Assume that there is a pulse on a scope with \$100\;mV\$ height, as shown in picture
This pulse is coming from a photomultiplier tube. How to determine the number of electrons from which this pulse is constructed?
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1\$\begingroup\$ We're going to need a LOT more information. In the first place, photomultiplers measure photons, not electrons. They do use electrons in the measurement process, but that's not actual thing being measured. \$\endgroup\$– Connor WolfMay 20, 2013 at 9:58
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\$\begingroup\$ @ConnorWolf: Thank you very much for your comment! What I need to know is the number of photoelectrons on the anode! That information is hidden in the pulse. If I am aware of that it is rather easy to "count" the number of photons in the photocathode. \$\endgroup\$– ThanosMay 20, 2013 at 10:00
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\$\begingroup\$ I believe @ConnorWolf is looking for information from the photomultipliers. With some sort of hint at the gain involved here no one can guess anything. What are they? Part numbers, data sheets etc... \$\endgroup\$– SpoonMay 20, 2013 at 11:50
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\$\begingroup\$ oops.. With OUT some sort of hint \$\endgroup\$– SpoonMay 20, 2013 at 12:05
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\$\begingroup\$ Is the photomultiplier connected to anything besides the oscilloscope (like a power source)? If it is, we'll need to see the schematic of those connections to give an answer. \$\endgroup\$– The PhotonMay 20, 2013 at 14:37
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2 Answers
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Assuming 1MOhm input resistance of the oscilloscope you can compute the current from the voltage and integrate it to get the charge. Divide by the charge of an electron and you get the number of electrons.
For the lower pulse that would give about -200nA * 15ns = -3e-15 C ~ 20000e.
Note that I'm not an expert w.r.t. photomultipliers, so I don't know whether this is of the right magnitude and whether there are some caveats of this measurement method.
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\$\begingroup\$ It's more likely a 50 Ohm system, but integrating the charge is the right approach. \$\endgroup\$– mngMay 20, 2013 at 14:25
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\$\begingroup\$ Thank you very much for your answer! I believe it's the right method but I was trying to make the calculation using the detector's capacitance(\$C=\dfrac{Q}{V}\$)... The thing is that for a \$50\Omega\$ scope input, the current is \$4mA\$ and I am not sure if this is a typical value... In other detectors a few \$nAmps\$ are more than enough to destroy the detector, but this may be due to the difference in type. I'll give it another look just to be sure! Anyway, thank you very much! \$\endgroup\$– ThanosMay 20, 2013 at 15:08
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\$\begingroup\$ The TDS2022 does not have an internal 50 ohm termination (some higher end scopes do have a switchable one), so the measurement would not be of a consistently 50-ohm system unless an external terminator is used at the scope input. (The manual says 1 M ohm in parallel with 20 pF) \$\endgroup\$ May 20, 2013 at 19:58
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PMT pulse heights generally go up to 1 or 2 volts. Here is an example of an FEU-84 at ~1400V, looking at an LED flasher. The oscilloscope is a 300 MHz Tek.
Historically, people have often used a triangular approximation for the total charge, which will give you a fairly good estimate. Looking at the image above, let's say we have a pulse width of 40 ns and a pulse height of 700 mV, which is a current 'height' of 700 mV / 50 \$\Omega\$= 14 mA. The area of a triangle is base times height over two, so the total charge is 280 pC.
One wrinkle: if you are interested in knowing the number of electrons that hit the anode, you need to look at the base circuit. There may or may not be a termination resistor on the PMT side of the output cable. If there is no resistor, the 'scope is seeing the entire signal from the anode. If there is a resistor, then the 'scope is only seeing half of the signal. This assumes that cables and terminations are consistent, e.g. 50 \$\Omega\$.
