PID woes, and a wild fix

I’m nearly done with the multi-year, many-times-delayed construction of my PID box, and while I’ve successfully crammed everything into a 4” cubed box (no easy feat), upon plugging it in, the PID worked fine, but the SSR wouldn’t fire. I tried changing outputs (1 to 2, hot and cold), but to no avail. Upon examining the leads with a multimeter, I was further baffled: no voltage reading above some 20mV, no current, nothing. After some research, it turns out that, indeed, I’d purchased the near-useless Dwyer Love 32B 33 model - two 5A relays, no logic signal out. Woof. While this is still a solid PID, it’s useless for the power I need (minimum 1,000W, sake of argument), so how can I fix this issue?

Well, the cleanest solution would be to simply buy a new unit, but they’re $90, and then I’d have a useless PID. Well, what about trying to pass a logic signal through the relay? An interesting idea; here are the pitfalls: it’s an AC relay, so can it handle a DC signal, even if the amperage is low? What would be the power source; could I cram a small power supply into my already-cramped box? Could the SSR even take that amperage?

Let’s cross the power supply option off, as best case scenario I’d have to cut the thing open to wire it into the power line in the box. So, what could give me a logic signal that doesn’t require a second cable into the box? That’s right, a battery.

Next problem: can the PID take a DC signal? Assuming the 3A, 240VAC SPST relay in mine is like most others, it should be able to handle a sub-50 mA DC signal, so we have to design around that. I couldn’t find anything suggesting the SSR wouldn’t be able to handle that, but for the sake of power conservation, I deferred to the Arduino’s 20 mA signal as a target. As for the battery itself, 9V seems simple enough.

Thanks to good ol’ E&M, we know V=IR, so for a 20 mA signal, that’s 450 ohms, which is good, since P=IV, and hence the circuit should draw .18W, whereas the common resistor I found was rated to .25W. It all checks out.

Finally, where to place the resistor? Well, the SSR needs to see the full signal voltage, so we need only place the resistor in parallel with the SSR terminals; that way we get a signal of virtually 20 mA assuming small internal resistance, and all 9V (we just need 3V, but can go into the 30s). And there you go! That’s one way to get a logic signal out of a relay-based PID without using a power supply. Neat!