# PN Junction Diode

1. Jun 22, 2006

### meldave00

During the processing of a PN junction diode when the Ptype material and the Ntype material are joined together. The holes tend to diffuse from the Ptype material to the Ntype material and conversly the electrons from the Ntype material diffuse from the ntype material to the ptype material. This diffusion eventually stops because the minority carriers that diffused over now create an Efield that stops the diffusion of charge and hold them in place. This is called the depletion region of the PN junction diode.
The Efield produces by the minority carriers in turn produces a built in potential called Vbi.

My question?
1). If I put a voltage meter across a PN junction diode. I get 0 V. Why don't I get Vbi????

2). What if somehow I was able to probe at the edges of the depletion region. Would I then be able to measure vbi instead of 0V?

regards,

David

2. Jun 23, 2006

### Hammie

From what I remember, I was told that if you could probe inside of the diode you could measure this voltage. I think it can also be measured indirectly by observing the forward voltage drop accross the unit.

3. Jun 23, 2006

### Gokul43201

Staff Emeritus
The voltmemter measures the difference between the mean values of the potential energies of the charge carriers at the two points. On one side, this is the mean potential energy of a lot of holes and a tiny number of electrons; on the other, it is the mean PE of a lot of electrons and a tiny number of holes. At equilibrium, the mean energies on both sides must be equal and hence, you won't measure any potential difference.

If you could measure the potential energies of only one type of charge carrier (ie: electron/hole) and compare between both sides of the junction, you will see an energy difference proportion to Vb. (Well, that's essentially the definition of the barrier potential - it is the energy needed by each type of charge carrier to cross over)

Don't believe so, for the same reason as above.

Here's a typical picture : http://www.techfak.uni-kiel.de/matwis/amat/semi_en/kap_2/illustr/pn_junction1.gif

Last edited: Jun 23, 2006
4. Jun 23, 2006

### antonantal

I will try to explain it from an electromagnetic rather than quantum point of view:

1)To simplify things a bit I will consider the spatial charge from the depletion region to be distributed entirely on the edges of the region: all the positive charges are in the N side on the edge of the depletion region and all the negative charges are in the P side on the edge of the depletion region. Ofcourse you will have an E field between the 2 edges which is the sum of the E fields produced by each edge alone. But each edge produces E fields in both the left and the right side of it. So although you have an E field between the edges, the sum of the E fields from one metallic contact to the other will be 0 and so will be the voltage.

2)In this case you would be able to measure Vbi because you will measure the voltage resulting from the sum of the E fields from each edge, only in one direction (constructively).

5. Jun 23, 2006

### meldave00

PN Juntion Diode

Thanks for the reply. Its food for thought. I will think about what you said. I appreciate it.

6. Jun 25, 2006

### vanesch

Staff Emeritus
I'm not 100% sure myself that the following is entirely correct, but I'm inclined to say that what a voltmeter measures, is the electrochemical potential (usually \mu) of the charge carriers in its test pins, and not really the electrostatic potential, because after all, every voltmeter is based upon the *displacement* of charges.

Now, as much as in a PN junction, the *electrostatic potential* is not constant, but makes the known jump in the depletion layer (where there is hence an electrostatic field), normally, the electrochemical potential is constant across the junction for both kinds of charge carriers (which is nothing else but the thermodynamic condition of equilibrium).

If you make contact with the test rods of your voltmeter, a new equilibrium will arise across this contact, which will put the electrochemical potential equal on both sides of the contact, and this for the two contacts. Hence, the electrochemical potential at the two testrods is the same, and the voltmeter will read 0.

Now, of course, for the voltmeter, if the two testrods are about of equal composition, one can subtract the "thermodynamic" part of the electrochemical potential, and hence obtain the purely electrostatic potential, so it can be said that the electrostatic potential between the testrods (ONCE THE CONTACT IS MADE) is equal too.

But there has been introduced an extra electrostatic field at the two contacts (voltmeter rod - semiconductor) which compensates the electrostatic field in the PN depletion layer, simply by the requirement of equal electrochemical potential of charge carriers at the contact.

