.

Since diffusion constant depends strongly on temperature, diffusion current will depend on temperature. However, mobility depends also on temperature, so drift current will depend also on temperature. The result is that the final electric field, on my knowledge, has no a big dependence on temperature.

I don´t understand the energy level diagram. Neither before the migration nor after, when the potential forms a smooth curve between the junction.

DaTario

Since the built-in electric field push electrons from P-side to N-side, an electron in the P-side has more energy than an electron in the N-side in the same way as a book on the table has more energy than a book on the floor. Thereby, you must draw the energy level in P-side higher than in the N-side. In the transition zone, the connecting curve is actually smooth, but their exact shape is something complicated. You will not lost anything relevant drawing it as a line. To work with holes, you must have a lot of imagination since they need more energy for staying in the N-side than for staying in the P-side. I solve this problem, putting my book head down.

Do holes produce current? How so?

DaTario

In a philosophical sense, holes cannot carry current because they are not material. However, the unoccupied states allow a current to flow. This current is, obviously, made from electrons. An electron linked to a silicon atom can hop to an unoccupied state in other silicon atom. This is equivalent to the unoccupied state hoping from the second atom to the first. By hoping and hoping, an electron can move in response to a field or a concentration gradient. However, it must be noted that this is not a "free electron". The electron who hopes is an electron in the valence band. This is the way in which the valence band can participate in conduction.

There are a lot of good, but complicated reasons, to imagine that these hoping valence electrons are actually holes. It is conceptually possible to expel holes from Solid State Physics in the same way in which some people want to expel electric and magnetic fields from electromagnetics and work only with "action-at-a-distance". However, it has been proven that the hole is a very useful concept and it is better to work with it than without it. It is so useful that it is common to forget that actually it is an abstract concept and that the physical reality behind it are the electrons of the valence band.

Lydia Alvarez

 

[DaTario]

Since the built-in electric field push electrons from P-side to N-side, an electron in the P-side has more energy than an electron in the N-side in the same way as a book on the table has more energy than a book on the floor. Thereby, you must draw the energy level in P-side higher than in the N-side. In the transition zone, the connecting curve is actually smooth, but their exact shape is something complicated. You will not lost anything relevant drawing it as a line. To work with holes, you must have a lot of imagination since they need more energy for staying in the N-side than for staying in the P-side. I solve this problem, putting my book head down.

Thank you for this answer. But let me exploit you a little bit more. Why exactly the electron in the P side has more energy than in the N-side?

You said that P side (N side) has initially more holes (free electrons) than free electrons (holes). Nevertheless, they are neutral at t=0 (time of the junction), but this structure provides conditions to some charge diffusion process which spoils this neutrality. Expand this argument, please.

Best wishes,

DaTario

P.S. I've got a PhD in quantum optics, so I have some familiarity with QM concepts and methods.

 

[LydiaAC]

Hello DaTario:
In the junction region you have a big electric field. This field is zero in the P-side far from the junction, begin to increase while approaching to the junction, stay big a little in the N-side and then began to decrease while going far from the junction in the N-side.
An electric field in a point is defined as the force a 1C charge would suffer if it were put in that point. The electric field in the junction points from N to P, so a positive charge put in the junction will suffer a force directed from N to P. The electron has a negative charge, so its force will be opposite to that on a positive charge, so it is directed from P-side to N-side. This is only about the definition of electric field and the shape of the electric field in the junction.

