PN junction: basic electrostatics


by Dr_Pill
Tags: basic, electrostatics, junction
Dr_Pill
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#1
Jan25-13, 10:11 AM
P: 40
Hi,

I'm stuck with the electrostatics of the PN junction.



I just can't see why the electric field has to be negative in the depletion regio.

The two equations give positive values for the Electric field in the intervals, so how can it be negative??
In the equation for the n regio, there are two minus signs, so its positive, in the p-regio one minus sign but x is negative, so again positve.
This is driving me insane.

And why is the electric field negative in the n-regio, its positively charged.

I have also some questions about the potential difference across the regio, but let's explain this first :)
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marcusl
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#2
Jan25-13, 10:47 AM
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In the n region, mobile electrons diffuse leftward across the junction leaving fixed positive crystal ions to the right of the origin. Likewise, mobile holes from the p side diffuse rightward, leaving negative ions behind. The E field points from positive to negative charges. It is maximum at the origin where the charge separation is greatest. It decreases linearly to either side (in this abrupt junction model) as the charge separation tapers to zero far from the junction.
Dr_Pill
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#3
Jan25-13, 10:52 AM
P: 40
Quote Quote by marcusl View Post
In the n region, mobile electrons diffuse leftward across the junction leaving fixed positive crystal ions to the right of the origin. Likewise, mobile holes from the p side diffuse rightward, leaving negative ions behind. The E field points from positive to negative charges. It is maximum at the origin where the charge separation is greatest. It decreases linearly to either side (in this abrupt junction model) as the charge separation tapers to zero far from the junction.
But why is it negative?

The equations for the electric field are positive in the both the n and the p region.


And if it is negative everywhere, how can it be pointed from + to -?

marcusl
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#4
Jan25-13, 11:26 AM
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PN junction: basic electrostatics


Quote Quote by Dr_Pill View Post
But why is it negative?
The E field points from + charge to - charge. Since +q is located at +x and -q at -x, E points from +x to -x. As a vector, it must be negative.
[tex]\vec{E}=-E\hat{x}[/tex]
Quote Quote by Dr_Pill View Post


The two equations give positive values for the Electric field in the intervals, so how can it be negative??
In the equation for the n regio, there are two minus signs, so its positive
No, you can't just count signs. You have to think a little! In the n region, x is between 0 and x_n so (x_n - x) is always positive and the field is negative.
Dr_Pill
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#5
Jan25-13, 11:41 AM
P: 40
Quote Quote by marcusl View Post
The E field points from + charge to - charge. Since +q is located at +x and -q at -x, E points from +x to -x. As a vector, it must be negative.
[tex]\vec{E}=-E\hat{x}[/tex]

No, you can't just count signs. You have to think a little! In the n region, x is between 0 and x_n so (x_n - x) is always positive and the field is negative.
Yeah.My brains are a bit foggy(been a long time since I studied physics&math)
It's just, in the lecture notes that I have, the field is positive! The lectures notes that I have, are not about pn junction, but about np junction however.

Should I scan it? Then you can take a look, it pisses me off :)


I am still also a bit confused how the electric field can be negative in the n-regio, since the n-regio is positively charged.


Wait: is the electric field negative because it's pointing from right to left?
marcusl
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#6
Jan25-13, 11:47 AM
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So your confusion is actually about something entirely different than what you scanned and posted here? If it doesn't match what is here, I suggest that you ask your professor about it.
Dr_Pill
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#7
Jan25-13, 12:01 PM
P: 40
Quote Quote by marcusl View Post
So your confusion is actually about something entirely different than what you scanned and posted here? If it doesn't match what is here, I suggest that you ask your professor about it.
No, that was another confusion :)
Don't have a professor, not in school.

I'm learning on my own and I got several books, however the lecture notes I got from a friend are slightly different: there it is a np junction and all the diagrams are reversed etc...
marcusl
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#8
Jan25-13, 12:11 PM
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Quote Quote by Dr_Pill View Post
Wait: is the electric field negative because it's pointing from right to left?
Yes.

In my experience, using too many sources ends up being confusing. As you have discovered, it's easy to get further confused by differing approaches, notations, etc. Try picking one book to use (in my self-study I steer clear of lecture notes because they are rarely as detailed or carefully constructed as texts). Consult others only to clarify a point of confusion, and take care when you do :o)
Dr_Pill
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#9
Jan25-13, 01:47 PM
P: 40
I think I have another question.

Suppose you take a surface that comprises both p and n regio's, then the total charge is zero, so according to gauss law the electric field is zero in the pn junction.

What is wrong with this reasoning?
marcusl
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#10
Jan25-13, 04:14 PM
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Did you review Gauss's law in your E&M text before writing this post?
Dr_Pill
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#11
Jan26-13, 06:12 AM
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Quote Quote by marcusl View Post
Did you review Gauss's law in your E&M text before writing this post?
Wait, is it because the charge density isn't equal on both sides?
All the charges summed up equals zero, but the charge densities are different.

So, if you make a pn junction with Nd=Na, then u have zero field strenght?
marcusl
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#12
Jan26-13, 03:22 PM
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Your statement of Gauss's law is completely wrong. Please review it, then actually try to work the problem you posed. Come back later if you get stuck or have a specific question.
Dr_Pill
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#13
Jan31-13, 01:19 PM
P: 40
I reviewed some electromagnetics and I understand it now, guess I was mistaking electric field for electric flux and charge density for charge.
My excuses.
But I have another question that's probably stupid:
One of the boundary conditions for solving the electric field on the pn-junction with poisson equation, is stating that @ x=0 the electric field must be continuous.
Well, if the field is not continuous, it means that there is a presence of charge just @ your junction right since field lines always end on a charge.
So why exactly is this impossible?
Also: why does the electric field reaches a maximum value @ x=0, what is the physical meaning of this.

Thanks in advance.
marcusl
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#14
Jan31-13, 03:19 PM
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Quote Quote by Dr_Pill View Post
I reviewed some electromagnetics and I understand it now,
Excellent!
Quote Quote by Dr_Pill View Post
But I have another question that's probably stupid:
One of the boundary conditions for solving the electric field on the pn-junction with poisson equation, is stating that @ x=0 the electric field must be continuous.
Well, if the field is not continuous, it means that there is a presence of charge just @ your junction right since field lines always end on a charge.
So why exactly is this impossible?
You might be thinking of cases like a sphere in a uniform electric field where charges can accumulate on the surface. In the present case, the semiconductor crystal supports free passage of electrons and holes across the junction, so there can be no charge buildup.
Quote Quote by Dr_Pill View Post
Also: why does the electric field reaches a maximum value @ x=0, what is the physical meaning of this.

Thanks in advance.
The origin is where there is maximal charge separation to left and right, so the field is highest.


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