These Band Diagrams are great tools to understand interface characteristics between two materials. To keep things straight in my head, I line up the Fermi Levels for each material (impose thermal equilibrium) then think about the charge redistribution needed to make them line up. Keep in mind that the band structure of the bulk material is unchanged, and the band bending occurs only in the interface region. This works for metal/semiconductor, metal/metal and semiconductor/semiconductor interfaces.
Question: I understand that the two materials have different Fermi levels, but why must the bands bend "to make the fermi levels the same?" I keep hearing this but don't understand for metal-on-semiconductor context. EDIT: Never mind! See 8:29 for the answer to my Q.
Fermi level is the level that the state will always have 50% chance of begin occupuied. If Fermi level is not constant in the system, there will be net current flowing. However, we assume in thermal equalibrum. (You assume in steady state). No transfer of energy occuers. So if the fermi level is not constant, it violates the assumption.
At 10:28 are we sure it's a negative voltage that would be applied to decrease the barrier height and not a positive one? I'm used to seeing that height defined as (phi_i - Va) where Va is a negative bias so now I'm confused.
@@JordanEdmundsEECS The video is really helpful. But I have a question. If the negative charge increase. The bending of the n-type semiconductor should be more curved. Wondering why is the barrier height will decrease?
@jordanedmundseecs I think that if it was a p type semi conductor the Fermi energy would be near the valence band. On your drawing, it is near the conduction band.
Why at the 8:50 minute the semiconductor bands rise high? I think they should rise down so Ef of the metal (white line) will "stick" to Ef of the semiconductor (blue line). You drew that graph assuming the Ef of the semiconductor is greater than that of the metal.. then at 8:57 it seems you changed your mind
A remark on the bending of the vacuum level: As many here I was confused about why the vacuum level is not constant but bends. I thought a lot about it, because the vacuum level is the energy of an electron far away from the sample. This should be the same regardless of the material, voltage,… In fact, there is an ambiguity in the definition: 1. There is a vacuum level, often used in semiconductor physics, that is measured right above the surface of the material. This vacuum level shifts, as you said. 2.) There is a vacuum level as I explained above (which is measured from infinity). This vacuum level is an absolute reference. In the second case, the electron affinity/work function changes, and the vacuum level stays constant. Why? Because, if I apply a voltage to the sample, the energy of the electrons may decrease (or increase), which leads to the bending of valence an conduction band. Of course this changes the energy needed to take an electron from the sample and bring it far away (the vacuum energy). In the end you do not only have to overcome the ionisation energy (like in the case without applied external field) but you also have to overcome the energy of the electric field. Another effect is happening when we apply no voltage but e.g. make a metal-semiconductor junction or a p-n-junction: There is a potential difference (energy difference) between p- and n-side because of the electric field in the depletion region. This causes the work function to shift. The siping itself does also shift the work function, but this effect is negligible (as I was told). In short: I think the vacuum level is constant and the electron affinity/work function changes. It also leads to the same band diagram (at least vor E_c and E_v). What do you think about it?
u explained this for system of metal and semiconductor with same value of work function, i.e. fermilevel and vacuum level of both align with eachother. i want u to check that again.
The vacuum level is always the same distance from the conduction band, so it has to bend as well. That doesn’t mean that an electron will have more energy when it is freed from the semiconductor. Once it’s in vacuum, this band structure no longer applies.
Better explained than the textbook or the lecture in my Semiconductors class. Thanks!
These Band Diagrams are great tools to understand interface characteristics between two materials.
To keep things straight in my head, I line up the Fermi Levels for each material (impose thermal equilibrium) then think about the charge redistribution needed to make them line up. Keep in mind that the band structure of the bulk material is unchanged, and the band bending occurs only in the interface region.
This works for metal/semiconductor, metal/metal and semiconductor/semiconductor interfaces.
We love you prof.
