Transcript Undergraduate Admissions & College of Engineering

Diffusion #1
ECE/ChE 4752: Microelectronics
Processing Laboratory
Gary S. May
January 29, 2004
Outline
 Introduction
Apparatus & Chemistry
 Fick’s Law
 Profiles
 Characterization

Definition



Random walk of an ensemble of particles from
regions of high concentration to regions of
lower concentration
In general, used to introduce dopants in
controlled amounts into semiconductors
Typical applications:
 Form diffused resistors
 Form sources/drains in MOS devices
 Form bases/emitters in bipolar transistors
Basic Process
 Source material transported to surface by inert
carrier
F
DC 0
x  (D /  )
 Decomposes and reacts with the surface
 Dopant atoms deposited, dissolve in Si, begin to
diffuse
Outline
Introduction
 Apparatus & Chemistry
 Fick’s Law
 Profiles
 Characterization

Schematic
Dopant Sources



Inert carrier gas = N2
Dopant gases:
 P-type = diborane (B2H6)
 N-Type = arsine (AsH3), phosphine (PH3)
Other sources:
 Solid = BN, As2O3, P2O5
 Liquid = BBr3, AsCl3, POCl3
Solid Source
Example reaction: 2As2O3 + 3Si → 4As + 3SiO2
(forms an oxide layer on the surface)
Liquid Source
 Carrier “bubbled” through liquid; transported as vapor to surface
 Common practice: saturate carrier with vapor so concentration is
independent of gas flow
=> surface concentration set by temperature of bubbler & diffusion system
 Example: 4BBr3 + 3O2 → 2B2O3 + 6Br
=> preliminary reaction forms B2O3, which is deposited on the surface;
forms a glassy layer
Gas Source
 Examples:
a) B2H6 + 3O2 → B2O3 + 3H2O (at 300 oC)
b) i) 4POCl3 + 3O2 → 2P2O5 + 6Cl2
(oxygen is carrier gas that initiates preliminary reaction)
ii) 2P2O5 + 5Si → 4P + 5SiO2
Outline
Introduction
 Apparatus & Chemistry
 Fick’s Law
 Profiles
 Characterization

Diffusion Mechanisms
 Vacancy:
atoms jump from one lattice
site to the next.
 Interstitial:
atoms jump from one
interstitial site to the next.
Vacancy Diffusion
 Also called “substitutional” diffusion
 Must have vacancies available
 High activation energy (Ea ~ 3 eV  hard)
Interstitial Diffusion
“Interstitial” = between lattice sites
 Ea = 0.5 - 1.5 eV  easier

First Law of Diffusion
C
F  D
x
F  D
C
x

F = flux (#of dopant atoms passing through
a unit area/unit time)
 C = dopant concentration/unit volume
 D = diffusion coefficient or diffusivity
 Dopant atoms diffuse away from a highconcentration region toward a lowerconcentration region.
Conservation of Mass
C
F
  C 

 D

t
x x  x 
 1st
Law substituted into the 1-D
continuity equation under the condition
that no materials are formed or
consumed in the host semiconductor
Fick’s Law
C
 C
D 2
t
x
2
 When the concentration of dopant atoms is
low, diffusion coefficient can be considered to
be independent of doping concentration.
Temperature Effect
  Ea 
D  D0 exp 

 kT 



Diffusivity varies with temperature
D0 = diffusion coefficient (in cm2/s) extrapolated
to infinite temperature
Ea = activation energy in eV
Outline
Introduction
 Apparatus & Chemistry
 Fick’s Law
 Profiles
 Characterization

Solving Fick’s Law
 2nd
order differential equation
 Need
 Need
one initial condition (in time)
two boundary conditions (in
space)
Constant Surface Concentration
 “Infinite source” diffusion
 Initial condition: C(x,0) = 0
 Boundary conditions:
C(0, t) = Cs
C(∞, t) = 0
 x
 Solution: C ( x, t )  C s erfc 
 2 Dt



Key Parameters
 Complementary error function:
erfc( x)  1  erf ( x)
erf ( x) 
2
x
e


u 2
du
0
 Cs = surface concentration (solid solubility)
Total Dopant

Total dopant per unit area:

Q(t )   C ( x, t )dx

Q(t )   C ( x, t )dx
0
0
Q(t ) 

