0
$\begingroup$

Empirically, a photomultiplier tube (PMT) is said to present better signal-to-noise ratio (SNR) than an avalanche photodiode (APD). Some materials say that it is attributed to the higher gain and lower dark current of PMT.

The benefit of higher gain is confusing to me, since I suppose that noise (from incident photons and from the PMT detector) and signal are both amplified by gain. Why does higher gain result in better SNR?

| cite | improve this question | |
$\endgroup$
0
$\begingroup$

You have quoted that photomultiplier tubes have a higher gain and lower dark current than avalanche photodiodes and considered that noise and signal are both amplified by gain, but what is important is the lower dark current of photomultiplier tubes compared with avalanche photodiodes.

This means that there is less inherent noise associated with photomultiplier tubes, so for the same signal they will present a better signal-to-noise ratio than avalanche photodiodes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, Mick. The lower dark current of PMT indeed matters. Then does it mean that you don't think that higher gain helps to better SNR? That is exactly what makes me confusive. I have read something like "higher gain of PMT gives more amplified signal, but does not increase noise". But I cannot understand, my thinking is just aforementioned "noise and signal are both amplified by gain". $\endgroup$ – xxfzero Sep 27 '19 at 15:22
  • $\begingroup$ Consider, if you have a particular SNR at the input then at the output of an ideal amplifier (no internal noise) both the signal and the noise have been amplified, but the SNR is the same as at the input because both have been amplified by the same multiplication factor. But in a real amplifier there is some internal noise (interference, dark current, etc.) so at the output the signal has been amplified, but also the noise has been amplified plus extra noise has been added within the amplifier itself. $\endgroup$ – Mick Sep 30 '19 at 7:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.