This week's classes will focus on FETs, so you are on your own, so to speak, with problems 1-3, which involve things we have already covered (but might be pretty challenging). (i.e., don't wait thinking that we will cover them in class.)
These problems are tricky to design and to get the parameters in a range that works. I am interested to see your results for 1c and 2a, which are the easier parts. Please email me those as soon as you get something if you like. (Then we can see if the problem is going to work out okay. The voltage range should hopefully stay below the band bending.)
1. Consider a semiconductor with an energy gap of 1.5 eV, \( D_2 = 0.8 \times 10^{22} \, states/(eV-cm^3)\) and \( D_3 = 0.8 \times 10^{22} \, states/(eV-cm^3)\), \(D_n = D_p = 100 \, cm^2/sec\) and \(\tau_r = 10^{-10} \, sec\). Suppose you make an n-p junction with an area of 1 cm^2 where \(\mu\) is 0.2 eV away from the majority carrier band on each side. [1a) and b) are warm-up problems and should be pretty easy and quick to do. c) is where it gets more difficult.]
a) What is the total band bending in this case (the difference between E_c on the left and on the right)?
b) What are the majority and minority carrier concentrations on each side? (away from the interface region) [Don't be discouraged if the minority carrier density seems too small; that might be a natural consequence of the larger gap (1.5 eV).]
c) If you connect a battery to the junction, how much bias voltage would you need to get a current of 1 amp?
d) How much voltage would you need to get a current of 0.5 amps?
e) Which way does the current go (most easily)? Sketch a picture.
2. Suppose you get rid of the battery and instead have a resistor, R, in its place. Then you illuminate the junction with \(10^{19}\) photons per second per cm^2. Each photon has an energy of 1.5 eV.
a) If half the photons are absorbed, in the junction region, then how much current would you get for R=0 Ohms?
b) Which way does the current go through the resistor? Sketch a picture.
c) How much current would you get for R = 1 Ohm?
d) Is this difficult? Why? (discuss here) Would it be easier if you were asked to get an estimate of I to just one or two sig figs?
https://drive.google.com/file/d/0B_GIlXrjJVn4djlsd2ExZElKNHc/edit?usp=sharing
3. We can think of the resistor as a stand-in for something that we would like to provide power to.
a) For an illumination of \(10^{19}\) photons per second per cm^2, what value of R provides the most power? (assume 50% absorption in an electron-hole pair generating process in the junction region) What value of V, the voltage across the resistor, does that correspond to?
b) For an illumination of
\(.25 \times 10^{19}\) photons per second per cm^2, what value of R provides the most power? (assume 50% absorption, as before) What value of V, the voltage
across the resistor, does that correspond to?
[Hint: You may not be able to do something elegant here. Focus on getting an actual value of R (and V). Imagine you are working for a company or group and they just want to know which resistor to put in the circuit (what impedance to put in the circuit). They want a number in Ohms accurate to 5 or 10%. Don't let them down.]
These problems may not all be do-able in the usual straightforward manner that you may be used to. Please discuss where you get stuck here.
You might want to do 5 before 4, though you would have to read 4 first.
4. Consider a semiconductor with an energy gap of 1.0 eV, \( D_2 = 0.5
\times 10^{22} \, states/(eV-cm^3)\) and \( D_3 = 0.2 \times 10^{22} \,
states/(eV-cm^3)\). Suppose this semiconductor is doped so that \(\mu\) is 0.35 eV above the valance band edge. Suppose you grow, evaporate or in some way create an insulator 20 nm thick (a material with a large band gap, say 5 eV) on a surface of this semiconductor, and that on top of the insulator you evaporate a metal layer (roughly 100 nm thick). Then you attach a wire to the metal and to the back of the semiconductor.
