Transcript The silicon substrate and adding to it - Rose

The silicon substrate and
adding to it—Part 1
 Explain how single crystalline Si wafers are made
 Describe the crystalline structure of Si
 Find the Miller indices of a planes and directions in
crystals and give the most important direction/planes in
silicon
 Use wafer flats to identify types of Si wafers
 Define
 Semiconductor
 Doping/dopant
 Resistivity
 Implantation
 Diffusion
 p-n junction
 Give a number of uses of p-n junctions
 Calculate
 Concentration distributions for thermal diffusion
 Concentration distributions for ion implantation,
and
 p-n junction depths
Silicon—The big green Lego®
silicon
substrate
silicon
substrate
Bulk micromachining
Surface micromachining
Three forms of material
Grains
Crystalline
Polycrystalline
Amorphous
Silicon wafers
Polysilicon (in surface
μ-machining)
Glass and fused quartz,
polyimide, photoresist
Creating silicon wafers
• Creates crystalline (cristalino) Si of
high purity
• A “seed” (semilla) of solid Si is
placed in molten Si—called the
melt—which is then slowly spun
and drawn upwards while cooling it.
• Crucible and the “melt” turned in
opposite directions
The Czochralski method
• Wafers cut from the cross section.
Creating silicon wafers
Grains
Polycrystalline silicon
(American Ceramics Society)
Photo (foto) of a monocrystalline silicon ingot
It’s a crystal
a
a
a
Cubic
Body-centered
cubic (BCC)
Unit cells
a - lattice constant, length of a side of a unit cell
Face-centered
cubic (FCC)
It’s a crystal
The diamond
(diamante) lattice
Miller indices
The Miller indices give us a way to identify different
directions and planes in a crystalline structure.
Indices: h, k and l
• [h k l ]  a specific direction in the crystal
• <h k l >  a family of equivalent directions
• (h k l )  a specific plane
• {h k l }  a family of equivalent planes
How to find Miller indices:
1. Identify where the plane of interest
intersects the three axes forming the unit
cell. Express this in terms of an integer
multiple of the lattice constant for the
appropriate axis.
2. Next, take the reciprocal of each quantity.
This eliminates infinities.
3. Finally, multiply the set by the least common
denominator. Enclose the set with the
appropriate brackets. Negative quantities are
usually indicated with an over-score above
the number.
Te toca a ti
Find the Miller indices of the plane
shown in the figure.
1. Identify where the plane of interest
intersects the three axes forming the unit
cell. Express this in terms of an integer
multiple of the lattice constant for the
appropriate axis.
4
3
Respuesta: (2 4 1)
2
2. Next, take the reciprocal of each quantity.
This eliminates infinities.
1
c
a
1
How to find Miller indices:
b
1
2
2
3. Finally, multiply the set by the least common
denominator. Enclose the set with the
appropriate brackets. Negative quantities are
usually indicated with an over-score above
the number.
For cubic crystals the Miller indices represent a direction vector perpendicular to a plane with integer
components. Es decir,
[h k l] ⊥ (h k l)
¡Ojo! Not true for non-cubic materials!
Non-cubic material example
Quartz is an example
of an important
material with a noncubic crystalline
structure.
(http://www.jrkermode.co.uk/quippy/adglass.html)
Miller indices
What are the Miller indices of the shaded planes in the figure below?
a. (1 0 0)
b. (1 1 0)
c. (1 1 1)
Te toca a ti:
Find the angles between
a. {1 0 0} and {1 1 1} planes, and
b. {1 1 0} and {1 1 1} planes.
Wafer types
Si wafers differ based on the orientation of their crystal
planes in relation to the surface plane of the wafer.
Wafers “flats” are used to identify
• the crystalline orientation of the
surface plane, and
• whether the wafer is n-type or p-type.
<1 0 0> direction
(1 0 0) wafer
Relative position of crystalline planes in a (100) wafer
Orientations of various crystal directions and planes in a (100)
wafer (Adapted from Peeters, 1994)
It’s a semiconductor
(a) Conductors
(b) Insulators
(c) Semiconductors
The “jump” is affected by both temperature and light  sensors and optical switches
Conductivity, resistivity, and resistance
Electrical conductivity (σ) 
• A measure of how easily a material conducts electricity
• Material property
Electrical resistivity (ρ) 
• Inverse of conductivity; es decir ρ = 1/σ
• Material property
By doping, the resistivity of silicon can be varied
over a range of about 1×10-4 to 1×108 Ω•m!
L
L
R

