SEMI-INFINITE BOUNDARY CONDITIONS .1 m odel : e rror f unctIons

dz

dy yz

z dz

dy y

= −

= − 2

1 2

which is easily integrated to give

ln

.

z y A

z Ae ^y

= − + ′

= ⁻

2

2

Moreover, integrate z one more time to get C(y) z dC

dy Ae C y A e dw B

y

y w

= =

( )

^{= +}

−

∫

− 2

2 0

where w is just a dummy variable of integration. Putting in y x= / 4Dt leads to the final important result

C x t( , )=A

∫

0^{x Dt}e^−w2dw B+ .

4

/ (11.6)

Therefore, the partial differential equation can be solved as an ordinary differential equation, Equation 11.6, with the dimensionless variable y x= / 4Dt, and it gives a specific solution for two integration constants A and B. It should be noted that the partial differential equation was solved simply by making a variable substitution—a similarity variable—to give an easily integrated ordi- nary differential equation: no advanced mathematics beyond calculus! Of course, the right choice of the variable y was critical.

11.3 SEMI-INFINITE BOUNDARY CONDITIONS

11.3 Semi-Infinite Boundary Conditions 387

Now the Gaussian error function of y, erf(y), is defined as

erf y( )= ²π

∫

^0ye^−w2dw (11.8) which is just a number for each value of y, given by 2/ π times the area under the Gaussian curve, e^−w2, from 0 to y, as shown in Figure 11.3. The error function was defined this way, so that the error function of infinity would equal one, that is, erf (∞) = 1, since

∫

0^∞e^−w2dw⁼ π/2. The complementary error function of y, erfc(y), is simply 1 − erf(y) and

erfc y erf y e dw e dw

erfc y e dw

w y w

w

( )

^{= −}

( )

^{= −}

( )

⁼

− −

∞

∞ −

∫

∫

1 2 2

2

2 2

2

0 0

0

π π

π

∫∫

^{+ 2}π

∫

⁰e^−w2dw

y

resulting in

erfc y e ^wdw

( )

^{= 2}π

∫

y^{∞ − 2} (11.9)

as shown in Figure 11.3. Therefore, the solution to the semi-infinite boundary condition problem of boron diffusing by solid-state diffusion into silicon, Equation 11.7, can be written as follows:

C x t C erfc x

S Dt

, .

( )

^{= }

 



4 (11.10)

This same equation applies to the diffusion of carbon from the surface of a piece of iron or steel to increase its hardness; ; or the diffusion of a gas such as hydrogen into a thick piece of polymer ; and the diffusion of ¹⁸O into aluminum oxide, Al2O3. The latter can be used to measure the diffusion coefficient of oxygen in aluminum oxide by analyzing the ¹⁸O content as a function of depth below the surface and comparing it to Equation 11.10. When discussing diffusion in solids, introductory books on materials science and engineering frequently simply introduce Equation 11.10 without any background—the black-box approach—and use it to calculate the concentration of carbon in a piece of steel as a function of time and distance. Hopefully, Equation 11.10 is no longer a black box!

−3 −2 −1 0 1 2 3

0.0 0.2 0.4 0.6 0.8 1.0

exp (−w2)

exp (−w²)

w

πerf (y) 2

πerfc (y) 2

y

FIGURE 11.3 Plot showing relation of the error function of y, shaded area, and that of the complementary error function to the Gaussian curve.

11.3.2 d

^IffusIonof

b

^oronInto

s

^IlIcon

The diffusion of dopants into silicon at low concentrations, in the order of parts per million, is used to form electronic devices such as diodes and transistors on the surface of single crystal silicon to make integrated circuits. For example, boron could be diffused into the surface of silicon that was doped n-type^* with, say phosphorus, to form a p-n junction diode somewhere below the surface.

Suppose that the silicon were doped to be an n-type semiconductor with 10¹⁷ cm⁻³ P atoms. For B diffusion in Si, D0 = 0.76 cm²/s and Q = 3.46 eV = 333.8 kJ/mol (Mayer and Lau 1990, 210). Take T = 1500 K (1327°C), then D B

( )

^{=0 76.} exp( 333,800/(8.314 1500))⁻ × ^{=2 37 10. × −12}cm s^{2 −1}. At 1500 K, the equilibrium constant for the reaction B H g2 6

( )

2B g

( )

⁺3H g2

( )

is Kp=5 47 10. × ⁻¹⁴. (Roine 2002) Therefore,

p B p H p B H Kp

2 3

2 2 6

1500 5 47 1014

( ) ( )

( )

⁼

( )

^{= . × − .} (11.11)

For silicon, ρ = 2.33 g/cm³ and M = 28.09 g/mol (Haynes 2013), so η(Si) = the number of Si atoms per cm³ = (2.33)(6.022 × 10²³)/(28.09) = 5.0 × 10²² cm⁻³. If Cs(B) = 10¹⁹ cm⁻³, then the mole fraction of boron, X(B) needs to be X(B) = 10¹⁹/5 × 10²² = 2.0 × 10⁻⁴. Because the mole fraction is small, it can be taken to be equal to the activity—partial pressure of B(g)—in the diborane–hydrogen atmo- sphere. To achieve a pressure of p(B) = 2.0 × 10⁻⁴, assume a carrier gas of hydrogen having a fixed p(H2) = 10⁻² atm The pressure of p(B2H6) is then required to be

p B H p H p B K

p B H

p

2 6 2

3 2

2 3 4 2

14

2 6

10 10 5 47 10 0 18

( )

⁼

( ) ( )

=

( ) ( )

×

( )

⁼

− −

. −

. aatm which is experimentally feasible.

Appendix gives some properties of the error function, and there are many places where tables of the error function exist (Abramowitz and Stegun 1965, 310). In addition, there are many software packages and graphing calculators that have both the error function and the complementary error function as built-in functions. So a spreadsheet was used to calculate the concentration of boron as a function of time and distance, C(x, t)/CS for various times at 1500 K, with the results plotted in Figure 11.4. As can be seen from the figure, if the p-n junction were to be established about 1 μm below the surface, then a total time just short of 1000 seconds would be required for the diffusion.

Of course, the concentration could have been specified at x = 1 μm and the exact time required could have been calculated from the same program. For example, C/CS = 0.1, then by iteration in a spreadsheet of erfc(y) −0.1 = 0, gives y = 1.165 so

y x

Dt t D

x

= = = y

 

 = ×

 ×

 

− 

−

1 165 4

1 4

1 4 2 37 10

1 10 1 165

2

12

4

. ,

( . ) .  =

2

777 seconds

* Recall from Chapter 9 that an n-type dopant such as phosphorous has one more electron than silicon, and when it replaces a silicon atom in solid solution, it donates its extra electron to produce n-type electrical conductivity—conduction by a negative charge. In contrast, a p-type dopant such as boron has one less outer electron than silicon and accepts an electron producing a net positive charge to give p-type conduction. A p-n junction acts as a diode allowing current to flow in only one direction: one of the many devices critical in integrated circuits.

11.4 Infinite Boundary Conditions 389

the exact time necessary for the concentration at x = 1 μm to be C/CS = 0.1. Also note that x/ 4Dt ≅1 for this value of 0.1 CS to be reached. This suggests that an approximate—back of the envelope—value for the depth of diffusion in one direction is x/ 4Dt ≅1, which was the main point of Section 11.2.2.

11.4 INFINITE BOUNDARY CONDITIONS