Timeline for Why n+ contact of p type Ge is 700um thick compared to p+ contact of n type Ge only 0.3um?
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| Mar 18, 2019 at 18:41 | answer | added | ggs | timeline score: 0 | |
| Feb 12, 2019 at 14:14 | comment | added | Jon Custer | Start with physicstoday.scitation.org/doi/10.1063/1.3037792, then go look at Ortec and Canberra product literature for their detectors. | |
| Feb 12, 2019 at 4:03 | comment | added | ggs | Sir pl respond. | |
| Feb 9, 2019 at 11:22 | comment | added | ggs | Dear Sir, the following questions remains....1. Why n+ contact in n type Ge is not several hundred microns? More depletion layer shall come as per your answer. 2. Why we require both additional p or n type contact? in p type Ge, why we require additional p type junction also? crystal itself should work as p junction. Only n type contact should do the job. 3. Why Be Endcap is not used in p type coaxial Ge detector? We shall get high efficiency at both low and high energy. Pl dont get angry. | |
| Feb 9, 2019 at 11:16 | comment | added | ggs | Dear Sir, the following questions remains.... | |
| Feb 5, 2019 at 17:43 | comment | added | Jon Custer | You seem to believe that I exist to answer your question - that is pretty annoying. Look, I've given you plenty of information above about just how gamma detectors work, why you want a thick, lightly doped region to get maximum performance, and why lithium ends up being the dopant of choice. How much clearer do I need to be? | |
| Feb 5, 2019 at 14:59 | comment | added | ggs | Dear Jon Custer pl respond.... | |
| Feb 5, 2019 at 8:06 | comment | added | ggs | It is not clear why 700um Li contact is required? Is it mandetory requirement or due to technology used for doping? | |
| Feb 4, 2019 at 19:54 | comment | added | Jon Custer | Lithium is used in silicon and germanium detectors since its diffusion can be driven by biasing the device. There is no way (within reason) that one could diffuse a more normal dopant (B, P, As) over 100's of microns. But with lithium one can get a nice, even doping profile at fairly low levels across millimeters to centimeters. Now, add a highly doped, ion-implanted p++ contact, and you have a great volumetric sensor. Remember, the goal is to make as wide a depletion layer as you can to increase the active volume. | |
| Feb 4, 2019 at 16:00 | comment | added | ggs | So do you mean to say since Li is diffused, the thickness is more? And since implanted, p type thickness is less? Then Q remains, why Li is also not implanted? | |
| Feb 4, 2019 at 15:45 | comment | added | Jon Custer | If the p contact is doped much higher than the n side, then the depletion layer will grow much further into the n region. So, if the p is, say, $10^{18}$ and the n is $10^{15}$, then the n side of the depletion layer is 1000 times wider than the p side. | |
| Feb 4, 2019 at 15:31 | comment | added | ggs | Why same does not happen for p+ contact of n type Ge? | |
| Feb 4, 2019 at 14:34 | comment | added | Jon Custer | Because the n-type region is diffused lithium, and the p-type region is implanted boron. Under reverse bias the depletion region grows dramatically into the n-type region, resulting in a large detection volume. | |
| Feb 4, 2019 at 14:16 | history | asked | ggs | CC BY-SA 4.0 |