MOS Capacitor - Depletion Mode Explained

/ edmundsj
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MOS capacitor in depletion mode. What depletion mode means, what the band diagram looks like in the semiconductor as well as the full MOS structure, and what voltage we need to apply to put the MOS capacitor into depletion mode.
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.

[@andrewk2625]

Your lecture is clear to understand, not too complicated that everyone can follow and fun, it's so helpful and enjoyable. Thanks!!

[@alecadanglao8386]

Top notch dude really needed this refresher cuz i got exams in 3 hours T^T

[@JordanEdmundsEECS]

Hahaha I hope that went well xD

[@abdolw3682]

Thank you very much, you are life saver, so much hearts ♥️

[@mouwaten100]

a positive voltage at the gate attracts negatively charged ions at the other side, but those boron ions are already bounding with Si atoms. So when they move up they bring with them their silicon partner atomes, correct?

[@reageno.764]

when we are applying a positive or negative voltage, is it moving Ec,Ei, and Ev bands or is it moving the fermi level?

[@JordanEdmundsEECS]

Great question, you're exactly correct. If you apply the voltage to the capacitor, after everything settles there is no current flowing, so the fermi level must be constant everywhere (this is the definition of the fermi level). It's Ec, Ev, and Ei that bend to accommodate the voltage.

[@rajgandhi4042]

at 1:22 you mention boron atom give up their electron but how it has one e less than si.. Plz guide us

[@JordanEdmundsEECS]

I definitely misspoke, thanks for finding that! Boron atoms will accept electrons and become negatively charged, phosphorus ions give up their electrons and become positively charged.

[@komalsrivastava973]

How negative voltage applied on the semiconductor is going to drag the band diagram up? Please explain! Thanks!

[@JordanEdmundsEECS]

This is a common confusion, band diagrams are the *electron* energy, so there’s a negative sign in the potential energy because the electron has negative charge (-q*V).

[@iteshbharti8560]

which book should I follow for MOS ?

[@JordanEdmundsEECS]

Neamen is my personal favorite and very comprehensive.

[@shahrukh3531]

Learning from India 🇮🇳

[@rajgandhi4042]

At 6:56 you mention Ef is constant what that means 1)it can't move in band diagram of 2) bending of Ef will not take place?

[@JordanEdmundsEECS]

Exactly. EF is constant means that it is the same value throughout the entire band diagram. This will always be true if the system is in equilibrium. Bending of EF does not happen in equilibrium. It will only happen when you have current flowing, and even then it gets kinda complicated (in the form of quasi-Fermi levels).

[@rajgandhi4042]

@@JordanEdmundsEECS when we apply positive voltage to gate after undoing the intrinsic band banding, Ef of metal should move down and Ef of semiconductor remains at the same place. Is it what you want to say?

[@gersh1332]

Video is broken at 720p60 - video feed stops but sound keeps going.

[@apiyopao4586]

At 1:23 I think u made a mistake you should have said boron ions accept electrons instead of saying give up electrons and become negatively charged That's why boron is called acceptor impurity as it's trivalent. After all if an atom gives up electrons it will become positively charged ion and not negatively chraged ion

[@JordanEdmundsEECS]

You are absolutely correct, my mistake. Thanks for pointing it out!

[@divya4752]

Please make videos on mos amplifier

[@JordanEdmundsEECS]

Videos on this should be coming in the next few days :)

[@ThePreppar]

Very good, but not 'scaring away' but like charger repell each other :)

[@kenneth31251]

Can you write my solid state physics exam?

[@JordanEdmundsEECS]

XD someday.