2

C s Dt  1.13C s Dt
Represents area under diffusion profile
Example
For a boron diffusion in silicon at 1000 °C, the surface
concentration is maintained at 1019 cm–3 and the diffusion
time is 1 hour. Find Q(t) and the gradient at x = 0 and at a
location where the dopant concentration reaches 1015 cm–3.
SOLUTION:
The diffusion coefficient of boron at 1000 °C is about 2 ×
1014 cm2/s, so that the diffusion length is
Dt  2 1014  3600  8.48 106 cm
Q(t )  1.13Cs Dt  1.13 1019  8.48 106  9.5 1013 cm2
Example (cont.)
dC
dx
x 0
Cs
 1019
23
4




6
.
7

10
cm
Dt
  8.48 106
When C = 1015 cm–3, xj is given by
 1015 
x j  2 Dt erfc  19   2 Dt (2.75)  4.66 105 cm
 10 
-1
dC
dx
x  0.466m
Cs  x 2 / 4 Dt

e
 3.5 1020 cm 4
Dt
Constant Total Dopant
 “Limited source” diffusion
 Initial condition: C(x,0) = 0

 Boundary conditions:
 C ( x, t )  S  dose
0
C(∞, t) = 0
 x2 
 Solution: C ( x, t ) 

exp  
Dt
 4 Dt 
S
Example
Arsenic was pre-deposited by arsine gas, and the resulting
dopant per unit area was 1014 cm2. How long would it take to
drive the arsenic in to xj = 1 µm? Assume a background
doping of Csub = 1015 cm-3, and a drive-in temperature of 1200
°C. For As, D0 = 24 cm2/s and Ea = 4.08 eV.
SOLUTION:
 4.08
  Ea 


13
2
D  D0 exp 

24
exp

2
.
602

10
cm
/s



5
 8.614 10 1473 
 kT 
x 2j  10 8
5




S
1
.
106

10
  1.04  10 12 t ln 

 4 Dt ln 


t


 C B Dt 
Example (cont.)
t • log t – 10.09t + 8350 = 0

The solution to this equation can be
determined by the cross point of equation:
y = t • log t and y = 10.09t – 8350.

Therefore, t = 1190 seconds (~ 20 minutes).
Diffusion
Profiles
Pre-Deposition

Pre-deposition = infinite source
C(x)
Boron diffusion
(Cs ~ 1020 cm-3 )
Cs
Csub
(phosphorus, ~1015 cm -3 )
p-type
xj
n-type
x
xj = junction depth (where C(x)=Csub)
Drive-In


Drive-in = limited source
After subsequent heat cycles:
C(x)
area = S
Csub
xj
x
Multiple Heat Cycles
2

S
x
C ( x, t ) 
exp  
 4( Dt )
 ( Dt )eff
eff

where: ( Dt ) 
eff




n
Dt
i 1
i i
(for n heat cycles)
Outline
Introduction
 Apparatus & Chemistry
 Fick’s Law
 Profiles
 Characterization

Junction Depth

Can be delineated by cutting a groove and etching the
surface with a solution (100 cm3 HF and a few drops
of HNO3 for silicon) that stains the p-type region
darker than the n-type region, as illustrated above.
Junction Depth
 If R0 is the radius of the tool used to form the
groove, then xj is given by:
x j  R02  b 2  R02  a 2
 In R0 is much larger than a and b, then:
a2  b2
xj 
2R0
4-Point Probe

Used to determine
resistivity
4-Point Probe
1) Known current (I) passed through outer probes
2) Potential (V) developed across inner probes
r = (V/I)tF
where: t = wafer thickness
F = correction factor (accounts for probe geometry)
OR:
Rs = (V/I)F
where: Rs = sheet resistance (W/
)
=> r = Rst
Resistivity
s  1 / r  q(  n n   p p)
where: s = conductivity (W-1-cm-1)
 r = resistivity (W-cm)
 n = electron mobility (cm2/V-s)
 p = hole mobility (cm2/V-s)
 q = electron charge (coul)
 n = electron concentration (cm-3)
-3
 p = hole concentration (cm )
Resistance
w
t
length = L
area = A
R
rL
A

rL
wt
Sheet Resistance




1 “square” above has resistance Rs (W/square)
Rs is measured with the 4-point probe
Count squares to get L/w
Resistance in W = Rs(L/w)
Sheet Resistance (cont.)

Relates xj, mobility (), and impurity distribution C(x)
Rs 
1
xj
q  C ( x)dx
0


For a given diffusion profile, the average resistivity
( r = Rsxj) is uniquely related to Cs and for an assumed
diffusion profile.
Irvin curves relating Cs and r have been calculated for
simple diffusion profiles.
Irvin
Curves