So in the end if I consider the relaxation of both electrons and holes (e- to the CB edge and h+ to the VB edge) I don't need this assumption, isn't it? Because for each photon, the final energy that I can exploit is Eg. I ask you this because my idea was to compute a ratio between energy exploitable (so I should compute the E lost through relaxation) and total E available. But maybe I already know the energy exploitable from this thinking...

P.S. Professor Shih draw one graph like yours during the lecture, but only for electrons... In this sense, we talk about energy lost through relaxation, right? Because this energy is release from electrons to holes?

Excuse me Tian, Can I ask you one clarification about point 3.4? I am trying to estimate the amount of energy lost through the relaxation process. But when we consider electrons adsorbing energy from photons, can we assume that electrons at the beginning are on the limit of valence band Ev? Thank you!

So Tian, excuse me again, in general when we say that the Fermi energies align only one Fermi energy move, and in principle, we already know the value of Ef after alignment. I ask you this in general, because I don't want to follow this route for the assignment.

So, in the end, the key message is that with respect to the initial case figure 2.b only one energy EcA or EcB rise up summing up the potential profile.

Sorry Tian, I need EcA and EcB at infinite to plot the band diagram in Q2.8. Can I say that Ecb at infinite is the same value that we see in figure 2.b? I thought that as psi(inf)=0... Thank you in advance!

Tian excuse me again for the disturb, just one comment about 3.1, I get values for B of the order of 10^(-8)... Isn't it too small for spectral radiance? thank you as always!

Excuse me Tian, In Q2.7 should we consider nA0 as a constant right? So can we take one value between 10^14 and 10^20?

And just one clarification, why in eq.7 we always choose the minus sign in front of the RHS term? If psi0 is negative the derivative shouldn't be positive?

Sorry Tian

Can I ask you a question about point 2.6? To get eq. 7 I followed the same procedure as in HW4 Q2.1. However, I am a bit confused about the limits of the integrals. Indeed if I integrate from inf to x I have to choose psi(x=inf)=0 in order to get eq. 7. But for a p-n junction, for instance, psi is not equal to zero on both sides at inf. So this equation doesn't give me the potential profile but I have to reconstruct it by adding phibi, right? And then If I need for instance the band diagram I can simply sum it to Ec, isn't it? Thank you in advance and sorry for the disturb!

Sorry Tian, what do you mean with "abrupt junction method will fail"? Because I considered also A side as positively charged. Maybe my comment is silly because I still don't know how to proceed with this question

Excuse me Tian, In Q2.4 the equilibrium electron densities are far from the interface, after the depletion region?

Sorry, can I ask you one question about point 2.1? When I try to compute the Fermi energies, If I take Evac as zero reference I get negative values. Does it make sense? Because in the slides the values are often positive, right? Moreover for the n-doped semiconductor can I assume Nc circa Nv? Thank you in advance for your help!

Dear Tian,

When I derive eq 1.4 I don't get the minus sign in front of it. This is why I consider Na=Nd (concentration of dopants). Indeed from the plot 1.a I see at x=xn Nd=-a*xn>0 and at x=xp Na=a*xp>0. Is there another reason for the minus sign or should I consider Na=-Nd?

Thank you in advance! Best regards Alessandro Andreotti

PS: In Q1.5 it's written er=11.0 Did you mean 110 or 11 is correct?

One curiosity about the band diagram in Q2 what does the energy line Evac represent?

@ttian sorry Tian, for what concerns point 3.7 we have to draw 8 plots. But for the second 4 the first assumption (no solvent flow) is still valid Or we have to consider only deltaP=0 and solvent flow different from zero? Thank you!

Dear TAs,

Can I ask you a question regarding the derivation in Q3.1? In my derivation, I have one term that is int[-beta exp[-beta U(x)] g(x) dU]. At this point, can I take g(x) out of the integral? If I do so, I get this solution for w: =exp(-beta U(L/2)) x<=-L/2

=exp(-beta U(-L/2)) x>=L/2

But now can I generalize and say w=exp(-beta*U(x))? Otherwise, it doesn't make sense to have an expression only outside the membrane (of course if I can take out g(x) from the integral). Thank you in advance for your help and your time! Kind regards Alessandro

Dear TAs,

Can I ask you a question regarding the derivation in Q3.1? In my derivation, I have one term that is int[-beta exp[-beta U(x)] g(x) dU]. At this point, can I take g(x) out of the integral? If I do so, I get this solution for w: =exp(-beta U(L/2)) x<=-L/2

=exp(-beta U(-L/2)) x>=L/2

But now can I generalize and say w=exp(-beta*U(x))? Otherwise, it doesn't make sense to have an expression only outside the membrane (of course if I can take out g(x) from the integral). Thank you in advance for your help and your time! Kind regards Alessandro

Dear Tian, Can I ask you a question about eq. 16.5 of Lecture 16?

During the lecture Prof. Shih provided us with this expression for the solute flux N=-w*D/L*delta(C)+w*c*v
Why eq.16.5 has the term D/L that multiplies c*v too? In other words, the solute carried by the solvent isn't independent from diffusion?

And sorry, for what concerns the term delta(c), on the assignment it is written that it is c1-c2. But in this case, when c1>c2 the first term on the right hand side is negative but the diffusive flux is from reservoir 1 to 2. So shouldn't it be delta(c)=c2-c1? Thank you in advance! Best regards Alessandro Andreotti

Dear Tian, I have one doubt about eq 11.7 of the lecture notes. In this step, you simply converted Ci,o into Mi,o. But the factor 1000 shouldn't be at the denominator? Cio= #/m^3 => Cio= mol x Na/m^3 => Cio= Mio x Na/1000 I ask you because I found the same expression also on the Wikipedia link you provided us this afternoon. Thank you in advance for your help! Best regards Alessandro Andreotti