If you like solar cells, try to make sure you are able to spend a lot of time on problems 6 and 7. Those are the problems that really help you understand solar cells. Especially problem 7.
Extra credit: Can some of you solve problem 1 soon and post what you get here? That might be helpful in contextualizing some of the later problems of this homework.
For these problems, assume a symmetric n-p junction doped to \(10^{17} cm^{-3}\) on either side. [and to a semiconductor for which \(E_g = 1 eV, \quad kT=.025 eV\) and \(D_c = D_V = 12 \times 10^{21} \frac{states}{eV*cm^3}, \quad B_c = B_V = 3 eV\)].
1. In the ideal junction approximation, I believe that we found that for a junction connected to a battery of voltage V_a, the current density will have the form:
\(J(V_a) = J_o (1 - e^{-eV_a/kT})\)
Based on the calculation we did of the diffusion current at x_d, what is a reasonable estimate for the value of J_o? What are the units of J_o?
2. (LED question) With a battery connected to the junction (in series),
a) how much current do you get through the junction for an area of 1 cm^2 and \(V_a = -0.4 V\)?
b) If 50% of the electrons involved in that current flow emit a photon, how much power is that? In what units would you like that power to be expressed?
c) how much radiated power (in photons) do you get from the junction if \(V_a = -0.5 V\)?
Discussion question. This is a question that you can discuss here starting as soon as possible. Hopefully this discussion will help other people with their homework.
Consider an n-p junction with no bias voltage. a) Suppose somewhere along "x" a photon excites an electron from the valence band to the conduction band. What do you think will happen to that electron? Does it matter where that occurs? b) Suppose there is a uniform flux of photons in the semiconductor and some of them excite electrons from the VB to the CB. What happens? c) What energy would you want that photon to be?
4. (Solar cell question) Suppose there is an incoming flux of 10^16 photons per second per cm^2 on a 1 cm^2 area junction. Suppose that 50% of them excite an electron from the VB to the CB. Suppose also that there is a wire connecting the far side of the n side to the far side of the p side.
a) What happens in the junction?
b) What happens in the wire?
c) What sort of assumption would you need to make in order to do a quantitive estimate of the current in the wire? What do you estimate that current to be?'
d) What would the current be for an incoming flux of 10^17 photons/(sec*cm^2) on a 1 cm^2 area junction (suppose that 50% of them excite an electron from the VB to the CB).
5. (Solar cell- capacitor question) Suppose there is an incoming flux of 10^16 photons per second per cm^2 on a 1 cm^2 area junction. Suppose that 50% of them excite an electron from the VB to the CB. Suppose also that there is a capacitor in series with the junction. (no battery or resistor, just the capacitor).
a) After a long time, what is the equilibrium charge of the capacitor? How come? Explain your reasoning.
b) Do a sketch showing the junction, the circuit, and which side of the capacitor has positive and negative charge.
c) After a long time, what is the equilibrium charge of the capacitor for a flu of 10^17 photons per second per cm^2?
d) extra credit opportunity: Derive and solve* the time-dependent differential equation that tells you the capacitor charge as a function of time (Q(t)). (Do you need to add something to the problem to get a time scale?) *Perhaps the solution can only be done numerically because the non-linear relationship between I and V. Maybe just getting the differential equation would be challenging enough.)
6. Consider a junction (of area A) for which the current-voltage relationship without illumination is: \(I(V_a) = I_o (1 - e^{-eV_a/kT})\) where \(I_o\), which is associated with a bias voltage induced diffusion current, is \(I_o = 3 \times 10^{-11}\) coulombs/sec. (extra credit: Comment on whether or not this value of \( I_o\) seems reasonable based on your understanding of biased junction diffusion current (and problem 1).)
Now suppose additionally that the junction is illuminated by a flux of photons such that \(8 \times10^{16}\) photons/second are absorbed in the depletion region and each of those photons excites an electron from the VB to the CB which then is pushed over to the n side by the electric field in the depletion region.
a) suppose there is a wire connecting the n-side to the p-side. If we make the simplifying assumption that each of those excited electrons contributes to the current, then what is the current through the wire?
b) Consider that case where there is a capacitor in the wire (in series with the junction). What charge and voltage would the capacitor reach in steady state? (Asymptotically) Let's say C= 10 Coulombs per Volt.
c) What charge and voltage would the capacitor reach for \(10^{15}\) photons per second absorbed ?
d) What charge and voltage would the capacitor reach for \(10^{16}\) photons per second absorbed? How come this is not 10x as large?
