High-pressure oxidation In an oxidation system, increasing the gas pressure incr

Question

High-pressure oxidation

In an oxidation system, increasing the gas pressure increases the parabolic reaction rate (by increasing Ns). Of course, the furnace become more complicated because the tube must sealed up in order to pressurize it above normal atmospheric pressure.

In experiments, it’s been found that the parabolic rate is directly proportional to pressure – doubling the pressure doubles B.

Determine the time that would be required to grow 4.0 m of oxide using wet oxidation at 1150°C in a normal atmospheric pressure oxidation tube. The silicon wafer is initially bare and has a (100) orientation.

To speed this up, we might consider increasing the pressure. By what factor would the pressure have to be increased (i.e. by what factor would be B have to be increased) in order to grow the 4 m in just 5 hours?

t (atmospheric) = ___________

factor = _____________

Explanation / Answer

1)   the parabolic rate is directly proportional to pressure – doubling the pressure doubles B.

so( height of oxide layer/time)^2 is proportional tp P

so( h/t)^2 = kP (say)

h/t =   (kP) ^1/2

so t= h/(kP)^1/2  

2)    now t1   =   h1 (kP1)^1/2

         t2        = h2(kP2)^1/2

t1/t2 =   h1/h2 ( P1/P2)^1/2     ----- (1)

suppose 1 m oxide film is formed in 1 hour at atmospheric pressure

So here t1= 1 h , t2 = 5 hours

              h1 =    1 m

            h2 = 4*10^-

             P1= 1 atm

P2         =     ?

Put these values in (1) equation

1/5   = ( 1/10^ - 6 ) ( 1/P2)^1/2

1/5*10^ -6    = ( 1/P2) ^1/2

2 * 10^-7    =   1/P2) ^1/2

squaring      4*10^ - 14   = 1/P2

So P2 = 2.5 * 10^ 13

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