Let's try to do a lot of this via discussion here:
1. a) Describe a picture of how a solar cell could work (which you could compare and contrast to an LED if you like). What processes are important; what currents are generated? In what regions of the junction?
b) What might be the influence of different biases (+, -, magnitude) that could arise? Consider, for example, connecting the junction +- infinity regions to each other via a capacitor (or a resistor).
c) if you are familiar with photosynthesis, compare, or contrast, its process(es) with those of a solar cell.
2. a) What does MOSFET stand for?
b) sketch a MOSFET and describe how it works.
c) what is the difference between "normally on" and "normally off"?
(discuss here)
3. extra credit: discuss here. What does the C in CMOS stand for? What is an advantage of having both hole and electron FET's? (discuss here)
3. Come up with another problem to add to this assignment. (and present it here as a comment)
4. For our usual p-n junction parameters mu=5000, Nd=10^17, etc.
( z) Would someone please post our usual parameters here.)
a) what value of the recombination time (which we have been calling $\tau_p$ on the right side, where diffusing holes are the minority carriers) would make the width of the diffusion region 5 or 10 times larger than $x_0$? (width defined by 1/e..., feel encouraged to ask about that or explain it in detail here.)
5. (optional, xc) For a "Scientific American" type article, do an illustration of a p-n junction under forward bias that includes all important regions: (depletion, recombination or diffusion, and, hmm, what should we call that outer one?)
Discuss here: the best names for each region? (This is an audience with no preconceptions regarding the names of the regions.)
a) Is "depletion" an ideal choice? Can we do better?
b) recombination vs. diffusion vs other choice??? what is most accurate, descriptive or helpful.
c) what about the outer region.
PS. You can choose to focus on photon emission (LED), or phonons, i.e., whereby the energy of recombination turns into heat (or both).
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ok here's my extra problem for number 3. i don't really know the answer to it, but it's something that i am curious about.
ReplyDelete3. Given what you know about how solar cells work, describe what things you could do to make the solar cell most efficient, ie, generate the most current for the same amount of incident photons. Do this in terms of doping levels, mobility, or whatever other quantities you deem appropriate...
I think one thing they do is actually link a lot of devices together. Something about how the most current is generated in the region right around the p-n junction, so the most efficient configuration is a system of small devices which are mostly junctions.
ReplyDeleteKelsey, I think of that as more in the realm of circuit design. While it is important, it doesn't really address what makes each p-n junction (in the chain) more of less efficient. I think Brian's question is more related to the basic question of what would tend to make a single solar cell more or less efficient?
ReplyDeleteI agree its an interesting question. I'm still trying to figure out how a solar works, though.
ReplyDeleteAt a basic level, you want a current resulting from the excitation of electron-hole pairs by incident photons.
I think that we're going to be looking at a heavily doped junction with a forward bias. That gives us a minority carrier current. The majority carrier current is unlikely to be significantly affected by the excitations--there's so many carriers already, a few more won't make much difference. A relatively small fluctuation in the minority carrier current will have a noticeable effect, though.
If it is the minority current that we care about (what a series of assumptions I'm making!), then we want it to extend out from the junction as far as possible. If the junction is heavily doped on one side and lightly doped on the other, the minority carrier current should be stronger at the edge of the depletion zone on the lightly doped side (since its inversely related to the doping of the other side), and therefore extend further from the edge of the depletion zone (since it decays exponentially from the boundary value).
It would be good if the compound has a high mobility, of course, since that will also increase the current.
Brian - That's probably a really important question in industry, but I don't know anything about solar cells at all yet. Maybe I just missed something in class... But anyway, after some discussion on 1a, that's definitely a question that would be cool to address.
ReplyDeleteAnyway, for 1a, my current picture of a solar cell is essentially* the reverse of an LED. By my understanding, an LED works by emission of a photon of energy $E_g$ every time a recombination occurs; a solar cell, then, could use incoming photons from the sun (of energy $E_g$ or somewhat higher) to kick electrons back up to the conduction band and therefore generate extra carriers.