You're a life saver
I have an exam and it all makes sense now
Thanks 🙏
Question: I understand that the two materials have different Fermi levels, but why must the bands bend "to make the fermi levels the same?" I keep hearing this but don't understand for metal-on-semiconductor context.
EDIT: Never mind! See 8:29 for the answer to my Q.
Fermi level is the level that the state will always have 50% chance of begin occupuied. If Fermi level is not constant in the system, there will be net current flowing. However, we assume in thermal equalibrum. (You assume in steady state). No transfer of energy occuers. So if the fermi level is not constant, it violates the assumption.
THANK YOU
Dude i fucking love you, all your videos are so helpful
Keep doing more videos please. thet are very good. Thanks 😊
At 10:28 are we sure it's a negative voltage that would be applied to decrease the barrier height and not a positive one? I'm used to seeing that height defined as (phi_i - Va) where Va is a negative bias so now I'm confused.
Va is assumed to be a positive voltage here. A positive voltage applied to the p-side should reduce the barrier height.
@@JordanEdmundsEECS The video is really helpful. But I have a question. If the negative charge increase. The bending of the n-type semiconductor should be more curved. Wondering why is the barrier height will decrease?
@@簡廉羲 yeah you are right instead of negative it must be a positive pole to reduce the curve
It should be positive voltage to reduce the barrier
@jordanedmundseecs
I think that if it was a p type semi conductor the Fermi energy would be near the valence band. On your drawing, it is near the conduction band.
Why at the 8:50 minute the semiconductor bands rise high? I think they should rise down so Ef of the metal (white line) will "stick" to Ef of the semiconductor (blue line). You drew that graph assuming the Ef of the semiconductor is greater than that of the metal.. then at 8:57 it seems you changed your mind
The same doubt🥲
A remark on the bending of the vacuum level:
As many here I was confused about why the vacuum level is not constant but bends. I thought a lot about it, because the vacuum level is the energy of an electron far away from the sample. This should be the same regardless of the material, voltage,…
In fact, there is an ambiguity in the definition: 1. There is a vacuum level, often used in semiconductor physics, that is measured right above the surface of the material. This vacuum level shifts, as you said. 2.) There is a vacuum level as I explained above (which is measured from infinity). This vacuum level is an absolute reference.
In the second case, the electron affinity/work function changes, and the vacuum level stays constant. Why? Because, if I apply a voltage to the sample, the energy of the electrons may decrease (or increase), which leads to the bending of valence an conduction band. Of course this changes the energy needed to take an electron from the sample and bring it far away (the vacuum energy). In the end you do not only have to overcome the ionisation energy (like in the case without applied external field) but you also have to overcome the energy of the electric field.
Another effect is happening when we apply no voltage but e.g. make a metal-semiconductor junction or a p-n-junction: There is a potential difference (energy difference) between p- and n-side because of the electric field in the depletion region. This causes the work function to shift. The siping itself does also shift the work function, but this effect is negligible (as I was told).
In short: I think the vacuum level is constant and the electron affinity/work function changes. It also leads to the same band diagram (at least vor E_c and E_v). What do you think about it?
Sir can you plz help... I want notea on ptype Schottky diode
what is the difference between one said junction (p+n''''''n+p) and schottky diode
Aren't there 2 barriers? One on each contact to the semiconductor?
Hello sir, what about p-type semiconductor? How the depletion region will be? And where the electric field direction will be? Thank you
I loved it!
u explained this for system of metal and semiconductor with same value of work function, i.e. fermilevel and vacuum level of both align with eachother. i want u to check that again.
Nice! Love from Brasil
Nice
Love from INDIA
Thank you.
Great content! But at the end when bending was performed, I don't think the vaccum level Eo will bend ... please correct me if I am wrong...
The vacuum level is always the same distance from the conduction band, so it has to bend as well. That doesn’t mean that an electron will have more energy when it is freed from the semiconductor. Once it’s in vacuum, this band structure no longer applies.
Are u sure the fermi level will be same through out the combination? Hmm
At thermal equilibrium yea
Thnx soooooo much
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