I never actually understood what exactly is a Hole? Could you explain it clearly.
It's simply the absence of an electron.
Because of historical accident, before the discovery of the electron, current was assumed to flow from positive to negative.
It's a bit error prone to have electrons moving one way and current moving the opposite way - so we imagine the movement of the absence of electrons, which go in the same direction as current
But that don't seem to explain the whole story. If movement of holes were actually movement of negatively charged electron in the opposite direction then the Hall effect shouldn't have shown positive hall voltage. But as far as I know, the holes behave totally like positive charges, in every respect. This is what bothering me for years!
Consider a string of hydrogen atoms. If you take away one electron on a particular hydrogen atom you would get a string of something like: ... H, H, H+, H, H ...
If the current moves in the right-direction the positive-charge (not the hydrogen cation!) moves in the left-direction. As you can see the hole is the positive charge induced by the un-screened nucleus.
If B is along +ve z and either velocity of holes is in +ve x or velocity of electrons is in -ve x, Hall voltage is +ve.
If B is along +ve z and either velocity of electrons is in +ve x or holes in -ve x, Hall voltage is -ve
I don't think thats correct. Look at the following illustration I sketched
If the movement of holes were actually movement of electrons, then should always find -ve hall voltage (Right Figure) . Shouldn't we?
I understand your problem but not the solution. Apparently the others are unfamiliar with the the Hall effect in P-type materials in violation of classical elecrtromagnetism. Perhaps someone can enlighten us both.
Maybe you should take into account that electrons at the upper end of the band have a negative effective mass. Hence a hole (the absence of one of these electrons) has a positive mass.
I agree with the critic.
O.K. I guess you mean that the holes are not just the absence of electron but absence of electron with negative effective mass (Someone somewhere has already told me this). But if that is the case then; when the electric Field is applied in upward direction (the right side figure of my sketch some posts above), then electrons having negative mass should have moved in the same direction (upward). O.K yes, That could produce and hence explain the +ve hall voltage, but the absurd thing is that, that would result in current in downward direction (opposite to direction of applied field).
But we always see the current in the direction of applied field, don't we?
In post#6 you have a contradiction.
This is because you are trying to show the current in three different ways
That is not a good idea.
You have shown a conventional current. This current is in one direction only, regardess of the movement of its carriers and is one of the determining vectors of the Hall voltage.
This conventional current arrow may be made of all holes, all electrons or a mixture.
Regardless the arrow is still the same.
So the two important vectors in your diagram are the magnetice field and the current (with its associated electric field of course).
This leads to build up of negative charge on the left and positive on the right which in turn makes the polarity Hall voltage, developed across the section negative on the left and positive on the right.
Are you also aware that this voltage develops its own electric field so that the final electric field is no longer parallel to the current direction but the vector sum of the two fields?
I think this is a fine example of trying to work in terms of charge carriers rather than a (ficticious) entity called conventional current brings difficulties. All the other equations of physics are adjusted to be in sync with this.
Yes, I have considered that positive and negative charges are deflected in opposite directions; but since their direction of motion is also opposite; they are finally deflected in the same same direction, as shown in the figure (post 6).
Either you missed it or I am in serious trouble with the Left-Hand Rule.
But your current vector is the same in both diagrams.
The LH rule expects this vector (as you have it) irrespective of the polarity of the carriers.
This is a sign convention thing. You also have to start introducing -ve signs into the vector cross product if you must use charge carrier direction. Don't forget that the current vector also defines a direction in space, ahich you have chosen by using conventional current (wisely in my opinion).
