Questions about doping, number of carriers

[lucasdekam]

Hi, I’m going through the lecture about doping and devices again and I came across some things I don’t yet understand.

• First of all, why can we assume that n_D and n_A are zero at ambient temperatures?

• Don’t n_e and n_h also include electrons and holes added by impurities (i.e. extrinsic charge carriers)? If yes, then why can we still make use of the law of mass action n_e n_h = n_i^2 from intrinisic semiconductors?

• How do n_e \approx N_D - N_A and n_h \approx N_A - N_D follow from the simplified charge balance equation? (I feel like this is the most important)

Edit: given the law of mass action I think I kind of get this one now. You can rewrite the charge balance equation to
\frac{n_e^2}{N_D - N_A} - \frac{n_i^2}{N_D-N_A} = n_e
where the second term is (almost) zero, assuming that |N_D - N_A| \gg n_i, so n_e \approx N_D - N_A. I’m still confused why the law of mass action applies though.

• Why are the obtained expressions for E_F independent of E_c and E_v? Those do appear in the exponentials.

• What can we learn from the E_F against 1/T plot? The only thing I can think of is that at low temperature it goes to E_D as all donated electrons go to the lowest available level.

• Why does the blue line for n_h in the n vs 1/T plot stop so early? (perhaps not so important)

• Why is it important to calculate the amounts of charge carriers and consider these intrinsic/extrinsic limits? In the book, they don’t really consider this (I think?), but still discuss of doping and pn-junctions et cetera.

(Maybe a nice idea for the q&a is to make these plots for p-doped semiconductors)

[Kostas]

@lucasdekam

First of all, why can we assume that nDn_D and nAn_A are zero at ambient temperatures?

Well, lets first of all look at how n_D is defined:

n_D = N_D \frac{1}{e^{(E_D-E_F)/kT} + 1}.

So we see that n_D is essentially controlled by E_D - E_F. We also previously established that the Fermi level is far away from the conduction/valance band such that E_c−E_F \gg kT. Additionally, we established that the donor states are weakly bound such that E_D \approx E_C. Combining those gives us E_D−E_F \gg kT which then gives us n_D = 0. Similar argument can be used to show n_A = 0.

@lucasdekam

Don’t nen_e and nhn_h also include electrons and holes added by impurities (i.e. extrinsic charge carriers)? If yes, then why can we still make use of the law of mass action nenh=n2in_e n_h = n_i^2 from intrinisic semiconductors?

Yup, n_e and n_h include the contributions from the dopants. Why can we still use the law of mass action? Well, why not ? When deriving this, we only assumed E_G \gg k_BT , so it doesn’t constraint this to a specific type semiconductors (doped or pristine). What reduces the above equation to only apply to intrinsic semiconductors is the charge balance equation n_e = n_h, which then reduces the law of mass action from n_e n_h = n_i^2 to n_h = n_e = n_i.

@lucasdekam

How do ne≈ND−NAn_e \approx N_D - N_A and nh≈NA−NDn_h \approx N_A - N_D follow from the simplified charge balance equation? (I feel like this is the most important)

You seem to have figured this one out

@lucasdekam

Why are the obtained expressions for EFE_F independent of EcE_c and EvE_v ? Those do appear in the exponentials.

It does depend on those quantities. E_F depends on E_G which is E_c - E_v.

@lucasdekam

What can we learn from the EFE_F against 1/T1/T plot? The only thing I can think of is that at low temperature it goes to EDE_D as all donated electrons go to the lowest available level.

You are correct that at zero temperature E_F goes to E_D. Furthermore, 1/T is not really good for this, but if you look at the plot of E_F vs T, then at low temperatures (but not zero, in the extrinsic regime), your E_F would be linear with T:

E_F = E_G - kT\ln[N_C/(N_D-N_A)], \textrm{ for } N_D > N_A

where the slope tells us something about the concentration of dopants. If you keep increasing the temperature, we would enter the intrinsic regime and so the slope of the line would change to:

E_F = \frac{E_c + E_v}{2} - \frac{3}{4}kT\ln(m_e/m_h).

which would tell us about the effective masses of conduction/valance bands.

@lucasdekam

• Why does the blue line for nhn_h in the nn vs 1/T1/T plot stop so early? (perhaps not so important)

Because the plot needs to be improved upon

@lucasdekam

Why is it important to calculate the amounts of charge carriers and consider these intrinsic/extrinsic limits? In the book, they don’t really consider this (I think?), but still discuss of doping and pn-junctions et cetera.

I’m not sure I understand the question. Are you asking why is it important to define different regimes? The reason for that is that different regimes reduce our equations to simplified forms and thus give us nice analytic solutions. If we didn’t do that, then we would be forced to solve the charge balance equations numerically all the time, which is not fun.

