Here is our first quiz. As soon as you feel ready, please take this quiz online by posting your responses as a comment to this post (and writing your name in your comment so i know who you are). The sooner you do this the better, moreover, it is okay to update your responses (answers) with a 2nd comment later (and i can even delete your first comment if you request that). I hope that makes sense.
Anyway, here is pop quiz #1: (PS. Let me know if you would rather have gotten this at the beginning of class? (just wondering) I thougth ths might be less stressful and offer more oppoptunities for learning.
1. What 3 essential "components" do you need in order to make a minimal model of a semiconductor?
2. What does the center energy of a band of states in a crystalline solid primarily depend on?
(what is the most important thing?)
3. What does the bandwidth of a band of states in a crystalline solid primarily depend on?
(what is the most important thing?)
4. Suppose a model solid is constructed from a periodic array of identical finite square wells. Each well is 10 eV deep, has a width d less then a, the periodic spacing, and, in isolation, each well would have a ground state at -9 eV and a 1st excited state at -5 eV.
Sketch what the bands for this model solid might tend to look like. Where are they roughly centered? How wide are they? Which one is wider?
(You can get extra credit if you notice and explain the subtle things associated with the difference in sign of the overlap integrals that play a role in the respective bandwidths for these bands and how that effects the graph of E vs q.)
5. What is exp{E/KT} for T = room temperature and:
a) E= 1 eV
b) E= 2 eV
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This is David Matthews. Here goes:
ReplyDelete1. To be a semiconductor, a material must have two bands of "allowed" electron energies with a gap on the order of 1 eV between them - much larger and it'll be an insulator. The less energetic of these bands, the valence band, must be full at T=0, while the conduction band must be empty. You can also have some thin energy bands in the gap from dopants.
2. The energy of a particular band centers around the energy of the corresponding orbital in just one isolated atom.
3. The width of the band (the deviation from the central energy) comes from the phenomenon of energy level splitting, which occurs when you introduce more atoms to the system. Bands corresponding to states with more degeneracies will be wider.
4. This is tough to do with ASCII!
5.
a)e^(1/(8.617*10^-5*273)) = 2.89*10^18
b)e^(2/(8.617*10^-5*273)) = 8.34*10^36
------
All answers are subject to change when others post their answers!
Actually, I'm not sure about the last sentence from #3. Scratch that.
ReplyDeleteHere's my best guesses:
ReplyDelete1. The three crucial components of the basic model of a semiconductor l are a full energy band (valence band), an empty energy band (conduction band), and a small energy gap between them. This basic configuration composes a semiconductor in that it is an insulator in the ground state at T=0, but additional thermal or light energy can excite electrons into the conduction band, in which state the material is a conductor.
2. Based on the calculation done in class, the center of an energy band of states in a crystalline solid is roughly determined by the level of the equivalent energy band of a single atom, or possibly molecule, that composes the crystal.
3. By the same calculation, the bandwidth depends on the overlap integral of the system--how neighboring atoms/molecules are interacting with each other. If the overlap integral is large, and neighboring atoms/molecules have strong interactions, then the single-atom model is not a good approximation, and the energy band will be wide. Interactions break degeneracy, so that what was one energy level becomes a spread.
4. I'll play with this tomorrow.
5. according to WolframAlpha:
a) e^[(1 eV)/([8.617*10^(-5) eV/K]*[293 K])] = 1.59*10^17
b) e^[(2 eV)/([8.617*10^(-5) eV/K]*[293 K])] = 2.53*10^34
Oops, I used the wrong temperature... 273 K is a bit cold for room temp. Here's my new quiz, you can delete my first posts.
ReplyDelete1. To be a semiconductor, a material must have two bands of "allowed" electron energies with a gap on the order of 1 eV between them - much larger and it'll be an insulator. The less energetic of these bands, the valence band, must be full at T=0, while the conduction band must be empty. You can also have some thin energy bands in the gap from dopants.
2. The energy of a particular band centers around the energy of the corresponding orbital in just one isolated constituent - one atom or molecule, depending on the material.
3. The width of the band (the deviation from the central energy) comes from the energy level splitting that occurs when you consider multiple atoms in a system instead of just one. Energy levels that were previously equal will split slightly and, once many atoms are introduced, will create a "band" of energies around the original.
