What Happens When PN Junction is Shorted?

[Bhargava2011]

Hi friends, here's my question.
The two terminals of a PN junction are at different potentials so, if I connect a them with a wire i.e. short them, What would happen? Would the potential difference between the two ends becomes zero or what? As far I'm understanding, there would be a flow of electrons from the P region to the N region because of potential difference. But as soon as the PD becomes less, the barrier potential becomes low and therefore the diffusion process will start, in which electrons would move from N to P and so on...
But, this would constitute a current without any battery, etc.
And why can't we measure inbuilt potential difference of the PN junction in Lab using voltmeters.
I'd greatly appreciate any help.
Thank you :)

 

[Bassalisk]

Hi friends, here's my question.
The two terminals of a PN junction are at different potentials so, if I connect a them with a wire i.e. short them, What would happen? Would the potential difference between the two ends becomes zero or what? As far I'm understanding, there would be a flow of electrons from the P region to the N region because of potential difference. But as soon as the PD becomes less, the barrier potential becomes low and therefore the diffusion process will start, in which electrons would move from N to P and so on...
But, this would constitute a current without any battery, etc.
And why can't we measure inbuilt potential difference of the PN junction in Lab using voltmeters.
I'd greatly appreciate any help.
Thank you :)

From my view, if you short it out, a small current will flow until that potential difference is gone. After that, diode will behave just like a wire.

 

[Bhargava2011]

From my view, if you short it out, a small current will flow until that potential difference is gone. After that, diode will behave just like a wire.

Thank you Bassalisk.
But, after the potential difference is gone i.e. the barrier potential becomes zero then the diffusion of electrons would start again since there is no electric field now that could oppose there flow.
Again, a pd would be created and it would lead to an another current of electrons moving from low potential to higher potential and so on...
So, we have a continuous current!
I know it won't work like that but, I'm not able to reach to any other conclusion.

 

[Bassalisk]

Thank you Bassalisk.
But, after the potential difference is gone i.e. the barrier potential becomes zero then the diffusion of electrons would start again since there is no electric field now that could oppose there flow.
Again, a pd would be created and it would lead to an another current of electrons moving from low potential to higher potential and so on...
So, we have a continuous current!
I know it won't work like that but, I'm not able to reach to any other conclusion.

It does make sense what you are talking. Interesting. I will see this through and come back to you. If cabraham or sophiecentaur come by, they will answer it for sure.

 

[Bhargava2011]

It does make sense what you are talking. Interesting. I will see this through and come back to you. If cabraham or sophiecentaur come by, they will answer it for sure.

Thanks Bassalisk, I'm eagerly waiting for the solution. And cabraham or sophiecentaurc, sounds like they are the experts here.

 

[jim hardy]

there's something called a "Depletion Region" where holes in P region get filled by electrons from N region.. and vice versa

http://en.wikipedia.org/wiki/Depletion_region

When enough charge (a minute amount) has moved to complete that process, current ceases.

The potentials are internal and they cancel so external voltage is zero.

You MIGHT be able to measure the charge with an electrometer - i don't know - would make an interesting lab experiment.

Something related happens in a vacuum tube diode. Shorting its terminals will result in a small current - electrons are shaken off the hot cathode and will migrate to the plate. It's called (if i remember right from 1961) contact potential and can produce a couple tenths of a volt.
But in a vacuum tube diode there is an external energy source, the cathode heater.

hope I'm right - keep me honest, centaur?

old jim

 

[Bhargava2011]

The potentials are internal and they cancel so external voltage is zero.

Could you explain it in details. I didn't get how the potentials can cancel each other.

