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- Path: sparky!uunet!dove!enh.nist.gov!rev
- From: rev@enh.nist.gov
- Newsgroups: sci.physics
- Subject: RE: photon 'detectors' - how reliable?
- Message-ID: <21JAN93.18141260@enh.nist.gov>
- Date: 21 Jan 93 23:14:12 GMT
- References: <wwadge.727584610@csr> <MERRITT.93Jan21103409@macro.bu.edu>
- Sender: news@dove.nist.gov
- Organization: NIST
- Lines: 64
-
- >In a previous article, merritt@macro.bu.edu (Sean Merritt) wrote:
- >>In article <wwadge.727584610@csr> wwadge@csr.UVic.CA (Bill Wadge) writes:
- >>> Is this possible? Can one really build a device so sensitive that it will
- >>> detect 100% reliably every photon that enters it, and never
- >>> go off by accident?
- >>>
- >>> Seems unlikely, but then I'm not a physicist ...
- >>
- >>What your asking about is the "quantum efficiency" of the detector.
- >>If this is a device that uses a method of converting photon
- >>flux into an electrical current(photo-electric device) if we
- >>ignore noise:
-
- [description of detector responsivity deleted]
-
- >>the parameters here are i and Popt. It is easy to see that you
- >>can't hope to get 100% qe, infact real devices seldom do better
- >>than 25%. I think the best photomultipliers are something less
- >>than this. If you consider quantum noise or S/N it just lowers
- >>the qe more. Although it is much more mathematically complicated.
- >>
-
- Depending on the photon energy this is not always the case. Detectors that
- have quantum efficiencies greater than 1 are available, but you need to work in
- the extreme ultraviolet or soft x-ray region. These are not photoemmitters,
- but semicondutor devices where the incident radiation creates electron-hole
- pairs in a diode. For example, silcon has an electron-hole pair energy of
- about 3.63eV (I think). If a photon with 7.26eV is absorbed, you can get two
- electron-hole pairs, and the quantum efficiency is 2, or 200% depending on how
- you want to put it.
-
- >>Usually the qe is specified interms of wavelength, and a function
- >>called "responsivity" R = R (lambda).
- >>
- >>
- >>qe(lambda) = [R(lambda)*h*c]/(e*lambda)
- >>
- >>the units of R are [area/power] power is in watts or lumens.
- >>somettimes itis given in [volume/power].
- >>
-
- Well, we just leave it as QE(lamda) in units of electron/photon.
-
- But returning to the original question, even though you can get more than one
- electron out of your detector for every photon going in, you don't actually,
- detect every photon and no others. As was mentioned there is noise, usually
- from thermal excitation of electrons. Also there are loss mechanism that come
- into play so not every photon gets detected. There are reflections off the
- front surface (and any other boundaries inside) of the detector, there are dead
- regions in the detector (SiO2 absorbs photons but does not result in any signal
- in a photodiode), and there is always the possibilty that the photon will just
- go through the detector without interacting with it at all.
-
- To my mind, the "perfect" detector can't be made, but it is possible to
- calibrate detectors by knowing their quantum efficiency and that allows you to
- do science as if you did see every photon coming your way, even when you might
- only "really" see one in every ten.
-
- +-----------------------------------------------------------------------------+
- + From the desk of: | rev@enh.nist.gov OR rev@nbsenh +
- + Rob Vest |---------------------------------------------+
- + National Institute of | Why are all the good quotes taken by +
- + Standards & Technology | some other .sig file?? +
- +-----------------------------------------------------------------------------+
-