Why does a photomultiplier tube produce a negative output pulse?

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The standard output waveform of a photomultiplier tube has a "negative" pulse, like below. Why is the amplitude negative?
Example-of-an-acquired-waveform-of-the-PMT-output-and-the-electric-signal-induced-on-the.png
 
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And the electrons are negative !
 
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hutchphd said:
And the electrons are negative !
Best answer so far. :wink:
 
sophiecentaur said:
Best answer so far
I think @phinds was better. It's not impossible to design a circuit where the pulse is positive, even though the electrons are negative. A PMT is but one element in a circuit.
 
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Agreed but, pour a load of electrons on something and its Potential will go in the negative direction. The OP was a bit too simple, of course, but electrons are at the coal face, here.
 
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hutchphd said:
And the electrons are negative !
The conventional current flows from ground, down through the current to voltage resistor to the anode. That is why the pulse is negative.
 
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Vanadium 50 said:
A PMT is but one element in a circuit.
But it does make a practical difference where ground potential is chosen. These were developed in the 1930's. I think I'll stick with door #1...
 
hutchphd said:
But it does make a practical difference where ground potential is chosen. These were developed in the 1930's. I think I'll stick with door #1...
I have used a PMT tube with grounded cathode as part of a TV camera. But grounded cathode not useful for DC outputs referenced to ground.
I think the PMT concept was invented by Philo Farnsworth as part of his Image Dissector TV camera.
 
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Hamamatsu says PMTs were invented by Iams et al. in 1935 and the modern multidynode tubes in 1936 by Zworykin.

As far as the negative signal, had it been positive and someone said, "oh, it's common to have an odd number of inverting amplifiers in the base", we all would have nodded sagely.
 
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When I nod sagely it means I've dozed off...just give me a little jab..
 
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I took the clue for the 'no output' level of zero. The question wording refers to the tube itself afaics. It's clearly been connected into a circuit that establishes a bias voltage of 0V. Then, along come the negative charges.