How to sketch the band diagram of intrinsic semiconductors including the fermi level with the electric field present verses distance? Its not a homework question.
Er.. your question is very vague. What exactly do you not know? A search on the energy band diagram of intrinsic semiconductor would have given you plenty of results. Did you try it? If you did, where exactly are you having difficulties? And what "electric field"? Zz.
Zapper, I know the energy band diagram of intrinsic semiconductor where Fermi energy level lies in the middle of conduction band and valence band. If we consider that characterstics, its the graph which is plotted w.r.t energy. What I have asked is to sketch the energy band diagram of intrinsic semiconductor which includes fermi level with uniform electric field verses distance. I googled but I didn't found the relevant sketch. Thanks!
I can only guess at what you are asking, because you are still not explaining it clearly. Are you asking for the situation where a perpendicular electric field is applied to the surface of the semiconductor, and you want the effect on the semiconductor bands due to this external field from the surface and into the bulk of the material? If it is, then you should be searching for "bend bending" diagram. Zz.
please clarify me with respect to your second question. The question i have posed was asked in an iit interview. But i still do not find the answer
This is getting sillier. Now it is *I* who have to clarify what I thought you were asking? 1. Uniform E-field. I assume you know what that means. 2. Surface of semiconductor is perpendicular to this uniform E-field. Again, I assume you know what this is. 3. E-field affects the bands in the semiconductor. OK so far? 4. Is this what you are asking? I think if I don't get a definite answer to what you are asking after this, I'm done. Zz.
The energy bands are related to the electric potential through E=q*V where E is the energy, q the electron charge. If your Electric field [itex]E_{field}[/itex] is constant through the device, then your potential, and thus your energy Bands will be linear in position. That is [itex]qV=-q\int^x_0 E_{field} dx'=-qE_{field}x=E_c + Const.[/itex] where [itex]E_c[/itex] is your conduction band, and the valence band is just [itex]E_v=E_c-E_g[/itex] where [itex]E_g[/itex] is the bandgap. So, just draw a straight line with a slope [itex]-qE_{field}[/itex] to get the shape of your bands with respect to position in your semiconductor.
Let me pose the Question in this way: Sketch the energy band diagram (E versus x) including Fermi level of an intrinsic semiconductor under uniform electric field in x-direction.
Just draw to lines )one for conduction band, one with the valence band)with a slope -qEfield, seperated by a distance Eg. Then, since for an intrinsic case, the fermi level Ef is almost nearly right at the midgap, just draw a dotted line in between your bands with the same slope. You have to be careful though, since this is a quasi fermi level. By definition, when you apply a voltage (and hence create an Efield) you are taking the system out of equilibrium and putting it into a steady state. So, to get an exact answer, we need to know what is on either side of your device (n+p junction, metal-semiconductor interface, oxide and semiconductor interface, etc.). Then you would draw a constant fermi level at you boundries, and the difference in the two sides of your device wll give you the applied voltage time q (or -q*d*Efield, where d is the total depth or length of your device).
Because the Energy of the bands is equal to [itex]qV+Const.[/itex]. So, the conduction band for example is [itex]E_c=qV+Const.=-q\int_0^x E_{field} dx'+Const.=-qE_{field}x+Const.[/itex] for a constant electric field, which I have denoted by [itex]E_{field}[/itex]. The Valence band is just taken by [itex]E_v=E_c-E_g=-qE_{field}x -E_g+Const.[/itex] where [itex]E_g[/itex] is just the bandgap energy. So the slope of the bands is [itex]-qE_{field}[/itex]