Schottky Diode Part 2 - Depletion Region and Capacitance

[@jasoncornell2474]

NOOOOO... I need part 3. Great explanations but part 3 sounded VERY useful. Explaining the V-I Curve.

[@saramillan4400]

I wish my professor explained this well - I'm learning more from your videos than from class thank you so much for posting these!

[@anantjain1108]

++ plus plus * infinity

[@asawirshakeebasmaail5114]

Thnx so much , u can't imagine how helpful this video is U saved my life الله يعطيك العافية

[@AvigailKolobov]

nice videos, you made it very easy to understand. Thank you!

[@JordanEdmundsEECS]

Thanks :)

[@ramiromillanfraile7399]

Very good video!! This is going to help me for my exam, thank you!

[@CannonballCircuit]

Hey Jordan, I thought that the Fermi levels were in the forbidden zone between the valence and conduction bands, meaning that no electrons could be present there at steady state?

[@orgoon7697]

Extremly helpful, thank you

[@dannchan00]

May I know which video is the continuation to this video, tq

[@karlc1980]

Thank you for the video. Around minute 1 you show that electrons will move from the n-type semi-conductor to the metal through diffusion, until E_F's become equal. Will this in general happen with any n-type semi-conductor/metal combination, or will this only work in certain rare cases? I seem to understand that this happens exactly when E_F in the semiconductor, before contact, is higher than E_F in the metal (so actually E0-EF,metal > E0-EF,semic.). So could it just as well happen that the E_F initially was lower in the n-type semiconductor than in the metal? Then the electrons would flow from the metal to the semi-conductor?

[@JordanEdmundsEECS]

Yup, exactly. This actually works not only for any metal/semiconductor combination, but even for metal/fluid interfaces and arbitrary semiconductor/semiconductor interfaces. It’s super powerful. Which Fermi level is higher dictates where electrons flow from.

[@williamp3677]

Is there a part 3 ?

[@jasoncornell2474]

Yeah. He said "in the next part"... but I can't find it. I need help understanding the I/V curves.

[@matthewchau8907]

So in a n-type, the metal's work function has to be greater than semiconductor's? What happens when the work function of the semiconductor is larger?

[@usdesk]

The band gap bending will be inverted. This happens when you join a p-type semicondcutor.

[@renatoberaldo2335]

how about the characterist CV curve of schottky contact.. do you have one ?

[@renatoberaldo2335]

This redcuction occurs at interface ??

[@musg5336]

Hello Jordan. Hope all is well with you. I have a question with regards to the capacitance. How is the thickness just the depletion region. The depletion region consists of positively charged ions and the rest of the distance on the n side is are the electrons. The width should be the distance between the two right? If yes, the how is it xn.

[@JordanEdmundsEECS]

The relevant distance is the separation between your two ‘seas’ of *mobile* charge carriers. There are a bunch of mobile charge carriers in the metal (and there is a ‘depletion region’ in the metal, but it’s a fraction of an atomic length), and there are a bunch of mobile charge carriers outside your depletion region, which has length xn. The separation between these two determines your capacitance. You just treat the ions as if they are a dielectric, because they cannot move to conduct current.

[@concernedhuman1518]

@@JordanEdmundsEECS Hello Jordan. Nice video. I have a question: why do we use an AC voltage with mV amplitude while performing C-V measurements in Schottky or PN junctions? Thanks and regards!

[@usdesk]

Where is the part 3 class .?

[@prashantupadhyay5110]

Around 5 30 min, u talked about charge nuetrality. Can u talk about it more charge nuetrality level and Fermi level

[@ShailendraPandeyOwm]

Thanks sir

[@ashlynnundlall]

How do we use figure 9.3 in Neaman ? How do you get Vbi from it ? They say look for the intercept but I don't know what those arrows mean ?