Additionally, suppose that, by design (i.e., choice of metal, doping in the semiconductor...) the Fermi energy of the metal is the same as the chemical potential (mu) of the semiconductor. Additionally, suppose that the minimum energy to excite an electron from the conduction band edge of the semiconductor to the vacuum is 4 eV.
a) What is the density of holes in the semiconductor?
b) What is the minimum energy to excite an electron from the valence band of the semiconductor to the vacuum?
c) What is the work function of the metal? Can that be determined from what you are given? If so, how?
d) As a warm-up for the next part, sketch the position of the Fermi level of the metal and of the semiconductor band edges, E_c and E_v, as a function of x, where x is the distance from the metal-insulator interface.
e) Insert a battery in the wire and do another band sketch for the case where the gate voltage is 0.5 eV higher than the voltage at the back end (far from the interface) of the semiconductor. You can assume that voltage drop (and variation of mu) occurs primarily in the insulator.
f) Do you see anything possibly occurring in the vicinity of the interface that might be interesting?
5. Suppose we regard the metal-insulator-semiconductor structure of the previous problem as a capacitor, where one plate is the metal and the other is the semiconductor. Applying a voltage across them should lead to a charging in which there is equal and opposite charge on the two sides. One could calculate C, but for this problem let's take C as given. Recall that Q=CV.
a) What sign of V would lead to a negative charge in the semiconductor?
b) Where might that charge reside in the semiconductor?
c) Assuming the same doping level, etc as in the previous problem, how thick a depletion layer would you need for a given voltage V?
d) Do a sketch (graph) as a function of x (distance from the interface) showing the space charge density for a voltage that depletes the valance band near the interface and also,
e) mu and the band edges as a function of x.
(possibly one more problem to follow. )
These problems are. I might add one or two more problems.
ReplyDeleteFor number 2, is it supposed to be photons/(sec - cm^3)?
ReplyDeleteI don't think so. Can you visualize it? This is a large area junction, area = 1 cm^2.
DeleteHow come? (not that I can see)
ReplyDeleteI think you are making it more difficult than you need to.
ReplyDeleteThe problem stipulates the half the photons are absorbed. in the junction region. Hmm, i better edit that in.
ReplyDeleteDo we need the electron/hole doping levels for #1?
ReplyDeleteyou can calculate them.
DeleteI don't think you mean N_a. Maybe you are referring to a value of n or p (where it is the minority carrier)?
DeleteI get 5x10^-3 as well, does seem ridiculous
DeleteI think it is fine. Just go ahead and use that.
DeleteThey are really the same thing. It would just effect voltage polarity. I would not draw the I-V curves any differently.
ReplyDeleteThat's possible. I doubt there is an easy way at all. It is hard any way you do it.
ReplyDeleteI am getting the impression that for a band gap this big (1.5 eV = 60 KT) the equilibrium minority carrier concentrations are very small, much less than 1 carrier/cm^3 (e.g., for problem 1). Is that what you are getting?
ReplyDeleteFor problem 1, what is causing the bending of the bands? Different doping levels on each side or an applied voltage?
ReplyDeleteDifferent doping levels.
DeleteHmm. I am able to find the initial doping level based on what we were given, but I'm confused on finding the bending of the band without knowing the new doping level (n+n_i)
DeleteThis problem is really easy. I think you are not realizing that. It can be done in about 15 seconds.
DeleteI mean part a is. You know were mu is on each side relative to the bends and we assume the mu is constant. Draw a picture and you will see...
Deleteyou don't need the doping level to do 1a.
DeleteI see now. I was over complicating the problem. Thanks for reminding me that mu is constant
Deleteyes, I think so. Given the capacitance....
ReplyDeleteit is relatively easy.
What is the applied voltage for problem 2?
ReplyDeleteIn problem 2 there is a resistor and an n-p junction. Do you mean the voltage across the resistor?; or the voltage across the np junction? or some other voltage?
DeleteI dont understand how to get part c in Problem 2 . I get part one to be 1.6A
ReplyDeleteReally? I get o.8A, since only half of the photons are being absorbed (one per electron).
DeleteYou're right though, I'm not sure how to find \( V_R \) either.
DeleteWouldn't 0 ohms give the highest power output in problem 3a? Since 0 Ohms would give the highest current
ReplyDeleteIt would give the highest current, but remember that \( P=I^2 R = IV \).
DeleteThe resistor is a stand-in for something you want to power. For example, a battery you want to charge or a light bulb or a motor. R is the impedance of that.
DeleteThis is one example of what you might call an "impedance matching" problem. To get use-able power from a solar cell, impedance matching is very important and, as you'll see, a little non-trivial since the best value of R depends pretty strongly on the illumination intensity.
