MOSFET Band Diagram Under Applied Bias
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- เผยแพร่เมื่อ 27 ก.ค. 2024
- / edmundsj
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In this video, I discuss the MOSFET Band Diagram under Applied Bias, or how applying a voltage between the gate and the semiconductor changes the result we got in the last video. I show how to solve for the surface potential and the oxide voltage drop and explain everything in terms of the total amount of band bending.
Understanding the MOSFET band diagram immediately allows you to understand the threshold voltage, which we will discuss in the next video.
This is part of my series on semiconductor physics (often called Electronics 1 at university). This is based on the book Semiconductor Physics and Devices by Donald Neamen, as well as the EECS 170A/174 courses taught at UC Irvine.
Hope you found this video helpful, please post in the comments below anything I can do to improve future videos, or suggestions you have for future videos.
These are great set of lectures. I keep coming back to these for a quick, yet insightful review and just appreciate how beautiful device physics is :)
Very neatly explained and very clear. Glad to watch many more videos from you, sir !!
Thank you! This was very useful for helping me derive the threshold voltage. It is much appreciated!
very nice explanation, thank you!
Very helpful. Much appreciated
Life saver playlist. Thank you so much sir🙏
great video sir.... neatly explained
Great video again but Jordan what If the applied voltage (Vg) is negative?
Excellent question. In this case, the total amount of band bending will be reduced, and if you apply a large enough negative voltage it will bend the other way (this is called “accumulation”).
How can we solve for Phi-S using the quadratic without knowing Cox?
It’s assumed that you know Cox (since as the designer, you know the materials you are using and the geometries involved), I’m glad someone asked so I could clear that up.
Ah, thank you very much! I was thinking only C = Q/V (but we would also know how to find C using the Area of the insulator as well as the dilectric constant of the insulator itself, correct?)
You are an awesome teacher! I can't stress enough how awesome your videos are. Much appreciated sir. xD
Stupid question, but I cant find explanation for this... Does Fermi Energy at some distance x, represent potential energy of all particles at that x? And that is the reason we define work function from vacuum energy to Fermi Energy, as it is the work necessary to overcome potential energy (electrical & chemical) difference? Is work function just energy needed to overcome electric potential difference, and is vacuum energy not representing chemical potential energy?
It's actually a very deep question. The fermi energy represents what's called chemical potential, or the tendency of particles to diffuse in space. The work function is the amount of energy it would take to rip an electron from the fermi level into vacuum (overcoming that potential difference).
Essentially very similar to a MOS Capacitor then?
Exactly the same. I actually made this video prior to the other one xD
But shouldnt vacume energy always be zero, by us bending it, arent we changing its value? This is very confusing :|
The vacuum "band" here is only relevant when we talk about the semiconductor itself (it's just as easy to excite an electron from the valence band to vacuum regardless of how much it bends). Once we get to vacuum, the energy is now zero and these bands no longer mean anything :p That's my interpretation, anyway.
It was not a good explanation of the band bending near the surface. It seemed that you applied Vg after bending. However, you should have explained how those electrons accumulated near the surface...
You generally do apply Vg after the bands bend. The applied voltage can either undo the band bending or make it more severe. Physically the electrons come from whatever source you decide to attach.
at 6:38, its |Vfb|+ Vg= |Vox|+|Phi(s)| , its Voltage flatband instead of phi ms, phi ms is the bend at the semiconductor only oxide not included.