1. Solution 2.5
Problem text: Rapid Thermal Annealing.(see chapt 6) has gained interest. The process allows wafers wafers that have a high concentration to be heated rapidly minimizing diffusion. Explain the desirability of such a process based on the discussion of phase diagrams and solid solubility
The above is the solution the publisher/author gives to the problem.
It seems OK. We can add the following page

2. Fys 4310 Sol 2.5
We see from Fig. 2.4 (and 2.3) in text book that all solubility values increase with temperature (at leat up to 1250 C).
Thus a high temperature treatment may cause a large concentration to be dissolved in the material. That could possibly lead to smaller contact resistance, large emitter doping and optimal emitter injection efficiency. (In the last case, it has to be considered at very high doping concentration the efficiency decreases with concentration , .also a lower series resistance of devices can be achieved.
This was certainly what the text book author had in mind. But it can be discussed whether this high doping is really achieved. Even if the phase diagram indicates so.
To acieve highly doped electrically active material region, we must have that
The atoms enter into substitutional positions and is dissolved.
They do not segregate out by cooling but remain as singleseparate atoms.
For 2) the cooling has to be fast such that the state of the system at the higher temperature gets frozen in. The point about heating fast is only to minimize diffusion. We could ask ourselves: what is the reason for a lower solubility with decreasing temperature? A free energy consideration is useful: The free energy is minimized in equlibrium. Relative to the entropy term, the internal energy gets more important at low temperatures. The enrgy is lowered by the system creating for example Si-As compounds thus As-As bonds . In order for this to happen the time has to be sufficiently long to nucleate the phase by diffusion. With fast cooling that won’t happen..
For 1) the time at high temp must be long enough for the system to be near equilibrium what the distribution of dopant atoms is concerned. First we need to create vacancies and they have to diffuse to where the dopant atoms are Thus we need some time, but if the time is too long the dopant profile will broaden by diffusion. Sometimes that’s undesirable as for example emitter doping. ß

3. FYS 4310 Sol 2.12
Problem text: A melt contains 0.1 atomic percent phosphorous in silicon. Assume the well mixed approximation and calculate the dopant concentration when 10 % of the crystal is pulled, when 50 % is pulled and when 90 % of the crystal is pulled.
The above is the solution the publisher/author gives to the problem.
It consider the dopant concentration at the surface of the solid! (which of course is OK)
The target of the problem could also be understood as that in the liquid. See next page

4. > restart;
Consider the situation when a fraction x0 is solidified., Consider the solidification occurring in one dimension. The length of the solidified rod is Lx, The total length when all is solidified is L0.
We call the fraction solidified for x1. x1 = Lx/L0.
FYS 4310 Sol 2.12
The total number of dopant atoms in the system is conserved,
We take the sum of atoms in the solid and the liquid and divide by L0 and the cross section an get
> eq0:=int(Cs(x),x=0..x1)+CL*(1-x1)=C0;
We could solve this equation, eq0, to find Cs(x), We have done that already and the expression is given in 2.23 in the book,
So we don’t do it again here. We just find from above the expression for CL explicitly.
> eq1:=CL=solve(eq0,CL);
We can put in for Cs from equation 2.13 in the book..
> Cs:=x->k*C0*(1-x)^(k-1);
Then the expression for CL becomes
> eq1;
Note this could be written directly from .2.13’ and
the definition of the segregation coefficient

5. >From Table 2.1 gets k for phosphorous, k=0.35
> subs({C0=0.1*atprcnt,x1=0.1,k=0.35},eq1);
> subs({C0=0.1*atprcnt,x1=0.5,k=0.35},eq1);
> subs({C0=0.1*atprcnt,x1=0.9,k=0.35},eq1)

6. Fys 4310 sol 2.13
Problem text
A boule of Si is pulled from a melt that contains 0.01% phosphorous(P) in the melt.
(a) what concentration of phosphorous (P) would you expect at the top of the boule (x = 0)?
(b) If the boule is 1 m long and it has a uniform cross section, at what position ( or x value) would you expect the concentration to be twice as large as it is at the top?
(c) Now consider the melt to contain gallium as well.( Ga is a p-dype aceptor dopant for Si, but it is not commonly used) The concentration of Ga in the melt is such that at the top of the boule (x = 0), the concentrations of gallium and phosphorous are exactly equal. If the concentration of gallium half way down the boule (x=0.5) is twice that of the phosphorous, what is the segregation coefficient (k) for gallium.?

7. Fys 4310 sol 2.13
Fys 4310 sol 2.13

8. Fys 4310 sol 2.13