Hey, I love your videos. They are extremely helpful for my semiconductors course. We're also using Neamen. I was wondering if, in the future, you could share text references to accommodate people like myself who like to learn through a combination of books and UA-cam videos. Thanks! Keep the excellent videos coming, they are much appreciated!
If anyone confused about the direction of the band bending, I was too, until I've found out that it depends on the fermi level of the metal. So if there's a "low work-function" metal (fermi level above the p-type's), you'll get the band bending shown here. If the metal is "high work-function" (fermi level beneath the p-type's), you'll get upwards band-bending. Also, for the latter you'll be in accumulation mode to begin with, i.e. at any voltage less than the flat band voltage (including 0). You'd need to overcome the flat band voltage to get into inversion. I hope I'm not terribly wrong.
the band diagram discussed dring 6:30 is p-Gate on n-Substrate, it got me confused for quite a while. However, Great Job on doing this video. Thanks a lot.
I don't really get why there is band bending without any applied voltage. And between 6:22 and 7:20 you talk about these work function as if you already talked about, but I can't find where, can you help me?
From 10:50 you mention one thing wrong that if phi Ms is positive, line moves up when we move from left to right but it should be opposite. Please answer what is actually the answer
Sure! I believe that is correct, phi_ms refers to the difference in the Fermi levels before you bring together the semiconductor and the metal. A positive phi_ms means the Fermi level of the metal is higher energy (semiconductor is lower energy), so when the Fermi levels meet electrons will jump from the metal to the semiconductor, and you will end up with an E-field that points to the right, which yields the energy profile in the video.
@@JordanEdmundsEECS your definition of phi Ms is wrong. It is energy required to move electron from Fermi energy to free space. And according to this phi Ms in this case will be negative. I am referring sung-mo-kang and In that it is mentioned like this what I am saying
You are correct about the sign, phi_ms in this case should be negative. The metal Fermi energy is higher in energy, but its work function is smaller. Phi-ms is defined in terms of these work functions, see Neamen p. 383.
thank you! these videos are great! I'm new to semiconductors and I need it to understand my project. I have a question though why the Fermi level never changes? the nature of the Fermi level is the probability so shouldn't it change with applying a voltage?
It does change, check out the definition and derivation of Fermi level, for example, it depends linearly on temperature But on itself is a function derivativated from fermi-dirac distribution
It’s just a way of visualizing what’s going on. Imagine you take your hand, grab the edge of the band structure, and physically drag it up. This is going to cause everything to start slanting.
I have two questions that I found myself hard to understand or confused all the time. I try to make a statement from my understanding and Dr. Jordan please correct me or guide me with proper principle. 1)The band diagram is drawn in respect of electron energy, so applying positive voltage (containing excess holes) at a particular region will drop any band diagram of that region but apply -ve voltage (containing excess electron) will increase any of the band diagram? 2)I thought the Ef is a constant at equilibrium (contact) , means it doesn't change and form a straight line for all materials (oxdie,metal and SC ) contact. Hence by applying voltage potential, it can be shifted?
Your assumption regarding the energy bands being drawn in terms of electronic potential is correct. More -ve voltage means we go up in energy in the band diagram and vice versa. Regarding your 2nd question, on the application of voltage, the built-in potential developed in equilibrium condition gets reduced or increased depending on the polarity of the bias.
By applying a voltage on the gate we are shifting the Fermi level of the gate metal (or polysilicate). This creates a difference between the Fermi level of the gate metal and that of the semiconductor. Difference in fermi-level between the two encourages all the subsequent charge distribution phenomena and the band bending. Applying external bias takes a system out of equilibrium so fermi levels can change.
Thanks Jordan for your semiconductor lecture series. They are really good. Can you pls confirm the energy band bending in accumulation mode for n-MOS (p-type substrate). Shouldn't it be upward bending in the band diagram @6:12 in video
Great question! What matters here is the *relative* voltage of the body and the gate, so applying a positive voltage to the body is the same as applying a negative voltage to the gate.
Thanks :D currently I use an iPad Pro (12.9”) running Duet Pro at a resolution of 2048x1536 at 60fps, the software I use is SketchBook from Autodesk and OBS Studio for the screen capture.
"..wait how did i draw it" ... *restarts*
You just saved my semester. Thank you very much.
he saved my degree
Wow what an explanation we need cool teachers like you in our Universities
Hey, I love your videos. They are extremely helpful for my semiconductors course. We're also using Neamen. I was wondering if, in the future, you could share text references to accommodate people like myself who like to learn through a combination of books and UA-cam videos. Thanks! Keep the excellent videos coming, they are much appreciated!
he is following the book from Neamen called semiconductor physics and devices. Just google it the pdf is online
If anyone confused about the direction of the band bending, I was too, until I've found out that it depends on the fermi level of the metal. So if there's a "low work-function" metal (fermi level above the p-type's), you'll get the band bending shown here. If the metal is "high work-function" (fermi level beneath the p-type's), you'll get upwards band-bending. Also, for the latter you'll be in accumulation mode to begin with, i.e. at any voltage less than the flat band voltage (including 0). You'd need to overcome the flat band voltage to get into inversion. I hope I'm not terribly wrong.
