What is the polarity of a Pn junction diode

[Ahsan Khan]

Hello all

Textbooks teaches us that in the formation of Pn junction in between P and N type semiconductor a region is formed where free electrons of N diffused in P side and exhaust by the holes of P near the junction. Such a region becomes devoit of free charge careers and their are only ions in these region viz negative ions in P side positive ions in N side of junction. This forms the depletion region. And due to negative ions on P side and positive ions on N side a fictious battery is formed with negative terminal on P side and positive terminal on N side. And this the inbuilt voltage and will remain so no matter we use diode in forward biasing or in reverse biasing.

This direction of polarity is in compliance with theoretical details we deal but while dealing with circuit problems texts are giving solutions by considering P side to be positive and N side negative.

I am attaching the pictures attempting to use kirchhoffs law in diode problems assuming P as positive.

Any help will be highly appreciated.
Thanks a bunch :-)

 

[Ahsan Khan]

Do I need to be more clear?

 

[NascentOxygen]

Do I need to be more clear?

Yes, just the small matter: what is your question??

Your attachments make this appear to be in the nature of homework exercises. So you need to provide your attempt at the exercise in question.

In order for an ordinary diode to conduct current, an external voltage must make the P side more positive than the N side, by approximately 0.5v for a silicon diode.

 

[Ahsan Khan]

Yes, just the small matter: what is your question??

My question is that the P side should be negative and N side should be positive that is the potential barrier of diode should be such that P side(being negative) should be at lower potential and N side (being positive) should be at higher potential. In other words potential must increase as one goes from P to N. And this makes kirchhoffs law to say that under forward bias the external battery and potential barrier both are acting in same direction, this is something I am finding not good.

If order for an ordinary diode to conduct current, an external voltage must make the P side more positive than the N side, by approximately 0.5v for a silicon diode.

Yes such should be the case but that is coming possible only if we take P side of junction diode at higher potential and N side at lower potential only then will kirchhoffs law will condition the current to be flow only when applied voltage is more than a minimum value.

I request you to actually apply kirchhoffs law on the problems I gave, you will find they took P at higher potential and N at lower potential.
What I understand from text is that P should be at lower potential and N must be at higher. And here is the contradiction.

 

[NascentOxygen]

The barrier potential has a polarity with P negative and N positive. To cause forward conduction, the applied voltage needs to oppose this and overcome it, this means making the P material more positive than the N.

 

[Ahsan Khan]

The barrier potential has a polarity with P negative and N positive. To cause forward conduction, the applied voltage needs to oppose this and overcome it, this means making the P material more positive than the N.

Yes this should be but the workout problems(I attached pics in first post) use the reverse assumption( P side Positive and N side negative) in applying kirchhoffs laws. Just try to actually apply the law please.

 

[Ahsan Khan]

Why they make diode equivalent circuit with P side positive(at higher potential i. e replace it with bigger plate of battery) and N side negative(at lower potential I. e replace it with small plate of battery)

 

[NascentOxygen]

The barrier potential is not a voltage you can measure by connecting a voltmeter to a floating diode. Think of it as a hill that electrons must overcome in order to pass through. So 0.5v of the external voltage drops across the diode when it forward conducts.

So when it is conducting current you will measure 0.5v across it (with P more positive than N), in addition to a voltage of similar polarity due to bulk resistance of the material. The 0.5v you see with a conducting diode can be modeled as a voltage source, with P more positive than N because that's the polarity you measure with a voltmeter across the conducting diode.

 

[Ahsan Khan]

So should I say as one goes from P to N potential decrease? As the examples of the book is assuming.

 

[NascentOxygen]

So should I say as one goes from P to N potential decrease? As the examples of the book is assuming.

That is what the voltmeter shows for a forward-biased conducting diode.

 

[Ahsan Khan]

So 0.5v of the external voltage drops across the diode when it forward conducts.
The 0.5v can be modeled as a voltage source, with P more positive than N because that's the polarity you measure with a voltmeter across the conducting diode.

Wait!
How can be P side more positive than than N side. When the P side near the junction has negative ions and N side has positive ions.

 

[NascentOxygen]

Wait!
How can be P side more positive than than N side. When the P side near the junction has negative ions and N side has positive ions.

