Let's try focusing all questions and discussion*, on HW and everything else, as comments to this one post. That way if you want to see what everyone has posted, all you have to do is look here. (This will be the active "comment post" for one week starting now and going until next Tuesday's class.)
* The one exception at this time would be your quiz response. Post that as a comment to the quiz post itself rather than here. Everything else, here!
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I am having a particularly difficult time with the quantum mechanics of the first homework and quiz and I'm not sure if I will be able to finish and understand them fully by Thursday. Also, I have not taken Physics 139. I know it is the first assignment, but none the less, I am concerned that not completing the assignments will negatively impact my grade and my understanding for the midterm and final. I was hoping you could either alleviate my worry or allow me to turn in the homework after your office hours on Thursday so I can have a chance to ask you questions about the problems I do not understand. -anon
ReplyDeleteThis is probably a common concern I think. Basically you should not worry about that. Because there are people of varying backgrounds, aspects of the 1st HW will not be appropriate for everyone. This will not effect you adversely.
ReplyDeleteThis will not be an issue for future HW's which will tend to build on what we cover in this class and not rely so much on in-depth background and understanding of quantum theory and quantum notation.
I'm struggling with 9 as well. Would the localized bound state refer to the lowest energy state? Which in this case would similar to intrinsic Si except with one electron in the excited state from the Phosphorus??
ReplyDeleteHmm. Your question is outstanding and shows a lot of insight and intuition. It makes me realize how difficult and subtle this problem is.
ReplyDeleteEssentially what was once the excited state of the P has become part of a delocalized band of states in the crystal. (and that is the conduction band). The localized trap, or bound state, that forms around the P atom has a larger length scale, maybe 5 Angstroms instead of 0.5 Angstroms. It looks like the ground state of an H atom, only bigger (i.e., it has a larger Bohr radius). It arises from the attraction between the +1 phosphorous core and a conduction band electron, and, technically speaking, it is formed from a linear combination of the unbound conduction band states which can be combined to make a localized "ground state" orbital in much the same way that a combination of free states (momentum eigenstates) can be constructed to make the g.s. of H or of any 1D quantum potential such as the 1d harmonic oscillator, delta function of square well.
i've gone through the whole homework assignment, but for some reason the first problem is still giving me some trouble.
ReplyDeletesince we are dealing with FINITE potential wells here, the wave function actually extends(horizontally) outside the potential well, correct? and when, for example, problem 1b asks for "the four lowest energy states," the only idea that i had was this:
state one: wells #2-4 in ground state, well #1 in 1st excited state
state two: wells #1, 3, & 4 in ground state, well #2 in 1st excited state
etc...
but this would assume that the problem means four lowest energy states, not counting the state where each well is in the ground state...
so i'm a little confused as to what these are supposed to look like. any help would be great
In a problem like this, the lowest energy state is always going to be the ground state, where the wavefunction peaks above each well. Since the wells are finite, the tails of the exponential decay will extend slightly beyond the edge of the well. These tails are what contribute to the overlap integral.
ReplyDeleteThere's some simple rules to determine what the excited states look like. First think about what the excited states of a single finite well look like. We're sketching the excited states of a set of finite wells some arbitrary distance apart; if this distance goes to zero, the wavefunction of the system must collapse to that of a single finite well.
Therefore, you can use node counting to figure out how the wavefunction will behave. The ground state is the 0th excited state, and so has 0 nodes--it never changes sign. The 1st excited state has 1 node--it should change sign once. The 2nd excited state has 2 nodes, meaning it should change sign twice. This rule works for all excited states.
The last rule is that the nodes be distributed symmetrically among the potential wells. A single node is in the middle. They don't have to be evenly distributed, but they do have to be symmetric.
That, at least, is how I learned to sketch wavefunctions. Good luck!
ok thanks kelsey!
ReplyDeleteIn an attempt to offer a completely qualitative visualisation, you might imagine a lumpy string which you can twist.
ReplyDeleteThe more twists in the string, the more the lumps alternate their orientation - the higher the energy.
The ground state could have all the lumps in one orientation. In order to twist the string of these lumps into the first excited state configuration, we require that the twisting creates only ONE node... along the length of the array, the lumps change orientation ONCE. The second excited state requires TWO nodes - ie. a sequence of lumps pointing up, then a sequence of lumps pointing down, followed by a sequence of lumps pointing up again.
Among the many caveats, symmetric states alternate with ANTI-symmetric states such that an antisymmetric state has an orientation that can be mapped onto itself by a combination of folding AND flipping.
Louis Steele
ReplyDeleteQuestion about HW 2a #4c and #5
#4c: It says calculate the intrinsic carrier density of the model where Nv=2Nc, but I can't figure out how that is possible. I thought with an intrinsic semiconductor Nv=Nc. Am I wrong?
#5: Which two models are we supposed to be comparing? The Nv=Nc model to the Nv=2Nc model?
^_^
I am a little disappointed that:
ReplyDelete1) there have been so few questions or comments here
2) that no other student has responded to this inquiry (from Louis) yet.
I am optimistic that we will do better in the future.
Regarding 4c, what intrinsic means is no doping. In the absence of doping there is a natural relationship between holes and electrons which serves as an essential mathematical constraint. In general, a material, or a model, will not have Nc = Nv. Nc and Nv are model parameters and there is no reason they should be equal. What they refer to is density of states. That is distinct from "occupied states".
When the density of states in the VB is greater than that in the CB, how can the number of holes still be equal to the number of electrons? These are "thermal" carriers, right? What controls the thermal carriers?
Regarding 5: yes, exactly.
For an intrinsic semiconductor, we require that n=p. The number of electrons in the conduction band must equal the number of holes in the valence band, since it is the excitation of the electrons that creates the holes.
