# Why there is NO current in a PN junction under equilibrium

Tags:
1. Oct 24, 2015

### Amal Thejus

1. The problem statement, all variables and given/known data
Its said that there is no Net current in a PN junction under equilibrium as the drift and diffusion currents cancel each other.
We are talking about a Step junction at equilibrium.
2. Relevant equations

Its just a qualitative analysis.

3. The attempt at a solution

The only way i can think of it is as the Positive charged area in the transient region attracting the electrons in the N region, towards it. Now the particle flow due to diffusion and drift in the N region are in the same direction.

Now the negative charged area in the transient region repells the electrons in P region in effect attracting the holes towards it. So here also the direction of particle flow due to diffusion and drift are in the same direction.

#### Attached Files:

File size:
24.5 KB
Views:
45
• ###### IMG_20151024_122632.jpg
File size:
28.3 KB
Views:
45
2. Oct 24, 2015

### Roy_1981

Your attempt is pretty much accurate. At the junction, Electrons from N-region and holes from P-region diffuse to the opposite/reverse doped region due to concentration gradient of electrons and holes across the junction. (Electrons cross over and occupy the holes on the p-side, neutralizing each other for a while and the regions on both sides of the junction get depleted of charge carriers). As the electrons (and holes) cross over/keep pouring into to the P-side (N-side) it gets negatively charged (positively charged). This build-up of negative charge on the p-side (and positive side on the n-side) leads to a creation of an electric field in the direction from N to P region at the junction/transition/depletion region. This electric field repels or opposes diffusion and when it reaches a critical point/value completely cancels diffusion (i.e. the number of electrons diffusing from N to P is cancelled the number of electrons drifting under the electric field back from P to N side and similar for holes).

3. Oct 24, 2015

### Prannoy Mehta

The reasoning is apt. It's not only due to repulsion but also the attraction by the N junction atoms. The N junction is neutral, it is not charged. Suppose it was formed by doping Silican with Arsenic. The arsenic behaves like a positive charge and the extra electron could be thought of as a negative charge. Both keeping the diode neutrol. We a PN junction is formed the electrons from the N side are attracted towards their free positive charge. As soon as this occurs, P junction has atoms which are fixed which can be thought of as non mobile negative charges which repel this new electron, and the non mobile positive charge in the N side attracts it. (The more technical term would be a electric field, I did not use it over here.)

There is this textbook NCERT Grade XII Physics Part 2, if you don't mind terrible fonts, and blue color, this is a very good book for introduction of semiconductors and electronic devices up till transistors.It will be written as Chapter 6, it is the 14th chapter in the book. The band theory is not explained in a very detailed manner, but is sufficient.

Hope that helped.

4. Oct 24, 2015

### Amal Thejus

Thank you so much.

5. Oct 24, 2015

### Amal Thejus

I checked the NCERT textbook. Its clearly written there. Thank you so much

6. Oct 24, 2015

### Amal Thejus

Thank you so much