Why there is NO current in a PN junction under equilibrium

AI Thread Summary
In a PN junction at equilibrium, there is no net current because the drift and diffusion currents cancel each other out. Electrons from the N-region and holes from the P-region diffuse across the junction due to concentration gradients, leading to charge carrier depletion on both sides. This diffusion creates an electric field that opposes further diffusion, ultimately balancing the currents. The N-region remains neutral due to the fixed positive charges from dopants like arsenic, while the P-region has non-mobile negative charges that repel incoming electrons. This interplay of attraction and repulsion maintains equilibrium in the junction.
Amal Thejus
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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.
 

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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).
 
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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.
 
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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:
 
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:
 
Thank you so much:smile:
 
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