That has been your question from the start.

See those negative ions in the depletion region? To get electrons past them you need to give the electrons some energy (from an external source) to overcome the repulsion.

Perhaps do some reading about the depletion region on wikipedia.

 

[Ahsan Khan]

My earlier concepts(like [1]positive charge cause positive potential and negative charge cause negative potential [2] potential decreases in the direction of electric field) and the theory of textbook(which is saying [1] their are negative ions near P side and positive ions near N side, [2] their is electric field from N side to side) make me difficult to accept that potential decreases as one goes from P to N.

I know I must be missing something and I am feeling really good that I am being helped. It will more helpful if you give answer in more open way so that a poor fellow like me can know it.

Regards :-)

 

[Ahsan Khan]

That has been your question from the start.

See those negative ions in the depletion region? To get electrons past them you need to give the electrons some energy (from an external source) to overcome the repulsion.

Perhaps do some reading about the depletion region on wikipedia.

Yes electrons do need some extra energy to past them but it depends from which side(N to P or P to N) you want electrons to pass. In our circuit we want electrons to move from N side to P side. According to you the polarity of junction is positive at P side and negative at N side. And it is well known fact that in low to high potential direction electrons movement is favoured.

So it will difficult for electrons(if P side has more potential than N side) to pass from P side to N side as electrons have tendency to move from higher to lower potential. But this potential(P more positive than N) will aid the electrons in passing from N to P side as our case is. And one can notice in forward bias current(positive charge) moves from P to N and electrons are moving from N to P side. So your assumption that P is more positive tha N side should in fact aid the current.

Kindly make your own circuits and give direction of current.

I read in Wikipedia this

Departure of an electron from the N-side to the P-side leaves a positive donor ion behind on the N-side, and likewise the hole leaves a negative acceptor ion on the P-side.

The uncompensated ions are positive on the N side and negative on the P side.Regards

 

[AlexCaledin]

The uncompensated ions are positive on the N side and negative on the P side.

Those ions cannot move, so they form the embedded electric field across the junction; that field is applied to free carriers ("electromotive force") and directed from N to P - therefore P becomes the positive terminal of the "battery".

But, without forward current or pair generation near the junction, the "battery" is "disconnected" by the gap separating holes and electrons.

 

[Ahsan Khan]

Those ions cannot move, so they form the embedded electric field across the junction; that field is applied to free carriers ("electromotive force") and directed from N to P - therefore P becomes the positive terminal of the "battery".

I am also saying that field is from N to P, but the fundamental concept is that potential decreases in the direction of field which clearly means potential decreases from N to P, so P must be lower potential.
I don't know what is happening with me that why I can't grasp it or what is the matter how can people here at famous physicsforums can say that potential in the direction of field attaing higher values. :-(
Regards

 

[AlexCaledin]

The thing is, potential is somewhat tricky in a semiconductor... When electrons go from N to P, their potential energy increases as they "climb" over the junction - that's according to your potential. But in the P domain, they have to "fall down", across the energy gap, and all that additional energy is completely lost...

 

[Ahsan Khan]

The thing is, potential is somewhat tricky in a semiconductor...

I would like to know it in the simplest way, linking the issue in hand.

When electrons go from N to P, their potential energy increases as they "climb" over the junction - that's according to your potential.

When electrons go from N to P, what happens to potential in reality? Will like to keep in mind that the term potential is defined as the work done(by external agent without change in kinetic energy or negative of the work done by field) in moving POSITIVE charge between two points. So potential is conveniently defined for positive (test) charge, if it difficult for positive charge to move from side A to side B we say side B has higher potential. But if charge is negative(like electrons) if it is difficult to move from side A to side B, it would mean side A has higher potential.

When eleBut in the P domain, they have to "fall down", across the energy gap, and all that additional energy is completely lost...

I do not understand what you are saying here. Sorry for my dump mind. :-)

Regards

 

[Ahsan Khan]

I fear that it may be taken as arguments or at least considered repeating my own(false) misconceptions but my heart knows it is only due to hunt for the truth. I am not understanding how can potential be more at P than it is at N side when it is clear the field of depletion region is from N to P.

