In the case of the intrinsic semiconductor, the chance of finding the electron at the Fermi level is 50% while when you n-dope the semiconductor is less likely to find it in the middle and more likely to occupy a higher level. It is like moving the distribution towards higher concentration. The same argument could be applied for the p-doped. That's how I explain it intuitively in my mind. Thank you very much for your videos they are very helpful.
Hi Jordan, the doping introduced the doping level in the band gap (close to Ec or Ev), and based on the ionized energy, the ionized percentage, can we say that doping can change or narrow the band gap of a semiconductor? Or we need a Case-by-case discussion, the shallow/deep doping (doping energy level), and the doping concentration (lightly doped / heavily doped) to decide whether the doping can change the band gap in the energy-band structures? In most instances, the ionization energy of (III, IV, V) impurities is about ~ several meV compared with the intrinsic bandgap ~several eV, which is too small to be ignored. But the metal dopant's ionization energy (~0.xxeV) is comparable to the original band gap energy, how can we discuss these conditions? Can we call that condition: semiconductor alloys to tune the bandgap in the bandgap engineering area? also, can you talk more about 2D materials doping and their bandgap engineering? Many thanks. Best,
In the last video, you said that adding electrons by doping decreases the number of holes because the electrons and holes "collide." How is this possible if the electrons are part of the new Energy state near the conduction band, and the holes are only in Ev? Also, a quick google gives the definition for Fermi Energy (Ef) as "the highest occupied energy level of a material as 0K." How does this make sense in our band diagram, since electrons can still go up to higher energy states in the conduction band?
Shouldn't the Fermi level be in in between the donor level and the conduction band? At 0 k there are namely states which are occupied above the Fermi level.
Definitely, the temperature at which you get a reasonable amount of ionization is below 300K most of the time. I believe the energy gap between the conduction band and the donor level is ~0.045eV (for phosphorous). If you want to find out precisely how many carriers are still left in their donor states, you simply integrate the fermi function from Ec to infinity and then compare that to the value of the fermi function at the donor level. In general, the ionization percentage is going to depend both on the temperature *and* the fermi level.
The concentration of electron charge carriers n and hole charge carriers p is approximately the same, call it n_i. Therefore, we can say n * p = n_i * n_i = (n_i)^2
In the case of the intrinsic semiconductor, the chance of finding the electron at the Fermi level is 50% while when you n-dope the semiconductor is less likely to find it in the middle and more likely to occupy a higher level. It is like moving the distribution towards higher concentration. The same argument could be applied for the p-doped. That's how I explain it intuitively in my mind. Thank you very much for your videos they are very helpful.
Lol that's exactly how I think about it!
you are a good teacher and you speech is clear and audible. thank you!
What a nice explanation! Cleared all my concepts.
Thank you for your explanation. It helped a lot
Hi Jordan, your videos are excellent and all concepts are so clearly explained - thank you very much! What drawing software do you use for them?
Hi Jordan, the doping introduced the doping level in the band gap (close to Ec or Ev), and based on the ionized energy, the ionized percentage, can we say that doping can change or narrow the band gap of a semiconductor? Or we need a Case-by-case discussion, the shallow/deep doping (doping energy level), and the doping concentration (lightly doped / heavily doped) to decide whether the doping can change the band gap in the energy-band structures?
In most instances, the ionization energy of (III, IV, V) impurities is about ~ several meV compared with the intrinsic bandgap ~several eV, which is too small to be ignored. But the metal dopant's ionization energy (~0.xxeV) is comparable to the original band gap energy, how can we discuss these conditions? Can we call that condition: semiconductor alloys to tune the bandgap in the bandgap engineering area?
also, can you talk more about 2D materials doping and their bandgap engineering?
Many thanks. Best,
There is a little mistake. There is no "degree Kelvin" since Kelvin is an absolute scale. Therefore it is just Kelvin.
I think this might be the most common comment on this video xD You are correct, thanks :)p
In the last video, you said that adding electrons by doping decreases the number of holes because the electrons and holes "collide." How is this possible if the electrons are part of the new Energy state near the conduction band, and the holes are only in Ev? Also, a quick google gives the definition for Fermi Energy (Ef) as "the highest occupied energy level of a material as 0K." How does this make sense in our band diagram, since electrons can still go up to higher energy states in the conduction band?
Shouldn't the Fermi level be in in between the donor level and the conduction band? At 0 k there are namely states which are occupied above the Fermi level.
This is beautiful
Sir i got a question. Is Nc "effective density of states" change with doping?
Best solid state videos :)
Sir, I think we need atleast 150 K (not 300K) for all the electrons to reach to E_c from E_d since the energy gap is roughly 2kT.
Definitely, the temperature at which you get a reasonable amount of ionization is below 300K most of the time. I believe the energy gap between the conduction band and the donor level is ~0.045eV (for phosphorous). If you want to find out precisely how many carriers are still left in their donor states, you simply integrate the fermi function from Ec to infinity and then compare that to the value of the fermi function at the donor level. In general, the ionization percentage is going to depend both on the temperature *and* the fermi level.
Ef doesn't change , its the Ec and Ev changing which give an impression that Fermi level changed. am i correct?
No, the fermi level does actually depend on the doping.
Please can you do a full derivation of equation 4.11 of Neaman. 4th Edition. Their are some other derivations by people but I don't understand them.
The best in the world.
It should be group 13( Boron family) and group 15( Nitrogen family) rather than group 3 and 5 which belong to the transition metals like Sc and V
So, can I say that all the excess electrons of the donor atoms get leveled at Ed at T=0 K ??
Exactly! At T=0K, the electrons won't have any energy to jump up into the conduction band, and they will stay stuck at the energy level Ed.
thanks :)
How can I prove Mathematically n*p=n^2 ?
The concentration of electron charge carriers n and hole charge carriers p is approximately the same, call it n_i. Therefore, we can say
n * p = n_i * n_i = (n_i)^2
Jordan please tell me, what is reason behind the origin of band gap?
@ 2:22 Room-temperature 300 K??? Now I know where this global warming is coming from :-)
Horseshoes and hand grenades xD
excuse the pedantry but kelvin is not a degree scale
#mostCommonComment 🤣
Yo you are the fucking goat
Samij aav.