ReplyDeleteHowever, that does not mean that the number of states available to be occupied at the upper edge of the valence band (N_v) and the number of states available to be occupied at the lower edge of the conduction band (N_c) must be equal. Those values depend on the density of states function, which cannot be assumed to have the same value at different energies (and judging by the sketches drawn in class, rarely does).
That's why its interesting to calculate mu in an intrinsic semiconductor--it has to shift depending on the ratio between N_v and N_p to preserve the relationship n=p.
Subject: Help with HW 2a.
ReplyDeleteOne student has a pretty strong background in Stat Mech and quantum, but had to miss all our classes so far and thus is unfamiliar with our particular methods and approximations.
Can anyone explain how to do the problems of HW2a, including how one goes from the real Fermi function to the approximate form we use (and the justification for that) and how one gets from the real density of states to Nc and Nv (and the definition of those), etc. Thanks.
I couldn't attend thurs class so I don't know if I fully understand the doping... What I understood from Tues is that when you change the material of the semiconductor (or dope it), the energies of the bands also change. In class we solved for n which is the number of electrons in the conduction band, and p which is the number of holes in the valence band. n=Nce^[-(Ec-mu)/kT], where Nc=kTDc, and Dc is the density of states in conduction band. and p=Nve^(-mu/kT), where Nv=kTDv, and Dv is the density of states in the valence band. n and p are pretty much saying the same thing, but it's just looked at differently. The holes in the valence band are electrons in the conduction band, so we can set n=p. But Nc and Nv can be different, so that would just change mu in our equations.
ReplyDeleteWhat I wasn't sure about in Hw 2a is when you say we dope it with 10^16 carriers/cm3, or 10^16 donors/cm3, that just changes whether that is =n or p right? Then we solve for either Nc or Nv, and plug it into n=p, which should give us a mu right?
And I also don't understand what doping level is in #2. And in #4, when you change the ration of Nv to Nc, does that just add a fraction into n=p?
If anyone can answer those, it would be very appreciated. And Zack if we could may be meet on Tues april 13 before class (I have a break from 10am-12), please reply with a good time for you, it would be nice to talk those over with you since I had to miss class. I'm flexible so whatever works. Thanks a lot.
Emma
ReplyDeleten only equals p before the doping takes place. If it's doped with donors then it'll have more electrons in the conduction band at room temperature so now n will be the number of donors and you can use the equation you have for n, along with the given value of Dc, to solve for mu. If it's doped with acceptors then p becomes the number of acceptors and you can use the equation you have for p, along with the given value of Dv, to solve for mu.
Hey Mike.
ReplyDeleteSo, we started with the integral that we used some last quarter in 155:
N = Integral[D(E)*f(E,T),(E,-infinity,infinity)]
total # of electrons as integral of density of states (# states available) and Fermi-Dirac distribution (weighting that determines how many states are occupied).
We split this into an integral over the energy range of the valence band and an integral over the energy range of the conduction band, since those are the states we care about in the semiconductor model.
We then approximated the Fermi-Dirac function by saying, as in 155, that at finite T there is a small deviation from 1 at Emu.
valence band: f(E,T) ~ 1 - Exp[(E-mu)/(k_B*T)]
conduction band: f(E,T) ~ Exp[-mu/(k_B*T)]
With this approximation, the electron density (decrease in occupancy of valence band) and hole density (occupancy of conduction band) can be determined:
n = Integral[D(E)*Exp[(E-mu)/(k_B*T)],(E, conduction band)]
p = Integral[D(E)*Exp[-mu/(k_B*T)],(E, valence band)]
By approximating D(E) as having constant values over the two bands (D_v and D_c), setting the energy of the valence band as 0, and keeping only the largest terms of the integrands, we get simpler expressions:
n = k_B*T*D_c*Exp[(E_gap-mu)/(k_B*T)]
p = k_B*T*D_v*Exp[-mu/(k_B*T)]
Then we make the model simpler and say we only care about the density of states at the upper edge of the valence band and the lower edge of the valence band, since those are the only states with any real possibility of being involved in electron excitations at room temperature, and introduce new notation:
N_c = k_B*T*D_c density of states at T at edge of conduction band
N_v = k_B*T*D_v density of states at T at edge of valence band
This (slightly) simplifies the form of the carrier density expressions:
n = N_c*Exp[(E_gap-mu)/(k_B*T)]
p = N_v*Exp[-mu/(k_B*T)]
For an intrinsic carrier (no doping) we require that n = p, since the holes are resulting from the excitation of the electrons, so it is possible to solve those expressions to obtain:
mu_i = .5*(E_gap + k_B*T*Log[N_v/N_c])
For doping with a donor like phosphorous, we assume that the electron carrier density is the density of the dopant. For doping with an acceptor like boron, we assume that the hole carrier density is the density of the dopant. I think that's everything needed for the homework.
Oh, and T is always room temperature so k_B*T = .025 eV.
Kelsey:
ReplyDeleteThanks for the thorough and excellent comments--this has clarified a number of questions I have had!
Mike
Kelsey:
ReplyDeleteI tried calculating mu using n = Nc*Exp[(E_gap - mu)/kT] but got a value for mu larger than E_gap. Is the correct relation n = Nc*Exp[-(E_gap - mu)/kT] instead, for this would give me a value of mu less than E_gap.
Mike
Yes. My bad.
ReplyDeleteHow will the density "row" (that we used in class today) change once we start having different values for Nv, Nc, Na and Nd? And how will we calculate x0 when it will have two different values on either side of the junction? I guess I don't really need to know this but I am curious.
ReplyDeleteCaleb