Dear Naveen Bagadi ,

I would like to second prof. Mitdank.

When you apply a voltage V on a uniform n-type semicondcutor, it will be dropped uniformly inside the materials such that the electric field E will be constant through out its thickness d. Simply the magnitude of E= V/d .

So the electric field profile will a horizontal line with a height from the x-axis = E=V/d.

If you want to get more about this topic please follow the book:Book Electronic Devices

Best wishes

Hello Naveen,

If you apply a constant electric field E to a material, a potential gradient gradV is generated (E = - grad V). The band edge (here band of conductivity) is shifted with the local potential change delta V. The slope of the band edge Ec(x) gives the electric field E.

With Regard

R. Mitdank

Your question needs to be clarified. If you are applying a voltage "difference" between two points of your uniformly doped semiconductor later, tjen the explaination given by Rüdiger Mitdank is of course the right answer,

But, if you apply a unique voltage to the layer (and possibly another voltage to another different layer of the same device) the answer would change and depend on the global configuration of the voltage sources and the precise structure of the device you are working with.

Dear Naveen Bagadi ,

I would like to second prof. Mitdank.

When you apply a voltage V on a uniform n-type semicondcutor, it will be dropped uniformly inside the materials such that the electric field E will be constant through out its thickness d. Simply the magnitude of E= V/d .

So the electric field profile will a horizontal line with a height from the x-axis = E=V/d.

If you want to get more about this topic please follow the book:Book Electronic Devices

Best wishes

Hello. As a simplified answer (verified by answers of Rüdiger Mitdank and Abdelhalim abdelnaby Zekry), if a semiconductor is doped uniformly, and have Ohmic contacts, it behaves like a conductor, with the conductance of mu_e*q_e*n_D (for n-type) or mu_h*q_e*n_A (for p-type). Therefore the potential drops with a constant slope (that is E-field) across the material. Here, mu is the mobility of electrons or holes, q_e is the elementary charge (of a single electron), and n is the number of charges due to the ionization of donor (D) or acceptor (A) dopings.

I provide you a link to get articles giving more informations about your question.

https://www.researchgate.net/project/Improvement-of-the-performances-of-the-silicon-solar-cell-by-integration-of-the-electric-field-source-in-the-photovoltaic-system

But if there is non-zero electric field inside a uniform doped semiconductor, then why there is no electric field outside the depletion region in the n region and p region of the PN diode.

Dear Naveen Bagadi ,

You have to pay attention that when you solve for the electrostatic field there will be no electric field in the neural regions outside the space charge region in the transition region. When current flows in these neutral regions it must be driven by the existence of an electric field.

I explained this matter in details in my book: Book Electronic Devices

Please read it thoroughly!

Best wishes