The concept of mean free path can help you to understand semiconductor process t

Question

The concept of mean free path can help you to understand semiconductor process theory better. For example in ion implant, ions are accelerated to high energy to shoot at wafers. The ions impinged into the wafer surface become dopants in silicon. A) can you imagine what will happen to the ions if they collide with gas molecules not pumped out in the implant chamber before they get to the wafer? Is it desirable for ion implant? (Part A not graded) B) if you do not want this collision to happen, you need to pump down the chamber until on average, an ion will not collide with a single gas molecule in the entire length of its travel until it hits the wafer. The entire length of its travel is say 10m.

Question?

Can you estimate the level of pressure the chamber has to be under before on average the collision will not occur?

Explanation / Answer

Solution:

Part A) When the ions are accelerated they have necessary kinetic energy to impinge into silicon wafer. But upon collision with gas molecules, ion loses its much of the energy and direction due to which the doping process does not take place.

Part B) The ions motion is directed towards the silicon wafer. Normally atmospheric air in the chamber contain mostly nitrogen N2 and oxygen O2 gas which has comparable or greater size than the dopants ions like phosphorous (n-type dopant) or boron (p-type dopant).

The mean free path of the gas molecules affect path length of the dopant ions (10 m): if the mean free path of gas molecules is less then there are more chances of gas molecules colliding with each other and the dopant ion and if the mean free path large then then there is less chances of collision with themselves and the dopant ions. Thus if the mean free path of the gas is set to 10 m then an ion will not collide with gas molecule until it hits the silicon wafer .

Now the pressure exerted by the ideal gas is given by

P = nMvrms2/3V      --------------------------------------------------------(1)

M = molar mass of gas; vrms = root mean square speed of the gas molecules

V = volume of the gas and n = number of moles of the gas

NA = Avogadro’s number = 6.02*1023 mol-1

R = gas constant = 8.31 J/mol.K

d = diameter of gas molecule ~ 3Ao = 3*10-10m

But we have

vrms = (3RT/M)    --------------------------------------------------------(2)

thus pressure P becomes

P = nM(3RT/M)/3V

P = nRT/V

V = nRT/P

We also have the expression for mean free path of the gas molecules

= 1/[2**d2*(N/V)]

= V/[2**d2*N]

but V = nRT/P nad N = number of molecules = n*NA                    

= (nRT/P)/[2**d2* n*NA]

P = (RT)/[2**d2*NA*]

Now plugging the values R, T = room temperature = 300 K and d = diameter of gas molecule and NA and mean free path = 10 m

P = (8.31 J/mol.K)*(300K)/[2*3.141*(3*10-10m)2*(6.02*1023 mol-1)*10m]

P = 3.25*10-3 N/m2

P = 3.25*10-3 Pascals

or

P = 3.25*10-3 Pa*(1atm/1.05*105Pa)

P = 3.10*10-8 atm

P = 3.25*10-3 Pa*(760torr/1.05*105Pa)

P = 2.36*10-6 torr

Thus the pressure of the chamber should be less than 3.25*10-3 Pa or 3.10*10-8 atm or 2.36*10-6 torr.

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