 

[jim hardy]

the Wilipedia link has a drawing showing the gradient inside the silicon - a picture is worth a thousand words

think of an analogy: two capacitors in series, both charged to maybe 0.1 volt, connected + to +

between their negative leads there is zero volts

 

[Bhargava2011]

the Wilipedia link has a drawing showing the gradient inside the silicon - a picture is worth a thousand words

think of an analogy: two capacitors in series, both charged to maybe 0.1 volt, connected + to +

between their negative leads there is zero volts

I think you are talking about this picture

And I got the concept of the capacitors but, I don't see any analogy. Here, it is clearly shown that the two regions p and n are not at the same potential, the n is at a higher potential and the p at a lower one. So, there is a potential difference between the two terminals.
Please point out if I'm making any mistake in the analysis.
Thank you.

 

[maverick280857]

Current will not flow!

You MIGHT be able to measure the charge with an electrometer - i don't know - would make an interesting lab experiment.

Its called the junction potential, and NO, you cannot measure it. First of all it is established across the depletion width at equilibrium. We are not talking of conduction here but rather equilibrium attained during the formation of the depletion width itself. The depletion width is formed (by diffusion across the heterogenenous metallurgical junction) in consistency with Poisson's equation. (Another viewpoint is that it is formed so that Maxwell's equations remain satisfied.)

Remember that the assumption made here is that there is no electric field outside the depletion width! This is an idealization, but it is a reasonably good model at low voltages and for reasonably long devices (with lengths greater than the diffusion length -- the so called long base assumption, a term which comes from BJT terminology).

So, please understand that any act of measuring the junction potential even under the assumption that the contacts are Ohmic and ideal will disturb equilibrium. The diode in this condition is not a voltage source. The potential difference across the depletion width cannot be measured without also accounting for the contact potential at the contacts. A careful calculation will tell you that the contact potential difference exactly equals the junction potential in magnitude with the polarity reversed. So since the 'electrode measurement' will measure the algebraic sum of the contact potential difference and the junction potential, the result will be 0.

The fact that the junction potential cannot be measured directly (and can only be inferred) is very well explained in most standard books on semiconductor physics and devices. Please see for instance the book Operation and Modeling of the MOS Transistor by Yannis Tsividis, or the book Solid State Electronic Devices by Ben Streetman and Sanjoy Banerjee.

PS (to the Original Poster Bhargava2011): Please do go through the books I've suggested, and do not rely on Wikipedia over a standard textbook on the subject -- there are many mistakes!

 

[Bhargava2011]

A careful calculation will tell you that the contact potential difference exactly equals the junction potential in magnitude with the polarity reversed. So since the 'electrode measurement' will measure the algebraic sum of the contact potential difference and the junction potential, the result will be 0.

I'd try to refer the books you've mentioned.
But, could you please explain in detail about the contact potential and the junction potential and their sum. I think its the main point here. If I'll understand it, my problem would be solved.
Thank you so much.

 

[es1]

Thank you Bassalisk.
But, after the potential difference is gone i.e. the barrier potential becomes zero then the diffusion of electrons would start again since there is no electric field now that could oppose there flow.

But if the potential difference is gone what pushes the charge over the barrier. The height of the barrier might be zero but if there is no field to push over it then there will still be no current.

 

[maverick280857]

I'd try to refer the books you've mentioned.
But, could you please explain in detail about the contact potential and the junction potential and their sum. I think its the main point here. If I'll understand it, my problem would be solved.
Thank you so much.

The full explanation is somewhat tedious to write down and will not be illuminating without a diagram. I know that Tsividis has it, and I shouldn't deprive you of the thrill in figuring it out yourself (since he explains every detail carefully anyway). If you are interested in device physics, you should definitely check out his book...its a classic.

 

[jim hardy]

""A careful calculation will tell you that the contact potential difference exactly equals the junction potential in magnitude with the polarity reversed. ""'

So Maverick

do i understand there's an effect from dissimilar affinity for electrons between the silicon and the metal leads? Like in thermocouples?

It's that simple?

I'd have thought that right at the outside edges of depletion region there'd be a gradient due to immediate proximity of excess charge, which at a few depletion region widths would fade. Once you're a few microns away from depletion region, distances to its two sides + and - don't look much different so cancel.. That's where i thought the equal and opposite potentials were located, right in edge of D R..

learn something every day maybe i'll know domething one day.

old jim

 

[Bhargava2011]

But if the potential difference is gone what pushes the charge over the barrier. The height of the barrier might be zero but if there is no field to push over it then there will still be no current.

