# Boiler thermocouple voltage monitor

#### Frankchie

Nov 14, 2017
149
I was planning on using the ADC input of an esp8266 microcontroller to monitor the voltage of my gas boiler's thermocouple. My main question is safety. How can I connect to the thermocouple without a safety issue. For example, if some failure happened that applies voltage to that thermocouple it could allow the gas valve to open even if the pilot light was off. Obviously that would be pretty bad.

I could monitor the voltage through a high value resistor so if voltage was inadvertently applied to the gas valve the current would be limited. But I'm not sure how much current the gas valve needs to open or how much current it needs to hold open. The typical thermocouple voltage is about 20mv and best I can measure the input resistance of the gas valve is less than 0.1 ohm. The largest resistor that the ADC can tolerate is about 50k, but that would drop the measurement to about 16mv when it should read 20mv. I think I can live with that. For added safety I would operate the microcontroller from a 4.8v volt battery rather than a 5v output wall wart. That would limit the maximum current to about 10ua upon worse case failure.

What you think? Is the above too high risk? Is there a better way to safely monitor this thermocouple voltage?
Thanks,
Frank

#### Alec_t

Jul 7, 2015
3,331
20mV/0.1Ω = 200mA. I doubt the valve will hold open with less than, say, 50mA. So if your ADC runs on 5V any resistor > 5V/50mA = 100Ω should be ok. A 1k resistor would limit any fault current to 5mA.
But how will you get access to both conductors of the thermocouple? If yours is anything like the one on my bolier it has a co-axial structure, with the inner conductor shrouded and hence inaccessible.

#### Frankchie

Nov 14, 2017
149
Thanks, Alec_t,
I just did some tests on a spare gas valve that I have and it drops out at 3mv. my proposed circuit suggests a worse case fault only provides about 10uv across the gas valve (4.8v into a 0.1/50,000 ohm voltage divider) . So just about anyway you look at it should be safe.

But in this case is "should" good enough given the potential consequences? I guess a dual failure like the resistor shorting and the ADC pin supplying 4.8 V could be problem, but I think the chances of a resistor shorting when operating at such a small load is infinitely small. The ADC fault is much more likely since a program malfunction could conceivably make that ADC pin an output pin, but it still should be unlikely.

My guess is that the probability of the above dual fault is less likely than the existing probability of the gas valve simply sticking open (it actually has two internal valves that would have to stick open). So maybe I can live with that risk.

Of course there are lot's of factors that I have not considered. I don't know what I don't know. (water leaks? somehow excessive heat? earthquakes?). So my inner self is still saying find another way. I'm hoping that somebody here knows a better way.

BTW, I have a little screw in adapter inserted between the thermocouple and the gas valve that exposes the inner conductor.

Thanks, again,
Frank

#### Frankchie

Nov 14, 2017
149
A related question: Are there DC current sensors that don't require insertion into the circuit? Hall effect sensors? If so, that sounds pretty safe for this application, although they would have to operate at pretty low current levels.

Thanks,
Frank

#### Alec_t

Jul 7, 2015
3,331
3mV/0.1Ω = 30mA. I'm surprised it can go as low as that before the solenoid drops out. We live and learn.
Although 200mA in a single conductor should provide a strong enough magnetic field to be detected easily by a Hall effect sensor, the inner and outer conductors of the thermocouple (assumed co-axial) would be carrying equal current in opposite directions so their fields would cancel. You might strike lucky, however, and be able to detect the field produced by the solenoid of the valve. Worth a try.

#### Frankchie

Nov 14, 2017
149
Alec_T,
I'm going to buy a Hall effect sensor and give it a try. My adapter gives me access to the inner conductor of the thermocouple.
Thanks, again,
Frank

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