This subject is being over-analyzed. IMO the Hall Effect experiment doesn't adequately describe what a hole is. A hole is not a particle. Yes, I said it! A hole is an absence of a particle. In semiconductors the absence of an electron within (not at the top) the DOS is manifested locally on an atom. Thus, it may seem like a hole is a particle, but in reality it is not. The positive charge of the hole comes from the nucleus. Holes are created by adding localized p-type defect states near the VBM, which pulls electrons out of the VB and this creates holes in the VB. The p-type defect gets neutralized while an atom in the parent lattice gets ionized . This is why holes are localized around an atom. In metals, a hole has a different meaning because as the temperature is raised you are technically creating holes in the VB, but the holes are not localized to a particular atom, which is why it's impossible to describe current in terms of hole-flow in metals.
I hope this helps
Yes you have shown the electron and hole flow in opposite directions.
Your diagram in post 6 is both correct and compatible with vin300 statement in post 5.
The Lorenz force pushes both carriers to the same side. This is fairly easy to show mathematically.
As a result the Hall voltage across the section will be either positive or negative on the left depending upon whether the flow is holes or electrons.
This situation when there are both carriers present (semiconductors) is much more complicated. It is not true to say that just because there are more holes or electrons the left side will end up more positive or more negative. This is because the Lorenz force equation has to be modified to take into account the diffeent mobilities of holes v electrons. This can lead to the minority carrier determining the Hall voltage polarity.
You can have the equations if you want them, but it is a lot of algebra.
Its a great relief to hear that.
Now, I am at point at asking what I intended to ask.
In some material Hall Voltage is found to be positive. This is explained by saying that the current is constituted by holes (Left figure, Post 6).
But the contradiction arises when we study about the nature of holes. If we are to say that holes are actually absence of electrons (and there is no such thing as positively charged hole) and that the motion of holes is accomplished by transfer of electrons in the opposite direction, then we are forced to model the current flow in the material by Rightside Figure in Post 6. And now to the arousal of great contradiction the Hall Voltage is Found to be -ve.
Please for gods sake, let be get through this. I am dying.
First let us look at the model you are employing in post#6.
The lattice is electrically neutral. So if we take a negative electron from point A and shove it over to point B, point B becomes –1 negative and this leaves a corresponding +1 positive charge at point A, supplied by the positive charges in the area that were formerly balancing the -1 charge on the electron we have moved.
However, once the hole and electron have separated (ie the pair has been 'created') there is nothing in the model to continue to link a particular hole to a particular electron. If the electron moved on again to point C it would not leave another hole at B, just a neutral point.
Equally the hole could move by ‘displacing’ a different electron, or if you like a different electron could occupy point A which would again be neutral, leaving a +1 hole somewhere else in the lattice.
Now we know that all charges, holes and electrons, moving as part of the current, are subject to the leftwards Lorenz force.
But all other electrons in the lattice are not subject to this force as they are not moving as part of the current. And there are plenty available.
So a leftwards movement of the hole at point A can be accompanied by a rightwards movement of electrons without violating any laws.
This, of course, is movement at right angles to the current flow so no work is required from the electric field causing the current ie there is no interference with it.
Note also my earlier comment about the different ‘mobilities’ of holes and electrons.
Now to address the title question of your thread.
A simple explanation of holes is positional. This is because electrons are regarded as point particles. As such they have a location in space and it makes sense to talk about electrons moving one way and 'holes' moving the other.
However when an electron moves it does not leave behind ‘nothing = absence of electron’
It is rather like pigeon holes in a letter rack and letters or pages in a book and writing. Just because there are no letters in a particular pigeonhole or no writing on a particular page does not mean that the pigeon hole or page does not exist. They are still there, available for letters or writing.
So if you imagine the solid block as the rack or book and the pigeonholes or pages as the wave states and the letters or writing as the electron perhaps you can see the analogy.
Remember the analogy is not perfect, just an aid to visualisation.
Further electrons and holes are not point particles. They have no location in space. In solids the wavefunctions extend across the entire block of material.
Some are filled some are not, but all are present.
This is how the electron can actually ‘move’ – It jumps from state to state. This is a quantum thing and the quantum Hall effect is much more complicated than the classical smooth one.