There is something called diffusion current. The electrons diffuse into the p region which gives the current. But, my concepts are not clear enough so, not absolutely sure!

 

[jim hardy]

""There is something called diffusion current. The electrons diffuse into the p region which gives the current. But, my concepts are not clear enough so, not absolutely sure! ""

Mr Bhargava,
i too am struggling to make my mental picture a clear one.

I do not understand what happens right at edge of depletion region.
Seems to me electrons diffuse into p region, attracted by p donors' affinity for electrons, until the electrostatic pull from the ions they left behind just balances the tug from p donors.
(of course it's a symmetric push from n and pull from p donor affinities balanced by coulombic attraction but i simplify for sake of brevity)
Hence the depletion region width is calculable from doping density and a few other numbers.
Seems there's a wall of charge at each edge of depletion region so why a gradient on only one side?

i am puzzled.




This link seems to support what Maverick said about the careful calculation;
that where the metal leads join the silicon is the location of the equal and opposite voltages required to satisfy Kirchoff and not at edge of the depletion region like i thought.
http://ecee.colorado.edu/~bart/book/book/chapter3/ch3_3.htm


So be advised this reply that i prepared last night probably has a glaring mistake.
Maybe we can tweak it into something useable for a teaching tool.

Yes that Wikipedia link was not a good choice.

I think this subject suffers from unclear teaching, same as "why does a rocket go".

The reason there's no external voltage is there's equal and opposing potentials inside and outside the depletion region.
Were this not so, a diode would deflect a voltmeter.

Here's a more scholarly writeup
http://ecee.colorado.edu/~bart/book/book/contents.htm
this section goes to to your question:
http://ecee.colorado.edu/~bart/book/book/chapter4/ch4_3.htm

and page 21 of this link gives a picture that paints for me the mental image i need to work it in my head:
http://www.scribd.com/doc/6661380/The-Junction-Diode2

look at the picture on p21 entitled "Diodes - The Depletion Region"
let us do a thought experiment on that picture.

1. Connect a wire from the left side of p region to right side of n region, so that the n and p regions are at same potential.
No current can flow through that wire because the depletion region prevents current flow between n amd p regions.

2. Imagine yourself very small, small enough to walk among the atoms in the silicon crystal.
Imagine also you are carrying a unit positive charge in one hand.
Imagine also you are equipped with a meter that can integrate the product of Force acting on your Unit Charge X Distance traversed. This would be the work done in moving the charge.

Recall what is potential - potential is work done in moving a unit charge from one place to another.
Absolute potential is work to move a charge from infinity to where you are, which is impractical because you can't get out there to make the measurement.
So we always pick some place for a reference and measure potential with respect to that point, hence voltage is a potential "Difference".

So let's call your [edit ] force X distance meter [ /edit] a "Potential Meter". And let's say potential increases when we have to supply the work.

3. Imagine yourself inside the silicon at the left side of p region where wire connects.
You see the crystal structure, orderly atoms with an occasional missing electron.
You set your "potential meter" to zero here.

4. Walk toward right approaching depletion region, and watch your potential meter.
We are approaching a region that has excess electrons, which attract our positive charge, so we are extracting work not supplying it.
Potential meter is decrementing because we are being pulled not pushed.
Note reading when you reach the p edge of depletion region. Negative.

5. Enter depletion region, continue to right. Now we are moving away from negative charge toward positive, so force on our our unit positive charge is other direction, pushing against our travel. Instead of extracting work we are having to supply it. Potential meter is now incrementing.

6. Continue moving right. Note meter reading when you reach n edge of depletion region.
Positive.

7 . Continue moving right through n region. We are moving away from a region of positive charge so force is once again in direction of our travel. Meter resumes incrementing. Note meter reading at right edge of N material.

