# Solution to Exercise 3.1.1

 How far off from perfection is a 1000 Wcm Si crystal (at 300 K)? The resisitvity given is about the best (i.e. highest) value that Si crystal growers can achieve on a routine base. Consider what level of dopants corresponds to 1000 Wcm? How far away from perfection (= truly intrinsic behavior) are the crystal growers in terms of dopant concentration? For that you must know the intrinsic carrier density and resistivity at room temperature in Wcm. Calculate the carrier density with the numbers and relations provided and find some suitable value for the mobility (from the various illustrations in chapter 2). This exercise not only demands that we generate numbers from some general formulas (which is not as easy as it looks), but also gives an idea of how close we can get to the real numbers with our simple models.

First some numbers from the literature. According to "Semiconductor Materials", the intrinsic electrical conductivity of Si at 300 K is
3.16 µS/cm. The NSM archive has rather similar numbers.
[S] = "Siemens" is a quaint German measure of conductivity, it is simply 1/W
This translates into a room temperature resistivity rof r = 1/s = 316 000 Wcm.
Alternatively, numbers for the intrinsic carrier density found in the sources given above or in arbitrary books are somewhere in between 1.00 · 1010 cm– 3 or 1.38 · 1010 cm– 3.
Lets see if we can get numbers like this by calculation:
The carrier density is given by

ne = Neeff · exp –   EC  –  EF
kT

Neeff  can be estimated from the free electron gas model in a fair approximation to

Neeff  = 2 · æ ç è ö ÷ ø

The dimension of this Neeff  is

[Neff ]  =  kg3/2 · eV3/2 · eV– 3 · s– 3  =  kg3/2 · eV– 3/2 · s– 3

That is a bit strange. Nevertheless it is right - try to do something about the kg! If you have problems of figuring out how to get the proper dimension m– 3, use the link.
Inserting numbers (me = 9,109 · 10– 31 kg; k·T = 1/40 eV, h2 = (4,1356 · 10– 18)2 eV2s2 = 1,71 · 10– 35 eV2s2), we obtain

 Neff = 4.59 · 1015 · T3/2 cm–3 = 2.384 · 1019 cm–3 T = 300 K = 2.384 · 1025   m–3

The intrinsic carrier density thus is

ne = 3.22 · 1019 cm– 3 · exp –   EC  –  EF
kT
=  3.22 · 1019 cm– 3 · exp –   0.55 eV
0.025 eV
=  9 · 109cm– 3

This is just a little bit smaller than than the values given above; rather amazing, considering that the free electron gas model is just a very simple approximation.
Now we can see what kind of mobility m we would get with ni = 1 · 1010 cm– 3 and a conductivity s = 3.16 µS/cm = 3.16 · 10– 6 W– 1cm– 1
We had the simple law s = 2eµni (the factor two takes into account that we have holes and electrons), and thus µ = s/2eni. This gives us

µ  =  3.16 · 10– 6
2 · 1,602 · 10– 19 · 1 · 1010
W– 1 · cm– 1 · C– 1 · cm3

With [W]= [V/A] = [V · s/C] we have

µ  =  986 cm2 · s– 1 · V– 1

as an expected result. The unit [cm2 · V– 1 · s– 1] comes from the original definition of µ, which was (drift) velocity divided by field strength.
Looking around a bit we find tabulated values of, e.g., 1400 cm2/Vs, which is just off by a factor of two - and that we do not take seriously. So, what have we learned so far?
1. It is not so easy to really calculate the intrinsic properties. Getting the right order of magnitudes is already pretty good. This is due, of course, that we have approximations a plenty, coupled with lots of exponentials which are quite sensitive to small changes in the argument.
2. If we accept an intrinsic carrier concentration for one kind of carrier at room temperature around ni = 1 · 1010 cm–3, we would need a dopant concentration that is at least an order of magnitude smaller if we want to claim truly intrinsic properties. That means we demand

Ndop  £  1 · 10– 9 cm– 3  £  20 ppqt

Find out what ppqt means yourself.
The minimum doping concentration Nmin achievable (corresponding to the maximum resistivity rmax of 1000 Wcm or the minimum conductivity smin of  1 · 10– 3 W– 1 · cm– 1) must be about

Nmin  =  316 000
1000
»  300 ni  =  3 · 1012 cm– 3

In the "master" curve for resistivity vs. doping, we find a value between 5 · 1012 cm– 3 and 1 · 1013 cm– 3, so again we are close enough.
The final answer thus is: We are still at least a factor of 100 away from "perfection" with respect to unwanted doping.
And of course, we can not make any statement about the perfection achieved with respect to impurities that do not influence the carrier concentrations

To index

Density of States

Exercise 3.1.1

Ohms Law and Materials Properties

Doping and Mobility

Exercise 3.4.1

Mistake in Picture

© H. Föll (Semiconductor - Script)