I would suggest making two copies, ideally by hand, and keeping one so you have something to study from for the midterm on Friday.
For all these problems, please feel free to ask questions and postulate answers here. It is expected that you may wonder what is being asked. Figuring that out is part of the goal of the assignment and an opportunity to deepen your understanding of the physics. Problem 5 is especially important and difficult I think.
Note on scattering times, May 1st. The scattering times I originally put in this HW assignment seem a bit too short. Let's use instead \(\tau_t = 10^{-12}\) sec and \(\tau_r = 10^{-9}\) sec. I think that this will give us more realistic values for the length scales and mobilities that depend on these times, as well as other quantities.
Extra Credit. May 2. For x greater than x_d, use the n(x) that you get below to calculate the upper Ef as a function of x. Sketch a graph of Ef as a function of x for the entire junction.
1. Write a paragraph describing the equilibrium state of an n-p junction. Include a 3-part figure showing charge density, electric field, electric potential as a function of x. Explain the origin of depletion.
2. For the symmetric \(10^{17} cm^{-3}\) doping case from the previous homework,
a) use \(E_c (x)\), as calculated in the depletion approximation, to calculate n(x). Plot n(x).
b) Combine that with the electric field as a function of x, also calculated in the depletion approximation, to get the drift current associated with electrons in the depletion region. Assume the the transport scattering time is \(\tau_{t}= 10^{-12}\) seconds.
c) What is the peak value of this current?
3. For the symmetric \(10^{17} cm^{-3}\) doping case from the previous homework,
a) use \(E_c (x)\), as calculated in the depletion approximation, to calculate n(x).
b) Use that n(x) to calculate the diffusion current associated with electrons in the depletion region. (Assume the the transport scattering time is \(\tau_{t}= 10^{-12}\) seconds.)
c) What is the peak value of this current?
4. When you apply a voltage across a n-p junction, you can assume that far from the interface the Fermi energy will be higher on one side than the other by an amount \(e V_a\) (where V_a is the applied voltage).
a) Suppose you ground the p side. What sign of voltage on that n side will make the Fermi energy higher on the n side than on the p side?
b) Suppose that you assume that in the depletion region the fermi energy bifurcates into two values separated by an energy \(eV_a\). Suppose that in this non-equilibrium situation the upper Fermi energy determines the electron density in the conduction band. Calculate n at \(x=x_d\). How much bigger is it than what you would get with just one Fermi energy?
5. For the symmetric \(10^{17} cm^{-3}\) doping case from the previous homework:
a) For zero bias voltage, what is n(x) at x=x_d. (This is on the p side of the junction.)
b) Suppose that we have bias voltage of -0.1 Volts on the n side with the p side grounded. Additionally, assume a bifurcated Ef. What is n(x_d) in this case? By what factor is it enhanced?
Presumably n(x) returns to its equilibrium value in some way as x increases? To model that one needs a continuity equation. If n is independent of y and z , then one can write that a continuity equation for electrons in the conduction band as* \(dn(x)/dt = -(-1/e)dJ(x)/dx - (n(x) - n_{eq})/\tau_r\) where \(\tau_r\) is the recombination or relaxation time for electrons in the conduction band and \(n_{eq}\) is the equilibrium value of n(x) (that is, the value n would have it Ef were not bifurcated).
c) Describe what each of the 3 terms in this equation represent physically.
We would like to find a "steady state" solution, that is, one where n is independent of time.
d) In that case, use the continuity equation to determine n(x) for n greater than x_d. You can assume that the only current in this region is diffusion current, which is related to n(x) via: \(J_n(x) = kT (e\tau_t/mm^*) dn(x)/dx\). (You can use \(\tau_t = 10^{-12}\) sec and \(\tau_r = 10^{-9}\) sec.)
* I think this should be (-1/e) not 1/e and originally written.
6. With reference to problem 5,
a) Over what length scale does n(x) approach equilibrium for x greater than x_d?
b) Calculate the diffusion current in the limit that x approaches x_d from above. (That is, what is J_n at x = x_d?)
