Transcript MS-630 Lecture 6 IC processing

IC Processing
Initial Steps: Forming an active region
Photoresist
is chemically
removed in
acid, or
stripped in
an O2
plasma
Si3N4 is etched away using an F-plasma:
Si3dN4 + 12F → 3SiF4 + 2N2
Or removed in hot phosphoric acid
After stripping
photoresist,
field oxide is
grown. Field
oxide provides
insulation
between
adjacent
junctions
N and P wells are formed
Photoresist mask is applied, and active ions implanted by ion
bombardment. Typically, 150-200 keV accelerating energy
After implantation, ions
are diffused into
substrate to form wells
After well formation, additional N and P layers are formed in respective N and P
wells, then a layer of polysilicon is deposited. Polysilicon is electrically
conductive and used for gate voltage connections.
Insulating layers of SiO2 are grown around the gate, followed by N or P
bombardment for form the NMOS or PMOS source and drain regions.
After forming gate, source and drain
regions, Ti film is deposited by
sputtering to act as electrical
interconnect
Ti is reacted with Nw to form TiSi2 where it contacts silicon (black
regions) or TiN elsewhere. Then, it is coated with photoresist and
etched, followed by deposit of another insulating SiO2 layer.
Another coat of photoresist followed by
etching exposes gates for connections
A barrier region of TiN is applied, followed
by thin-film application of W, which
undergoes CMP to provide a flat surface
with exposed contacts
Finally, aluminum is sputtered on wafer,
masked and plasma etched. Additional
interconnect layers may be added the
same way.
SEM photograph of interconnects formed in an integrated circuit.
Conductive metals are carefully chosen to provide right conductivity (or
resistivity) and dielectric properties
Photolithography
K1 ~ 0.6-0.8 and
K2 ~ 0.5.
NA is the numerical
aperture number,
NA=n*sin(a)
where n=1 and a is
the angle formed by
the point light source
and the aperture
width
Example
Estimate the resolution and depth of focus of an
excimer laser stepper using KrF light source (l =
248 nm) and NA=0.6 Assume k1 = 0.75 and k2 =
0.5.
Solution:
R = k1*l/NA = 0.75(0.248/0.6) = 0.31 nm
DOF = ± k2*l/NA2 = ±0.5(0.248/(0.6)2) = ±0.34 mm
Shrinking device size drives need
for finer replication methods:
Typical Photoresist Problems
Wet and Dry Etching
Wet Chemical Treatment
Etching Challenges
Dry Etching
Wet vs Dry Etching
Thin Films and Diffusion
Diffusion is not constant across cross section, and continues with
every subsequent high-temperature step; hence, we use charts as
below to calculate surface concentrations, Cs, from average
conductivity,
Effective diffusion-time, (Dt)eff, is the sum
of the diffusivity and time at each step:
(Dt)eff= D1t1+D1t2(D2/D1)=D1t1+D2t2
Effective diffusivity is:
DAeff=Do+D-(n/ni)+D=)n/ni)2 for N-type
DeffA=Do+D+(p/ni)+D++(p/ni)2 for P-type
Values are tabulated, as in table 7.5
Diffusion Data
Example
Figure 7-17 Dopant surface
concentration vs. effective
conductivity for various substrate
concentrations, CB
Chemical Vapor Deposition (CVD)
Typical thin-film problems
Sputtering
Physical Vapor Deposition (PVD)
Suggested exercises
Do Problem 2.1 in Silicon VLSI Technology
Look over example problem (7.3) and
examples on page 390 and 412.