How this creates a potential (or a current), though, I'm not exactly sure; my first guess is that generation events will affect carrier density and therefore drift current. What doesn't make sense to me about that is that there wouldn't be any preference for extra carriers to go one way or the other. The carrier density would be different, but the gradient would still be the same - each side of any particular point would just have the same amount of carriers added to it. I thought that maybe the increased carrier concentration in the part exposed to the sun would diffuse into the unexposed parts, but then the extra of one type would cancel out with the extras of the other type.
My second guess, then, comes out of searching for some reason for newly generated carriers to move in the same direction as others of their type and create an actual current. As charged particles, my first thought is an electric field like the one we have in the depletion region; if a generation event were to happen in that field, then the hole and electron would each be pushed away from each other, each creating a charge current in the same direction and adding to our total drift current. Then, as these carriers are kicked out of the depletion region, we'd end up with extra holes being injected into our p-type side at $-x_0$ and extra electrons in our n-type side at $x_0$. But this, then, would increase the majority carrier concentrations on each side; didn't we say it was minority carrier currents that were important? If anything, it seems like this would reduce the minority carrier concentration by encouraging more recombination events. Perhaps, though, there's some relation that I'm forgetting about that would make things work out there.
Anyone have a better model that would explain how a current gets created?
(*: is the word "just" ok if it has extra syllables?)
Kelsey - I think (but I'm not sure) that your idea falls the same way my first guess did, in that the minority carrier concentration would indeed be increased, but it would be by the same amount everywhere so gradient that's driving the actual current would be the same.
ReplyDeleteAs for how to make it work better, your approach sounds like you would try to maximize the total area covered by the depletion region in each microscopic device and then, by stringing them together, in the overall macroscopic device. Since it's in or near the depletion region that everything interesting happens, this approach makes sense to me since more photons will be interacting with the interesting parts of your solar cell!
Number 2:
ReplyDeletea.) MOSFET = Metal-Oxide Semiconductor Field-Effect Device
b.) i basically took my description and picture of a MOSFET from this website, which gives a pretty good basic description of the device:
http://www.techpowerup.com/articles/overclocking/voltmods/21
c.) "normally on" refers to a device that has an "electron channel" between the two n+ regions (see the pic in the link i posted above) BEFORE an external voltage is applied to it. "normally off" then refers to a device that needs to have a voltage above a certain "gate voltage" applied to it, before the electron channel can form and charge flow from the source to the drain can happen. (pg. 301 in Streetman describes these two concepts)
i guess i'll take a stab at the EC problem (the first number 3):
ReplyDeleteCMOS = Complementary Metal-Oxide Semiconductor
This means you use both n-type and p-type materials for the creation of your FETs. From what i have read (which isn't much), these are the most popular kind of transistors, especially in computer technology. they use almost no power when not engaged, so that is a plus i guess. i'm not sure exactly why, but my guess is that it has something to do with the gate voltage of "normally off" FETs, which you could basically use as an on-off switch, by just varying the amount of voltage applied. i'm curious to see what other people have to say about this one...
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ReplyDelete4.) Usual parameters for the p-n junction:
ReplyDeletemobility = 5000 cm^2/V*s
Na=Nd = 10^17 cm^-3
Nv=Nc = 5*10^18 cm^-3
Egap = 1.0 eV
ΔE = 0.1 eV
Permittivity of Si: 11.8*(8.85*10^-12) C/(V*m)
i need some clarification on number 4:
ReplyDeletewhat is the "diffusion region"?
would that be the region between x0 and the quantity L that we defined in class on thursday?
and i don't get how x0 is defined by 1/e... what does that mean?
