I love how this teacher insists on concepts. It shows how much he cares about driving ideas home and have his students understand the concepts. Kudos to you, Sir.
I asked the question below, not quite thinking a lot. I just starred a bit more at my own question and answered it myself. The equation used was actually valid for both reverse and forward bias. The photodiode is just not used in the forward bias mode but in the reverse bias mode. This is truly an outstanding teaching.
Around time stamp 35 minutes, I have an issue with the formula. The Diode current I=Is(exp(V/KT/q)-1) is a forward bias current due to minority carriers in both N and P regions where the externally applied voltage (forward) reduces the potential barrier. However, in the photo-diode, it seems like we're applying reverse biased voltage where we are expanding the potential barrier to accelerate generated e's in the i-region. Why are we using the same formulas of current in two situations completely different? (one forward bias and the other reverse bias) to want to subtract the P-current?
You put reverse bias say 5 v the formula simpley reduces to I =-Is becz of the negilagable exponential term. So here in General sockley equation is used which is for both F.B and R.B but it preety fair to consider I =-Is for R.B and I=-Is-Id for photo detector case.
Good explanation! You have mentioned that electron-hole pair generated in p-region or n-region may not experience force, they wander freely. I believe that your assumption is wrong. Even the electron-hole pairs generated in p and n regions experience force from external voltage. For example, minority carriers (electrons) in p-region experience repulsive force from voltage or negative terminal of the battery. However, the recombination rate for the electron-hole pairs generated in p or n-regions is very high compared to the electron-hole pairs generated in depletion region. The electron-hole pairs generated in p or n-region have to travel from one end to another end in the semiconductor but electron-hole pairs generated in depletion region have to travel only a half distance so recombination rate is very low for the charge carriers generated in depletion region. Therefore, the charge carriers generated in depletion region will carry more electric current. The more the charge carriers are generated in the depletion region the greater is the minority current.
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I love how this teacher insists on concepts. It shows how much he cares about driving ideas home and have his students understand the concepts. Kudos to you, Sir.
I asked the question below, not quite thinking a lot. I just starred a bit more at my own question and answered it myself. The equation used was actually valid for both reverse and forward bias. The photodiode is just not used in the forward bias mode but in the reverse bias mode.
This is truly an outstanding teaching.
Hats offf!!! How the teacher in department of physics can have so much knowledge about the Electronics and communication.
This is all device, semiconductor physics.
Physicists are all-rounders.
Sir, you are phenomenal in explaining the concepts
bruh I watch this like watching a movie, it's both joyful and helpful ☺☺
Awsome explainations
Excellent
Great way to illustrate
thank you prof
I am really blessed to watch your lecture sir. excellent teaching thank you sir.
Excellent lecture
Thank you, very helpful.
Wonderful ♥️
very nice lecture
Very nice potential sir
excellent
Good!
Thanks Professor.
Ultimate
So nice thanks
Around time stamp 35 minutes, I have an issue with the formula. The Diode current I=Is(exp(V/KT/q)-1) is a forward bias current due to minority carriers in both N and P regions where the externally applied voltage (forward) reduces the potential barrier. However, in the photo-diode, it seems like we're applying reverse biased voltage where we are expanding the potential barrier to accelerate generated e's in the i-region. Why are we using the same formulas of current in two situations completely different? (one forward bias and the other reverse bias) to want to subtract the P-current?
You put reverse bias say 5 v the formula simpley reduces to I =-Is becz of the negilagable exponential term. So here in General sockley equation is used which is for both F.B and R.B but it preety fair to consider I =-Is for R.B and I=-Is-Id for photo detector case.
very helpfull sir🙏
Is it possible to know what is the textbook used in this course?
I use SOLID STATE PHYSICS ,by Kittell
Good explanation! You have mentioned that electron-hole pair generated in p-region or n-region may not experience force, they wander freely. I believe that your assumption is wrong. Even the electron-hole pairs generated in p and n regions experience force from external voltage. For example, minority carriers (electrons) in p-region experience repulsive force from voltage or negative terminal of the battery. However, the recombination rate for the electron-hole pairs generated in p or n-regions is very high compared to the electron-hole pairs generated in depletion region. The electron-hole pairs generated in p or n-region have to travel from one end to another end in the semiconductor but electron-hole pairs generated in depletion region have to travel only a half distance so recombination rate is very low for the charge carriers generated in depletion region. Therefore, the charge carriers generated in depletion region will carry more electric current. The more the charge carriers are generated in the depletion region the greater is the minority current.
Индусы- это лучшее, что создали для российского образования
Excellent