edits: Tuesday 2 PM: Problems 3 and 8.
1. Sketch a picture that shows the conduction and valence bands of an n-p junction as a function of x, the distance from the interface. Show how the bands bend upward in what we call the "junction region", and then level off after that. (FYI, technically, the conduction band in the n region should never be below the valence band in the p region for an ordinary junction. Let that "constrain" your drawing.)
2. For an n-p junction functioning as an LED, do a sequence of drawings illustrating the journey of a single electron from the n side to where it emits light.
3. For an n-p junction functioning as an LED, suppose there are \(10^{18}\) electrons per second crossing the junction and that 80% of those are emitting light.
a) How many milliwatts of light would that be? {I think that to calculate this one would need to know the energy per emitted photon. Let's make that 2.0 eV.}
b) Draw a circuit with a battery and an n-p junction in "LED configuration". (What is your level of confidence regarding the battery polarity?)
c) What is the current in this circuit? What is the difference between the current on the left side of the battery and the current on the right side of the battery?
d) If the relationship between current and voltage for this circuit is given by: \( I(V) = I_o (e^{qV/KT}-1)\), where \(I_o = 6.8 \times 10^{-18}\) coul/sec and KT=.025 eV, then (numerically?) find the voltage of the battery. (What voltage must the battery put out to create and maintain this current?) { I think for this calculation you can ignore the "1" and just use the equation: \( I(V) = I_o e^{qV/KT}\).}
e) Sketch a graph of \(I\) vs V from V= 0 to 1 Volt. What is \(I\) at V= 0 ? What is \(I\) at V= 1 Volt ?
f) extra credit: Suppose you want a power output (of light) 10x less than in parts a-d. What current would you require? What voltage would that correspond to? Based on your results, would you rather use a voltage source or a current source to operate a diode?
4. a) What is the difference between a n-p junction LED and an n-p junction LASER?
b) What do LED and LASER stand for?
5 a) Does the width of what we call the junction region depend on doping levels of the n and p regions?
b) (if yes) which way?
c) How come?
6. Consider an n-p junction (solar cell) illuminated by light (E&M waves) with a frequency f, where \(hf = 2.0\) eV. Let's say that there are \(10^{19}\) electrons per second excited from the valence band to the conduction band in the junction region due to this illuminating light.
a) What is the wavelength, frequency and color of this light?
b) How much energy must be lost by the electromagnetic field per second in order to create these excitations?
7. Consider an n-p junction (solar cell) illuminated by light
(E&M waves) with a frequency f, where \(hf = 2.0\) eV, as in the previous problem. Let's say
that there are \(10^{19}\) electrons per second excited from the valence
band to the conduction band in the junction region due to this
illuminating light and that this leads to a current coming from the junction and passing through a resistor in series with this n-p junction.
Suppose that each excited electron contributes exactly one electron to the
current.
a) What would that current be? Draw a picture of the circuit including the n-p junction and the resistor.
b)
Suppose the resistor has a resistance of 0.5 Ohms. What is the power
converted to heat in the resistor in this case? (In units of eV/sec,
Joules/sec or Watts.)
c) Make a table showing power converted to heat in the resistor for R=0.5, 1.0 and 1.5 Ohms. What is the voltage across the resistor in each case? (add that to your table)
d) How do these values compare with the amount of energy per second lost by the electromagnetic field?
e) Is there anything troubling about your table? Which part of it concerns you? Graph power dissipated in the resistor as a function of R from about R = 0 to 2 Ohms. Also show in that graph the power lost by the E&M field (from the previous problem).
f) What problem do you see? How might you resolve that? Are there any assumptions made earlier in this problem that you might now question?
8. For an illuminated n-p junction and a resistor in series, as in the previous problem, one can perhaps write the relationship between current and voltage as: \( I(V) = - I_{sc} + I_o (e^{qV/KT}-1)\) where KT has a fixed value of KT=.025 eV (which corresponds to room temperature). \(I_o\) is difficult to derive or explain. Let's use \(I_o = 10^{-18}\) coul/sec. This "diode equation" shows the relationship between the Voltage across the n-p junction and the Current coming out of (i.e., going through) the junction.
a) Draw the circuit. What is the relationship between the current coming from the n-p junction and the current going through the resistor?
b) What is the relationship between the voltage across the n-p junction, V, and the voltage across the resistor, \(V_R\)?