A
A
Conductivity, resistivity, and resistance
Te toca a ti
Find the total resistance (in Ω) for the MEMS snake
(serpiente) resistor shown in the figure if it is made of
• Aluminum (ρ = 2.52×10-8 Ω·m) and
• Silicon
1 μm
100 bends total
Respuesta:
• Al: 509 Ω
• Si: 1.3 GΩ !!
100 μm
1 μm
Entire resistor is 0.5 μm thick
Doping
(a) Phosphorus is a donor – donates
electrons
(b) Boron is an acceptor – accepts
electrons from Si
 Charge carriers are “holes.”
Phosphorus and boron are both dopants.
P creates an n-type semiconductor.
B creates a p-type semiconductor.
Doping
Two major methods
• Build into wafer itself during silicon growth
• Gives a uniform distribution of dopant
•  Background concentration
• Introduce to existing wafer
• Implantation or diffusion (or both!)
• Non-uniform distribution of dopant
• Usually the opposite type of dopant (Es decir, si
wafer es p-type, el otro es n-type y vice versa)
• Location where dopant concentration matches
background concentration se llama p-n junction
p-n junction
Uses of doping and p-n junctions:
• Change electrical properties (make more or less conductive)
• Create piezoresistance, piezoelectricity, etc. to be used for sensing/actuation
• Create an etch stop
Doping
Often implantation and diffusion are done
through masks in the wafer surface in order
to create p-n junctions at specific locations.
How do we determine the distribution of
diffused and/or implanted dopant?
Mass diffusion:
Mass “flux”
Concentration
gradient
dC
j  D
dx
Diffusion constant
Compare to
D  D0e

dT

q  k
dx
dV
 
dx
Ea
k bT
Frequency factor and activation energy for diffusion of dopants in silicon
Doping by diffusion
C
dC
j  D
dx
time
x
At t = 0, or C(x, t = 0) = 0
Conservation of mass (applied to any point in the wafer)
C ( x, t ) j
C

 D 2
t
x
x
2
Need
C(x = 0, t > 0) = Cs
• 1 initial condition
• 2 boundary conditions
C(x → ∞, t > 0) = 0
erfc( ) is the complementary error function:
Solución 
 x 

C ( x, t )  C s erfc
 2 Dt 
erfc( ) 
2


e
x
2
d
Appendix C
Doping by diffusion
x
xdiff  4Dt
Diffusion of boron in silicon at 1050°C for various times
Diffusion length  rough estimate of
how far dopant has penetrated wafer
Doping by diffusion
Total amount of dopant diffused into a surface per unit area is called the ion dose.

2 Dt
 x 
C
erfc
dx

Cs


Q(t )   s

 2 Dt 
0
C(x = 0, t > 0) = Cs
Q = constant
Cs
  x2 
Q

C ( x, t ) 
exp
Dt
 4 Dt 
time
Gaussian distribution
x
Doping by implantation
Distribution is also Gaussian, but it is more complicated.
 ( x  RP ) 2
C ( x)  C P exp 
2
2

R
P




• CP – peak concentration of dopant
• RP – the projected range (the depth of peak concentration of
dopant in wafer)
Doping by ion implantation
• ΔRP – standard deviation of the distribution
Range affected by the mass of the dopant, its acceleration energy,
and the stopping power of the substrate material.
Peak concentration  C P 
Qi
2 RP
Doping by implantation
Doping is often (de hecho, usually) a twostep process:
1st implantation – pre-deposition
2nd thermal diffusion – drive-in
If projected range of pre-deposition is
small, can approximate distribution with
Typical concentration profiles for ion implantation
of various dopant species
  x2 
Q

C ( x, t ) 
exp
Dt
 4 Dt 
Replace with Qi
Junction depth
C
implanted
dopant
background
concentration
p-n junction
x
Te toca a ti
A n-type Si-wafer with background doping concentration of 2.00×1015 cm-3 is doped
by ion implantation with a dose of boron atoms of 1015 cm-2, located on the surface of
the wafer. Next thermal diffusion is used for the drive-in of boron atoms into the wafer
a 900°C for 4 hours.
a. What is the diffusion constant of boron in silicon at this temperature?
b. What is the junction depth after drive-in?
Hints:
• Assume that the distribution of ions due
to implantation is very close to the wafer
surface
• Useful information:
− kb = 1.381×10-23 J/K
− eV = 1.602×10-19 J
Te toca a ti
A n-type Si-wafer with background doping concentration of 2.00×1015 cm-3 is doped
by ion implantation with a dose of boron atoms of 1015 cm-2, located on the surface of
the wafer. Next thermal diffusion is used for the drive-in of boron atoms into the wafer
a 900°C for 4 hours.
a. What is the diffusion constant of boron in silicon at this temperature?
b. What is the junction depth after drive-in?
a.
D  D0 e

Ea
k bT
Set = Cbg
  x2 
Q
b. C ( x, t ) 

exp
Dt
 4 Dt 
Replace with Qi

Qi

x j  4 Dt ln
 C Dt
 bg



