diffusion #2 ece/che 4752: microelectronics processing laboratory gary s. may february 5, 2004

Diffusion #2

ECE/ChE 4752: Microelectronics ECE/ChE 4752: Microelectronics Processing LaboratoryProcessing Laboratory

Gary S. May

February 5, 2004

Outline

ObjectivesObjectives Double DiffusionsDouble Diffusions Concentration-Dependent DiffusionConcentration-Dependent Diffusion Diffusion in SiliconDiffusion in Silicon Lateral DiffusionLateral Diffusion

Objectives

Discuss the concept of double diffusions, an important part of how we fabricate our CMOS transistors in the lab.

Introduce some “second-order” diffusion effects.

Outline


After p-well Diffusion

C(x)

x

Csub

pn

xj0

NA(x)

After NMOS Source/Drain n+ Diffusion

C(x)

Csub

n+

pn

ND(x)

NA(x)

xj1xj2

Notation: p-well Pre-dep

Boron doping Pre-Dep: tpp @ Tpp => Dpp, Cspp

tpp = p-well pre-dep timeTpp = p-well pre-dep temperatureDpp = p-well diffusion constant at pre-dep

temperatureCspp = surface concentration for p-well pre-

dep

Notation: p-well Drive-in

tpd @ Tpd => Dpd

tpd = p-well drive-in time

Tpd = p-well drive-in temperature

Dpd = p-well diffusion constant at drive-in temperature

Notation: n+ Source/Drain Pre-dep

Phosphorus doping Pre-Dep: tnp @ Tnp => Dnp, Csnp; Dp1

tnp = n+ source/drain pre-dep time Tnp = n+ source/drain pre-dep temperature Dnp = n+ source/drain diffusion constant at pre-

dep temperature Csnp = surface concentration for n+ source/drain

pre-dep Dp1 = boron diffusion constant at source/drain

pre-dep temperature

Notation: n+ Source/Drain Drive-in

tnd @ Tnd => Dnd; Dp2

tnd = n+ source/drain drive-in time

Tnd = n+ source/drain drive-in temperature

Dnd = n+ source/drain diffusion constant at drive-in temperature

Dp2 = boron diffusion constant at source/drain drive-in temperature

Profile: After p-well Diffusion

NA x Sw

Dt w

----------------------x–

2

4 Dt w------------------exp=

Sw2Cspp

-------------- Dpptpp=where:

Dt w Dpptpp Dpd tpd+=

= well dose

= well “Dt”

There is a pn-junction xj0 where NA(x) = Csub

Profile: After n+ Source/Drain Diffusion

ND x Ssd

Dt sd

-----------------------x–

2

4 Dt sd-------------------exp=

Ssd2Csnp

-------------- Dnp tnp=

Dt sd Dnptnp Dndtnd+=

where: = source/drain dose

= source/drain “Dt”

BUT: now the well profile has changed to…

New Well Profile

There is a pn-junction xj1 where ND(x) = NA(x)

There is a new pn-junction xj2 where NA(x) = Csub (where xj2 > xj0)

NA x Sw

Dt eff

------------------------x–

2

4 Dt eff--------------------exp=

Dt eff Dpptpp D pdtpd Dp1tnp Dp2 tnd+ + +=where:

= overall effective “Dt”

Example

Suppose we want to design a p-well CMOS diffusion process with a well depth of xj2 = 2.5 m. Assume the n-type substrate doping is 1015 cm-3.

Example (cont.) If we start with a boron pre-dep with a dose of 5 × 1013 cm-2,

followed by a 1-hr drive-in at 1100 oC, what is the initial junction depth (xj0)? Neglect the depth of the pre-dep. The B diffusivity at this temperature is 1.5 × 10-13 cm2/s.

SOLUTION:

NA x Sw

Dt w

----------------------x–

2

4 Dt w------------------exp= where: Sw = 5 × 1013 cm-2

(Dt)w = (1.5 × 10-13 cm2/s)(3600s) = 5.4 × 10-10 cm2

NA(xj0) = 1015 cm-3 =>x j0 4– Dt w

Dt w 1015

51310

--------------------------------------ln

12---

= = 1.24m

Example (cont.) Find the necessary (Dt)eff for the p-well to reach the desired

junction depth of xj2 = 2.5 m.

SOLUTIONSOLUTION (This must be solved by iteration!!!): (This must be solved by iteration!!!):

where: x = xj2 = 2.5 m NA(xj2) = 1015 cm-3

Sw = 5 × 1013 cm-2

NA x Sw

Dt eff

------------------------x–

2

4 Dt eff--------------------exp=

=> (Dt)eff = 2.46 × 10-9 cm-2

Example (cont.)

What is the approximate p-well drive-in time needed if all steps are carried out at 1100 oC?

SOLUTION:

tpdDt

ef f

Dpd---------------- 2.46 9–10

1.513–10

------------------------=

= 1.64 × 104 s = 273.3 min

Example (cont.)

If the n+ source/drain junction depth required is xj1 = 2.0 m, what is the p-well doping at the source/drain junction?