If this(P higher potential) is true I need a clear comprehending explanation for this. For I know electrons(negative charge) only feel difficulty in going to a region of low potential do the explanation that since electrons have to work hard in going from N to P will mean P has more potential is wrong for obvious reason(electrons require hard work to ho to rather low potential). Further how can we forget the fact that potential decreases on the direction of field an field is from N side to P side this further let me not accept that potential of P is higher than that of N.

Regards

 

[jim hardy]

while dealing with circuit problems texts are giving solutions by considering P side to be positive and N side negative.

There's the rub.

P and N describing material refer to doping.

Positive and Negative in circuits refer to externally applied potential across a device.

A diode whose P side is made positive with.respect to its N side by more than its barrier potential might conduct.

A diode by itself held in your fingers has no positive or negative side.

That unfortunate terminology confuses a lot of beginners. Maybe we should have called semiconductor doping D and R for Donor and Receptor, instead of P and N.
P and N almost rhyme with Positive and Negative,
but one applies to an internal property
and the other to an external connection..

its that simple, i think.

 

[Ahsan Khan]

The only thing I want to know with reason is why potential decreases as one goes from P to N side of junction diode?

I am having hard time going through circuit problems of diode where they apply kirchhoffs law. In applying kirchhoffs law for diode they take a drop of 0.7V as they move from P to N, while concepts says their should be a rise of 0.7V as one goes from P to N side.

As a matter of fact P side has negative ions and N side has positive ions potential at P should be negative and that at N is positive that is N should be at higher potential. So as one goes from P to N their should an increase or rise in potential. I am not taking about the applied potential( forward or reverse biasing polarity) I am talking about the self potential of diode(which alone however I know do not let any current pass in closed circuit) due to depletion region.

I request everyone interested in giving me help to actually apply kirchhoffs law in the circuits given on attached pictures before having said, what they assume, what I am asking.

 

[AlexCaledin]

Let us talk as mere tech guys. Potential is voltage measured with a voltmeter connected to certain wires. Wires have free electrons in (almost) equilibrium, that's why two ends of a wire have (almost) the same potential. The N-side of a diode, having free electrons, is like a wire. Its potential is determined by the external circuit connected to the N terminal of the diode - not by any ions.
Now let the external circuit provide the potential -0.6V (a silicon diode, P grounded.) There will be a considerable current across the junction, depending on the junction parameters and temperature. Often the external circuit keeps that current at a certain value, and it's in that case that the diode has its own "potential", called forward voltage drop, 0.5 ... 0.8 volt, N negative.

 

[jim hardy]

I request everyone interested in giving me help to actually apply kirchhoffs law in the circuits given on attached pictures before having said, what they assume, what I am asking.

would you post legible figures? Those in first post are too small and out of focus for me.

 

[Dr Ross]

The fictitious battery concept works if you imagine the EMF in parallel (+ on N side, - on P side) with the diode but ONLY at the junction region. An external battery connected - to the diode's P side + to N, will find the junction EMF opposing it and the region depleted of charge carriers. Reverse the battery (forward bias) lowering energy level differences and current flow is encouraged. Does this help?

 

[Ahsan Khan]

My question can be simplified if you see the below picture.

 

[NascentOxygen]

My question can be simplified if you see the below picture.

There is no picture below.

 

[Ahsan Khan]

Here it is. :)

 

[NascentOxygen]

I think we are going in circles.

 

[thankz]

try page 148 of sedra&smith microelectronics 1st ed ,that will give you the answer.

now off the top of my head I'm going to say it's because the electron orbital position (?am I saying it right) is jumping to a higher energy band during ion donation across the pn junction and that just happens to be .7 volts.

 

[Dr Ross]

Ovalis, it helped me to consider the electric field as a sort of vacuum that sucks electrons in its direction. It has the Potential to attract electrons to the positive symbol of the battery for example. The text you shared reads "our initial assumption [diode as battery] was wrong". There is emf at the pn junction but it is electrostatic. Benjamin Franklin mislabeled electrical polarities much to the confusion of subsequent generations and the concepts are hard to put into words anyway. Nascent and Alex gave (imo) helpful, clear, and simple information. About the Kirchoff solution, I suggest changing the sign on your diode junction's emf.