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The general instruction when using a silicon detector is to either apply voltage only in atmospheric pressure or in high vacuum. Not in between!

I can't find a physical answer to it. Why is it so important?

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  • $\begingroup$ [Citation needed] I've never heard this, and I've been in the detector business for a zillion years. $\endgroup$ – Carl Witthoft Feb 18 '14 at 12:30
  • $\begingroup$ I also haven't heard of this, but I've never really worked in between atmosphere and high vacuum. $\endgroup$ – Chris Mueller Feb 18 '14 at 13:05
  • $\begingroup$ @CarlWitthoft : Thank you very much for your comment! I like your approach, because I can't find a reason on why the pressure should be bellow a certain limit. $\endgroup$ – Thanos Feb 18 '14 at 14:41
  • $\begingroup$ @ChrisMueller : Thank's for your comment! I want to use it on $10^{-1}$ Torr. It's not atmosphere and definitely it's not high vacuum! $\endgroup$ – Thanos Feb 18 '14 at 15:27
  • $\begingroup$ I on the other hand have come across this phenomenon, in an undergrad lab in fact, where we had to turn off the detector only while pumping down to vacuum. Failure to do this would result in some unhappy equipment and an unhappy grade. $\endgroup$ – user10851 Feb 18 '14 at 16:40
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I just realized what the problem is. It actually doesn't have anything to do with the detector. When working in vacuum systems you have to worry about the dielectric breakdown of the air as the pressure is reduced. It turns out that the breakdown voltage hits a minimum around $\sim 1$ Torr depending on the species of the gas (see the curves below). This phenomenon is known as Paschen's Law; you can read more about in this Wikipedia article.

If you are applying high voltage across your Si detector when this breakdown occurs, you will surely ruin your detector. It looks like you will be OK at $0.1$ Torr, but I make no guarantees :).

Paschen Curves

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  • $\begingroup$ Well, ok, but who uses a Si detector which isn't in a properly passivated & insulated package? $\endgroup$ – Carl Witthoft Feb 18 '14 at 16:55
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    $\begingroup$ @CarlWitthoft We typically remove all packaging for our in vacuum Si photodiodes in order not to throw away any photons. $\endgroup$ – Chris Mueller Feb 18 '14 at 16:58
  • $\begingroup$ @ChrisMueller : Thank you very much for your answer! Very interesting answer!!! However, there is something I don't understand... My Si detector has only one cable, where the voltage is applied. Where will the spark occur? Between the detectors frame(i.e. ground) and the cable's central core? $\endgroup$ – Thanos Feb 19 '14 at 10:56
  • $\begingroup$ @Thanos The cable must have two leads even if one is the shield of the cable. The Si acts as a current source which drives current from one lead to the other. The breakdown would most likely happen between the place where these two leads connect to the photodiode, since that is where the air is acting as the dielectric. Typically the shield of the cable will connect to the diode case which is connected to the back of the Si while the central lead will have a very fine gold wire which connects it to the front of the diode. $\endgroup$ – Chris Mueller Feb 19 '14 at 12:20
  • $\begingroup$ Exactly! That's my question! There isn't air inside the detector or the cable, is there? The only air that exists is between the detector case(which is in 0 potential) and the vacuum chamber(which is also in 0 potential). So where would the dielectric breakdown occur? $\endgroup$ – Thanos Feb 19 '14 at 13:01

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