Transcript Lecture 5

Diffusion #2
ECE/ChE 4752: Microelectronics
Processing Laboratory
Gary S. May
February 5, 2004
Outline
 Objectives
Double Diffusions
 Concentration-Dependent Diffusion
 Diffusion in Silicon
 Lateral Diffusion

Objectives

Discuss the concept of double diffusions, an
important part of how we fabricate our CMOS
transistors in the lab.

Introduce some “second-order” diffusion
effects.
Outline
Objectives
 Double Diffusions
 Concentration-Dependent Diffusion
 Diffusion in Silicon
 Lateral Diffusion

After p-well Diffusion
C(x)
N A (x)
n
p
Csub
x
xj0
After NMOS Source/Drain n+ Diffusion
C(x)
ND (x)
NA (x)
n+
n
p
Csub
xj1
xj2
Notation: p-well Pre-dep
Boron doping
 Pre-Dep: tpp @ Tpp => Dpp, Cspp
 tpp = p-well pre-dep time
 Tpp = p-well pre-dep temperature
 Dpp = p-well diffusion constant at pre-dep
temperature
 Cspp = surface concentration for p-well
pre-dep

Notation: p-well Drive-in
 tpd
@ Tpd => Dpd
 tpd = p-well drive-in time
 Tpd = p-well drive-in temperature
 Dpd = p-well diffusion constant at drive-in
temperature
Notation: n+ Source/Drain Pre-dep


Phosphorus doping
Pre-Dep: tnp @ Tnp => Dnp, Csnp; Dp1
 tnp = n+ source/drain pre-dep time
 Tnp = n+ source/drain pre-dep temperature
+
 Dnp = n source/drain diffusion constant at predep temperature
 Csnp = surface concentration for n+ source/drain
pre-dep
 Dp1 = boron diffusion constant at source/drain
pre-dep temperature
Notation: n+ Source/Drain Drive-in
 tnd
@ Tnd => Dnd; Dp2
+
 tnd = n source/drain drive-in time
 Tnd = n+ source/drain drive-in temperature
 Dnd = n+ source/drain diffusion constant
at drive-in temperature
 Dp2 = boron diffusion constant at
source/drain drive-in temperature
Profile: After p-well Diffusion
2
Sw
–
x
NA x  = ---------------------- exp -----------------4  Dt  w
  D t w
where:
Sw
2Cspp
= -------------- D pp t pp

= well dose
 Dt w = Dppt pp + Dpd tpd = well “Dt”
 There is a pn-junction xj0 where NA(x) = Csub
Profile: After n+ Source/Drain Diffusion
2
Ssd
–x
ND  x = ----------------------- exp -----------------4 D t sd
  Dt  sd
where: Ssd
2Csnp
= -------------- D np tnp

= source/drain dose
 Dt  sd = Dnpt np + Dndt nd = source/drain “Dt”
 BUT: now the well profile has changed to…
New Well Profile
2
Sw
–x
NA  x  = ------------------------ exp ------------------4 D t e ff
  D t e ff
where:  Dt eff = Dppt pp + D pdt pd + Dp1t np + Dp2 tnd
= overall effective “Dt”


There is a pn-junction xj1 where ND(x) = NA(x)
There is a new pn-junction xj2 where NA(x) =
Csub (where xj2 > xj0)
Example
Suppose we want to design a p-well CMOS
diffusion process with a well depth of xj2 =
2.5 mm. Assume the n-type substrate doping is
1015 cm-3.
Example (cont.)

If we start with a boron pre-dep with a dose of 5 × 1013 cm-2,
followed by a 1-hr drive-in at 1100 oC, what is the initial
junction depth (xj0)? Neglect the depth of the pre-dep. The B
diffusivity at this temperature is 1.5 × 10-13 cm2/s.
SOLUTION:
2
Sw
–
x
NA x  = ---------------------- exp -----------------4  Dt  w
  D t w
where: Sw = 5 × 1013 cm-2
(Dt)w = (1.5 × 10-13 cm2/s)(3600s) = 5.4 × 10-10 cm2
15
NA(xj0) = 1015 cm-3 =>

 D t w 10  
x j0 = – 4 D t w ln ------------------------------------- 
13


5 10
1
-2
= 1.24mm
Example (cont.)

Find the necessary (Dt)eff for the p-well to reach the
desired junction depth of xj2 = 2.5 mm.
SOLUTION (This must be solved by iteration!!!):
2
Sw
–
x
NA  x  = ------------------------ exp -------------------4 D t e ff
  D t e ff
where: x = xj2 = 2.5 mm
NA(xj2) = 1015 cm-3
Sw = 5 × 1013 cm-2
=> (Dt)eff = 2.46 × 10-9 cm-2
Example (cont.)

What is the approximate p-well drive-in time needed if
all steps are carried out at 1100 oC?
SOLUTION:
 Dt  eff
2.46 10–9
t pd  ---------------- = -----------------------– 13
Dpd
1.5 10
= 1.64 × 104 s = 273.3 min
Example (cont.)
 If the n+ source/drain junction depth required is xj1 = 2.0
mm, what is the p-well doping at the source/drain
junction?
SOLUTION:
2
Sw
–
x
NA  x  = ------------------------ exp -------------------4 D t e ff
  D t e ff
where: x = xj1 = 2.0 mm
(Dt)eff = 2.46 × 10-9 cm2
=> NA(xj1 = 2.0 mm) = 9.76 × 1015 cm-3
Example (cont.)

