PN Junction Band Diagram
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- เผยแพร่เมื่อ 4 พ.ค. 2018
- / edmundsj
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In this video, I go over the band diagram of the P/N junction, which will allow us to find the electric field and the P/N junction width through what is called the "built-in potential" Vbi. This is an important quantity in circuit design as well and is where the "turn-on voltage" of a diode comes from.
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.
This is the video I was finding for a thousand years.
so glad I found this video, you explain it very clearly and have a good narrating voiv
e, thank you!
thank you so much i have an exam in like 6 hours and I thought I'm going to fail. now I will fail a little less
Same as me. 😞😞😞
Woohoo! I hope you did fail a little less.
@@JordanEdmundsEECS failed by 2 marls
Awesome thank you so much for the semiconductor videos!
Thank you so much my college prof explained none of this . So it was very difficult to grasp where where the 0.7 and such values came from .
This guy.
;)
Thank you so much,literally lifesaver❤
Great. This is as good as it gets.
thank you so much for such a nice explanation :)
You are welcome! Glad it was helpful.
Very interesting Video thank you ! is this sketch right because in the n-Type are more electrons normal marked with a big minus ... (p&n-Type reversed?)
Thank you very much sir.
so to differentiate forward bias band diagram with reverse bias band diagram is depending on the width of build-in potential. if forward bias, the width is small to indicate that the electron from n-type conduction band is required to have more energy to reach p-type conduction band. am i right? please correct me
Sehr gut !
I'm watching your video in Korea and thank you for this quality video.
And I have a question about how to get the number of -xp and xn on x-axis
can you explain how the difference formula comes at 2:37. kTln(Na/ni^2)
Some of the books says in the pn-junction when higher energy electrons in the conduction band on the n-side diffuse to the p-side to reach equilibrium, the entire band structure on n-side will shift down relative to p-side. I am not sure what is the exact mechanism on how losing high energy electrons causes the band structure on n-side to shift down? I really hope you can enlighten me!
Thank you
Appreciating the video! One question; How come the gap between the valence and the conduction band is the same for p-type and n-type. Seems counterintuitive.
Great question! The underlying band structure (valence band, conduction band, intrinsic Fermi energy, etc.) are only dependent on the properties of the material itself (silicon in most cases), and so the distance between those bands won’t change with your doping. The only thing that doping changes is the Fermi level E_F, which affects the concentration of electrons and holes. You can think of E_F (or equivalently doping) as the only knob you are able to tweak to control a semiconductor device.
Why is the efficiency of the AM0 spectra lower than the AM1.5 spectra, even the incident
power is higher (for single-junction cells)? Can you please tell me
Very good video. I have a question: is the depletion region voltage drop the same as V_bi? Why they have the same value (if that is the case)? Thanks in advance.
2 years late, but it is because the example shown is in equilibrium that we can say V_BI is the same as the voltage drop in the depletion region.
In other cases like fwd / rev bias, the voltage drop would vary according to the external bias applied, and V_BI remains constant throughout all cases.
Hope im right haha, all the best
Thanks
Great video ! one question, when you say the Fermi energy (value) is constant you mean Ef so at the equilibrium condition there's not flow of electrons. If there are some variations, it's only the value E_i that changes right ?
The Fermi-level is defined as when the product of the fermi function = 50%. This 50% is stated by a line that represents a 50% probability that an electron will have a charge of that corresponding value.
So if the Fermi-level is always 50%, and both energy diagrams are on the same scale, you can combine both diagrams by using their respective fermi-levels as the middle (since they both resemble 50%).
Sir why is the shape of the depletion region like that
Jordan, please correct me if I am wrong. When we fuse two materials together their fermi level or their electro potential energies align to reach equilibrium. Your choice of the word "constant" in describing fermi levels in two different material is confusing. Fermi levels are different in different materials and they align when they are fused to reach equilibrium.
Correct! The term “constant” only applies once the two materials reach equilibrium, at which point the fermi level is flat throughout the entire system.
Finally clear English!
hi, is this banddiagramme the same by OLED??
thank you
Yup, it's just OLEDs are made out of a different material (organic materials). Almost all the same physics are at play.
Hi, I have a question.
I understand that Ef is constant within the semiconductor including the junction because otherwise there should be a current flowing and it is obviously not true at equilibrium condition.
However, I don't understand why Ei should be different for both N-type and P-type semiconductors at the junction.
Especially, I believe Ei should be equal for both N-type and P-type semiconductors because it is an"intrinsic" property of the material.
I am confused because creating a PN junction actually changed the properties of the material.
Your intuition is correct, Ei is the same everywhere *relative to the valence and conduction bands*. It should always be roughly in the middle no matter how the bands bend.
@@JordanEdmundsEECS Thank you for your reply.
hmm I have a question.
I assume Ec, Ev and Ei are material properties. So I assume those energy levels should stay the same for both P and N semiconductors.
Is this true?
If that is true then, when we create a junction, the energy diagram bends and the energy level of Ecn and Ecp differ by the applied voltage, which contradicts my assumption that Ec,Ev, and Ei are equal for both p and n semiconductors.
We use the model that energy diagram bends, so I believe my first assumption is wrong. If it is wrong then could you tell me why Ecn,Evn differ from Ecp,Evp?
Maybe I am asking a very silly question..
@@trueroughly1691 you are question is not at all silly, apparantely no one in the internet is answering this, have been searching for a while. why the Ec's and Ev's are different.
