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?
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?
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.
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.
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.
@@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!
NOOOOO... I need part 3. Great explanations but part 3 sounded VERY useful. Explaining the V-I Curve.
I wish my professor explained this well - I'm learning more from your videos than from class thank you so much for posting these!
++ plus plus * infinity
nice videos, you made it very easy to understand. Thank you!
Thanks :)
Thnx so much , u can't imagine how helpful this video is
U saved my life
الله يعطيك العافية
Very good video!! This is going to help me for my exam, thank you!
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?
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?
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.
Extremly helpful, thank you
Is there a part 3 ?
Yeah. He said "in the next part"... but I can't find it. I need help understanding the I/V curves.
May I know which video is the continuation to this video, tq
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?
The band gap bending will be inverted. This happens when you join a p-type semicondcutor.
how about the characterist CV curve of schottky contact.. do you have one ?
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 ?
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.
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.
@@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!
This redcuction occurs at interface ??
Where is the part 3 class .?
Around 5 30 min, u talked about charge nuetrality.
Can u talk about it more charge nuetrality level and Fermi level
Thanks sir