4. This is tough to do with ASCII!
5.
a) e^[(1 eV)/(8.617*10^(-5) eV/K * 293 K)] = 1.59*10^17
b) e^[(2 eV)/(8.617*10^(-5) eV/K * 293 K)] = 2.53*10^34
This comment has been removed by the author.
ReplyDeleteI think Dave's and Kelsey's answers are excellent and hard to improve upon - but I will try. In the meantime I will also work on #4.
ReplyDeleteEthan
Hi Ethan,
ReplyDeleteYou don't have to improve on them. You can just address the quiz, in your own words, as if their comments weren't there. (This is for everybody.) I feel confident that the experience of articulating your answer, without explicit reference to external sources, is very valuable.
1) 3 components
ReplyDelete-> A valence band, the lower energy band and a near full band of electrons
-> A conduction band, the higher energy band with near empty band of electrons
-> Energy cap between the valence and conduction band such that incoming energy of Egap can excite the electrons from the valence to the conduction
2) From our derivation, the center of the band is a result of the intrinsic semiconductor to be at such a location that n=p
3) The bandwidth of a band is resultant of the interation between atoms in the periodic crystaline structure of the solid.
4) I think this is what you ment
5) KT @ Room Temp -> 1/40 eV
e^(1/(1/40))=2.35*10^17
e^(2/(1/40))=5.54*10^34
Here's a (rough) verbal sketch for #4:
ReplyDeleteThe lowest energy band will be centered around -9 eV, and the first excited energy band will be centered around -5 eV.
An electron in the first excited state has more energy, and is therefore less tightly bound. I would therefore expect the overlap integral to be larger for the first excited state, since there will be more interaction between an electron and neighboring wells. The first excited energy band should therefore be wider than the lowest energy band.
For the ground state, the overlap integral is negative, so the minimum of the energy band will be at q=0. Because the wavefunction changes sign at the central node of the first excited state, the overlap integral of the first excited state will be positive. The maximum of the energy band will therefore be at q=0. These energy gap between these two bands will therefore be largest at q=0 and smallest at q=+/-pi.
1. The energy levels and bandwidths of the valence band, energy gap, and conduction band.
ReplyDelete2. I am confused with what this question is asking. However, center energy to me seems to apply to fermi energy. And this depends on the chemical potential and temperature of the semiconductor.
3. The bandwidth of a crystalline solid depends on the structure of the crystal, namely its location in the periodic table.
4. Mimicking the graphical approach used in class(E vs. DOS) The graph would be flat up until the valence band which will jump to -9eV. The beginning is located, from the problem, at d. The center of this band roughly at a-(d/2). The graph will then "drop" to 0ev at a. This will remain flat until roughly a+(5/9)*d.(This band will be wider and shorter with the same area, however will still fall in the periodicy of a. So, the beginning of d changes, not the end) This will "rise" to -5ev. The center of this band located at 2a-5d/18.
The overlap of the integrals comes from the imperfectness of the square wells. The bands have a slight amount of area taken from the top and given to the bottom(essentially rounding the well). This will slightly widen both wells.
5. ~9.22E16(1eV @ room temp w/ KT~.0256(class))
~8.50E33(2eV)
1. A full valence band, empty conduction band, and an energy gap separating the two.
ReplyDelete2. The center of the bands depend on the energy levels of the individual atoms when they are separate; i.e. ground state, first excited state, etc. The center of the energy gap can be approximated by the Fermi energy of the system.
3. The way I'm thinking about the band width is that as the individual atoms are moved together their energy levels merge (much like the probability wave functions do when atoms are covalently bonded) this broadening the bands.
4. my little sketch
5. a) 9.218*10^16
b) 8.497*10^33
(I used room temperature as 298K (25C), so KT is about 0.0256)
1. a continuous range of allowed lower energies (the valence band), a continuous range of higher level energies (the conduction band), and a continuous range of energies separating these two bands, energies for which electrons are not allowed to have.
ReplyDelete2. the ground state energy of an electron in a single-well potential
3. the bandwidth is proportional to the overlap integral, t.
4. the valence band is centered at -9eV. The conduction band is centered at -5eV, it is shorter and wider than the valence band. The width of the valence band is proportional to the strength of the well and the spacing between each well.