[lucasdekam]

Thank you so much for taking your time to answer these carefully! It is extremely helpful. I have a couple of follow-up questions, but feel free to answer these later

@Kostas

We also previously established that the Fermi level is far away from the conduction/valance band such that E_c−E_F \gg kT.

Okay, but we previously also approximated the Fermi-Dirac distribution by a Boltzmann distribution, not as zero (otherwise there would be no conduction possible, of course). Is the difference that the E_D is just one state, whereas the conduction band is almost a continuum of states?

@Kostas

Why can we still use the law of mass action? Well, why not ?

Maybe my understanding of N_C and N_V is still off. Are they different for pristine and doped semiconductors? I’m thinking a (n-)doped SC should have more electrons in the conduction band compared to a pristine SC, so would N_C then be larger…? But I’m not sure what factor in N_C could cause the increase in that case. If N_C is the same for both doped and pristine SCs, I’d guess n_e = (N_C + N_D) e^{-(E_C - E_F)/k_BT} but then it seems like we’re counting N_D double in the charge balance for doped SC’s…

(sidenote: isn’t N_{C/V} = 1/\lambda_T^3, with \lambda_T the thermal wavelength that appears in statistical physics of an ideal gas? Is that related in any way? Just out of curiosity)

@Kostas

EFE_F depends on EGE_G which is Ec−EvE_c - E_v .

Hmm, can you double-check whether this is really not a typo in the lecture notes? I’d think
N_D - N_A \approx N_C e^{-(E_C - E_F)/k_BT}
(n-doped) leads to
E_F = E_C - k_BT \ln\frac{N_C}{N_D - N_A},
and I think Anton wrote this down in the lecture also. For a p-doped SC I get an E_V term. But maybe I’m missing something here.

@Kostas

EFE_F would be linear with TT … where the slope tells us something about the concentration of dopants

Thanks, now I also understand the discussion during the lecture.

@Kostas

Because the plot needs to be improved upon

So at T=0 or 1/T\to\infty, n_h should go to zero right? Then all holes are in their E_A state / all electron-hole pairs have been annihilated?

@Kostas

different regimes reduce our equations to simplified forms and thus give us nice analytic solutions.

Yeah maybe it was a bit of a vague question. I suppose I was looking for the bigger picture, and I think this describes it well. Thank you!

[Kostas]

@lucasdekam

Okay, but we previously also approximated the Fermi-Dirac distribution by a Boltzmann distribution, not as zero (otherwise there would be no conduction possible, of course). Is the difference that the EDE_D is just one state, whereas the conduction band is almost a continuum of states?

That’s a fair point. I recommend looking at Doping and Devices Ex. 1 subquestion 2, because that essentially explains why n_D = 0. But in short, n_D is so much smaller compared to other quantities like n_i and N_D - N_A, and therefore we neglect it completely in equations.

Now, why, when considering intrinsic semiconductors, did we not set n_e=0 and instead approximated Fermi-Dirac distribution with a Boltzmann factor? Well, despite n_e and n_h being small, they were the only contributions and thus we had to consider them.

@lucasdekam

Maybe my understanding of NCN_C and NVN_V is still off. Are they different for pristine and doped semiconductors?

N_C and N_V are kinda like properties of the valance/conduction band. They only depend on the effective masses and T.

N_V = 2\left(\frac{2\pi m_h kT}{h^2}\right)^{3/2}.

Nowhere in this equation there is any property which is connected to doping. So it should not change as we add/remove electrons/holes in the conduction/valance bands.

@lucasdekam

Hmm, can you double-check whether this is really not a typo in the lecture notes? I’d think
ND−NA≈NCe−(EC−EF)/kBTN_D - N_A \approx N_C e^{-(E_C - E_F)/k_BT}
(n-doped) leads to
EF=EC−kBTlnNCND−NAE_F = E_C - k_BT \ln\frac{N_C}{N_D - N_A} ,
and I think Anton wrote this down in the lecture also. For a p-doped SC I get an EVE_V term. But maybe I’m missing something here.

Yup, you’re right. It seems the lecture notes seem to assume E_G = E_C and E_V =0.

@lucasdekam

So at T=0T=0 or 1/T→∞1/T\to\infty , nhn_h should go to zero right? Then all holes are in their EAE_A state / all electron-hole pairs have been annihilated?

Yup, at T=0 n_h and n_e should go to zero.

[lucasdekam]

Thanks again for the help! I think I understand it now, and otherwise I’ll ask about it for tomorrow Doing the exercises was helpful as well indeed