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## Diffusion #1

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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
- Decomposes and reacts with the surface
- Dopant atoms deposited, dissolve in Si, begin to diffuse

Outline

- Introduction
- Apparatus & Chemistry
- Fick’s Law
- Profiles
- Characterization

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

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

- 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 high-concentration region toward a lower-concentration region.

Conservation of Mass

- 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

- When the concentration of dopant atoms is low, diffusion coefficient can be considered to be independent of doping concentration.

Temperature Effect

- 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 one initial condition (in time)
- Need 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

- Solution:

Key Parameters

- Complementary error function:
- Cs = surface concentration (solid solubility)

Total Dopant

- Total dopant per unit area:
- 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

Example (cont.)

When C = 1015 cm–3, xj is given by

Constant Total Dopant

- “Limited source” diffusion
- Initial condition: C(x,0) = 0
- Boundary conditions:

C(∞, t) = 0

- Solution:

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:

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).

Drive-In

- Drive-in = limited source
- After subsequent heat cycles:

Multiple Heat Cycles

where: (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:
- In R0 is much larger than a and b, then:

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

where: s = conductivity (W-1-cm-1)

- r = resistivity (W-cm)
- mn = electron mobility (cm2/V-s)
- mp = hole mobility (cm2/V-s)
- q = electron charge (coul)
- n = electron concentration (cm-3)
- p = hole concentration (cm-3)

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 (m), and impurity distribution C(x)
- For a given diffusion profile, the average resistivity ( = Rsxj) is uniquely related to Cs and for an assumed diffusion profile.
- Irvin curves relating Cs and have been calculated for simple diffusion profiles.

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