#4:
ReplyDeleteSo we know that $p(x)=p^0+[p^0(e^{eV/kT}-1]e^{-(x-x_0)/\sqrt{\tau_pD}}$ and we want $\lim_{x\to 5x_0}e^{-(x-x_0)/\sqrt{\tau_pD}}=0$
This is all find and dandy, but we have an exponential decay function here that never actually equals zero so all I have been able to come up with is $\tau_p=16x_0^2e^{\infty}/D$ which is just some abysmally high number that doesn't really mean anything to us...
Any thoughts? Maybe we can set a functional limit to p(x) S.T. $\lim_{x\to5x_0}p(x)=0$ so far as we are concerned.
~Lou
to brianv:
ReplyDeletex_0 is not defined by 1/e. but the diffusion region is.
brianv:
ReplyDelete4.) Usual parameters for the p-n junction:
mobility = 5000 cm^2/V*s
Na=Nd = 10^17 cm^-3
Nv=Nc = 5*10^18 cm^-3
Egap = 1.0 eV
ΔE = 0.1 eV
Permittivity of Si: 11.8*(8.85*10^-12) C/(V*m)
I think that might over-specified since you can't set all three things. For example, one can leave Nc and Nv unspecified and say that $E_c - \mu = 0.1$ eV and $\mu -E_v = 0.1$ eV. Is that what your $\Delta E$ refers to?
For 1b): The influences of different biases on what? A solar cell? Or just a p-n junction in general?
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ReplyDelete@Dave
ReplyDeleteI'm a bit unsure about that too. It seems to me that a reverse bias couldn't be generated directly within a solar cell, especially if the LEDiode comparison is valid (which it very well seems like from my understanding + what I have read).
lg_maguis said:
ReplyDeleteSo we know that $p(x)=p^0+[p^0(e^{eV/kT}-1]e^{-(x-x_0)/\sqrt{\tau_pD}}$ and we want $\lim_{x\to 5x_0}e^{-(x-x_0)/\sqrt{\tau_pD}}=0$
Actually we use:
$\lim_{x\to 5x_0}e^{-(x-x_0)/\sqrt{\tau_pD}}=1/e$
Using my version of what Zack posted (I didn't bother with the limit and used $ x-x^0 = 500nm) $ ) I got a $ \tau_p $ of 0.1 sec, which makes sense to me.
ReplyDeleteHere is a more specific problem, related to #1, that I think may be interesting for understanding a conventional view of photon to electrical energy processes.
ReplyDeleteConsider a p-n junction, initially in equilibrium, hooked up to a capacitor (intitially uncharged).
a) sketch a picture (they are in series).
b) Suppose that starting at t=0 and continuing after that, photons start hitting the p-n junction at a rate that produces 10^5 generation events per second.
(To make sense of that, assume the junction is thin in the z direction (about 1 micron) and 1 cm long in the y direction and that x is perpendicular to the interface (of the p and n regions, like always).
What happens?
c) Estimate and graph the energy stored in the capacitor as a fucntion of time.
(discuss your assumptions)
Wait, I forgot about the square root, and that seems way too long.
ReplyDeleteI actually got 5*10^-7 sec, which seems much better.
@ Zack
ReplyDelete1a)Click
1b)The striking of the photons would cause excitations that lead to a current with in the PN junction. This would lead to a build up of charges on one side of the capacitor.
1c) I would assume that the capacitor would charge like any other capacitor attached to source, and take infinitely long to charge, with it be approximately 99% charged at $5 \tau$
yes zack that's what my ΔE was referring to.
ReplyDelete@Zach
ReplyDeletealso as an updated to my 1c) I was thinking about this, and thought that at some point it *might* be possible for charge to build up on one side, and cause a reverse current at some thresh-hold. I dont know if this possible, but it seems reasonable to an extent ...
Any comments?
Ok I'm just trying to do a kind of stream of consciousness thing for number 4:
ReplyDeleteI think I'm correct in saying that the condition we are trying to satisfy is this:
(L-x0) = 5*x0, or L = 6*x0.
We know L = (Dτp)^(1/2).
