So far, so goodHoneywell lists the specs for the Q340 as: open circuit voltage ~30mv; resistance .02 ohms. That would make the theoretical short circuit current 1.5 amps. I suspect the normal operating current is not in that range, as of course the load will not be zero ohms. If the load is in the range of even just a few ohms, that would give operating current in the range of 1 ma.
Honeywell lists the specs for the Q340 as: open circuit voltage ~30mv; resistance .02 ohms. That would make the theoretical short circuit current 1.5 amps. I suspect the normal operating current is not in that range, as of course the load will not be zero ohms. If the load is in the range of even just a few ohms, that would give operating current in the range of 1 ma.
Seeing as it is a coil that you are measuring you may want to measure impedance. There are a lot of variables.
http://www.google.com/search?hl=en&q=resistance+of+a+solenoid&btnG=Search
Nah, it's DC.
With AC solenoids, yeah, it may get messy.
You have a link to your tester?I have a Baso tester that I can vary the mv output to see when the solenoid drops out, and also the t'couple to see what its output is and how long it takes to fall to the dropout range when the pilot is extinguished.
Is there any way a bad or shorted valve solenoid can damage a good thermocouple?
This is awkward, but...
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