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Thermal doping review example

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Cross section cut view that is not to scale

Silicon wafer

Oxide film

Dopant containing film.

Pattern a wafer and place an oxide film on top of the exposed silicon.

Thermal Doping Example

This section was protected by the mask

Dopant will diffuse into the unprotected silicon as function of time and temperature in the furnace

Spin-on Dopant film

sol-gel film

Sources of dopants


Solid Source

Cross section view of oven rack to hold wafers and solid dopant

Wafer side that will house the functioning device.

Solid wafer made of the dopant material

Side of wafer that will have the functional device

Wafer side that will house the functioning device.

sol-gel film

for educational use only. From, R.C. Jaeger, Introduction to Microelectronic Fabrication, 2nd Ed., Prentice Hall, 2002

Sources of dopants


Solid as vapor source

Solid dopant placed in platinum boat

Liquid as vapor source

Sources of dopants



Pure vapor source

Sources of dopants





Silicon wafer

Oxide film


Pattern a wafer and place an oxide film on top of the exposed silicon.

Place a dopant containing film on the wafer and heat for some time.

Silicon wafer


Oxide film


Region of interest


Silicon wafer


Oxide film


HEAT


DEGLAZE

thenCLEAN


HEAT


Mask t

hic

kn

ess (

mic

ron

s)

Diffusion time (hours)


Oxide film needed to be thick enough to mask diffusion process

for educational use only. Fig 3.7 p 53, R.C. Jaeger, Introduction to

Microelectronic Fabrication, 2nd Ed., Prentice Hall, 2002

If your furnace is at 1100 degrees C, it will be at least 3.5 hrs before the boron gets through the1 micron thick oxide protective cover.

1

Practical factors How thick does the protective oxide have to be?


Thermal Doping ExamplePractical factors How much dopant will dissolve in the silicon?

The real issue is how many dopant atoms will replace silicon atom.

Imp

uri

ty c

on

cen

trati

on

(a

tom

s/c

m3)

You can dissolve more P and As atoms into crystal than can substitute for silicon atoms.

N0

At 900 C and maximum Boron concentration (solubility)at the surface is about 1.1 x 10

20 Boron atoms/ cm3

1.1 x 10

20 Boron atoms/ cm3

=

Therefore,

Thermal Doping ExamplePractical factorsHow much does temperature influence the dopant transport

into the silicon?


B and P

As

D(T) = D0 e-

EA

k T[ ]

These plots can be modeled as exponential functions

86.2 x10-6

ev/ K o

Atom D0 EA

B 10.5 3.69 ev

Al 8.0 3.47 evGa 3.6 3.51 ev

P 10.5 3.69 ev

As 0.32 3.56 ev

D(1173) = D0 e-

EA

k (1173)[ ]

From the model, what is the diffusion coefficient for P at 900 C?

(900 C equals 1173 K)

D(1173) = P

10.5 e-

3.69

86.2 x10-6

(1173)[ ]

D(1173) = P

?Come on! Work it out, its good for you.

1.48 x10-15cm /sec

2

10.5 3.69 ev

00( , ) 2 /Q N x t dx N Dt

102 ( / )j Bx Dterfc N N

0( , ) ( / 2 )N x t N erfc x Dt

What are the model equations for the diffusion of dopant from an infinite source?

Concentration profile through the diffusion region as a function of distance and time.

Thermal Doping ExamplePractical factors

X =0 at the outside edge of the wafer

t =0 before the diffusion starts.

Total dopant that was added to substrate.

Distance into wafer were the concentration of the n and p materials is identical.

Junction depth

2( , ) ( / )exp ( / 2 )N x t Q Dt x Dt

02 ln( / )j Bx Dt N N

What are the model equations for the diffusion of dopant from a constant or fixed source?


Concentration profile through the diffusion region as a function of distance and time.

X =0 at the outside edge of the wafer

t =0 before the diffusion starts.

Concentration at the surface as a function of time? 2( , ) ( / )exp ( / 2 )N x t Q Dt x Dt Put X =0 and solve

for all values of time.

Distance into wafer were the concentration of the n and p materials is identical.

Junction depth

X /4(Dt)2

Fu

ncti

on

valu

es


The error function and its complement are popular functions because they are solutions to differential equations that deal with diffusion problems.

Values for the function are available from tables or plots like this one,

What is erfc and how do I use it.?

Practical factors

erfc( )

-xe2

or approximation functions like this one also found in common mathematics software packages.

[ ]1/2



The gaussian curve on the right is also often used as a substitute for the erf complement. For most of the model curves shown the plots have similar shape and functional response.


Thermal Doping ExamplePractical Problem

You have a n-type silicon wafer that has a resistivity of 0.36 ohm-cm. You want to use boron to form the base region in the wafer for an npn transistor.

You perform a solid-solubility limited boron “predeposition” at 900 C for 15 minutes followed by (after deglaze and clean) a 5 hour “drive-in” at 1100C.

Find the boron surface concentration , the junction potential and the dose.

(I) just after the “predeposition” step.

(II) just after the “drive in” step.


Find the boron surface concentration, the junction potential, and dose.

Get N from the solubility graph for Boron at 900 C.01)

2) Find the value for diffusion coefficient at 900 C.

900 C 1173 K

D(1173) = BD0 e- [ ]

EA

k (1173)

,BFor Boron, B, the model becomes

3) Find the number of boron atoms, N ( x, t ) when x = 0 and t = 15 minutes (900 seconds).

N ( x , t ) = N0

erfc1/2

[ ]x2

D T

t4

,B


(a) Boron surface concentration just after the “predeposition” step.

Practical Problem

Determine the number of Boron atoms that correspond to the same resisitivity. (dopant concentration vs resistivity plot)

1)


(b) Boron junction depth in the original resistivity of 0.36 ohm-cm n doped wafer.

2) Use the concentration profile model as a function of distance and time and solve for the junction depth distance.

(II) just after the “drive-in” step.

1)

Integrate the area under the concentration profile model for the pre-deposition or the “drive-in” process.

1)

(c) Boron dose for this process.


Use the concentration profile model as a function of distance and time and solve when x = 0.

(a) Boron surface concentration just after the “drive-in” step.

N ( x , t ) =[ ]

x2

D T

t4e-Q2

D tT [ ]

1/2



1)




N ( x , t ) =[ ]

x2

D T

t4e-Q2

D tT [ ]

1/2

1) Solve concentration profile model as a function of distance and time for junction depth.


(b) Boron junction depth, just after “drive in” step, in the original resistivity of 0.36 ohm-cm n doped wafer.



1)




N ( x , t ) =[ ]

x2

D T

t4e-Q2

D tT [ ]

1/2

1) Solve concentration profile model as a function of distance and time for junction depth.

(b) Boron junction depth, just after “drive in” step, in the original resistivity of 0.36 ohm-cm n doped wafer.


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