5. room temp = 295K
a) exp[1eV/(k*T)]=1.25*10^17
b) exp[2eV/(k*T)]=1.57*10^34
1. An (intrinsic) semiconductor has to have a full valence band and a nearby (Egap~<5eV) conduction band empty of electrons, all at T=0. If the gap is too large, it becomes an insulator. If the gap becomes nonexistent, or the bands overlap, we have a metal. So, in that region in between, we have a semiconductor.
ReplyDelete2. The center energy E0 comes from the unperturbed energy state, the integral, where H_0 = -hbar^2/2m (d/dx)^2 + V_0(x), where V_0(x) is the potential for a single well. Then, when we perturb this state by adding in the interactions from nearby wells, we get the band structure. There are basically a bunch more states now available (they're actually still discrete, but the spacing becomes so tiny they're essentially continuous) once we take the other wells into account.
3. The bandwidth comes from the overlap integral, , and is actually equal to 2 times this integral. Granted, this is only a first approximation. There are finer corrections to this if we don't ignore the denominator of the perturbation (), which we are in this approximation setting equal to one (this is equivalent to saying that = delta(m,n) (delta = dirac delta function), or that the overlap of wavefunctions from different wells is very small compared to the overlap of a well with itself).
4. Yeah, I don't think I have the means to hand-draw this in this space. Also, I'm not sure of the answer (this is why I'm copping out and saying I can't draw it here). Some people have mentioned E vs Density of States--how can we figure that out? Any comments would be much appreciated.
Regardless of my ignorance, I'll make a few stabs in the dark. So, from what was shown in the notes online, it seems like the bands are centered at E0, that is, the energy level for a single well. So, it seems like the bands should be centered at -9eV and -5eV, but I could be completely wrong here. The bandwidths are due to overlap integrals. So, the level with more overlap has a wider bandwidth. Here I'm stuck--I don't know what wavefunctions to overlap for the first excited state. Do we just replace the subscript 0's in the ground state bandwidth integral (given in #3 above), or do we do something else. If we do the former, then it seems that the ground state band will be wider, since it seems to have a less rapid decay outside the well, and so a higher amount of overlap. The first excited state seems to have a higher amount of curvature in the region exiting the well, and so it has a smaller "skin depth" (for lack of a better term), meaning less overlap and a smaller bandwidth.
5. Let's say room temperature is about 300K, at which point kT is about 1/40th of an eV (according to Fred Kuttner). Then, we have:
i. E=1eV: Exp[E/kT] = Exp[1/(1/40)] = Exp[40] = 2x10^17.
ii. E=2eV: We doubled E, so this should square our answer: Exp[E/kT] = 4x10^34.
Looking at other people's answers, I think my value for kT may have been a little low (it's closer to 0.026eV than 0.025eV (or 1/40)), but perhaps for our purposes my level of precision may be sufficient.
Oh, by the way Zack, MikeK = Mike Kozina.
ReplyDeleteAs a clarifier for #1, I said intrinsic because the charge is balanced (there are an equal number of holes as electrons) in this system, and zero free charge carriers at T=0. Only when there is energy kT do electrons get excited from the valence band and go into the conduction band, allowing current to flow in the valence band via the missing electrons (the holes) and current to flow in the conduction band (via the electrons). Extrinsic semiconductors have an intrinsic charge imbalance due to impurities, and there are inherently either more holes than electrons or vice versa.
ReplyDelete1) A semiconductor needs a full valence band, an empty conduction band, and a small energy gap between the two that electrons sometimes overcome.
ReplyDelete2) Center of a band is based on the energy of the orbital from just a single atom.
3) The bandwidth comes from the interactions of the many atoms involved.
4) I'll come back to this one
5) room temp= T =300K
k=1.38x10^-23J/K/(1.6x10^-19J/eV)
=8.6x10^-5eV/K
a) e^[1eV/(kT)]= 6.09x10^16
b) e^[2eV/(kT)]= 3.70x10^33
~Emma Pulleva
back to #4...
ReplyDelete4) I'm thinking the valence band will be centered at the lower energy of -9eV. The conduction band will be centered at the higher energy of -5eV, and it will be wider because that spaces the energy wells closer together, as in the excited states.
1) must have a valence band, conduction band, energy gap. the engergy gap should be withn 0-3ev.