1. Could the confusion be that you mixing n-type materials with p-type? I believe current is solely due to the particles in the conduction band, so it might depend whether you have n or p majority.
I did not read through the whole thread .... I would have to review the Hall's effect before coming with a valid answer
I read and re-read your reply for some hints, but please forgive me, I couldn't still get it.
This is explaining how conduction begins, Right?
This is explaining how conduction continues.
Same story repeats, I think. How can electrons be made to move rightwards?
I would like to put it this way
Be it n-type or p-type material,
1. the only thing that can move is electrons, right? (Am considering Classical Model)
2. If the current is in upward direction, No matter how you try to explain the phenomenum, The net electron flow must be in downward direction, Right?
3. If the elctron flows in downward direction, then Lorentz Force pushes it left, and Hence
-ve Hall Volatge, Right?
4.I must be wrong at least Somewhere, Right? But where (1,2,3,4)?
A hole is NOT merely an absence of an electron. By that definition, no charge would be a hole.
A hole IS the absence of an electron in a formerly charge-balanced material, giving the material a net positive charge.
Your problem is in number 2/3
The crux of it is here
The electrons that need to move rightwards are not the electrons that take part in the upward flowing current.
Consequently they are not the ones being forced leftwards by the Lorenz force.
(You are aware that the force you have labelled 'Thrust' is called the Lorenz Force?)
Don't forget there are no electrons moving directly leftwards. These are the ones whose trajectory has two components, viz the upward flow of current and the leftwards push, resulting in a path curving towards the left.
Yes, I am aware of that.
I am aware of that too.
Then which electrons are they? And which force causes them to move rightwards?
(I think I have reached to the root point, now. Wow!)
Ok, here's my view:
is the equation of motion. Due to friction, the acceleration will lead to an equilibrium velocity v that will show in the same direction as the acceleration a.
Case a: m>0, q<0 (electrons at the lower end of the band)
v is antiparallel E, j =q v is parallel E. If E is in x-direction and B in z-direction, then the electrons will be deflected in direction of -y.
Case b: m<0, q<0 (electrons at the upper end of the band with negative effective mass)
v is parallel E, j is antiparallel E. If E is in x-direction and B in z-direction, then the
electrons will be deflected in direction of -y, as in case a. However, as the direction of the current is reversed in comparison with case a, the hall coefficient will be of alternate sign.
case c) m>0, q>0 (holes):
v is parallel E, j is parallel E and holes are deflected in direction of -y. The Hall coefficient equals the one in case b.
In the band picture, the situation in a nearly full band is complicated by the fact that most levels are filled and "Pauli block" the mutual motion. In a completely full band, the electrons cannot react at all to the applied field and acceleration a=0. In a nearly full band, the mean acceleration of the electrons will be very small, but will on the mean follow case b. Alternatively, the situation can be described in terms of some holes of case c.
Thanks for that DrDu. I was hoping you'd come up with more. You are the one who has done the work, not I, so I'm a little embarassed to mention this. In case b) where electron mass is less than zero, j is antiparallel with E. The positive current is moving counter to the applied field! So we get perpetual motion due to negative resistance (which would be awfully nice to have...)
I think we might consider the equilibrium condition where an electric charge gradiant has already accumulated in the ±Y-direction so that there is an additional Ey field due to this charge.
The drift velocity, vx = (½d)max = k-1max, where d is a some nominal positive distance between collisions, so that k is positive and
sign(vx) = sign(ax). (I'm just repeating what you’ve established in this.)
kvx = max = qEx + q(vyBz)
vy=0 by conservation of charge in the equilibrium condition, so that
1) kvx = qEx. Now the electrons are not giving us free energy. Bummer.
kvy = 0 = may = qEy – vxBz so that
2) qEy = vxBz
Equations 1 and 2 are as far as I've gotten. Maybe the negative electron mass model works. I have no idea. ---or did I make a fundamental error somewhere?
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