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TimeDomain CVD, Inc. |
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At the Sheath Edge: Ions Falling InWe've mentioned several times that ions from the plasma bombard surfaces. How many ions are there? To answer this question we need to look in more detail at the sheath region. As the ions enter the sheath, they are accelerated by the field and their density must decrease to maintain a constant flux. If this decrease is too fast, the ion density will fall off before the electrons (which diffuse a bit into the region of positive potential) vanish, leading to a net negative charge, which would then increase the ion velocity. The stable solution must occur when the ions entering the sheath have enough initial velocity to ensure that the sheath remains positive. Let's examine this condition quantitatively. We'll first examine a collisionless sheath: that is, we assume that the ions can fall through the sheath region without colliding with a neutral molecule. Notation is defined below:
What does this mean for plasmas of interest? Let's look at some numbers: The ion current in deposition plasmas will vary from a few microamps to a few milliamps per square centimeter. If we multiply this by a typical ion energy of 100 volts or so, we find that a lot of power is dissipated as these ions strike the walls: e.g. in a high density (1E11) plasma the power delivered to the substrate surface by ion bombardment is 0.5 W/cm2, or 150 Watts to a 200 mm wafer. In the case of a collisional sheath (more appropriate for many CVD processes at > 1 Torr), the sheath velocity is reduced relative to the Bohm velocity by the ratio of the ion mean free path to the Debye length. The Debye length is the characteristic length over which the plasma electrons screen an applied potential; numerically it is approximately where the length is in cm, the electron temperature in volts (i.e. actually kT/q), and the density in number/cm3. For a 4 eV plasma with density of 1E10/cm3, the debye length is 0.014 cm or 140 microns. The mean free path of ions is similar to the neutral mean free path.
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