I am a student of masters study with a background only in electrical engineering, but I have studied Miller indices and crystallographic planes and directions, personally, to help myself understand the process of silicon etching which is required in my current course of study. But, I am unable to understand the following written in my text book:
"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.
Answer
Certain types of wet etching agents have differential reactivity on different crystal surfaces. Consider that the Si-Si distance and coordination environment will be quite different.
Let's start with an isotropic etch - it just eats silicon. Well, if you started with a pinprick (i.e., just a point and not a gap in the mask) you'd etch out a hemisphere. Normally, we etch through gaps in the mask, so you'll get rounded sides.
With an anisotropic wet etch you can get flat, angular side-walls defined by crystal planes.
For instance, potassium hydroxide (KOH) displays an etch rate selectivity 400 times higher in <100> crystal directions than in <111> directions.
OK, so let's consider that etch through a gap in a mask into a <100> surface. As we go down 400 atomic layers of Si, we'll clearly have some etching along the side-walls (e.g., an "undercut"). But also consider that depending on the size of the feature for the etch, you'll go down into the <100> but not completely uniformly. Different wet anisotropic etches will have different selectivity rates between <100> and <111> and thus you get different aspect ratios.
Incidentally, dry etching (e.g., RIE, plasma) can give more vertical side-walls.
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