Why the electron has more energy in the P side? Because if you want to move it from N-side to P-side you need to do work. You need to manage to get a force opposite to that exerted by the electric field and use it for moving the electron a distance from its position in the N-side to its position in the P-side. It is the same what happens when you lift a weight. The force of gravity point through the floor, so if you want to lift a weight you must use your muscular force to apply a force opposite to the gravity force and use it to move the weight the required distance. Roughly speaking, force times distance is work. You must work to lift a weigh and you must work to move an electron from the N-side to the P-side.
If you want to move the electron from the P-side to the N-side, you do not need to do anything. The electric field is the one who "works". It provides the force to move the electron a distance. It is the same when you drop the weight. You do not work, gravity does. This is the way in which we know that the P-side has more energy than the N-side.In P-side you have a lot of holes. Neutrality is maintained because there are also negative acceptor ions. In N-side you have a lot of electrons. Neutrality is maintained because there also positive donor ions. For considerations of neutrality a charge in a hole or an electron or a charge in a ion has exactly the same value. However, for transport considerations, you must remember that holes and electrons can move freely and ions cannot.
Diffusion is what happens always that you have a concentration gradient. Actually it is only consequence of random motion. Actually, the probability of a hole to go from the P-side to the N-side is exactly the same than its probability to go from the N-side to the P-side. However, probability must be multiplied for the holes available to choose this direction. If in the P-side there is a lot of holes and in the N-side there are only a few, most holes will go from P-side to N-side than holes going from N-side to P-side. The net effect is a particle flux from P-side to N-side. This flux would be large if the difference in concentration is large, small if it is small and zero if it is zero. You can write this as the formula J=-Ddp/dx, where dp/dx is the (unidimensional) gradient of concentration and D a proportionality parameter very dependent on temperature. The minus sign is because the flux go from the high concentration side to the low concentration side.
In a hypothetical junction recently formed the concentration of electrons and holes are step functions. This gradient will subject them to the diffusion process and they gradually will go to the opposite side. Why neutrality is lost? Because the neutrality of each side is maintained through the ions and since ions cannot move they are not subject to diffusion. Acceptor negative ions cannot follow the holes in their travel for compensate their positive charge anymore. Nor can do the donor positive ions with the electrons.
By itself a diffusion process finish when the concentration gradient is zero but in this case, the lost of neutrality creates a big electric field in the junction. The equilibrium is reached when the tendency of electrons and holes to diffuse is exactly opposed by the tendency of electrons and holes of being pushed by the electric field. Electrons and holes continue crossing the junction constantly but since there is not difference in the number crossing from P to N and from N to P, the net flux is zero.
I hope this is useful.
Lydia Alvarez

 

[DaTario]

Thank you again. I must say however that the definitions of work and positive energy were already familiar to me, but thanks anyway. My problem is in understanding the following:

P-side has lot of holes. So, outer electrons feel higher attractive forces to this network of atoms than in the N-side. Ok1 ?

Positive carriers (whatever they are) will consequently experience more attractive forces in the N-side, Ok2 ?

Electric potential is usually higher near positive charges, but electric energy is a different concept, and must be higher near repulsive regions. Therefore, the energy diagram which puts P-side higher than the N-side (both valence and conduction band) must be related specifically to positive charges, Ok3 ?

Finally for the moment, if we would like to write an energy diagram for the electron, the situation would be the opposite, i.e., P-side would be lower than the N-side, Ok4 ?

Best wishes

DaTario

 

[LydiaAC]

P-side has lot of holes. So, outer electrons feel higher attractive forces to this network of atoms than in the N-side. Ok1 ?

Not OK.
A hole has a positive charge that should attract an electron... if there were not the same number of negative ions accompanying the holes. There are actually charges, but at a very local level. Far from the junction, the P-material is globally neutral, so there are no attractive nor repulsive forces for electrons. Close the junction, you find that some holes have abandoned the material by diffusion, so you get an excess of negative ions. In this region electrons feel REPULSIVE forces, not attractive ones

Positive carriers (whatever they are) will consequently experience more attractive forces in the N-side, Ok2 ?

Not OK.
Far from the junction, N-material is globally neutral, so holes are not subject to an attractive nor repulsive force. Close the junction, we have lost many electrons, so we have many naked positive ions and the hole is subject to a repulsive force not a attractive one.
In summary, electrons in P-side and holes in N-side (minority carriers) are repelled from the respective side and into the opposite.

Electric potential is usually higher near positive charges, but electric energy is a different concept, and must be higher near repulsive regions. Therefore, the energy diagram which puts P-side higher than the N-side (both valence and conduction band) must be related specifically to positive charges, Ok3 ?

Not OK
Electric energy is electric potential times charge, so it is true that for the same potential, electrical energy is different for electrons and holes. It is true that is higher in the repulsive regions.
Since electrons are repelled from P-side and attracted to N-side, the energy of electrons in the P-side is higher. The "energy diagram which puts P-side higher than the N-side" is in fact for electrons.

