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Hi. I'm testing a PMT and I'm surprised by the following. With power

  1. Jan 31, 2012 #1
    Hi. I'm testing a PMT and I'm surprised by the following. With power supply (-900v on K) and with no charge between output electrode and ground, I measure -126 V same with scope (Zin = 1 M) or DVM (Zin = 10 M). I must precise that this experiment has be made in a room not completely dark. Can someone explain why such a high voltage ?
     
  2. jcsd
  3. Feb 1, 2012 #2
    Re: Photomultiplier

    Well, a photomultiplier, from my understanding, converts photons into current. When a photon strikes the first plate it releases an electron if the photon energy is greater than the work function of the plate. This electron is accelerated by an electric field to the next plate knocking off another electron, so now you have two electrons. These two electrons are accelerated and hit the next plate creating two more electrons. These four electrons hit the next plate and create 4 more electrons, and so on. You get multiplication by using a large number of plates. This generates a fairly large current from a very small number of photons. The photomultiplier is generally used with a current to voltage converter. Depending on the gain of this current to voltage converter your voltage could be very small or very large. You can estimate the number of photons striking the first dynode based on the gain of the current to voltage converter, the output voltage, and the number of dynodes in the PMT.

    I have never used a PMT though so I am not sure what voltages these things typically output. -126 volts seems very high. Maybe you're right...maybe the room isn't completely dark.

    -Matt Leright
     
    Last edited: Feb 2, 2012
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