ReplyDelete2)it depends on the atomic energy of a single atom/particle.
3) the width depends on the interaction of atoms in the solid, which is mathematically interpreted by the overlap integral t in the notes.
4) Since the band is centered at the atomic energy of a single particle. if you consider that single particle in the ground state of -9ev, and in a first excited state of -5ev.
These energies would be the center of the bands. the width is determined by the overlap intergral. the band centered at -5ev is wider because the particle has more kinetic energy therefore it interacts more with the lattice.
As a result the overlap integral t gets bigger, therefore the bandwidth is greater.
5) T =300K using a calculator i got this.
a) e^[1eV/(kT)]= 6.09x10^16
b) e^[2eV/(kT)]= 3.70x10^33
1. The essential parts to a semiconductor are the valence band, conduction band, and the band gap.
ReplyDelete2. The center of a band of states in a crystalline solid primarily depends on the state at which the particle is in, which means that the particle goes to the center of the band when it jumps to the next state, and deviates from the center from it’s interactions with the lattice.
3. From what I wrote for question 2, the band width depends on the interactions with the lattice, which we can determine by the overlap integral.
4. The band is centered at the energy state of a single particle, so the particle would be at -9 eV on the ground state, and -5 eV in the first exited state. The width is determined by the overlap integral. The particle in the band centered at -5 eV has more energy, which causes it to have a greater interaction with the lattice and therefore have a greater overlap integral and larger bandwidth.
5. Let T =300K
a) e^[1eV/(kT)]= 6.1x10^16
b) e^[2eV/(kT)]= 3.7x10^33
1. For something to be a semiconductor, it has to have a full valence band, a conduction band, and an energy gap between the bands on the order of ~1 eV.
ReplyDelete2. The center of a band of states is basically the ground state of a single particle, not counting the other interacting particles.
3. The band width comes from the interactions of the other particles, which split the energy levels into a continuous 'band' of energies.
4. As shown in class, the graph is flat except for two bars at -9 and -5 eV. The ground state at -9 eV is taller and thinner, while the bar at -5 is shorter but wider, with each bar having the same area.
5. @ T = 300 K,
a. e^(1 eV / (kT)) = 6.09 x 10^16
b. e^(2 eV / (kT)) = 3.7 x 10^33
BTW, this is Cameron Brown.
ReplyDeleteBrian Vollbrecht
ReplyDelete1.) Valence Band (highest filled energy band at T=0), Conduction Band (energy band into which electrons from the valence band can be excited), and an Energy Gap between those two, on the order of 1 eV.
2.) The center energy of a band depends on the number of electrons in the outermost shell of the atoms which make up the solid.
3.) The width of the energy bands in a solid depends on the separation distance between atoms in the solid.
4.) Here’s my picture: file:///Users/brianvollbrecht/Desktop/Physics%20156/Quizzes/quiz_01_scan.JPG
5.) a.) exp(E/kT) = 2.35 x 10^17; b.) exp(E/kT) = 5.54 x 10^34
1) A typical situation with semiconductor material is that the valence-electron levels of individual atoms produce two energy bands, called (i)valence and (ii)conduction bands, that are separated by an (iii)energy gap which determines the basic material properties. The electrons that form the covalent bonds are energetically placed in the valence band. When a covalent bond is broken due to some thermal energy delivered to a valence electron, this electron jumps up into the conduction band, leaving behind a hole.
ReplyDelete2) The atomic energy of the orbital from a single atom or particle.
3) The interaction of atoms with the lattice determined by the overlap integral of the system.
4) http://picasaweb.google.com/villanueva.bb/Phys156#5457623903889774034
5) kT = (8.62*10^-5 eV/K)*(298K) = 2.57*10^-2
e^(1eV/2.57*10^-2 eV) = 8.06*10^16
e^(2eV/2.57*10^-2 eV) = 6.51*10^33
Caleb Caldon
ReplyDelete1)A valence band that is full at T=0, a band gap on the order of 1eV, and a conduction band that is empty at T=0.
2)The energy of the corresponding energy level in the single atoms that make up the solid.
3)The inner product part of our derivation that depended mainly on the eigenstate in the individual atoms and the periodically repeating potential.