7. This problem involves the same junction as in the previous problem, but with a resistor in series with it instead of a capacitor. Let's assume \(8 \times 10^{16}\) photons per second absorbed creating a current due to illumination.
a) With a resistance of 1 Ohm, can you get a pretty good estimate of the current through the circuit (to about 5% accuracy or better) without too much work? How come? What is I? What is the power, e.g., \(I^2 R\), generated in the resistor in this case? What about for R =: 2 Ohms, 4 Ohms, 8 Ohms and 32 Ohms. Plot power generated in the resistor as a function of R. Is the relationship linear?
b) What value of R would give you a voltage across the resistor of 90% of the voltage that you would get with a capacitor (as in problem 6b). What is the power generated in the resistor for this case?
c) Is there a value of R that gives you the highest power generated in the resistor, (or would that just be infinity)? If there is such a value, what is it for this case?
d) extra credit. Make a table showing values of R, I, V and power generated. Cover an interesting range.
e) Special Extra credit: Compare the power of the incoming photons to the power generated in the resistor! Think about and discuss conservation of energy, where energy goes in this process, etc! (Send me a pdf or post here anything you get.)
8. (Optional problem) (This problem is kind of interesting and may be relevant to understanding solar cells (illuminated np junctions). It is an extra thing you can do if you have time.)
a) For zero bias voltage, what did we find for the depletion length for this junction at that doping? What is the total band bending?
b) If you apply a voltage of -0.4 eV to the n side, how much does that change band bending and the depletion length? (The size of the depleted region.) Why does it make the depletion region less wide?
For the discussion question:
ReplyDelete(a)If one photon excites an electron from the valence band to the conduction band, wouldn't the electron be "pushed" to the right? Because the photon would induce an electric field within the junction? I'm assuming that the photon would hit the junction directly, possibly in the depletion region since everything interesting takes place there.
(b)If there is a uniform flux of photons, then it would cause multiple electrons to be pushed towards the n-side of the junction.
(c)For the amount of energy, my guess that it would have to counter act the force of the diffusion current on the n-side of the junction, in order to successfully push electrons in that direction. I'm not sure its the diffusion current though...my only hunch is that in the case of an un bias junction, nothing is really moving because everything is somewhat balanced with each other, then again, I could be confusing myself.
Yeah, I assumed that the n-side is on the left and the p-side is on the right. I guess its just a matter of coordinate systems? I did mean it would move to the left, like in class, my bad. Your point for part c makes sense.
DeleteFor the first problem, using J(x)=kTμ[n(x_d)-n_eq]/L, I got J_o=[μkTn_eq]/L, which is about 7.5*10^-5 Amps/cm^2 where: kT=.025eV, μ=88 C*S/kg, n_eq=10^6 cm^-3 and L=4.7*10^-5 cm.
ReplyDeleteThe first J(x) should actually be J(X_d)
DeleteFor p=10^17 cm^-3, what is neq? I mean I think your value is reasonable, but is it really exactly 10^6 cm^-3?
Delete"do we also assume as we did for part b, that only 50% of the electrons in the calculated current emit photons?"
DeleteYes
I think by definition \(n_{eq} = kT D_c e^{-32}\) when Ef, for equilibrium, is 0.2 eV. Equivalently, I think it is \(10^{17} * e^{-24}\) for this case.
DeleteThis may seem like a trivial question but I'm trying to figure out what to use for my n(x_d) function for the first problem. Would I have to involve the enhancement factor to get the current?
ReplyDeleteIs that \(V_a\) you solved for the voltage that should be applied in order to have that relationship between the capacitor and diffusion current? In that case, that would make perfect sense. What still throws me off about the problem is where does that voltage come from if there is no applied voltage. Charge builds up on the plates, then eventually moves along the circuit, but there should be no voltage drop across the plates, right? so there should be no bifurcation of the chemical potential levels?
ReplyDeleteShouldn't there also be a 1/e term in your voltage calculation? It should come from the original exponent I believe.
ReplyDeleteHas anyone tried the unit analysis for I_o on problem 5? From the definitions in my notes, I_o=kTμA/L where μ=e(tao)/mm* and L = [kT(tao)(tao)/mm*}^1/2. But every time I plug these in, I get units of A*cm^3 instead of just A, is anyone else having this problem?
ReplyDeleteImagine that the junction is really thin and that each photon excites an electron hole pair in the junction region, that is, between -xd and xd...
ReplyDeleteSo, I'm a bit stuck with the calculations from #6? I have the understanding that the total current that we end up calculating is the sum of the current-voltage relationship without a bias and the current induced by the flux of photons? Is that right?
ReplyDeleteThen, what do we use for V_a? Or am I thinking too simply?
For number 6 it is helpful to find the value of voltage where the current stops flowing. You can do this by using the current equation and solve for voltage. Once you ha e that just multiply by the capacitance to get the charge.
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DeleteSorry about the deleted comments. Somehow my phone manages to re post the same comment everytime I refresh the page.
DeleteI used positive for the chare but either way is the same idea. Good posts Louis.
Did you draw a picture of the charged capacitor ?
ReplyDeleteHmm. I see what you mean, but I think we should just stick with that anyway. Maybe a larger valuable would have been more realistic, and also a larger flux of photons.
ReplyDeleteI think this gets you to about 0.5 volts or so for I=0 though. Is that true?
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ReplyDeleteBut if V changes, then I has to change too. Because the solution is on a curve that is a 1 to 1 relationship between V and I.
ReplyDeleteI think the current is pretty much unchanged at low values of R, but then it changes when R becomes larger.
ReplyDelete