Dear Minshen Lin ,

In a pn junction a space charge region is formed in a form of dipole with the positive donor ion in the n-side and negative acceptor ions in the p-side.

So, an electric field associated with dipole charge in the direction from from the donor ions to the acceptor ions. This is called a built in electric field Ei due to the contact difference of potential between the n-side and the p-side.

If one forward biases the p-n junction by applying a positive voltage on the p-side relative to the n-side, an external electric field opposing that of the built in electric field is appearing on the space charge region. The consequence of this is that the space charge region width will be decreased to account for the net decrease in the potential difference between the -side and the p-side.

So, forward bias will decrease the potential difference between the n-side and the p-side.

If one reverse biases the junction by lowering the potential of the p-side the overall potential difference across the junction increases and the electric field increases compared to the thermal equilibrium case.

And so the space charge region widens to accommodate the increase in the potential barrier as consequence of the reverse bias.

In summary

at equilibrium there is a built in potential barrier = phi the contact difference of potential, the electric field is equal to Ei

at forward bias V the potential barrier decreases to phi-V , The electric field will be less than Ei

at reverse bias V the potential barrier increases to phi+V, the electric field will be greater than Ei.

If you study the p-n junction theory in the book at link: Book Electronic Devices

you will get thorough insight in it and all these issues become clear to you.

Best wishes

Dear Minshen Lin ,

In a pn junction a space charge region is formed in a form of dipole with the positive donor ion in the n-side and negative acceptor ions in the p-side.

So, an electric field associated with dipole charge in the direction from from the donor ions to the acceptor ions. This is called a built in electric field Ei due to the contact difference of potential between the n-side and the p-side.

If one forward biases the p-n junction by applying a positive voltage on the p-side relative to the n-side, an external electric field opposing that of the built in electric field is appearing on the space charge region. The consequence of this is that the space charge region width will be decreased to account for the net decrease in the potential difference between the -side and the p-side.

So, forward bias will decrease the potential difference between the n-side and the p-side.

If one reverse biases the junction by lowering the potential of the p-side the overall potential difference across the junction increases and the electric field increases compared to the thermal equilibrium case.

And so the space charge region widens to accommodate the increase in the potential barrier as consequence of the reverse bias.

In summary

at equilibrium there is a built in potential barrier = phi the contact difference of potential, the electric field is equal to Ei

at forward bias V the potential barrier decreases to phi-V , The electric field will be less than Ei

at reverse bias V the potential barrier increases to phi+V, the electric field will be greater than Ei.

If you study the p-n junction theory in the book at link: Book Electronic Devices

you will get thorough insight in it and all these issues become clear to you.

Best wishes

Dear Prof. Abdelhalim abdelnaby Zekry ,

Thank you so much for your thorough analysis.

I've been thinking about this question for some time and came up with the idea that the chemical potential is omitted when I tried to equate the applied voltage with the line integral of the built-in electric field. The complete equation should incorporate the 'chemical potential field' (in relation to the gradient of the chemical potential or the gradient of the carrier concentration) into the line integral. So, in that way, the result of the integral is the electrochemical potential, which should be equal to the output voltage or the applied voltage.

I haven't examined the self-consistency of this idea. I hope we can have some discussions.

Best regards.

Dear Minshen Lin ,

The chemical potential of the material is normally represented by the Fermi level position in reference to the vacuum level. This is by definition the work function of the materiel phimat.

In this way when one contacts two materials with each other they will have a contact potential difference equal to difference in the work function and this what we call the contact difference of potential. It results from the the different chemical potentials of the two contacted materials.

At thermal equilibrium, the chemical potential of the two materials gets horizontal flat line and so the band bending occurs between the materials to keep the same levels of the the chemical potential relative to the vacuum level. In this way the Vacuum level will be a curved line reflecting the contact potential difference between the material. Simply stated, the chemical potential difference leads to the formation of an equal electrostatic potential difference between the two materials equal to the contact difference of potential.

One you apply a potential difference from outside, all what you apply to obtain the net potential difference is the superposition principle that is applied to combine the applied potential differences.

This is may be the simplest way to understands this subject.

Best wishes