4)The valence band is centered at -9eV. The conduction band is centered at -5eV. The conduction band is shorter and wider than the valence band.
5)
a)1.57*10^17
b)2.46*10^34
1) In order to have a minimum model of a semiconductor, the following components are necessary:
ReplyDelete- an approximately full valence band at T=0
- an approximately empty conduction band at T=0
- an energy gap between the bands on order of magnitude ~1 eV. Carriers must be excited with substantial amount of energy to jump energy gap into conduction band.
(not that you probably had any trouble figuring out who this is, Im Garrett Rogren)
1) You'll need a conduction band, a valence band, and quite a gap in energy so an electron wouldn't be tempted to just jump from just the valence band up to the conduction band without much, much persuasion (an energy gap).
ReplyDelete2) I can't answer this question very well since I don't have a very good definition of "center energy", but I'm assuming it is either the Fermi energy (Mu) or comes from the number of electrons in the outermost shell of the crystalline solid we're talking about.
3) We could either word this as the distance between the two potential wells is the most important thing in determining this since that affects how much overlap we get in our overlap integral, which a further discussion of overlap integrals can be found on this blog. Zack does a good job explaining some of it.
4)brb
5) Assuming room temperature is 300k,
1eV=6.09*10^16
2eV=3.70*10^33
2)the center energy of a band states primarily depends ground state of the particles in single atom and is not influenced by other atoms around it.
ReplyDelete3)The Bandwidth of a band of states is a result of multiple atoms with small inter atomic spacing forming split energy states(due to linear combination of atomic orbitals). When the distances between the atoms approach equilibrium, the band splits into two bands separated by and energy gap, which contains no electron energy states.
4)The valence band is centered at roughly -9 eV and the conduction band at -5 eV. The conduction band is wider, but shorter than the valence band, however the both of the same area(n=p).
5) T=290k K=8.617*10^-5 eV/k
a.exp{1 eV/KT}=2.39*10^17
b.exp{2 eV/KT}=5.72*10^34
(sorry i posted as 2 comments, im still getting used to this)
1. A semi Conductor needs a valance band that has electrons and electron holes filling it, a conduction band that holds excited electrons, and an energy gap between the two that the electrons must overcome to be part of the conduction band.
ReplyDelete2. The energy of a band is dependent mostly on a single atom and its particular orbitals and its ground state.
3. The bandwith depends on the interactions of many atoms together in the semi-conductor. We can then find numerical parts of the bandwidth from the overlap integral.
4. The two bands are centered on the ground state, -9eV, and the first excited state, -5eV. The width can be found if we use the overlap integral. Because the second band is centered at -5eV it has more energy than the first band, and so the overlap integral will be larger. Hence the second band will have a larger bandwidth.
5. T = 290K so kT=.024989
a. exp[1eV/kT]=2.39E17
b. exp[2eV/kT]=5.73E34
1. Valence band, Conduction band, and energy gap. An intrinsic semiconductor with have a full valence band and an empty conduction band at T=0K. At normal temperatures (i.e.: room temperature), there is sufficient thermal energy for a small number of electrons in the valence band to gain enough energy to jump the band gap and enter the conduction band. This is useful in that it allows a current to flow through a semiconductor at normal temperatures. \\\\
ReplyDelete2. The atomic energy of a given orbital for an isolated atom. \\\\
3. The interaction of the atoms in a lattice can be accounted for approximately by the overlap integral $t=<\Psi_0(x)|V(x-a)|\Psi_0(x-a)>$. From this overlap integral (interactions), one can find the width of a band to equal 4t. \\\\
4. The bands are centered at -9eV and -5eV. The width of the bands can be determined by the overlap integral $t=<\Psi_0(x)|V(x-a)|\Psi_0(x-a)>$. The width of the band centered at -5eV is wider because t is larger for that band. This is due to the fact that particles in the first excited state have more kinetic energy than those in the ground state. (see sheet HW \#1, part 1 for diagram). \\\\
5a) For E= 1 eV, \\
$e^{1eV/(8.617\times10^{-5}eV/K)(293K)}=1.6\times10^{17}$ \\
5b) For E= 2 eV, \\
$e^{2eV/(8.617\times10^{-5}eV/K)(293K)}=2.5\times10^{34}$ \\
lg_magius --> Louis Steele
ReplyDelete