What is the polarity of a Pn junction diode

[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.

 

[ovais]

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

 

[ovais]

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.

 

[ovais]

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.