7. a) For a junction as in problems 5 and 6, with an area of 1 cm^2, make a table showing the value of the current I (in coulombs per second) for \(V_a = -.1\), -.2, -.3, -.4, -.5 and -.6 Volts.
b) Graph the current as a function of V_a.
8. Explain the nature of and difference between the two scattering times in problem 5.
9. How come we are only talking about the electrons in the conduction band in the vicinity of \(x_d\)? What about the other side of the junction around -\(x_d\)? What happens there? How might this effect your table from problem 7?
10. \(e \tau_t/mm^*\) is often called mobility and calculated in \(cm^2/Volt*sec\).
a) How the heck do you get coulomb*seconds/kg to turn into \(cm^2/V*s\) ?
b) What value of \(\tau_t\) gives you a mobility of 1 \(cm^2/V*s\) for m*=1?
c) What value of \(\tau_t\) gives you a mobility of 1 \(cm^2/V*s\) for m*=0.2?
...
This comment has been removed by the author.
ReplyDeleteI think you mean $$10^{17} cm^{-1}$$ for numbers 1 and 2
ReplyDeleteWill material from this homework be on the midterm?
ReplyDeleteYes.
Deleteexcellent idea. yes!
ReplyDelete"I don't remember what the answers were for the previous homework, so in this problem when asked to use the /(E_c/) and other things I am not sure what to use. "
ReplyDeletea student
Solutions to the previous HW are linked in the HW4 post.
DeleteSo for that part I think I got a factor of around around 54. I would double check the signs because there is a lot of negatives about.
ReplyDeleteFor 4.b, when you refer to a situation in which bifurcation doesn't occur, are we still assuming an applied voltage?
ReplyDeleteyou are assuming one fermi energy set to give you 10^17 holes
DeleteI see the problem. You make a good point. For that applied voltage, the enhancement should actually be \(e^{e|V_a|/kT}\). Does that make sense? Sorry about the confusion and thanks for pointing that out.
ReplyDeleteSo it is bigger than 1 because the exponential argument is positive.
ReplyDeleteYes. true. then there would be a minus sign.
ReplyDeleteYeah, the way I thought about it was without the absolute value bars, just that the minus sign from the electron charge cancels out with the negative voltage to give a positive exponential, and so when the voltage is positive there's a negative exponential
ReplyDeleteWith the sign convention we are using for voltage applied, I think the general formula would be \(e^{-eV_a/kT}\), in which is a positive number. For this problem it is 54 I think.
ReplyDeleteFor 5 d), In class you said that the both terms on the steady state equation were positive, but then went back and changed one to negative because the solution must be exponential, not sin/cos. Now for this problem both terms are negative, which is equivalent to them both being positive, so it seems the solution must be sin/cos. I'm not sure how to continue the problem, how do we remove one of the negative signs to get an exponential solution?
ReplyDeleteThe solution is definitely exponential, not sin and cos. I think that there may be a sign error in the term related to the diffusion current. Finding that error by understanding and interpreting the terms is the recommended approach.
DeleteExtra credit to anyone who can explain all the relevant signs here.
DeleteHmmm, i would say that the continuity equation in 5 d) as was previously written with both terms negative was correct because both terms represent a way n decreases, but J(x) is also negative so when you plug it in you end up with one positive term and one negative term.
DeletePeter. What you say makes sense. There is a slight twist though I think. Because J is flow of current and electrons have negative charge, the actual direction of electron motion is opposite to the direction of current flow. Does that make sense? So to turn J into the flow of electrons one has to divide by -1/e, not just by 1/e.
DeleteI think that perhaps fixes the sign problem issue. Does it?
Yes I believe so, thanks!
Deleteyes
ReplyDeleteSounds good. I think what we learned is that a scattering time about 5x shorter would have been a better choice for Si. Good work!
ReplyDeleteI think it is at about 0.8 x_d? Does that sound right?
ReplyDelete