"I understand it is because of different of the work function difference of the metal and Semiconductor but why the band bend in certain directions (either up/down) that is not clear ?

Why band bend in MOS under zero bias , please explain in details as it is not clear in books. "

Even under Zero Bias, there is potential due to workfunction difference, this creates a small Electric Field which is defined by dEC/dx or dEV/dx as we move from the surface of the semiconductor at the oxide interface towards the bulk, this field slowly vanishes to zero, that is dE/dx = 0 which is straight line for bands. But at the interface dE/dx that is change in energy with respect to distance x from the interface is small variable value, which changes, hence the bands bend.

Now, why up and why down depends on the p-type or n-type semiconductor as the location of fermi level is fixed near to either at conduction band edge in n-type or valence band edge in p-type, the change in energy level at the interface is either going up or going down.

I hope it clarifies, if not, follow the curve of dE/dx for the energy band diagram.

Even if the books doesn't looks convincing, try to read it again by following the dE/dx variation.

https://nptel.ac.in/courses/113106065/Week%206/Lesson14.pdf

Please read this article, good explanation with decent diagrams.

Hi,

Let us consider a mos capacitor with p type substrate that have fermi potential of 0.2 .

we know that the work function/electron affinity of metal is 4.1 ev.

and for substrate (silicon) it is 4.15 ev.

we also know band gap of silicon is 1.1 ev.

therefore from conduction band the fermilevel of p type substrate is .75 ev far away .

hence the total workfunction of substrate with respect to free space is 4.9ev = 4.15+.55+.2

now when all three comes in contact with each other they lead to single material , hence it is for sure that due to conservation of energy there fermi level will line up.

but if we see the work function difference between metal and substrate it comes out to be -0.8ev.

this implies that , if we take free space as reference point (let say line of x=0 of xy plane ) , then metal is 0.8 less negative that substrate. or in another way we can say that positive 0.8 appears on metal with respect to substrate.

Now is positive 0.8 appears on metal , then we all know the positive 0.8 will drop some across oxide , and the rest of it will drop on substrate, and we all know very well that on applying positive potential on the gate with p type substrate electrons accumulate near surface , or we can say the fermi level near surface moves towards conduction band.

This is the same thing which is happening with mos (p substrate) under 0 bias.

The fermi level is moved toward conduction band , now in order to make fermilevel straight the conduction band is move toward fermilevel , hence we see band bending downwards.

I hope this is explanatory.

Thanks.

Hi,

If I try to answer in a simple way, we have metal and semiconductor with different work functions. As soon as we are making the contact the fermi level is aligning itself in a straight-line throughout the metal, insulator and the semiconductor. As there was a difference between the metal and semiconductor work functions prior to the contact formation the bands in the semiconductor will bend to arrange the alignment as soon as the contact is formed.

This mainly starts with surface and bulk interaction resulting in band bending. Also, additional parameters (like differences in work function and affinity and bias, and so on) determine more contribution of band bending direction. Please follow the link also to get more information.

https://science-problem-jp.blogspot.com/2022/07/how-does-band-bending-occur-when-n-type.html