Finally for the moment, if we would like to write an energy diagram for the electron, the situation would be the opposite, i.e., P-side would be lower than the N-side, Ok4 ?

Not OK.
The energy diagram for the electron is as I said before, higher in the P-side and lower in the N-side. If you want to write an energy diagram for the hole, you must do the opposite, P-side lower and N-side higher. This is the reason that for studying holes I put my book head down.

In summary.
The majority carriers, holes in p-side and electrons in n-side are many, but this has not an effect in the net charge. This is exactly the curiosity with semiconductors. You know that by putting excess charge in a conducting material it will migrate to surface and you will not have any improvement in conductivity. In a semiconductor you can add carriers without adding charge, so the carriers do not migrate to surface but stay in the bulk and conductivity can be modulated.
The charges in the junction are due to a lack of majority carriers. A lack of holes in the P-side allows for a negative charge to build up, a lack of electrons in the N-side allows for a positive charge to build up.
The charges in the junction are NEGATIVE for the P-side and POSITIVE for the N-side. The labels P and N signal the kind of carriers but it is their absence and nor their presence which is responsible for the existence of these charges.
Lydia Alvarez

 

[DaTario]

I am really well impressed with your kindness and patience with my deep ignorance in this subject. I must appologize for being so slow in getting the idea, but I really feel I am learning something important here.

Let me check just two points this time.

Although N-type has more electrons than holes at the beginning of the junction, it is the diffusion process that allows for a positive charge to build up in this region, the converse applying to the P-side, Ok5 ?

You said that the local charge in both types of semiconductors travels through the bulk because these charges are created by thermal effects that promote either valence electrons to holes or electrons in the donor's level "Ed" to the conduction band. Is it Ok6 ?

Best wishes

DaTario

 

[LydiaAC]

Hello DaTario:
In am very interested in the Semiconductor Physics but I am also very interested in the the teaching of Semiconductor Physics. I always say my students they must to ask honest questions because this is the only way in which I can detect where is the problem and look for an appropriate answer. I have found that must students prefer to repeat what is said in the book, give the appearance they understand everything. When they get a PhD, they continue doing the same, because they think that having doubts is not acceptable, in despite of the fact that PhD are excessively specialized and cannot provide a full understanding of "everything". Have you read "Adventures of a curious character of Richard Feynman"? He gives a description of what he found in his time about physics students in Brazil. I can say that this is exactly the same situation for students in my environment.

Although N-type has more electrons than holes at the beginning of the junction, it is the diffusion process that allows for a positive charge to build up in this region, the converse applying to the P-side, Ok5 ?

OK

You said that the local charge in both types of semiconductors travels through the bulk because these charges are created by thermal effects that promote either valence electrons to holes or electrons in the donor's level "Ed" to the conduction band. Is it Ok6 ?

I am not sure I can subscribe this. Maybe it is not enough clear. Electrons in the valence band are attached to individual atoms while electrons in conduction band are free to move through all the crystal. The unoccupied states in a almost full valence band, can allow valence band electrons to move hoping from an state to another but a process called "conduction by holes"

When they say "electrons" they are referring specifically to "electrons in the conduction band". Electrons are created when a valence band electron is promoted to conduction band or when the fifth electron in the impurity is promoted from Ed level to conduction band. In the first case you create also a hole and in the second you create also a positive ion.
Holes are the unoccupied states and they are created when a valence band electron is promoted to conduction band or when the unoccupied state in a impurity, in the level Ea, is filled with an electron of the valence band, leaving an unoccupied state in this. In the first case you create also an electron and in the second you create also a negative ion.

Creating electrons or holes, never alter global charge but it can change dramatically the number of "free charges" or "carriers", that is, charge that can move. By thermal processes, you are obliged to create electrons and holes by pairs. You cannot get a P or N semiconductor only with thermal processes. Impurities allows you to create electrons without creating holes, or holes without creating electrons. This is the way in which you can get a P or N semiconductor.

