# Biasing network for phototubes and photomultipliers

I have some doubts about typical biasing network for vacuum phototubes and photomultipliers.

Precisely, let's start by a simple photocathode.

1. I have seen some biasing network like this one:

In this circuit, which is the precise role of $$\R_L\$$? I'd say it is a current to voltage converter, but I was asking if it may have also another role. For instance, if the phototube were a diode, that resistor would be useful to establish the correct current along the diode. Has it this role also in this situation?

1. In this picture there is another type of circuit (like the previous one but with a DC blocking capacitor). In the image of the real tube, it seems that the anodic resistor and the DC blocking capacitor are already inside the tube. Is it correct, or should the customer add them externally to the tube?

2. Finally, this is a photomultiplier

I'd say that the logic behind this biasing network is simply that of using a high anodic voltage between anode and cathode (like in a simple phototube), but with also some voltage dividers in order to get smaller voltages for the intermediate electrodes. But I was asking if it is exactly a voltage divider: is the current across each intermediate electrode low, so that we may consider that network a voltage divider?

• The phototube is arranged with a cathode and anode. It's not effective operated oppositely. So I'm not sure about your question regarding diodes. The phototube doesn't amplify. So the current will be proportional to the light striking the cathode. The resistor does in fact convert that current to a voltage. The value you select here will make all the difference in the world, so it is very important you select a carefully designed value for the resistor, taking into account your signal and applied voltage.
– jonk
Jul 16, 2020 at 19:40

For circuit (1), the bias arrangement for a solid-state photodiode must bias anode at a more negative potential than cathode...so that no-light results in no-current (or perhaps just leakage current). It is also acceptable to bias with zero volts.
With infinite load resistor, no diode external current flows. Any internal diode capacitance simply charges up to some DC voltage. Response time is slow. A low-resistance load resistor speeds response time immensely.

This circuit (1) is DC-coupled, so it is not clear if photocurrent response is required to extend to DC:

• For very large RL (similar to internal diode resistance) output voltage is not a linear function of incident light level especially at high exposure.

• For small RL, voltage across RL is close to a linear function of incident light.

Circuit (2) appears designed for high-speed response. The internal bias resistor, and coaxial output (coupled with a series capacitor having no lead inductance) is designed to drive a 50-ohm load. Such an AC-coupled output is certainly meant for fast response time.
With this circuit, one might be cautioned to apply a high bias voltage +Va slowly, so that turn-on transient does not couple a huge spike to the external 50 ohm load RL. Such a transient could destroy a sensitive amplifier. Connecting a 50-ohm load after applying DC bias +Va could similarly discharge the internal 0.01uf capacitor as a huge transient into a 50 ohm RL. An internal large-value bleed resistor would be a good idea if +Va is large...this should be inside the shell, with the shell permanently grounded.

simulate this circuit – Schematic created using CircuitLab

Yes, the dynode voltage divider ideally is a simple voltage divider in circuit (3). Resistors are chosen large enough not to load down the high-voltage supply excessively, yet small enough that dynode photo-currents are a small fraction of DC resistor current. I recall something like 100k resistors in the dynode string. These photomultipliers are mostly used at very low light levels (yielding small photocurrents), but be aware that dynode currents are subject to gain.
The load resistor attached to anode is a small fraction of dynode resistors. Just like circuit (2) a permanent load resistor might be a good idea to discharge any DC anode voltage to ground. For example, a 1k resistor wired directly from anode-to-ground might help prevent a charged-up coax from destroying a 50-ohm preamp with a single-pulse turn-on transient. I'll bet more than a few preamps have been destroyed this way.

A more casual approach might caution a user, "Increase DC bias slowly to its final 1000V".

In this circuit, which is the precise role of RL?

It represents the impedance of the load, e.g. 50 ohms if you use a 50 ohm RF amp.

In the image of the real tube, it seems that the anodic resistor and the DC blocking capacitor are already inside the tube. Is it correct, or should the customer add them externally to the tube?

This is something you should look at the datasheet to determine. Do you have a link to where that picture came from? I am not entirely sure what it is showing.

But I was asking if it is exactly a voltage divider: is the current across each intermediate electrode low, so that we may consider that network a voltage divider?

Generally yes. Remember, the cathode is giving you one electron per photon absorbed, and then you adjust the bias voltage to get the final output current you want. There is a limit where the tube will saturate or be damaged if you try to draw too much current, but usually you have very, very low current out of a PMT.

• about RL, what if the amplifier has infinite input impedance (like many low frequency amplifiers)? Should we insert a resistor in parallel like RL? Jul 16, 2020 at 19:25
• @Kinka-Byo If you hook a current source with thousands of volts bias to the high impedance of an opamp, it'll raise the voltage until the opamp blows up and starts conducting. Usually if you're using an opamp with a PMT at all then you use a transimpedance configuration which provides a path to ground. Jul 16, 2020 at 23:55