MSE 630
1.
Homework #3 – Diffusion & Ion Implantation
Fall, 2010
A diffused region is formed by an ultra-shallow implant followed by a drive-in. The final
profile is Gaussian. Derive a simple expression for the sensitivity of x J to the implant dose
Q. Is xJ more sensitive to Q at high or low doses?
Answer:
1

 Q
2


x j  4Dt ln 
 Dt CB 

 12


1   Q


 4Dt ln 
2   Dt C 
dQ

B 
dx j
dx j
dQ
 xj 

1
 Q


1


 

 Dt CB   Dt CB 
2Dt
x jQ
2Dt
Q
xjQ
so that the junction depth is more sensitive to Q for small values of Q.
2.
Suppose we perform a solid solubility limited predeposition from a doped glass source
which introduces a total of Q impurities / cm2.
(a) If this predeposition was performed for a total of t minutes, how long would it take (total
time) to predeposit a total of 3Q impurities / cm2 into a wafer if the predeposition
temperature remained constant.
(b) Derive a simple expression for the Dt drivein which would be required to drive the
initial predeposition of Q impurities / cm2 sufficiently deep so that the final surface
concentration is equal to 1% of the solid solubility concentration. This can be expressed
in terms of Dt predep and the solid solubility concentration CS .
Answer:
(a)
Q
2CS
Dt

 Q t
3Q  9t
(b)
C0,t drive in 
Q

2

Q
 0.01CS
Dt
2CS
Dt predep

Dt predep
Dt drivein
 Dt drive in 
 0.01
2
200 
Dt predep
  
3.
An epitaxial layer is grown with uniform concentrations of
15
3
15
3
( B  2 10 cm , P  110 cm ) so that the net doping is
oxidized at 1000˚C for 60 minutes. Approximately (use 2 Dt as a
motion) sketch the resulting impurity profiles for both boron and
oxide which result from the redistribution that occurs.
B and P impurities
p-type. The wafer is
measure of the profile
phosphorus under the
Answer:
Phosphorus and boron have similar diffusion coefficients at 1000˚C, approximately
14
2 1
1 10
cm s . The profiles in the substrate will change over a distance of a few Dt ,
so that while the oxide grows the profile will change because of segregation to the growing
oxide over a distance
2 110 14  3600 1.2 105 cm  0.12 m
The surface concentration will tend towards the value in the oxide, so is related to the bulk
concentration by the segregation coefficient
C S 
   0.3
CB B
C S 
   10
CB P
so that an approximate sketch of the dopant distribution after oxidation is as follows:
1e16
2e15 Boron
1e15 Phos
7e14
0.12 microns
4.
Arsenic is implanted into a lightly doped p-type Si substrate at an energy of 75keV. The
dose is 1 1014 / cm2 . The Si substrate is tilted 7˚ with respect to the ion beam to make it
appear amorphous. The implanted region is assumed to be rapidly annealed so that complete
electrical activation is achieved. What is the peak electron concentration produced?
Answer:
From Fig. 8-3, the range and standard deviation for 75 keV arsenic are
RP  0.05m
RP  0.02m
The peak concentration is
CP 
Q

2 R P
1  1014

4
2 0.02  10

 2 1019 cm  3
Assuming all the dose is active, then the peak electron concentration is equal to the peak
dopant concentration.
5.
How thick does a mask have to be to reduce the peak doping of an implant by a factor of
10,000 at the mask/substrate boundary. Provide an equation in terms of the Range and the
Standard Deviation of the implant profile.
Answer:
*
*
We want to reduce the peak doping N P in the mask at range R P by 10,000 at the
mask/substrate boundary. We will use the equation which describes the profile of an implant
in a mask layer



* 2 
d

R
P
*
*

N (d)  N P exp
*2
2R


P


When
N* (d)
*
NP
we have
*
*
d  R P  4.3R P
 10
4