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