PN junction: basic electrostatics

• Dr_Pill
In summary: If the total charge is zero, then according to Gauss's law, the electric field should be zero. However, in reality the electric field is negative in the pn junction. This is because mobile electrons in the p region diffuse leftward, while mobile holes diffuse rightward. The net charge is negative in the pn junction, so the electric field is negative.
Dr_Pill
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 :)

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.

marcusl said:
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 -?

Last edited:
Dr_Pill said:
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.
$$\vec{E}=-E\hat{x}$$
Dr_Pill said:

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.

marcusl said:
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.
$$\vec{E}=-E\hat{x}$$

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?

Last edited:
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.

marcusl said:
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...

Dr_Pill said:
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 )

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?

Did you review Gauss's law in your E&M text before writing this post?

marcusl said:
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?

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.

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.

Dr_Pill said:
I reviewed some electromagnetics and I understand it now,
Excellent!
Dr_Pill said:
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.
Dr_Pill said:
Also: why does the electric field reaches a maximum value @ x=0, what is the physical meaning of this.

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

Last edited:

1. What is a PN junction?

A PN junction is a type of semiconductor junction formed by combining a p-type (positive) and n-type (negative) semiconductor material. It acts as a diode, allowing current to flow in one direction while blocking it in the other.

2. How does a PN junction work?

When the p-type and n-type materials are brought together, electrons from the n-type material fill the holes in the p-type material, creating a depletion region. This creates a potential barrier, preventing current from flowing in the reverse direction. When a forward bias voltage is applied, the barrier is overcome and current can flow through the junction.

3. What is the role of electrostatics in a PN junction?

Electrostatics plays a crucial role in creating the potential barrier in a PN junction. The depletion region is created by the separation of positive and negative charges, which results in the formation of an electric field that opposes the flow of current.

4. How does the doping of the semiconductor affect the behavior of a PN junction?

The doping level of the p-type and n-type materials determines the width and strength of the depletion region. Higher doping levels result in a narrower depletion region and a lower barrier for current flow. This affects the overall behavior and characteristics of the PN junction.

5. What are the practical applications of PN junctions?

PN junctions are essential components in many electronic devices, such as diodes, transistors, and solar cells. They are also used in rectifiers, voltage regulators, and signal amplifiers. Additionally, PN junctions are used in integrated circuits for the creation of complex electronic circuits.

Replies
2
Views
1K
Replies
3
Views
955
Replies
14
Views
2K
Replies
12
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
42
Views
8K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
10
Views
2K