Then, we have:
τp = [(6*x0)^2]/D
Then we have to get x0 in terms of the quantities we know.
Don't we need to know the voltage bias to do this?
If we know the voltage, then we can just use this equation for x0:
(x0)^2 = [ε(eΔϕ-eV)/(e^2)Nd]
...and plug it in to my above equation to find the value of τp that satisfies our condition.
...Thoughts?
My contribution to #3 about adding another question: What does it mean for something to be a logic gate and how are CMOS FETs helpful in them?
ReplyDeleteI'm not sure if this is too far off topic but I find it interesting.
Dave
ReplyDeleteI got a value for Tau_p that's of the order of 10^-11 seconds. Where did you get the value of 500nm and what did you do with it?
Brad -
ReplyDeleteI don't think charge can only build up on one side of a capacitor, as charges on one plate would repel like charges and attract opposite charges to the other plate; that's why I think the capacitor would just charge as if the photocell were replaced with any other EMF.
Also, I don't think there's any way that you'd get a reverse current, besides a tiny leak through the p-n junction that might happen if the light that the solar cell was exposed to dimmed.
Brian -
You just made me realize that I used the width of the depletion region without considering a bias. I totally forgot that affected it...
Caleb - I only used 500 nm because that's a nice round value that I thought would be between the 5 and 10 times $x_0$. Unfortunately, I forgot that a bias will change the length of $x_0$.
ReplyDeleteWhat bias voltage did you use?
Oh, and I put it into Zack's equation
ReplyDelete$\lim_{x\to 5x_0}e^{-(x-x_0)/\sqrt{\tau_pD}}=1/e$
though the limit seems sort of unnecessary.
Dave-I'm having trouble recognizing how Zack's equation is appropriate for this situation. Does anyone care to explain that if they understand it.
ReplyDeleteBrian-I agree more with your way of doing things but we still all face the issue of what value to use for the external voltage.
ReplyDeleteDoes anyone have any ideas?
Zack probably just wants us to pick a value and use it. Maybe pick a few and see how it depends on it.
ReplyDeleteDave-
ReplyDeleteI agree, I think the limit is just cumbersome notation. Since the function is not undefined at 5x0, but rather is perfectly well-defined, there is no need for the limit.
Caleb-
Zack's equation says that at x=x0*5, we want the magnitude to decrease to 1/e of its initial value(ie value at x0), a typical decay length. Since Zack mentioned either 5 or 10 times, we could do either--that is, solve for a $\tau_p$ that would result in x=10*x0 yielding 1/e.
Actually, there is one technical issue with the limit. Since we are looking at x-x0 in the argument, if we want the decay length 5 times x0, then we need to evaluate the exponential at x=6x0, since we are subtracting one x0. Likewise, if we want 10*x0, then we should evaluate it at x=11x0.
Caleb/Brian:
I think we can just pick a value for our external voltage. How about 0.8V? I think we used something of this order earlier on, so why not use it now?
Dave:
I read the text about the direction of the current formed by incoming photons, and couldn't understand why it said the current was in the same direction as the drift current. Your explanation, however, made a lot of sense to me--I think the gist of it was that the carrier density increases absolutely (so n goes up), but this increase is uniform in space, at least locally (ie, it could increase more in the region outside the depletion region, or maybe within--I'm not sure), and so dn/dx doesn't change. Since Jdrift goes like n and Jdiff goes like dn/dx, the former changes but the latter doesn't. I don't know if this is what you meant, or even if this is accurate, but this is now the picture I have...perhaps we can clear it up in class.
Oh, and I agree that an LED is the reverse of this--as you need photons of energy greater than Egap to excite electrons and generate a current and a change in the voltage difference, an applied voltage can cause electrons and holes to recombine (ie, the electrons drop down from the higher-energy conduction band to the valence band) and give off photons of energy Egap. I wonder what other processes are involved with the electrons falling back down to the hole state--I've read that there are other ways electrons can fall back down in energy than just fluorescing (releasing photons).
