I'm no photomultiplier expert, first off.
But i did grow up on vacuum tubes.
From Hamamatsu's tutorial
i think it is a mistake to allow any(significant) voltage to accrue on the anode. To do so would make it repel electrons.
http://www.hamamatsu.com/resources/pdf/etd/PMT_handbook_v3aE-Chapter2.pdf page 6 of 8
2.4 Anode
The anode of a photomultiplier tube is an electrode that collects secondary electrons multiplied in the
cascade process through multi-stage dynodes and outputs the electron current to an external circuit.
Anodes are carefully designed to have a structure optimized for the electron trajectories discussed previ-
ously. Generally, an anode is fabricated in the form of a rod, plate or mesh electrode. One of the most impor-
tant factors in designing an anode is that an adequate potential difference can be established between the last
dynode and the anode in order to prevent space charge effects and obtain a large output current.
So the idea would be toimmediately pull the charge off the anode and turn it into a voltage. I think such a circuit is called a "transimpedance amplifier".
I'd call it instead a "charge integrator". or "accumulator".
Anyhow, point being you keep anode voltage low, ideally zero, so as to avoid space charge effects.
A picture is always worth a thousand words.
Take a look here,
http://pessina.mib.infn.it/Biblio/Conferenze/NSS09 PMT cross talk study Orlando 09.pdf
pages 3 and 4 of 11 for a picture of the approach.
Page 4 shows a simplified schematic : charge cannot flow into the opamp, so the opamp must pull any incoming charge on around into the capacitor.
So your V=Q/C isn't C of the anode, it's C of your integrator where you accumulate the charge.
Blow up the circuit in yellow block, page 3
and observe the "preamp" has no input resistor. It yanks charge off the anode and into the capacitor C connected around opamp..
Upon arrival of a bucketfull of charge at anode, opamp will pull that charge into C by producing voltage Q/C. Voltage will decay to zero with time constant RC.
When enough charge is arriving at anode to make DC current instead of pulses your voltage will be I X R, R being that resistor in parallel with preamp's C.
That's what Duk meant by "dependent on readout..."
Maybe we'll get lucky and somebody with genuine photomultiplier and spectroscopy experience will chime in.
I'm just an old maintenance hand with only very basic and unrefined knowledge.
Only photomultipliers i ever saw up close were in Bell&Howell "Fimosound Specialist" 16mm movie projectors from 1950's.
But I hope this ramble helps you figure out your gizmo.
Good luck to you
old jim
PS: to anyone reading -- correct me if I've erred please, for i like learn from my mistakes.