(at least, this is how I always understood this apparent "paradox").

7. Jun 25, 2006

### Gokul43201

Staff Emeritus
Vanesch, I think I agree with most of what you are saying - though the differences in "language" are throwing me off a bit. I guess I have a different way of thinking about this (eg: electrostatic potential gradients make charge currents and chemical gradients make particle currents, which must "cancel each other" in equilibrium...).

In any case, do you believe the voltmeter will register a non-zero number when the probe tips are touching points near the edge of the depletion regions? Would you expect the electrochemical potential to have a gradient (at equilibrium) in/near the depletion region?

I'm still not seeing the paradox here, so either I'm terribly confused, or I just don't understand what exactly is being said.

8. Jun 25, 2006

### Gokul43201

Staff Emeritus
It sounds to me like antonantal is modeling the depletion region like a parallel-plate capacitor (where the field between plates add, but outside, they cancel each other). Or am I misinterpreting this?

9. Jun 25, 2006

### vanesch

Staff Emeritus
Electrostatic gradients only make electrostatic fields
All net particle movement is dictated by gradients of electrochemical potential, which equals (up to some constant) however the electrostatic potential for charge carriers in a material of uniform composition.
To be sure, the electrochemical potential is the derivative of the local Gibbs free energy to the number of charge carriers under question.
In thermodynamical equilibrium (such as you quickly reach when measuring with a voltmeter on a PN junction, as there's no reaction going on anywhere), by definition, the electrochemical potential is equal everywhere.

No, because by definition, at equilibrium, the electrochemical potential of the charge carriers is the same everywhere. So no matter where you measure, you will always read 0 V on the voltmeter (unless its sense pins are, say, at different temperatures or so).

There's no paradox of course. It is just that the electrostatic potential, integrated ONLY over the junction, equals some, say, 0.6 V, while if you try to measure it with a voltmeter, it reads 0V, which may seem somewhat puzzling.
But this is because you introduced yourself two new junctions when you applied the test rods to the semiconductor, which will compensate exactly those 0.6 V (if the testrods are made out of the same material - and there will be later junctions of they aren't).

So a voltmeter does measure electrostatic potential (when its testrods are of the same material) AFTER we take into account the contact potentials of its testrods with the thing under study ; OR we can say: the voltmeter measures the electrochemical potential of a type of charge carrier of the two points we contact (and then we don't have to take into account the contact potential because that's taken care off automatically).

10. Jun 26, 2006

### Gokul43201

Staff Emeritus
I'll read this again, in less of a hurry, but I think I see the source of the cross-talk. I thought you were concurring with anto (and hence refuting me) on question 2, that there will be a voltage measured by probing near the depletion region.

On the other hand, I believe you think I'm saying that there is no internal gradient in the electrostatic potential (which is not what I'm saying - in my first post, for instance I was refering to Fermi levels being equal throughout, not electrostatic PE).

But I'll read this again later.

11. Jun 26, 2006

### antonantal

You're right. The misinterpretation is on my side. At first I thought that the depletion region is wide enough so that the outer electric field doesn't vanish. But than I realized that if any E field would exist outside the depletion region the state would not be equilibrium, and the carriers would automatically move and compensate the field in order to reach equilibrium.

I also realized wich was the simple "electromagnetic" explanation to the fact that if you put a voltage meter across a PN junction diode you get 0 V. When the diode is not forward biased, the built-in potential stops the carriers in the neutral regions from diffusing and entering the depletion region. So since there is no carrier flow from the neutral regions to the depletion region, the neutral regions are practically acting as electric insulators for the depleted region. So it's like trying to measure the voltage across a charged capacitor which has it's terminals covered with insulator.

12. Jun 26, 2006

### antonantal

We are talking about measuring the voltage on the edges of the depletion region. The depletion region doesn't contain charge carriers (in the approximation of total depletion and neglecting the thermically generated carriers)

13. Jun 27, 2006

### vanesch

Staff Emeritus
How are you going to do that, to connect your voltmeter to "the edges of the depletion layer" ??