Suppose the source/drain dose (Ssd) is 5 × 1014 cm-2. What
is the surface concentration in the source/drain regions and
the source/drain diffusion (Dt)sd?
2
Ssd
–
x
ND  x = ----------------------- exp ------------------4 D t sd
  Dt  sd
where: x = xj1 = 2.0 mm
ND(x = xj1) = 9.76 × 1015 cm-3
SOLUTION:
(i) (Solving by iteration): (Dt)sd = 1.52 × 10-9 cm2
(ii) Surface Concentration:
14
Ssd
5 10
ND  x = 0  = ----------------------- = ------------------------------------ Dt  sd
 1.52 10–9 
= 7.24 × 1018 cm-3
Example (cont.)

The phosphorus source/drain regions are deposited and
driven in at 1050 oC. At this temperature, the phosphorus
diffusivity is 5.8 × 10-14 cm2/s. Ignoring the contributions
of the pre-dep, what is the approximate source/drain
diffusion time (tnd)?
SOLUTION:
 Dt sd
1.52 10– 9
t nd  --------------- = ----------------------–
14
Dnd
5.810
= 2.62 × 104 s = 436.8 min
Example (cont.)

If the boron diffusivity is 6.4 × 10-14 cm2/s at 1050 oC,
correct for the p-well diffusion time to account for the
extra diffusion during the source/drain drive-in. (Neglect
the contributions of pre-dep steps).
SOLUTION:
Dp2tnd = 1.68 × 10-9 cm2
( “Dt” accumulated by boron during source/drain diffusion).
=> Initial p-well drive-in may be reduced by this amount, or:
 Dt e ff – Dp2 t nd  2.46 10–9  –  1.6810– 9 
tpd  --------------------------------------- = ---------------------------------------------------------------–
13
Dpd
1.510
= 86.9 min
Outline
Objectives
 Double Diffusions
 Concentration-Dependent Diffusion
 Diffusion in Silicon
 Lateral Diffusion

Vacancies


When host atom acquires sufficient energy to
leave its lattice site, a vacancy is created.
Vacancy density of a given charge state (#
vacancies/unit volume, CV) has temperature
dependence similar to carrier density:
 E F  Ei 
CV  Ci exp 

 kT 
where Ci = intrinsic vacancy density, EF = Fermi
level, and Ei = intrinsic Fermi level
Vacancy-Dependent Diffusion
 If diffusion is dominated by the vacancy
mechanism, D is proportional to vacancy
density.
 At low doping concentrations (n < ni), EF = Ei,
and CV = Ci (independent of doping), so
 D (which is proportional to CV = Ci ), also
independent of doping concentration.
 At high concentrations (n > ni), [exp(EF –
Ei)/kT] becomes large, which causes CV and D
to increase.
Intrinsic and Extrinsic Diffusion
Effect on Diffusivity
C
D  Ds 
 Cs
F  D



g
C
x
Cs = surface concentration
 Ds = diffusion coefficient at the surface
 g = parameter to describe concentration
dependence

Diffusion Profiles
Junction Depth
x j  1.1 Ds t
 For g > 0, D decreases with concentration
 Increasingly steep box-like profiles result
 Therefore, highly abrupt junctions are formed
 Junction depth is virtually independent of
background concentration
x j  1.6 Ds t
g=1
x j  1.1 Ds t
g=2
x j  0.87 Ds t
g=3
Outline
Objectives
 Double Diffusions
 Concentration-Dependent Diffusion
 Diffusion in Silicon
 Lateral Diffusion

Concentration Dependence
 Boron,
 Gold,
arsenic: g ≈ 1
platinum: g ≈ -2
 Phosphorus:
g ≈ 2 (sort of)
Phosphorus
Diffusion
Phosphorus Diffusion (cont.)
 When surface concentration is low, diffusion
profile is an erfc (curve a).
 As concentration increases, the profile
begins to deviate (b and c).
 At high concentration (d), profile near the
surface is similar b, but at ne, kink occurs,
followed by rapid diffusion in tail region.
 Because of high diffusivity, phosphorus is
used to form deep junctions, such as the ntubs in CMOS.
Outline
Objectives
 Double Diffusions
 Concentration-Dependent Diffusion
 Diffusion in Silicon
 Lateral Diffusion

The Problem



1-D diffusion equation is not adequate at
the edge of the mask window.
There, impurities diffuse downward and
sideways (i.e., laterally).
In this case, we must consider a 2-D
diffusion equation and use numerical
techniques to get the diffusion profiles
under different initial and boundary
conditions.
Diffusion Contours
 Contours of constant doping concentration
for a constant Cs, assuming D is independent
of concentration.
Interpretation
 Variation at far right corresponds to erfc
distribution.
 Example: at C/Cs = 10–4, the vertical
penetration is about 2.8 µm, whereas the lateral
penetration is about 2.3 µm (i.e., the penetration
along the diffusion mask-semiconductor
interface).
Implications



Because of lateral diffusion, the junction consists
of a central plane (or flat) region with
approximately cylindrical edges with a radius of
curvature rj.
If the mask has sharp corners, the shape of the
junction near the corner will be roughly spherical.
Since the electric-field intensities are higher for
cylindrical and spherical junctions, the avalanche
breakdown voltages of such regions can be
substantially lower than that of a plane junction.