# Hamamatsu PMT

1. Feb 18, 2015

### miss_physics

Hello every one,
I would like to find out the output signal voltage due to a single photon detected by H9500 PMT. I was able to calculate the charge collected on the anode due to a single photon but How do I get the voltage?
I would appreciate any help with that
Thank you

2. Feb 18, 2015

### dukwon

Typical operating voltage across a H9500 (and many other Hamamatsu flat panels) is 1.1 kV. The anodes are kept at ground unless you reverse-bias the device.

What happens to the signal after the anode depends entirely on your readout electronics.

Last edited: Feb 18, 2015
3. Feb 18, 2015

### jim hardy

in other words,

what becomes of that charge?
Presumably it goes into a charge-to-voltage converter of some sort.

4. Feb 18, 2015

### miss_physics

I meant how can I calculate the output voltage resulting?
I think if I was able to calculate the capacitance of the anode, I can find out the value of the voltage. But how?

5. Feb 18, 2015

### jim hardy

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
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.

6. Feb 18, 2015

### jim hardy

7. Feb 19, 2015

### miss_physics

Thank you so much for your detailed reply which was very helpful. I will definitely check the links you posted and will try get my way around.
Thank you again very much

8. Feb 19, 2015

### jim hardy

Thank you for the kind words and feedback. Makes an old guy feel still maybe just a teeny bit useful.

old jim

9. Feb 20, 2015

### Baluncore

The traditional way of converting the pulse of charge to a voltage pulse is with a series resistor. If that resistor is in the negative feedback loop of a fast op-amp, the voltage on the anode will remain at a stable “virtual earth” potential while the low impedance op-amp output provides the signal voltage.

The pulse of charge resulting from a single electron will usually have a gausian profile. For the sake of simplicity, approximate it with a rectangular profile of the same area, then if you know the charge and the duration of the pulse you can convert that charge to a current. I = q / t. Knowing the series resistance, R, get the output voltage by ohms law; V = I * R.