14. Jun 27, 2006

### vanesch

Staff Emeritus
Well, as this is practically impossible, it all depends on what one ideally defines as measurement I guess.
I fully agree that there is an *electrostatic* potential difference across the depletion layer (in other words, there is a non-zero electric field). But now, the question is: how is one going to *measure* this potential difference ? And I answered this from the point of view of a classical voltmeter with two contacts, which you have to put somewhere in contact with the "thing" you're going to measure. I based myself on two hypotheses: that a standard voltmeter is going to measure a (eventually tiny) displacement of charges, and that the two contact pins are made of identical material, and that we are going to look at charge displacement in equilibrium. I then tried to point out that all charge displacement is finally ruled by the electrochemical potential of the charge carrier that is supposed to displace, and that in any equilibrium situation, this electrochemical potential is constant. So it is not possible, with a standard volt meter, to measure anything else but 0V in an equilibrium situation.
So I concur with you that *no matter how you connect your voltmeter to a part of a PN junction*, you will always read 0V on your voltmeter.

One can interpret that in 2 ways. One can say that a classical voltmeter measures not electrostatic potential, but electrochemical potential. And the electrochemical potential is constant across the PN junction, so it is normal that one reads 0.
Or, one can say, that the classical voltmeter measures the electrostatic potential ON ITS PINS. Both statements are equivalent, because the pins are made out of the same material, so there's just a constant difference between the electrochemical potential and the electrostatic potential *in the pins*. However, this implies that one has to take into account any electric field that will be created by charge movements between the pin material and the material under test, which will be responsible for an extra electrostatic potential difference between the electrostatic potential of the pin, and the one of the material under test.
So, it is somehow simpler to say that you measure simply electrochemical potential with a voltmeter. And then it is obvious that, no matter how you're going to plug your pins onto a PN junction, you'll always find 0 V.

Now, I've been thinking about the following situation, and I have to say I'm not sure about the answer:
imagine that we drill a small hole through the entire PN junction, and we drop an electron into the hole. Will it accelerate or not when moving through the hole ? It would only see the electrostatic field, so I'd be inclined to say that you would see the electron accelerate across the hole. But then I'm not clear about any charges building up along the wall of the drilled hole, which might compensate...

15. Jun 27, 2006

### Gokul43201

Staff Emeritus
Oh, if you drop an electron into a hole, they will recombine with a flash of light!

I second that.

Does it help to paraphrase your question as: "Will there be a dipole-like field ouside a diode?" Yes, there would, wouldn't you think?

16. Jun 30, 2006

### meldave00

PN Juntion Diode

Hi,

What is the difference between electrostatic potential and electrochemical potential?

regards,

David

17. Jul 2, 2006

### vanesch

Staff Emeritus
Maybe this Wiki article can enlighten you:
http://en.wikipedia.org/wiki/Chemical_potential

The electrochemical potential is the chemical potential of a charged species, like electrons, or ions. One compound of it is the electrostatic potential of course, but there are other, thermodynamical contributions (essentially entropy). It is the increase in Gibbs free energy of the system, if we add one more particle (or a mole of particles) of the said kind to the system, under constant pressure and temperature ; or it is the increase in internal energy of the system if we add one more particle of the said kind to the system under constant volume and entropy.

The electrostatic potential is the increase in energy when we add one particle of the said kind to the system, *keeping the microstate of all the others the same* (in other words, by not changing the state of motion of all the other particles in the game).

It is a property of thermodynamical systems that, in equilibrium, the chemical potential of all different species is constant across the system (this minimizes Gibbs free energy). It is not a property that the electrostatic potential is constant across the system.
The gradient of electrochemical potential takes into account the "desire for diffusion, for reaction, for...." on top of the mechanical coulomb force. It determines the ultimate average movement of the charge carriers, taking into account diffusion, reaction and all that on top of electric field determined drift.