An interesting issue that I read about in the text but I don't think has been mentioned here is the possibility that photons may just pass through the material or be reflected. I wonder what factors (photon energy, light intensity, properties of the semiconductor) affect the relative amounts of reflected, absorbed, and transmitted photons. I guess that can be my question for #3.
I found a copy of my old AP Bio book and looked up photosynthesis. From what I gather, there is a collection of pigment molecules called an antenna complex because they collect all the solar energy there. This energy excites electrons from either chlorophyll a or b (two slightly different pigments), and the electrons are apparently passed from molecule to molecule, losing energy to free protons all the while (the mechanism behind this is not explained...). These protons apparently create a concentration gradient, which is used to fuel the combination of adenosine diphosphate (ADP) with a phosphorus to creaet adenosine triphosphate (ATP) through the enzyme ATP synthase. ATP is used to run the Calvin cycle, which takes the energy from ATP along with CO2 to produce sugars. So, it seems the main idea is that light excites electrons, which go from molecule to molecule in a chain and (somehow...still unclear to me...) create a proton (H+) gradient. That gradient is used through the enzyme ATP synthase to make ATP, the molecule that fuels sugar production. I'm far-removed from bio, so don't take my word at all on this--if anyone has any corrections, please make them--I'd like to know what's going on here.
ReplyDeleteIt was said earlier (I don't remember who said it) that the solar cell/pn junction acts like an emf in the circuit with the capacitor, and so the capacitor will charge as if the junction were replaced with a battery.--This makes sense to me. In this system, we have no external bias, but the incident light will produce a voltage (in the reverse direction, ie in the direction of reverse-bias) and charge the capacitor. I wonder if we were to charge the capacitor in this way, and then remove the light, if the system would act like an LED, as the charge stored on the capacitor decayed off? Any comments?
Zack:
You said that Brian overspecified, but I'm not sure if he did, since he didn't specify kT. (This is a technicality, but without kT Nc,Na, and $\Delta E$ don't overspecify.)
I used Nd=Na=10^17/cm^3,Nv=Nc=5*10^18/cm^3,kT=0.025eV,$\epsilon$=12*8.85*10^-12C^2/N-m^2. From this, I found $\Delta E$ = 0.8eV and x0=73nm. So, we want our diffusion length to be five times this--ie, we want $\sqrt{D \tau_p}=5x_0$.
Doing this, I end up with $\tau_p = 10^{-11}$s, in agreement with Caleb. Dave--I don't know how you got the 10^-7s, but here's what I did explicitly:
$\sqrt{D \tau_p}=5x_0$
$25x_0^2/D = \tau_p$
$x0=73nm=7.3*10^{-6}cm$
$D=\mu kT/e = 5000cm^2/V-s*0.025eV/e=125cm^2/s$
$\tau_p = (7.3*10^{-6})^2cm^2/(125cm^2/s)$
$\tau_p = 10^{-11}s$.
If something here seems suspect, please let me know--wait, I just realized I didn't apply an external voltage, so there would be no current, and this would be meaningless. Ok, so that's a problem...
Instead, I'll assume a forward bias of 0.5V, so that $\Delta E$ is now 0.8eV-0.5eV=0.3eV. This has the effect of scaling my x0 by $\sqrt{0.3/0.8}$, yielding x0 = 45nm, and it would scale my $\tau_p$ by 0.3/0.8 (since $\tau_p$ is proportional to x0^2), giving 4*10^-12s.
So, I guess I'll say that $\tau_p$ is 4*10^-12s. Still, I seem to recall doing this calculation after class last Tuesday and getting something like you were getting, Dave (around a microsecond), or maybe I am misremembering?
I guess that's all for now...Sorry for the extreme disjointedness of this post, I sort of just added to it as I worked through the hw. As always, any comments are welcome!
