Name:________________________
PHGN/CHEN/MLGN435/535 Pre-Lab #7 – Solutions
Topics: Diffusion, Schottky contacts
Reading: Campbell 3.1-3.2, 3.4-3.5; 15.6
Diffusion doping is one of those places where students have become confused in the past.
Later in this course, we will be using estimates of diffusion depths and concentrations to
determine processing times and temperatures for transistor structures. Part of the goal of
this assignment is to try to help you think about how this works given our dopant sources.
1. a) We want to dope the n-type base of a PNP bipolar transistor. To do this, we pattern
a wafer, grow a 200nm wet oxide mask to act as an antidiffusion barrier, pattern a hole in
the oxide for the base, and spin on the P8545 n-type dopant you used in class. We then
do a pre-deposition anneal at 950C for ½ hour. Using the manufacturers data provided
on the web (see this page), what
Oxide
Oxide mask
would the sheet resistivity and
Diffusion
diffusion depth of the base region
doped base
be?
p-type Si doped at 1x1016/cm3
b) Using these two values, find the resistivity of the doped region and from this the
dopant concentration.
c) Now we want to drive in the dopant to increase the depth of the base, and to lower the
average doping concentration. To do this we strip the oxide and the spin on dopant layers
in an HF solution, and we anneal at 1100C for 12 hours in argon. Make the assumption
that all of the dopant introduced during the pre-deposition is located on the surface of the
wafer (that is, that the diffusion
Diffusion
depth you got in part a is small
doped base
enough to be negligible). In Fair's
vacancy model assume the
background p-type doping
p-type Si doped at 1x1016/cm3
concentration is 1x1016/cm3 . This
is much less than the intrinsic
carrier concentration so can be ignored. Also assume the simple theory of diffusion
presented at the beginning of chapter 3 applies (that the diffusion coefficient does not
change with time or depth). Give me a simulated plot of the P concentration as a function
of depth through the base region. Also determine the depth of the PN junction that forms
between the base region and the substrate (find the point where the n-type doping
concentration is equal to the p-type background concentration). Does this agree with what
you got in your plot?
d) Finally, we have to worry about that oxide barrier layer in the first step. Did any P get
through the oxide during the predep and introduce dopants in regions where we don't
want them? For this part, assume the P concentration at the surface of the oxide is
constant and at the saturation limit of 5x1019/cm3. Now plot the P concentration as a
function of depth through the oxide and Si region below the oxide. Is the concentration
in the Si high enough that we need to worry about it?
2. Assume a Schottky barrier is formed by evaporating copper onto silicon doped n-type
with a doping concentration of 7x1014/cm3. The built-in voltage for the Schottky barrier
is 0.5V.
a) What is a Schottky barrier?
b) Find the depletion width in microns you would expect for the diode.
c) Find the capacitance of a circular Schottky diode with a diameter of 150 microns and
no applied bias
Show your work for full credit. Also, note there is a subtlety with how to handle the
dielectric constant. You can find the dielectric constant for Si in the back of the book,
but since the expressions in your text are in SI units, the dielectric constant to use is this
value in the back of the book multiplied by the permittivity of free space (ε0).
3. Summarize two different ways you could use C-V measurements to get useful
information about a device (you could pick multiple types of devices- Schottky barriers,
MOS devices, etc., or pick one and look at multiple uses of C-V for that device).
Solution to 1:
Solution to 2)
a) A Schottky barrier is a type of metal-semiconductor contact which has rectifying
behavior, similar to a pn junction. A Schottky barrier is formed when the work function
(the minimum energy required to completely remove an electron from a material) of the
metal is larger than the workfunction of a doped semiconductor.
b) For the depletion width, we have Eq. 3.42 with V=0 so
2εε 0Vbi
2 x11.7 x8.85 x10 12 x0.5
W=
=
= 9.61x10 − 7 m = .961um (note all units were
qN
1.602 x10 −19 x7 x10 20
changed to MKS values so the answer came out in meters).
c) Now to get the capacitance we need use
C=
εε A
0
W
=
11.7 x8.85 x10 12 π (150 x10 −6 / 2) 2
= 1.9 x10 −12 F = 1.9 pF
−6
0.961x10
Note that again all calculations were done in MKS units. Also note that some of you will
be expected to do this type of calculation on your project reports to compare the
electrical determination of your oxide thickness to physical (profilometer) and optical
(ellipsometer) determinations.
3. C-V measurements can provide an array of information. For example:
•
•
•
For an MOS structure, if you know the area of your structure and the dielectric
constant (permittivity), you can determine its thickness from: C = εA/tox, where C
is the capacitance measured in the accumulation mode of the MOS.
For a Schottky diode or a p+n or pn+ diode, C-V data can provide both the builtin voltage, Vbi, and the doping concentration of the lower doped side (or
semiconductor side, in a Schottky diode).
Other answers/uses are also acceptable.