Great video, Jordan!
the band diagram discussed dring 6:30 is p-Gate on n-Substrate, it got me confused for quite a while. However, Great Job on doing this video. Thanks a lot.
Sorry, I was wrong above. It took me few times of watching this video to understand the 'intrinsic band bending'.
@@xltian4368 can u explain in detail what happened there? I didn't get it
Thanks for awesome lecture for my winter vacation👍
I don't really get why there is band bending without any applied voltage. And between 6:22 and 7:20 you talk about these work function as if you already talked about, but I can't find where, can you help me?
happened to me too.
The video we are looking for is "MOSFET Band Diagram Explained".
excellent approach to explaination .. life saver!
The video really helps a lot!! Thank you!
Shouldn't the bending direction of p-type mos capacitor in accumulation mode be up?
Thnx sooooo much
U re the only one that i can understand from
Thnx again ♥️♥️
Why i dont had a teacher like you???
At 10:03 where you said "It's looks more p-type", Isn't that n-type ? As earlier in the video you considered a body of p-type...
Thank you for saving me and my family from Semicons
XD my pleasure
From 10:50 you mention one thing wrong that if phi Ms is positive, line moves up when we move from left to right but it should be opposite. Please answer what is actually the answer
Sure! I believe that is correct, phi_ms refers to the difference in the Fermi levels before you bring together the semiconductor and the metal. A positive phi_ms means the Fermi level of the metal is higher energy (semiconductor is lower energy), so when the Fermi levels meet electrons will jump from the metal to the semiconductor, and you will end up with an E-field that points to the right, which yields the energy profile in the video.
In your explanation whose work function is more metal or semiconductor?
@@JordanEdmundsEECS your definition of phi Ms is wrong. It is energy required to move electron from Fermi energy to free space. And according to this phi Ms in this case will be negative. I am referring sung-mo-kang and In that it is mentioned like this what I am saying
You are correct about the sign, phi_ms in this case should be negative. The metal Fermi energy is higher in energy, but its work function is smaller. Phi-ms is defined in terms of these work functions, see Neamen p. 383.
Why is it bend downwards when no voltage is applied?🙃
Great explanation otherwise :D
It bends downwards when a positive voltage is applied. With no voltage, the bands are just flat.
What does "undo" mean?
Yes what it actually means by it? @Jordan Edmunds
perfect explanation !!!
thank you! these videos are great! I'm new to semiconductors and I need it to understand my project. I have a question though why the Fermi level never changes? the nature of the Fermi level is the probability so shouldn't it change with applying a voltage?
It does change, check out the definition and derivation of Fermi level, for example, it depends linearly on temperature
But on itself is a function derivativated from fermi-dirac distribution
At 8:16 what is concept of draging band? Plz explain
It’s just a way of visualizing what’s going on. Imagine you take your hand, grab the edge of the band structure, and physically drag it up. This is going to cause everything to start slanting.
@@JordanEdmundsEECS ??
Thank you very much, you are life saver, so much hearts ♥️
What is the difference between accumulation mode and forward biasing?
Great video, thanks!
What happens when V=0 ? would that also be considered Accumulation Mode?
I have two questions that I found myself hard to understand or confused all the time.
I try to make a statement from my understanding and Dr. Jordan please correct me or guide me with proper principle.
1)The band diagram is drawn in respect of electron energy, so applying positive voltage (containing excess holes) at a particular region will drop any band diagram of that region but apply -ve voltage (containing excess electron) will increase any of the band diagram?
2)I thought the Ef is a constant at equilibrium (contact) , means it doesn't change and form a straight line for all materials (oxdie,metal and SC ) contact. Hence by applying voltage potential, it can be shifted?
Your assumption regarding the energy bands being drawn in terms of electronic potential is correct. More -ve voltage means we go up in energy in the band diagram and vice versa.
Regarding your 2nd question, on the application of voltage, the built-in potential developed in equilibrium condition gets reduced or increased depending on the polarity of the bias.
By applying a voltage on the gate we are shifting the Fermi level of the gate metal (or polysilicate). This creates a difference between the Fermi level of the gate metal and that of the semiconductor. Difference in fermi-level between the two encourages all the subsequent charge distribution phenomena and the band bending. Applying external bias takes a system out of equilibrium so fermi levels can change.
Thanks Jordan for your semiconductor lecture series. They are really good. Can you pls confirm the energy band bending in accumulation mode for n-MOS (p-type substrate). Shouldn't it be upward bending in the band diagram @6:12 in video
it is the intrinsic band bending @6:12. It took me quite a while to figure out.
As you said semiconductor was grounded, how did you apply voltage on it ?
Great question! What matters here is the *relative* voltage of the body and the gate, so applying a positive voltage to the body is the same as applying a negative voltage to the gate.
thank you
Hello, thanks for the videos. By the way which tools (software & hardware) do you use for these videos?
Thanks :D currently I use an iPad Pro (12.9”) running Duet Pro at a resolution of 2048x1536 at 60fps, the software I use is SketchBook from Autodesk and OBS Studio for the screen capture.
nice video sir !! Thanks
ok lol
Your english accent seems like you are not a indian youtuber... lol
اللهي تنستر
full of stupid mistakes. I could be the stupid one though.
You are.