--
\(I_{sc}\) is the current due to the illuminating photons (when R=0). Let's use \(I_{sc} = 1.6 \, amps \).
c) Plot of I vs V, where V is defined as the voltage on the p-side minus the voltage on the n-side of the n-p junction. (Focus on positive V.) At what value of V does the current become zero? {By the way, in the equation for I(V), when qV/KT is greater than about 10 then the 1 can be ignored with no significant loss of accuracy.}
d) Plot the power dissipated in the resistor as a function of V. (This can be calculated as the product, \(V_R I\). Where is this zero? (Two voltages.)
e) Numerically estimate at what voltage the maximum of power dissipation, \(V_R I\), occurs. {A table might be helpful.}
f) Since I and V and are also related by R=V/I, estimate what value of R would allow you to achieve this maximum value? This is considered the ideal "load" to get the most power out of a solar cell.
9. extra credit. Do the same thing for the I-V relationship, \( I(V) = - I_{sc} + I_o (e^{qV/KT}-1) - V/(1 \, Ohm)\). The extra term represents one of several possible additions/corrections to the ideal diode equation used in the previous problem. In particular, it reflects an effect associated with current leakage (across the junction).
There are two questions labeled number 3. How would you like us label these?
ReplyDeleteI fixed it. Thanks.
DeleteThis comment has been removed by the author.
ReplyDeleteIn problem 4 you ask us about p-n junctions. I was under the impression that a LED was created using a n-p junction. Am I mistaken?
ReplyDeleteI changed that. Thanks.
Deletefor 3a. to get milliwatts should we just use 80% of 10^18 ev/s and convert to watts?
ReplyDeletei tried computing the answer to 3a.) your way and ended up with a result on the order of ~10^7 milliwatts. I suspect this is a little high and we're missing something, but I could be wrong.
DeleteHow did you get ~10^7 mW?
ReplyDeleteI did 0.8(10^18 eV/s) = 8e17 eV/s.
Then converting eV to Joules, I multiplied 8e17 eV/s by 1.602e-19 J/eV to get 0.12816 J/s.
Then 0.12816 J/s = 0.12816 W = 128.16 mW.
There is a 2 eV photon released for every electron, if I am not mistaken, you answer assumes that e = 1 eV
DeleteRight, I made the same bogus assumption and then something went horribly wrong when I did the dimensional analysis. We can now do the problem correctly and without assuming anything, since the 2.0 [eV] condition has been added.
DeleteI'm not sure how you're getting eV/s, we are given 10^18 e-/s (current). The only way I can see us getting a numerical result (not just our percentage multiplied by the symbolic power) if we are not given the energy of the energy gap is if we assumed the resistance to be that 0.8, but this is a super iffy move. How can we calculate power of the circuit without voltage or resistance? (P=IV=R*I^2).
ReplyDeleteI think that you just can't answer this question unless you are given the energy gap! The problem is incomplete.
DeleteI agree I think it is incomplete. Unless he just wants an answer in terms of the energy gap.
DeleteI think the problem works out fine. First calculate current and then figure out the voltage. And P=VI. I got 256 milliwatts.
DeleteYeah, we can now do the problem fully since the professor recently added the 2.0 [eV] condition to the problem. I concur with your answer
Deletepower = energy / time, we're given the amount of electrons/sec, the rate at which they turn into photons, and how much energy there is per photon.
DeleteFor number 3 in general: what does it imply when only 80 percent of the *crossing* 10^18 electrons are emitting photons. Does it mean (1) that 20 percent of them are coming back to their original n side state after crossing, (2) that 20% of the electrons never drop down and just stay in the conduction band on the p-side after crossing or (3) that they dropped down to the valence band on the p side, as Fermi would tell us, but then the energy is lost in heat, or some other form?
