Why there is NO current in a PN junction under equilibrium

In summary, the conversation discusses the concept of equilibrium in a PN junction and how the drift and diffusion currents cancel each other out. The attempt at a solution explains how the electric field at the junction plays a role in this cancellation. The conversation also mentions a recommended textbook for further understanding.
  • #1
Amal Thejus
14
1

Homework Statement


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.

Homework Equations



Its just a qualitative analysis.

3. The Attempt at a Solution [/B]
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.
 

Attachments

  • IMG_20151024_122639.jpg
    IMG_20151024_122639.jpg
    21.8 KB · Views: 527
  • IMG_20151024_122632.jpg
    IMG_20151024_122632.jpg
    21.4 KB · Views: 510
Physics news on Phys.org
  • #2
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 canceled the number of electrons drifting under the electric field back from P to N side and similar for holes).
 
  • Like
Likes Amal Thejus
  • #3
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.
 
  • Like
Likes Amal Thejus
  • #4
Roy_1981 said:
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 canceled the number of electrons drifting under the electric field back from P to N side and similar for holes).
Thank you so much. :smile:
 
  • #5
Prannoy Mehta said:
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.
I checked the NCERT textbook. Its clearly written there. Thank you so much:smile:
 
  • #6
Thank you so much:smile:
 

1. Why is there no current in a PN junction under equilibrium?

Under equilibrium, the concentration of free electrons and holes on either side of the junction is equal due to diffusion. This results in a depletion region at the junction, where there are no mobile charge carriers available to conduct current.

2. What is the role of the built-in potential in a PN junction?

The built-in potential is created by the diffused charges at the junction and acts as a barrier for the flow of current. It prevents electrons and holes from easily crossing the junction, maintaining the equilibrium state.

3. Can a PN junction under equilibrium conduct current if an external voltage is applied?

No, the built-in potential will still act as a barrier for the flow of current. However, if the applied voltage is greater than the built-in potential, the junction will be in a non-equilibrium state and current can flow.

4. How does temperature affect the current flow in a PN junction under equilibrium?

As temperature increases, more electrons and holes are thermally generated, leading to an increase in the number of charge carriers in the depletion region. This reduces the width of the depletion region, allowing for a small amount of current to flow.

5. Can a PN junction under equilibrium ever conduct current?

In theory, no, as the built-in potential will always act as a barrier for current flow. However, in practical applications, there may be some leakage current due to imperfections in the junction or external influences.

Similar threads

Replies
3
Views
749
  • Electrical Engineering
Replies
2
Views
908
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Electrical Engineering
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
Back
Top