MOSFET Band Diagram Explained Part 2

/ edmundsj
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In this video I show how to put together the complete MOSFET band diagram (MOS capacitor band diagram). I explain how to do this using the fact that the vacuum level is continuous, and explain why this must be the case. The central concept is one of band bending, and "amount" of band bending, described by voltages. Understanding this concept allows is the most direct and powerful way to understand MOSFET energy band diagrams.
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.

[@gagandeepiiitd4377]

Thankyou for this. Struggled with the topic everytime I read this from any source. Finally, I think I know what is really happening conceptually :)

[@rakshitbhatia3885]

Very informative and lucid explanation. Thankyou for great content. Keep up the good work.

[@talhasmathemetatix6775]

thankyou for such great content

[@muhammedzakaria9031]

Awesome work , thank you

[@Ashxxik-EC001]

Amazing Content sir, Thanks a lot!

[@vamosabv]

Thanks a ton for the wonderful videos. They really help. I have a question. If we assume, E and E as the fermi levels of metal and semi-conductor before they are brought together, we know that they should line up when they're brought together. Does E now increase to E or is the lined up Fermi level for the entire MOSFET in between E and E?

[@abj2873]

Thanks for this great informative video. I have a question about the boundary level between the semic. And the oxide. How are the negative charges gathered there? And where are the positive charges now? Thanks a gain :)

[@SimonBenjamin]

Great videos, thanks for creating all these tutorials. One thing that might be a bit confusing to the viewer; at 2:21 you assert that electron affinity (in the semiconductor) is “exactly the same concept” as work function; but of course the affinity is the gap from the vacuum level to the minimum of the conduction band, whereas the work function is to the fermi level. Both exist and differ in the semiconductor.

[@floriansteiner5666]

I fully agree, and that is why total bending = V_ox + Phi_si = | Phi_m - Xi_si | is not correct. the difference you're addressing is Ec-Ef so I think it should be total bending = V_ox + Phi_si = | Phi_m - [ q Xi_si + (Ec-Ef)] |

[@soumitrajoy7660]

Jordan, excellent endeavor. However, I think a better approach is to start with drawing a constant vacuum energy level. Then draw semiconductor band taking into account of how far the Ef would be from vacuum level. Repeat this for metal, and oxide layer, but assume that there are no junction formed between them yet (M,O,S are far apart from each other). Then consider a junction formed between them, and the Metal and Semiconductor is short circuited by an external wire. When short circuited (V_ms=0), electron would flow from low work-function material to high work function material, until their fermi levels are matched. The oxide interface which loses electron to achieve equilibrium will see an upward band bending. Likewise, the other oxide interface will see a downward band bending.

[@floriansteiner5666]

do you know a book where I could find that approach ?

[@floriansteiner5666]

what you can definitely find is the same approach as in this video, in for instance: TAUR und NING Modern VLSI Devices chapter 2.3 together with this video it gives a very good understanding of the situation.

[@vrtg025]

Man you are great

[@tusharrohilla7154]

Awesome work sir

[@JordanEdmundsEECS]

IDEOLOGY OF PHYSICS Awesome name sir.

[@tusharrohilla7154]

@@JordanEdmundsEECS thanks ... it's my u tube channel

[@ywk7282]

Thank you for the video I have a question. You said the reason why the vacuum level is continuois is because it must be free at the interface. Is it right for electrons to be free in a insulator? Wouldnt that be a defect? Thanks for reading

[@shanusingh98]

Gazab (Hindi, amazing) teaching style sir ty

[@floriansteiner5666]

Hello, first of all - great videos, thank you! A question though, for total bending = V_ox + Phi_si = | Phi_m - Xi_si | Xi_si would not be enough ?! out of your scheme that you drew: Xi_si should be replaced with Xi_si* out of ... q Xi_si* = q Xi_si + (Ec-Ef)

[@omarhadeed8248]

16k views should have been 16k likes!

[@JordanEdmundsEECS]

Daw, thanks ^^

[@marcellmurgas7823]

Hi Jordan, I love these videos... explaining thigs really good. I got one question though. In the previous video you say that it is not known where the Fermi level is in the oxide however in order to know how much is the offset in the vacuum level between the metal and the oxide layer in other words how much the oxide layer's energy levels tilt you need to know where the fermi level is, don't you? Thanks for the answer in advance.

[@JordanEdmundsEECS]

As the oxide doesn’t really have any charge carriers, the Fermi level really isn’t meaningful. You can figure out how much it will bend by treating the oxide like a capacitor

[@arnavrawat9864]

What's the degree of the mathematical equation describing the bent portion of the curve? I mean, in oxide it's linear, in semiconductor is it a exponential with -ve power? Or is it 1/r^2 type curve? Which mathematical law is the curve governed by?

[@JordanEdmundsEECS]

It’s linear, and it can be modeled as a carpacitor.

[@arnavrawat9864]

I'm talking about the vacuum level line's shape in the semiconductor sir. The oxide vacuum level line is indeed linear.

