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I am a student doing my Master's Degree having a background in Electrical Engineering. I have studied on my own millers indices, crystallographic planes and directions so to be able to understand the process of silicon etching, which is required in my studies. The problem is that I am unable to understand the following text that is written in my textbook (I quote from it):

"When Silicon is etched anisotropically, i.e. depending on crystal direction, and a {1 0 0} surface direction is used, cavities will result which have the same {1 0 0} orientation at the bottom and are determined laterally by {1 1 1} surfaces. The reason is that {1 1 1} surfaces are etched off using a minimal etch rate and this way becomes the most resistant"

Could someone please help me understand this concept?

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2 Answers 2

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You may want to check the chemistry stack, but as far as I can remember, it has to do with the number of atoms in the plane. Using a 111, you have more atoms per plane, therefore the remaining channel has more atoms to remove in the same amount of time, which takes longer, but gives a stronger base for deposited oxides and metals to bond to on an atomic level.

Edit: Here is a paper that may help you "see" the differences: http://users.encs.concordia.ca/~mojtaba/Chapter1.pdf

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  • \$\begingroup\$ Ok, I'll take your advice and post this question in the chemistry stack exchange forum and see what kind of responses I end up with. \$\endgroup\$
    – Ghosal_C
    Commented Nov 17, 2014 at 0:43
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{1 1 1} planes in this case can be considered to be non-etching planes. As you know the {1 1 1} planes are sidewalls in the resulting structures with the bottom being (100) parallel to the top surface. Only if a { 1 1 1} has an exposed edge will an etch proceed along that plane. I imagine it like a ex-foliation (de-layering). If you etch long enough you will get an inverted pyramidal pit with a sharp ("zero" radius) at the bottom.

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