I have a question about the dead layer properties of HPGe gamma semiconductor detectors. I found this on Wikipedia

As of 2012 HPGe detectors commonly use lithium diffusion to make an n+ ohmic contact, and boron implantation to make a p+ contact. Coaxial detectors with a central n+ contact are referred to as n-type detectors, while p-type detectors have a p+ central contact. The thickness of these contacts represents a dead layer around the surface of the crystal within which energy depositions do not result in detector signals. The central contact in these detectors is opposite to the surface contact, making the dead layer in n-type detectors smaller than the dead layer in p-type detectors. Typical dead layer thicknesses are several hundred micrometers for an Li diffusion layer, and a few tenths of a micrometer for a B implantation layer. https://en.wikipedia.org/wiki/Semiconductor_detector

I have some problems understanding this. First of all I can't imagine how a lattice of Ge with Li doping would look like, since Li has only 1 valance electron (Does it mean that 3 holes are created?). If this is the case, why exactly Li is chosen? Moreover, I don't understand why one of the layers is thicker. My guess is that this has something to do with the lithium diffusion process, but that is just a guess.

I would appreciate your help to understand the properties of those layers.

  • 1
    $\begingroup$ First, almost all elements have some sort of level in the gap for Ge and Si - don't fixate on the 'explanation' used for dopants. Lithium has a beautiful shallow donor level in both Si and Ge, shallower even than the typical dopants. It is used for thick diffusions because the diffusivity is strongly enhanced by electric fields, so you can easily get the hundreds of microns of diffusion (with field) that would take forever with typical dopants. $\endgroup$
    – Jon Custer
    Oct 12, 2016 at 16:18
  • $\begingroup$ Thank you for your answer. I don't understand is why Li is used for the n region? Germanium has 4 valance electrons and Li only 1. It should create holes?! $\endgroup$ Oct 12, 2016 at 16:50
  • 2
    $\begingroup$ You are too focused on valence electrons vs the general concept of atomic levels fitting in to band structure. Just look in any decent semiconductor book - it will have a large figure with the energy levels in the gap for many elements with many different atomic electron configurations. $\endgroup$
    – Jon Custer
    Oct 12, 2016 at 17:35
  • 1
    $\begingroup$ I think that I understand what you mean (I hope). One of the Li levels is 33mV away from the conduction band, closer than P which is 45mV. This makes it an electron donor. B on the other hand is 45mV from the valence band, which makes it acceptor. However, why that thick level of lithium (not used for $\gamma$ detection) is needed. Isn't it possible to use P or As? $\endgroup$ Oct 12, 2016 at 20:24
  • 1
    $\begingroup$ Yes, at room temperature the thermally-induced electron-hole pair creation will lead to a large current that you don't want. So, you cool the detector. $\endgroup$
    – Jon Custer
    Oct 17, 2016 at 17:42


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.