I guess I did overspecify- I thought that by this point, we all knew kT so well that it wasn't really necessary to list it, but we don't need Nv or Nc. That said, I would like to update my previous post:
ReplyDeleteΔE = 0.8 eV (i realized that i confused this with the difference between Ec and mu earlier - it was too small)
and...
e*V = 0.5 eV
With the voltage defined in this way, I found τp = 5.6x10^-12 s.
for L-x0 = 5x0.
Mike: I think your question about light being reflected and passing through the material is a really interesting one. Another interesting thing to consider is how deep within the material is light is exciting electrons? It seems like there would be in ideal region in material for the ehps to generate so they do not recombine before the free electron is used for conduction. Does this make sense?
ReplyDeleteAnother thing I was thinking about was the arrangement of light exciting electrons in a solar cell. Obviously the wavelength and the energy is very important, but if the light is polarized in a certain way could it have a higher probability of interacting with the lattice to excite more electrons?
There's also the question of what the wavelength of the incident light is. With white light, only some small fraction of the photons that hit the solar cell are going to be a wavelength that corresponds to the energy gap. Ideally, you'd want monochromatic light of the exact wavelength so that its energy is equal to the band gap.
ReplyDeleteActually, a good setup for a solar cell might be to arrange a network of p-n junctions of different materials with slightly different band gaps, so that each is gathering energy from a different range of wavelengths. Otherwise I think you'd be collecting only a very small fraction of the incident energy.
I think that the photons will pass through the solar cell, but only if it does not have enough energy to kick an electron up to the next energy level. Or it may be that it just doesn't hit any of the electrons in the material. Photons with High-energy are also not going to get absorbed into electron, but rather they will bounce off. So unless a photon is around the same energy as the band gap, then they don't contribute to the current, but could rather lower the efficiency.
ReplyDeleteI was wondering with the solar cell whether the generation current is primarily contributed by the generation of electron-hole pairs in the transition region, and is the transition region the diffusion length?
ReplyDeleteRegarding 3:
ReplyDeleteC in CMOS stands for complimentary. CMOS is a way of wiring symetric MOSFETs to create a logic gate. CMOS are usually used for common(drain, source, or gate) amplifiers. This configuration results in a lower heat output, much smaller wafers, use less power, and are more versatile as a logic tool.(With help from Kelsey E. and Julian)
It's probably to late for people to respond to this, and it might be a bit out of the scope of this class, but my question for 3 would be: what, or how would solar cells effect the efficiency of solar sails?
ReplyDeleteMy extra question would be to calculate how much energy takes to produce a solar cell, and compare that to how much energy a solar cell could produce in its life. compare and see if it's worth it.
ReplyDeleteFor number 3:
ReplyDeleteOk, so this isn't a very structured question, but I thought it was an interesting idea. So we're looking at solar cells and it a very tiny nutshell, photons with E>E_g can create electron-hole pairs that create a current due to the electric field in the depletion region which can lead to stored energy. I was thinking it would be interesting to create a relationship between the energy gap and emission spectrum of the sun to get a functional relationship between the energy gap and how much energy you could potentially get from the sun. For example, if the energy gap is 1eV, then only photons with energy greater than or equal to 1eV will create electron-hole pairs, and if you know the intensity spectrum of the sun, you could get some idea how much energy you could store. If you were to reduce the energy gap to 0.7eV, given the intensity spectrum of the sun, how much additional energy will you get from the photons with energy 0.7<hf<1.0?
~Lou
I got something really small for $\tau_p$ too. I did the following:
ReplyDelete$$\tau_p=16x_0^2/D=16x_0e/\mu kT$$
$$=16\epsilon\Delta V/\mu kTN_a=2.5\times10^{-12}s$$
Where $e\Delta V-E_{applied}=0.3eV$
This just seems way too small.....
~Lou
$\tau_p=16x_0^2/D=16x_0^2e/\mu kT$
ReplyDelete$=16e\Delta V/\mu kTN_a=2.5\times10^{-12}s$
The post above was hard to see