[AussieKenDoll]

The internal resistance and the bandwidth of a photomultiplier tube of an atomic absorption spectrophotometer are 5MΩ and 2kHz respectively. During a particular operation, it generates a current of 1.0 pA. If the photomultiplier tube is operating at a temperature of 25 C.

I. The noise in the voltage signal. Hint: V(RMS) = [4kTR(f)]^1/2

noise of the signal, V(RMS) = 1.283 x 10^-5 V

To find signal, I used the formula V=IR
and V= 1.0 X 10^-12 X 5 X 10^6 = 5 X 10^-6 V
 So V=sqrt(2) x V(RMS)
V(RMS) = 3.535 X 10^-6

so S/N = 0.2756

Is this calculation correct?

[Enthalpy]

Hi AussiKenDoll,

I have difficulties connecting the figures you cite with what I know about photomultiplier tubes (PMT), but I'm not an expert neither.

Could the figures characterize not the PMT but the circuit that uses it? A typical PMT "response" time is in the ns, the corresponding bandwidth >100MHz or more.
The internal impedance is huge and it depends essentially on the output current. Normally it is neglected.
When the main noise isn't the darkness current, then the PMT's noise is shot noise, which is bigger than the same current usually creates, because the charges arrive as packets, like in an avalanche diode. But as soon as you detect a few photons within a response time, the shot noise is smaller than the signal, in other words, the response slowness cumulates several charge packets.

The circuit's load resistor, which could be 5MΩ if it exists, adds noise. The circuit's bandwidth could be 2kHz if designed so. Then your computed resistor noise is correct.

The amplifier that follow adds noise, but for 5MΩ and 0-2kHz you might hope to add less than 12µV. This is far from trivial! Filtering the low frequencies away would help a lot.

If bunches comprise 10⁶ electrons, then shot noise would be 25pA, which only means that 2kHz bandwidth lets see individual photons. The other way round, 1pA and 10⁶ electrons per photon mean 1 detected photon every 0.16s, much slower than the instrument's reaction time. That noise is important too.

I suppose the instrument is not designed to operate with a current that low, which needs tremendous shielding against stray light. Instead of measuring a mean current, the designer would then put a threshold or the anode current and count the detection events.

And why the rms vs peak? This sqrt(2) applies to sine waveforms. I don't know if the spectrophotometer uses some sort of sine modulation, then OK, but one should check if the 1pA is rms, peak or peak-to-peak.

Whether the noise is "in" the signal or added to it? I rather heard that the signal is the unpolluted component.

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A nice doc about photomultiplier tubes there
https://www.hamamatsu.com/resources/pdf/etd/PMT_handbook_v3aE.pdf
Time response is in paragraph 4.3.1, page 62 noted 48
Signal-to-noise is in paragraph 4.3.7, page 87 noted 73

and Hamamatsu's catalogue
https://www.hamamatsu.com/resources/pdf/etd/PMT_TPMZ0002E.pdf

==========

In case you or someone else is interested, I propose to reduce much the photosensitive area of the photocathode of an existing PMT by depositing elsewhere a metal with big work function to hopefully reduce to peanuts the dark current just in liquid nitrogen, easier than superconducting detectors and as sensitive, there
https://www.scienceforums.net/topic/93628-observe-a-falling-charge-radiate/?tab=comments#comment-910916 last section
would be nice for other uses, like single-photon fast communications, possibly encrypted quantically.

I also propose to replace the dynodes by a semiconductor diode where the photoelectron arriving with few keV produces a bunch of electrons, even faster than in dynodes, with the preamplifier closer to the source, there
https://www.scienceforums.net/topic/104919-photocathode-resonates/?tab=comments#comment-985283
also sensitive to single photons if cooled a bit or if the circuit reacts quickly, like 50Gb/s.

[Enthalpy]

PMT have dark currents rather in the 1nA-10nA range.

Even if making a zero on a 1nA dark current, and if the dark current results from primary electrons multiplied by 10⁴, the shot noise over 2kHz is 80pA, worse than many other sources. So 1pA would not be measurable.

At least a part of the dark current is emitted at the dynodes or leaks at the anode. As these currents are less amplified by the dynodes, a circuit that puts a threshold to count pulses can be sensitive only to the primary electrons, hence reject some dark current.

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Assuming that 5MΩ is a load resistance, its noise (which isn't important here) can improve by using the resistor in the feedback of a transimpedance circuit, as is done with photodiodes.