ReplyDeleteMy guess is (3). Anyone have any other ideas or votes?
I vote (3), since I think (1) would require some extra energy we can't account for in our model, and the problem specifically states that all of the electrons are transitioning, so (2) can't be right either.
Delete3 for sure.
Delete3, these aren't ideal circuits and some energy will be dissipated in the form of heat as the electrons do work to make it over the hump and drop to the VB
DeleteI have a question about 6B and 7D. Does anyone know how to approach them?
ReplyDeleteFor 6B, isn't the energy gained by the electrons in the n-p junction also the energy lost by the electromagnetic field?
DeleteWe know how many electrons are zooming through the junction region every second and we know each electron's energy. Find the total energy "gained" by all those electrons every second. The electromagnetic field must lose that much energy every second to conserve energy.
ReplyDeleteTo the professor: for problem 9, what do you mean "do the same thing for the I-V relationship"? Do you mean redo all of problem 8 except with the new equation?
ReplyDeletefor 3e should we use the equation I=(Io)*e^(qv/kt) or should we keep the -1 in the equation for this problem?
ReplyDeleteGood question. What do you think? What arguments would you present for each choice?
Deletewell I wasn't exactly sure why we could just ignore that -1 in the first place. Would you mind explaining that?
DeleteI did not neglect the -1 term in part 3e) and got reasonable results. In fact, looking at the equation I = Io * (exp{qV/KT} - 1), if you want I = 0 exactly, that -1 term has to be included. Hope that helps.
DeleteHow big is e^{qV/KT}? Thinking about that is a good place to start.
DeleteFor the graph, what sets the vertical scale? What is the largest value you get for I?
@Evan. True, if you want zero exactly...
DeleteBut is there a noticeable difference between zero exactly and 10^{-18} amps (on the scale relevant in this problem}?
@Professor Certainly not. But is there a good reason for negating the -1 term if it's not too trivial to work with? For example, is there a practical reason we should avoid working with zero exactly in our model?
DeleteAck, I meant if it is trivial to work with. My bad.
Deletecould you explain why the electron flow on our diagram of the solar cell is the opposite direction than the electron flow in the LED diagram again ?
ReplyDeleteIn one the electrons are pushed up the hill by the battery; the the flow is up the hill.
DeleteIn the other there is not battery. The electrons appear on the hill and slide down to the bottom. Hence opposite flow.
For getting maximum power dissipation, the only way i can see getting a "maximum" (a minimum because its negative) is by using (current as a function of v)^2 * R, and solving for where the derivative = 0. Is this the right way to go about it?
ReplyDeleteThat is a tricky point, but I think that both I and V are negative (or have the same sign) so that power is positive. IV might be easier to work with, though i am not sure.
DeleteI think you are asked to do it numerically. The closed form method you suggest may be too difficult. Guess where it is. Calculate some IVs to actually find it (roughly). Get an actual number in volts, not some complicated expression.
Vr=-V (voltage across resistor is equal and opposite the voltage across the junction)
DeleteTherefore power dissipated across the resistor is positive (or you could just take the absolute value of P=V*I).
To find the maximum power dissipation I thought, around what voltage does the exponential start to pick up. Then I used the ideal diode equation I(V) to find the current. Then finally take the product of the two, as Zack suggests a table is helpful.
rethinking that, i used I(v) and multiplied it by -v, found where the derivative = 0 (to find the maximum), i came up with something that sounded incredibly correct in comparison to the graphs
DeleteNice, to repeat the ideas: we define the voltage across the junction to be positive, which gives us a negative voltage across the resistor, and as such we get a double negative for the power across the resistor P=IV.
DeleteWith this in mind, I have a slight feeling that, on number 9, the correction term (V/1Ohm) should be positive. Since it is supposed to account for a 'current leakage', I imagine it should be decreasing the magnitude of the overall current. So knowing the equation is for a negative current, decreasing the magnitude would mean we'd want a positive term.