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Photomultiplier Tube - Gain and Secondary Emission

  1. Sep 26, 2011 #1
    I'm studying about photomultipliers right now, but I'm having troubles understanding the concept of gain and secondary emission ratios...

    1) Gain : If I were to take a measurement of the output signal produced by photoeletrons and relate it the the high voltage of the PMT, I'd expect a linear relationship since the higher the PMT voltage is the faster it will accelerate the electrons which will cause the cascade at the dynodes to produce even more electrons thus resulting in larger output signals. Would that be correct? Gain = C(HV)^a where C=const. and a=# of dynodes

    2) Regarding the secondary emission ratio: defined as the number of secondary electrons produced per one primary electron. So from my understanding this ratio must be also related to the # of dynodes, as well as to gain. Which means that G = r^a where r=emission ratio and a=# of dynodes?

    I also noticed that there is also a capacitor value for each PMT, does that relate to the above somehow?

    Thanks for your time and help!
  2. jcsd
  3. Jun 6, 2012 #2
    1) Yes, the fact that the anode current output would be larger with a higher gain would be correct. In fact, you can directly calculate the peak of the anode current output by using the relation Ipeak=G*N*e/t
    G=gain of PMT
    N= number of photoelectrons
    e= electronic charge
    t=width of pulse
    However, this will be true only in the linearity range of the PMT. As you go on increasing the gain of the PMT, the anode current will saturate and reach the value of the bleeder current/ voltage divider current.

    2)Yes, the relation is correct.

    3) Capacitor value for each PMT, or capacitance value for each resistor in the voltage divider network?
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