Dear Firas,

welcome,

Your question is a basic question and concerns the energy band diagram of the p-n junction at equilibrium. It is so that when contacting tow materials their Fermi level will be consatnt through out the p-n junction as per your drawing.

Since the n-type material conduction band edge is near the Fermi level and the valence band edge will be near to the the Fermi level in the P-type material as per figure. And also because of equal bandgap for the material one gets that the that the conduction band edge in the P-type will be higher than the conduction band edge in n-type side. This is clear from Figure given in the question.

And so a potential energy barrier will be formed across the p-n junction. This potential barrier is the contact difference of potential between the two sides of the p-n junction.

More information can be found in the book: Book Electronic Devices

Best wishes

The answer related to trivalent and pentavalent.

Trivalent and pentavalent.

In n type semiconductor is formed by trivalent impurities and the majority charge carriers are electrons are -ve charged therefore it lies below forbidden energy gap

we can clearly determine from carrier lifetime.

Thank you mr.Sundara Siva Kumar v

But I am sorry for my question:

N -type semiconductors material is formed by trivalent or pentavalent ?!

Why the energy band diagram of n- type material in semiconductors silicon are lower than the energy band diagram of p- type material?

Because

The trivalent impurities exert lower forces on the outer-shell electrons than the pentavalent impurities

That mean :

The lower forces in n- type material mean that the electon orbits have greater energy than the electron orbits in the n- type materials.

Dear Firas,

welcome,

Your question is a basic question and concerns the energy band diagram of the p-n junction at equilibrium. It is so that when contacting tow materials their Fermi level will be consatnt through out the p-n junction as per your drawing.

Since the n-type material conduction band edge is near the Fermi level and the valence band edge will be near to the the Fermi level in the P-type material as per figure. And also because of equal bandgap for the material one gets that the that the conduction band edge in the P-type will be higher than the conduction band edge in n-type side. This is clear from Figure given in the question.

And so a potential energy barrier will be formed across the p-n junction. This potential barrier is the contact difference of potential between the two sides of the p-n junction.

More information can be found in the book: Book Electronic Devices

Best wishes

Dear Firas Nadheer ,

This is because the Fermi level in the two sides should be aligned (horizontal) in equilibrium . As Efn of n-side is near the C.B, and Efp of p-side is near the V.B. the p-side is raised or the n-side is lowered.

Mathematical Proof:

The current is proportional to the gradient of Fermi level so that Jn =e.n.un.grad(Ef), Jp =e.p.up.grad(Ef). In equilibrium, there is no current flowing across the p-n junction. Therefore, Efn=Efp=Ef (horizontal line)

Thank you very much Dr. Muhammad Hamza.

For about your useful information.

Thank you very much Dr. Abdelhalim Zekrey.

Again, Dr. Abdelhalim and Dr. Muhammad thank you very much for you and I respect you very much.

But my deep question is:

In n- type materials,

why Efn ( fermi level ) was near to conduction band Ec ?

And

In p- type materials,

Why Efp ( fermi level ) was near to

balance band Ev ?

Do you agree with me that the reason it :

The trivalent impurities exert lower forces on the outer-shell electrons than the pentavalent impurities

That mean :

The lower forces in n- type material mean that the electon orbits have greater energy than the electron orbits in the n- type materials.

Firas Natheer: "The trivalent impurities exert lower forces on the outer-shell electrons than the pentavalent impurities -- That mean: The lower forces in n- type material mean that the electon orbits have greater energy than the electron orbits in the n- type materials."

Well, you seem to have a completely unsuited starting picture in your mind, because you talk about atomic properties as if those atoms were not bound in a solid. But the situation changes dramatically when many atoms are bound together: For example, look at the change of the energetical positions of the silicon 3s and 3p states (i.e., its outer electrons), starting from being well-defined levels for single atoms but becoming bands for the crystalline solid, eventually forming the characteristic band gap of about 1.1 eV (as shown in the graphics on this page: www.tf.uni-kiel.de/matwis/amat/mw2_ge/kap_4/illustr/t4_2_1.html -- where "Leitungsband" means conduction band, "Lücke" means gap, and "Valenzband" is self-explaining).

As you can see, this band gap of 1.1 eV has nothing to do anymore with the original energetical position of the atomic levels, but everything that is relevant for the electronic and optical properties of this solid follows just from the electronic properties of this band gap. So, how can you talk about energies of electron orbits in this context -- or what did you have in mind, more specifically?

Now to your questions about the position of the Fermi level Ef: "In n-type materials, why Ef was near to conduction band Ec? And in p-type materials, why Ef was near to balance band Ev?"

This follows from the role of Ef as determining the energy position at which an electronic state, if present, were half-occupied -- which follows from the Fermi distribution. For electronic states higher in energy than Ef, the occupation probability is lower (approaching 0), whereas for lower energy states it is higher (approaching 1). In n-type materials, the energetical position of the extra electrons present at donor atoms is close to the conduction band, hence the Fermi level is also close to Ec. In p-type materials, the empty acceptor states are close to the valence band, hence also Ef is found there. In general, the position of Ef depends on temperature, but for silicon at room temperature this general picture still holds.