If you can read Spanish I have an article on semiconductors

http://insting.mxl.uabc.mx/~lydia/Conductividad_carta.pdf

I have never translated it into English because I think there is a lot of resources in English and I concentrate in generate resources in Spanish.

Lydia Alvarez

 

[DaTario]

No problem in reading spanish, thank you again.

Concerning this part of your saying:

Creating electrons or holes, never alter global charge but it can change dramatically the number of "free charges" or "carriers", that is, charge that can move.

By thermal processes, you are obliged to create electrons and holes by pairs.

You cannot get a P or N semiconductor only with thermal processes.

Impurities allows you to create electrons without creating holes, or holes without creating electrons.

I agree with the first, third and fourth sentences. But regarding the second sentence, I disagree for I believe the electronic transitions:
a) Ed to conduction band
b) valence band to Ea

occur also due to thermal reasons, and therefore it is not the case of saying that thermal effects produce only electrons and holes by pairs.

Best wishes

DaTario

 

[LydiaAC]

Hello DaTario:
You are right, you need some thermal energy to ionize impurities. However, it is enough with some 50 K to full ionize all of them, and once this has happened, you have one carrier for each impurity ion and this will not have a variation with temperature.
At room temperature you can assume "full ionization" and carriers generated from impurities are not thermal dependent. On the contrary, carriers generated from valence electrons being promoted to conduction band increase and reduce strongly as the temperature varies, so they are frequently called "thermally generated carriers".
It is true, as you say, that actually, all carriers are in some sense "thermally generated carriers". However, it is common practice to call that way only those which were generated in pairs.
Are you from a Spanish speaking country or you studied the language?
Lydia Alvarez

 

[DaTario]

I am from Brazil, Rio de Janeiro. And thank you so much for these explanations.

Best wishes

DaTario

 

[DaTario]

Dear Lydia,

Intending to explore just a little more, could you explain, in order to reach completeness in this thread, the mechanism through which the junction works as a rectifier, allowing current to go one direction and forbiding the other.



DaTario

 

[LydiaAC]

Hello DaTario:

The key is in the "barrier" caused by the electric field. This electric field pushes electrons from P side to N side and holes from N side to P side. So, electrons in the N side and holes in the P side (majority carriers) need to have more energy than this barrier in order to pass through the junction. On the other hand, electrons in the P side and holes in the N side (minority carriers) have no problem with the barrier and pass freely through it. This is important: the barrier affects majority carriers but no minority carriers.

Majority carriers are very sensitive to changes in the barrier. If the barrier becomes higher, flux of majority carriers decreases; if the barrier becomes lower, flux of majority carriers increases. You must remember that the equilibrium was reached when flux of majority carriers (by diffusion) was exactly the same that the flux of minority carriers (by drift). Since the flux of minority carriers does not change with changes in the barrier, you can see that a barrier higher than equilibrium will result in a net flux of minority carriers and a barrier lower than equilibrium will result in a net flux of majority carriers.

Then, if you add an external voltage to your diode, and put the positive terminal at the P-side and the negative terminal at the N-side (the same that the internal voltage) the barrier will increase. Then, holes will flow from N side to P side and electrons will flow from P side to N side The net result is a conventional current from N side to P side or an "inverse current". This current is very small, because it is caused by minority carriers which are very few. This current is also practically independent from the applied voltage, since the flux of minority carriers is not affected by changes in the barrier.

On the contrary, if you put your external voltage with the positive terminal in the P-side and the negative terminal in the N-side (reversed from the internal voltage) the barrier will be reduced. Then holes will flow from P-side to N-side and electrons will flow from N-side to P-side. The net result is a conventional current from P-side to N-side (direct polarization current). This current can be very high because the majority carriers are very much. Also, this current is very dependent of applied voltage. For very small voltages is small but for a little higher voltages, increase very swiftly.

It is important to note that the barrier is not a problem for minority carriers and the reason for the small inverse current in a diode is the scarcity of this carriers and not the barrier. If you manage to inject a lot of minority carriers in a material, you can have a very high current through an inversely polarized junction. This is what you do in a bipolar transistor.

I hope this is useful.

Lydia Alvarez