These are really good questions. Who can answer or address in some way Alex's questions?
ReplyDeleteSome thoughts on 3: What is the equation for power? Putting R=0 into that equation, what power do you get?
ReplyDelete\( P=I^2 R = IV \), making P = 0 when R = 0.
DeleteOther thoughts on 3: Are we only looking at photocurrent? What might be affected by changing the resistance in the circuit?
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteI have a question about 2c. We have a photocurrent that creates a voltage across our resistor. This voltage also appears across the diode and creates an opposing current as we discussed in class. However, by superposition this opposing current now reduces the current across the resistor and therefore reduces the bias voltage across the diode. This then reduces the drift current due to biasing which then increases the the voltage across the resistor from what it was previously (but the voltage is less than what it was initially). This process then repeats forever.
ReplyDeleteI believe this would initially produce a damped oscillatory response but that at steady state for \(t \rightarrow \infty \) the current would not be \(I = I_{ill} - I_0 e^{qV_a/kT}\) as we discussed. Does that sound right?
This is a great question. Your analysis is right on, but the equation for I is a self-consistent equation that takes all that into account. So the answer is the current would be that.
DeleteThat equation is saying, taking both illumination and bias into account, here are the stable points where the system will find a steady state. Each possible stable point is characterized by an I and a V. You might want to not put the subscript a on the V, as that might confuse people.
This was my issue exactly wasn't sure how to ask it exactly. For the V in that equation are we using IR? if so with current is used? and whats I_0? Also I would think V = 0 if theres no resistor which would mean $ I=I_{ill} - I_{0} $ which differs from the last part with no resistor.
DeleteAlso how do I do Tex? XD
\(I_o\) is in the first video above (Illuminated Junction). It is just 10 minutes long. That gives an expression for J_o and if you multiply by Area then you would have I_o. It depends on a number of things; I thin those were all "given". If not, we should make up some reasonable values here.
DeletePS. There is only one current. It is the same everywhere in that circuit. Even though it seems like current is created in the junction, in steady state it flows through it, the same amount in as out. So the current through the R and through the junction are the same.
I was thinking we use \(V = I_{ill} R\). I am still pondering this problem so I may not be right. \(I_0\) was the prefactor we found last week it is discussed in the video that was posted today. If \(V = 0\) then \(I = I_{ill}\) since there is a \(-I_0\) in the actual equation that is not written. You can ignore it for \(qV \gg kT\).
DeleteYou can enter TeX using \ followed by ( and closing with \ followed by ): \ (example\ )
Here are my thoughts for number 3. We have power: \(P = V \cdot I\). We can we graph P vs. V and it will should look something like this with different scaling: http://wolfr.am/1qg7G0G. This picture should be cut off so that \(P \ge 0\). We can see that there is a critical point for the power and this is where it is maximum.
ReplyDeleteNow it is non-trivial to solve exactly where this point is. I believe the maximum point is called a first order root in literature and you have to use newton's method iteratively to get a good approximation of where it is. It's been a while since I took calculus and the derivatives are all very messy and hard to work with, but since we are physicists here I think we can just use wolfram alpha's max() feature (or whatever is similar in Matlab or Mathematica) and assuming the function we give as an argument is properly solved for it will tell us what voltage we are looking for.
I am too poor at math to do it, but I found a good resource for Newton's Method if anyone is interested in going that route: http://www.ms.uky.edu/~rbrown/courses/ma113.f.12/l30-newton.pdf
I don't think you need to use anything so sophisticated as newton's method. If you have the function and you can see what it looks like, then maybe just try calculating a few values and see where it is biggest from that. Just by guessing and trying a few values of V. Does that make sense?
DeleteHere is a thought regarding problem 3 and related to the discussion above. What if one graphed I = -(1/R) V and I= (the equation for an illuminated junction) both as a function of V on the same graph. I think the stable point for the system would correspond the the point where those two graphs intersect. Each different value of R leads to a different stable point. Does that make sense???
ReplyDeleteI don't know if this helps you actually calculate anything - perhaps not --, but it may help with visualization and conceptually.
Delete