High resistivity silicon detector

Question

How is the resistivity of the silicon detector is related with the depletion of the detector bulk? In a thesis I read that to achieve low full depletion voltage, a high resitivity silicon detector is required. Why?

Answer 1

To create the depletion region, you pull charged carriers out of the depletion layer, leaving behind the (fixed) charged donor or acceptor ion sites. In high-resistivity material there is less doping so there are less of those ions (and also less free carriers before you pull them out) which means:

1. You need less voltage to pull the carriers out of a depletion layer of certain width.
2. You end up with less charge remaining in the depletion layer (p-charge remaining in the n-material in the cathode and n-charge remaining in the p-material of the anode, of course!)
3. Therefore the electric field in the depletion region also will be lower.
4. Conversely, when using a given voltage you can create a larger-width depletion layer in low-doped material than in high-doped material.

Since there is only a finite region you want to deplete before you call it "full depletion", it is 1) and not 4) that is relevant for your question.