the silicon substrate and adding to it—part 1 explain how single crystalline si wafers are made ...
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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®
Bulk micromachining Surface micromachining
silicon substrate
silicon substrate
Three forms of material
Crystalline Polycrystalline Amorphous
Grains
Silicon wafers
Polysilicon (in surface μ-
machining)
Glass and fused quartz, polyimide,
photoresist
Creating silicon wafers
The Czochralski method
• 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
• Wafers cut from the cross section.
Creating silicon wafers
Photo (foto) of a monocrystalline silicon ingot
Polycrystalline silicon(American Ceramics Society)
Grains
It’s a crystal
Cubic Body-centered
cubic (BCC)
Face-centered cubic (FCC)
a
Unit cellsa - lattice constant, length of a side of a
unit cell
a
a
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.
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.
a b c
1
2
2
1
1
2
3
4
Respuesta: (2 4 1)
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 non-cubic 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, andb. {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) wafer
<1 0 0> direction
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
A
L
A
LR
By doping, the resistivity of silicon can be varied
over a range of about 1×10-4 to 1×108 Ω•m!
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
100 μm
1 μm
1 μm
Entire resistor is 0.5 μm thick
100 bends total
Respuesta: • Al: 509 Ω• Si: 1.3 GΩ !!
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:
dx
dCDj
dx
dTkq
dx
dV
Mass “flux”
Concentration gradient
Diffusion constant
Compare to
Tk
E
b
a
eDD
0 Frequency factor and activation energy for diffusion of dopants in silicon
Doping by diffusion
Conservation of mass (applied to any point in the wafer)
C
x
time
2
2),(
x
CD
x
j
t
txC
dx
dCDj
Need
• 1 initial condition• 2 boundary
conditions
At t = 0, or C(x, t = 0) = 0
C(x → ∞, t > 0) = 0
C(x = 0, t > 0) = Cs
Dt
xerfcCtxC s
2),(Solución
erfc( ) is the complementary error function:
deerfcx
22
)( Appendix C
Doping by diffusion
Diffusion of boron in silicon at 1050°C for various times
x
Dtxdiff 4
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.
)(tQ ss CDt
dxDt
xerfcC
2
20
x
time
C(x = 0, t > 0) = Cs
Dt
x
Dt
QtxC
4exp),(
2
Q = constant
Cs
Gaussian distribution
Doping by implantation
Distribution is also Gaussian, but it is more complicated.
Doping by ion implantation
2
2
2
)(exp)(
P
PP R
RxCxC
• CP – peak concentration of dopant
• RP – the projected range (the depth of peak concentration of dopant in wafer)
• 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.
P
iP
R
QC
2Peak concentration
Doping by implantation
Doping is often (de hecho, usually) a two-step process:
1st implantation – pre-deposition2nd 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
Dt
x
Dt
QtxC
4exp),(
2
Replace with Qi
Junction depth
x
C
implanted dopant
background concentration
p-n junction
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?
Dt
x
Dt
QtxC
4exp),(
2
Replace with Qi
Set = Cbg
DtC
QDtx
bg
ij
ln4
Tk
E
b
a
eDD
0a. b. = 1.248×10-18 m
2/s
= 0.83×10-6 m