[@JordanEdmundsEECS]

Ah! My mistake. The vacuum level just follows the curvature of the underlying material, since the distance between the material and the vacuum level at any point in a given material must be the same (given by the electron affinity). It’s linear in the oxide and quadratic in the semiconductor.

[@ayushsrivastava1486]

Sir if the direction of electric field is from metal to oxide why bending is toward metal. Please expaexplain the concept....thank you.

[@JordanEdmundsEECS]

Sure, I believe it is that there are two minus signs (to get the energy from the electric field you first use the relation that Efield = -dV/dx and then the relation that Energy = -qV), so since the E-field is pointing to the right (positive) you expect the energy to increase moving to the right.

[@ayushsrivastava1486]

@@JordanEdmundsEECS thank you so much sir....love from india😃

[@tejaswininivarthichandrash597]

In the last video you showed the Fermi levels of metal and semiconductor had a large distance and you told that we'd have to make them be at the same energy level at the equilibrium ... But you explained nothing about that in this video.... Can you please tell how the Fermi levels are same here ? They had a difference in the last video

[@Joe_vanni]

there is a difference between the fermi levels when a voltage is applicated (no equilibrium).

[@coldspine3697]

God bless

[@nihalagrawal4646]

Why u have drawn all vacuum levels at different level

[@JordanEdmundsEECS]

Because the vacuum level just represents how much energy you need to strip an electron from the semiconductor, but it must also be continuous (otherwise if an electron was walking along on the surface of one material and crossed over to the other it would suddenly become a bound electron, which is nonsense). These requirements together *force* the vacuum level to bend.

[@monalisamallick9426]

could you please explain how the holes are moving away from the junction between oxide and semiconductor at 5:25 (kzbin.info/www/bejne/gXfbZ62umZxsZ8k ) ? Is it because of electric field and the majority carriers?

[@michellezheng3611]

Please see Mueller and Kamins device electronics textbook (pp.382-383) for a clear explanation. 1) Metal (e.g., aluminum) work function is less than the p-silicon's work function. 2) So when the parts of MOS capacitor are brought into contact, electrons will transfer from metal to p-Si. But how? 3) The oxide is a dielectric so it will not provide a path of this electron transfer, However, there is always some path for charge carrier to flow between the metal and the silicon. For example, the gate electrode and the silicon may be connected together. This is a key point - please see Mueller and Kamins textbook at p. 383. 4) Note that there is no external voltage applied to the metal. Simply connecting the metal and the silicon will cause a transfer of electrons from the metal to the silicon. 5) This electron transfer is equivalent to holes moving away from the p-Si and into the metal. These holes will spread out in the metal and form a thin layer at the metal - oxide interface. At the p-Si - oxide interface, the depletion of holes will cause the silicon in that area to become more intrinsic. 6) Important: since there is no external introduction of charge carriers, the MOS capacitor must be at thermal equilibrium, so the Fermi level must be flat for metal and silicon. 7) The flat Fermi level along with the depletion of holes at the p-Si - oxide interface cause the energy bands of the silicon to bend down at the interface. This is somewhat similar to the band bending in a pn junction, except here in a MOS capacitor, that there is no charge carrier transport between the metal and the p-Si through the oxide. I hope this helps you better understand the MOS capacitor's band diagram at thermal equilibrium.

[@Dempsey-180]

@@michellezheng3611 I also have the same question, this really helps! Thanks!

[@rishivarrey1176]

don't you think all vacuum levels should be on the same level

[@JordanEdmundsEECS]

Rishi Varrey Yes, initially I did think that because that makes intuitive sense (the “vacuum” is just external to the semiconductor so why should it’s level change?). However, the vacuum level is really only meaningful in a *relative* sense. It’s not a single energy level across the whole world, but just represents how hard it is to yank an electron away from the material. In order for that to be true, it has to be continuous across a material, which means it has to bend with the silicon.

[@sifatmunim1776]

@@JordanEdmundsEECS How do I figure out relative vacuum level positions? Here it shows that vacuum level for the metal is lower than that of the semiconductor. But from you last video, work function of the metal is greater than electron affinity of the semiconductor. So, if fermi levels are same, shouldn't the vacuum level for the metal be higher than that of semiconductor? I find it confusing.

[@rajgandhi4042]

We don't have to consider electron affinity of semiconductor, indeed we have to consider work function of semiconductor which is greater than metal which was also mentioned in last video

[@rajgandhi4042]

Sir has given just example what happen to bending of oxide when work function of metal is greater..

[@rajgandhi4042]

Sir I don't understand the concept of surface potential plz explain

[@matitrattore98]

This is MOS CAPACITOR band diagram, NOT A MOSFET. I don't know how you name the stuff in your country but in the rest of the world a MOSFET is a transitor, not a capacitor. You can use it as a capacitor of course, but then please change the name of your video

[@yakovhammer5739]

Love ur vids. I hate hearing u swallow it is hell in my ears.