I submit that your potential meter MUST read zero at three places:
far left where we started,
middle of depletion region,
far right where we ended.

and i submit readings at edges of depletion region are equal and opposite.

My model does not agree with the idealization Maverick pointed out in bold above, 'no field outside depletion region'.
That's doubtless so at distances greater than depletion region's width where distances to positive and negative regions are about the same.
What really goes on at the outside edge of depletion region?
If there's no gradient there, it seems to me it violates Kirchoff.

But i could be wrong.

Different explanations click for different people.

Here's another related PF thread
https://www.physicsforums.com/showthread.php?p=2658119#post2658119

old jim

i think i should repeat my thought experiment, but start my walking tour inside the wire not in the silicon. I thnk i will find a region with the potential gradient i need right where the wire joins the silicon, at extremities of the silicon not extremities of depletion region..
it seems logical an undepleted region may not support a gradient for charge would move. That could force the gradient to appear instead in another region where there's dissimilar electron affinities in proximity, like where the electron rich metal leads attach (as Maverick said) .

Bhargava - i post my thought experiment above uncorrected, with apology, please do not think me discourteous..
What if I'm wrong on both ?

maybe some of the more mathematically facile denizens like you and Maverick and Centaur and Yungman will help me out.

old jim

 

[es1]

There is something called diffusion current. The electrons diffuse into the p region which gives the current. But, my concepts are not clear enough so, not absolutely sure!

This is a little sloppy in terminology but the minority carrier diffusion current gets balanced by the drift current in steady state.

For a quick proof, assume the drift current did not match the diffusion current in steady state. Then there would be a build up of a charge which would mean that the width was changing. This is a contradiction, so they do match.

 

[Bhargava2011]

Jim, i think it would be better to refer books mentioned by Maverick. We may get a clear picture there. And here are some links which might be helpful http://www.youtube.com/watch?v=IMoJUqDlSQs"

 

[Bhargava2011]

This is a little sloppy in terminology but the minority carrier diffusion current gets balanced by the drift current in steady state.

For a quick proof, assume the drift current did not match the diffusion current in steady state. Then there would be a build up of a charge which would mean that the width was changing. This is a contradiction, so they do match.

hmm...I didn't think about that. I'll refer some books, anyways thanks :)

 

[es1]

I would recommend walking through the derivation of the Shockley Diode equation. Once you understand where the equation comes from you'll understand why it also has the point 0,0.

The textbooks do it best. Pages 430 to 439 have the basic derivation in the link below.

http://books.google.com/books?id=vC...&q=Shockley diode equation derivation&f=false

 

[jim hardy]

thanks , guys

i had a feeling something was "fishy" about my thought experiment. Shoulda looked closer at Poisson...

firstly i didn't go completely around the circuit.
secondly i was oblivious to "contact potential" at wire-to-silicon connection.

My bifurcated brain is wrestling right now with two more coccepts
a.. if i were to represent a diode as not two bars of dissimilar silicon but two semicircles,
then join them at both ends not just the middle;
that'd make two p-n junctions and voltage drop around the loop would be zero
b... that's not physically unlike a thermocouple where there's only a net emf if junctions are at different temperatures

when I've come to grips with what is fundamental difference between
dissimilar metals in a thermocouple; and
dissimilar silicons in a diode;

i will believe i have a good mental picture that i could explain to somebody without much math background beyond algebra.

somehow the "affinity for electrons" is involved in both.

Thanks for your patience and your help.
I spent a lot of my career (as an engineer in a maintenance environment) helping non-mathematical thinkers understand how stuff works. The old fire-horse in me senses a need for nuts-and-bolts explanation of this effect.
The people who keep wheels of industry in good repair are not your physicists and mathematicians but everyday folks who need such explanations - it makes them more effective troubleshooters.

And that's the mark of a good educator or manager, achieving extraordinary results with ordinary folks.
It's a tide that lifts all boats.