@@user-ge8hj9br6w I think I figured it out. It is because of the electric field created within the depletion region. There is an electric potential difference between P and N and there should be a relative energy level difference because PE at P and N is different due to the depletion region PE difference.
So take that into account, and from our premise that the PN junction is at an equilibrium state (otherwise there is a current but we assume no current), the energy level should be *relatively different* to satisfy both conditions. So Ei is naturally there as it is but what lifted up Ei of P-type semiconductor is the electric potential within the depletion region.
You know, PE is always relative. Jordan answered my question but at that time I just could not understand lol. Maybe I am wrong but I am going to the Embedded field so.. yeah I will stop here lol
@@trueroughly1691 I totally agree.I too found out that after commenting, yes its the potential difference conditions that should satisfy, there is no way Ec at p side and Ec at n side can be at same level after a potential(Vbi) or energy difference qVbi is created at the junction. Ecn and Ecp should adjust accordingly, Point energy make little sense here i assume, its the energy level difference that matters.
Since my background is in biology, I have no clue how to get the value of Nd (donor) and Na (accptor) from the p-type or n-type semiconductor. Did u explain this in your previous videos that i missed the point ?
Another question is, let say i'm doing experiment on a newly unknown material that is not discovered yet, how can i tell whether this material is N type or P type?
I really appreciate that u can answer these questions, Dr. Jordan, thx in advance.
This is something that you are usually given. When I want to figure out the doping of the Silicon wafers I work with, I do 4-point probe measurements and use the known mobility of electrons/holes in Si. If you don’t know the dopant type you can use the hot probe method.
@@JordanEdmundsEECS For your knowledge, I'm actually using biomolecule (liquid form) as a material to replace of the conventional Si doped material or other material in diode construction. Can hot probe method used for biomolecule? will they degrade in high temperature or the water will evaporate quickly?
Thx Dr. Jordan for the suggestion and it leads me to a direction, I'm trying to figure it out what type are them. I will look into it, if I have any question, I will ask again.
Dear Jordan at 2:58, why you have started sketching the band diagram for n type bit below in comparison to p type? I was looking for the reason behind it. Why we are not keeping the Ec and Ev lines for N and P type at the same level. Please do clarify my doubt.
Electron affinities are different when you dope Si with either P or B, from a chemical point of view. Although not a good reasoning, also the equilibrium condition for the Ef to remain constant along the depletion region and bulk material, requires the Ec and Ev of both p and n-type semiconductors to be arranged in this way, as it's shown in the video.
@@brunoaramburu740 Thanx for replying. I have figured out the reason.
Why we use Ei - Ef in p type and Ef - Ei in n type
I am not entirely sure why the Fermi energy is constant. I would have thought that the intrinsic Fermi energy is constant since it is the same for both materials. Can you help me understand that a bit more?
Your intuition is correct, the intrinsic Fermi energy is the same everywhere *relative to the valence and conduction bands*. As those bands bend, so too must the intrinsic Fermi energy, so that it’s at the same relative distance. The Fermi energy in contrast you can think of as the “equilibrium energy” for the entire system, by definition when the Fermi energy is constant, the system is in equilibrium. This means there is no net flow of electrons.
@@JordanEdmundsEECS okay, got it. Thanks.
Sir please make video for the band diagram in forward biasing
Can you make a video of band diagram of biased pn junction
I have no intuition about why E_F would be the same value in both halves. Can you explain why that is the case?
OK, so after thinking a lot, I realize the answer must be that this is only true in equilibrium. When the P-N junction is first created, the Fermi Levels, Ef,n and Ef,p, start out naturally different, and I think that would also mean that Ec and Ev are the same for both (uncertain here). However, there is a relationship between the Fermi Level and the probability of finding an electron. Due to the odds of finding an electron being higher in the N-type area, they will naturally drift over to the P-type area (this is modeled as diffusion current) and as this happens, the Fermi energies of the two haves start to become equal as any inequality will result in an equalizing drift. The shifting of the fermi levels more or less "Drags" the energies of the valence and conduction bands with them, creating the potential barrier. @Jordan can you confirm or elaborate on if I've understood this correctly?
I wish this were stated somewhere explicitly. I have struggled to find it.
You have understood this exactly as I currently understand it. This way of understanding things can also be extended to heterojunctjons, and you can use it to predict what their (rather strange) band structure looks like.
@@xephyr417 thank you for the nice explanation but I am still stuck at the point where Ec and Ev are the same. They should be same right ? Because the material is same. I am confused about this.
@@0xVikas That's my question too. Because Ei is an intrinsic material property of semiconductors, it should remain the same for both n-type and p-type semiconductors.
I can imagine that Ec and Ev can vary because the material is somewhat "ionized" at the depletion region(though I am not really sure about this because this is just my assumption). Therefore material properties Ecn, Evn, Ecp,Evp can be changed because the material itself is different from intrinsic semiconductors as it is electrically non-neutral.
However, I still cannot understand why Ei should vary because Ei is the intrinsic property of "undoped semiconductor"
@@trueroughly1691 maybe even Ei varies just like how Ec and Ev are changed, due to the precense of non uniform charges but yes, I am still not sure about this.
You have not discussed the behaviour of minority carriers I.e. holes on n_side and electrons on p_side at the time of junction formation.
So this example assumes that the p-type and n-type semiconductors both have the same band gap? That's not always going to be the case right?
Huh. That's a really great question. Yes, it does assume that. I never actually thought to question that assumption. In principle there is no reason they have to be *exactly* the same, but as long as you don't add too many dopants and don't disturb the underlying crystal lattice too much, you should have the same bandgap.
Thanks