What is the polarity of a Pn junction diode

[Ahsan Khan]

Ovalis

Its ovais :-), Hehehe

it helped me to consider the electric field as a sort of vacuum that sucks electrons in its direction.

I don't know why you change a well known fact that electric field sucks positive charge(not the negative charge particles like electron) in its direction. It is very well known fact that force on positive charge is in the direction of electric field and that on negative charge in opposite direction.

Nascent and Alex gave (imo) helpful, clear, and simple information.

All are saying same thing without explaining me how can the side(P side) with negative ions act as higher potential while the side(N side) with positive ions act as lower potential. Field is from positive side to negative side, right?

In Pn junction positive charge is on N side and negative charge is on P side, right?

So field must be from N to P side, right?

Potential decreases in the direction of field, right?

As field is directed towards P, P side should have low potential than N side.

But text is showing reverse if you actually applied kirchhoffs law, in those circuit you will notice it.

Which is where I feel contradiction between concepts I learn and how the things were solved.

About the Kirchoff solution, I suggest changing the sign on your diode junction's emf.

If you read what I wrote above, my concepts shows junctions emf with sign on P side negative and N side positive. But book uses reverse thing.

I want you teach me reasons why I should change the sign of my diode junction? I must be missing something otherwise.

Regards, thanks a bunch :)

 

[Jony130]

Please read this
http://www.tutorvista.com/content/physics/physics-iv/semiconductor-devices/p-n-junction.php#forward-biasing

Or look at this figure

A p-n junction is said to be forward biased, if the positive terminal of the external battery Va is connected to p-side and the negative terminal to the n-side of the p-n junction. Here the forward bias (external battery Va) opposes the potential barrier VB and so the depletion layer becomes thin.
And because this "depletion region battery" has 0.7V, so to overcome this potential barrier the external battery (Va) voltage must be larger than "depletion region battery" and must be connected in the opposite direction to "depletion region battery".

So only if this is the case ( if the positive terminal of the external battery Va is connected to p-side and the negative terminal to the n-side of the p-n junction) the forward diode current can flow.

 

[NascentOxygen]

@Jony130 See all those black ⊕ charges in the N doped material? Shouldn't they be (-)?

 

[Ahsan Khan]

I don't know what I should say. Jony130 send the link which contains some theory and diagrams. The figure showing barrier voltage VB with negative polarity(small plate) on P side and positive polarity(large plate) on N side. But just adjacent to it where diode symbol is their in the circuit they show the reverse polarity.

Before posting any content one should make sure if the content is authentic or not.

For every topic we have on Internet a bulk of information as well as misinformation.

In my previous post i tried very hard what I wanted to know. That how come the side(P) with negative ions can show higher potential as compared to the side(N) with negative ions. Please refer that post again.

Thank you

 

[Ahsan Khan]

Note how the picture contradict itself. In big diagram it shows negative terminal(small plate) of VB with P side(right side) but in small diagram adjacent to it showing symbol of diode it shows positive terminal on P side(right side).

How does the polarity reverse if one use symbol instead of two pieces of P and N semiconductor?

 

[Jony130]

Vb - battery on the diagram represents/show the polarity of a depletion region. And Va is a external battery.
So on the left (small diagram) where the diode symbol was used Va is also a external voltage source. Right??
So to force diode into conduction we need to connect an external voltage source (Va in the diagram) in the opposite direction to the "depletion region battery" (Vb on the big diagram). Why? because only when Va oppose Vb, Va is able to overcome this potential barrier (Vb) and current through the diode starts to flow.
Do you understand what I just wrote?

Also read this
http://www.allaboutcircuits.com/vol_3/chpt_2/6.html

 

[Dr Ross]

Ovais, you're getting confused a lot about a little thing. The majority carriers are the big thing. You obsess about the little thing: the barrier at the junction. By the way: Potential decreases linearly as the distance from the field point increases.

 

[Ahsan Khan]

I think I got it. If I am true now, I was missing the fact that unlike outside the circuit, inside the battery positive charge flows from lower potential to higher potential! Or negative charge moves towards lower potential!

That is inside the cell the plate towards which the flow of electrons is difficult will be the higher potential(statement 1)While outside the cell, negative charge likes to move towards higher potential.

We must therefore apply opposite rules(that electrons move towards higher potential, electrons move to higher potential only outside the circuit) as infact inside the battery electrons move(or like to move) towards lower potential. If we consider the potential barrier as a fictious battery, we can notice that electrons find it difficult to move toward P side(as it has negative ions) than as per statement 1, P side will have higher potential. I feel the difficulty I had lies in my assumption that positive charge ALWAYS like to go from higher potential to lower potential while it is actually reverse INSIDE the battery, and we are looking at diode as a battery.

Am I right?

 

[Jony130]

No, wrong again. In semiconductor we have mainly two types of a current:
1 - Drift current
http://www.ittc.ku.edu/~jstiles/312/handouts/Drift Current.pdf
2 - Diffusion current
http://www.ittc.ku.edu/~jstiles/312/handouts/Diffusion Current.pdf

And when the diode is forward biased the diffusion is the dominant current.
http://www.ittc.ku.edu/~jstiles/312/handouts/The pn Junction Diode.pdf
http://www.ittc.ku.edu/~jstiles/312/handouts/The pn junction in forward bias.pdf

 

[Ahsan Khan]

Images showing contradicting polarity of potential barrier VB.

 

[Jony130]

The polarity shown across the diode symbol indicates the polarity of a external sources, in this case the diagram shows the polarity of a Va voltage source.

 

[Ahsan Khan]

The polarity shown across the diode symbol indicates the polarity of a external sources, in this case the diagram shows the polarity of a Va voltage source.

You mean to say the polarity of diode depends on the polarity of sources? Here P side is connected to positive of battery(forward biasing) that is why P of potential barrier is positive?
If one apply kirchhoffs law in the diagram of diode symbol he/she will find that in this case applied voltage VA is reverse of potential barrier VB. That's OK.

But according you when we connect diode in reverse bias(P with negative of Battery) potential barrier VB will still be opposite to applied voltage VA, you can draw a circuit(for reverse bias and apply your rule that just because P is connected to negative it is negative emotions or lower in potential) and verify this. But textbook says in reverse bias applied voltage is along the potential barrier that is its field is in same direction as that of VB.

I mean if one stick to the rule that P is positive as it is connected to positive terminal of battery forgetting the fact that diode also has its own potential due to charges the depletion region, it will cause a lot of problem in explaining why in reverse bias the applied voltage VA and potential barrier voltage acts in same sense.

Regards

 

[Jony130]

Stop mixing the external voltage source with the barrier voltage (Vb). The barrier voltage (Vb) will never changes its polarity.

But according you when we connect diode in reverse bias(P with negative of Battery) potential barrier VB will still be opposite to applied voltage VA,

How can it be ??

it will cause a lot of problem in explaining why in reverse bias the applied voltage VA and potential barrier voltage acts in same sense.

Why ?? I do not see any problem. When diode is reverse biased by any external voltage source (Va) the electric field created by this external source (Va) is in the same direction as a Vb electric field. So this two field adds together and this is why depletion region increases his thickness. Notice that the positive battery terminal (Va) attracts N-type majority carriers, electrons, away from the junction and the negative battery () terminal attracts P-type majority carriers, holes, away from the junction. This increases the thickness of the nonconducting depletion region.