Britannica Sources used in this answer

How does a photomultiplier tube turn flashes of light into electrical signals?

Created from Britannica articles using AI. AI can make mistakes, so verify using Britannica articles.

detector output connected to a measuring circuit — radiation measurement Figure 1: (A) A simple equivalent circuit for the development of a voltage pulse at the output of a detector. R represents the resistance and C the capacitance of the circuit; V(t) is the time (t)-dependent voltage produced. (B) A representative current pulse due to the interaction of a single quantum in the detector. The total charge Q is obtained by integrating the area of the current, i(t), over the collection time, t_c. (C) The resulting voltage pulse that is developed across the circuit of (A) for the case of a long circuit time constant. The amplitude (V_max) of the pulse is equal to the charge Q divided by the capacitance C.

A photomultiplier tube converts flashes of light into measurable electrical signals through a two-stage process of conversion and amplification.1

First, when a light photon strikes the photocathode, it may eject a low-energy electron into the vacuum. The probability of this conversion, known as quantum efficiency, depends on the wavelength of the incident light.1

Because this initial pulse of electrons is too small to measure, the tube employs a series of electrodes called dynodes to amplify the signal. Each dynode is held at a successively higher positive potential, which accelerates the electrons. When an electron strikes a dynode, it triggers the emission of several additional electrons through secondary emission. This process repeats across the series of dynodes, multiplying the original electron count by a factor of up to 1,000,000. Finally, the amplified pulse is collected at an anode, where the resulting charge indicates the intensity of the original light flash.21