SOLUTION:NA x Sw

Dt eff

------------------------x–

2

4 Dt eff--------------------exp=

where: x = xj1 = 2.0 m(Dt)eff = 2.46 × 10-9 cm2

=> NA(xj1 = 2.0 m) = 9.76 × 1015 cm-3

Example (cont.) Suppose the source/drain dose (Ssd) is 5 × 1014 cm-2. What

is the surface concentration in the source/drain regions and the source/drain diffusion (Dt)sd?

SOLUTION: ND x Ssd

Dt sd

-----------------------x–

2

4 Dt sd-------------------exp=

where: x = xj1 = 2.0 m ND(x = xj1) = 9.76 × 1015 cm-3

(i) (Solving by iteration): (Dt)sd = 1.52 × 10-9 cm2

(ii) Surface Concentration:ND x 0=

Ssd

Dt sd

-----------------------5 1410

1.52 9–10 -------------------------------------= = = 7.24 × 1018 cm-3

Example (cont.) The phosphorus source/drain regions are deposited and driven in

at 1050 oC. At this temperature, the phosphorus diffusivity is 5.8 × 10-14 cm2/s. Ignoring the contributions of the pre-dep, what is the approximate source/drain diffusion time (tnd)?

SOLUTION:

tndDt

sd

Dnd--------------- 1.52 9–10

5.814–10

------------------------= = 2.62 × 104 s = 436.8 min

Example (cont.) If the boron diffusivity is 6.4 × 10-14 cm2/s at 1050 oC,

correct for the p-well diffusion time to account for the extra diffusion during the source/drain drive-in. (Neglect the contributions of pre-dep steps).

SOLUTIONSOLUTION::

Dp2tnd = 1.68 × 10-9 cm2

( “Dt” accumulated by boron during source/drain diffusion).

=> Initial p-well drive-in may be reduced by this amount, or:

tpd

Dt eff Dp2tnd–

Dpd---------------------------------------

2.469–10 1.68

9–10 –

1.513–10

-----------------------------------------------------------------== 86.9 min

Outline


Vacancies When host atom acquires sufficient energy to leave its

lattice site, a vacancy is created. Vacancy density of a given charge state (#

vacancies/unit volume, CV) has temperature dependence similar to carrier density:

where Ci = intrinsic vacancy density, EF = Fermi level, and Ei = intrinsic Fermi level

kT

EECC iF

iV exp

Vacancy-Dependent Diffusion

If diffusion is dominated by the vacancy mechanism, D is proportional to vacancy density.

At low doping concentrations (n < ni), EF = Ei, and CV = Ci (independent of doping), so

D (which is proportional to CV = Ci ), also independent of doping concentration.

At high concentrations (n > ni), [exp(EF – Ei)/kT] becomes large, which causes CV and D to increase.

Intrinsic and Extrinsic Diffusion

Effect on Diffusivity

Cs = surface concentration

Ds = diffusion coefficient at the surface

= parameter to describe concentration dependence

x

CDF

ss C

CDD

Diffusion Profiles

Junction Depth

For > 0, D decreases with concentration Increasingly steep box-like profiles result Therefore, highly abrupt junctions are formed Junction depth is virtually independent of background concentration

tDx sj 6.1

tDx sj 1.1

tDx sj 1.1

tDx sj 87.0

= 1

= 2

= 3

Outline


Concentration Dependence

Boron, arsenic: ≈ 1

Gold, platinum: ≈ -2

Phosphorus: ≈ 2 (sort of)

PhosphorusDiffusion

Phosphorus Diffusion (cont.) When surface concentration is low, diffusion

profile is an erfc (curve a).

As concentration increases, the profile begins to deviate (b and c).

At high concentration (d), profile near the surface is similar b, but at ne, kink occurs, followed by rapid diffusion in tail region.

Because of high diffusivity, phosphorus is used to form deep junctions, such as the n-tubs in CMOS.

Outline


The Problem 1-D diffusion equation is not adequate at

the edge of the mask window. There, impurities diffuse downward and

sideways (i.e., laterally). In this case, we must consider a 2-D

diffusion equation and use numerical techniques to get the diffusion profiles under different initial and boundary conditions.

Diffusion Contours

Contours of constant doping concentration for a constant Cs, assuming D is independent of concentration.

Interpretation

Variation at far right corresponds to erfc distribution.

Example: at C/Cs = 10–4, the vertical penetration is about 2.8 µm, whereas the lateral penetration is about 2.3 µm (i.e., the penetration along the diffusion mask-semiconductor interface).

Implications Because of lateral diffusion, the junction consists

of a central plane (or flat) region with approximately cylindrical edges with a radius of curvature rj.

If the mask has sharp corners, the shape of the junction near the corner will be roughly spherical.

Since the electric-field intensities are higher for cylindrical and spherical junctions, the avalanche breakdown voltages of such regions can be substantially lower than that of a plane junction.

diffusion #2 ece/che 4752: microelectronics processing laboratory gary s. may february 5, 2004

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