During a friend's gas fire service, the technician replaced a thermostat
and left the old one. I saw it, picked it up and have been puzzling for
days how it works..
It doesn't have a single tube with a bulb to heat at one end and an
actuator at the other. Those I understand. It has two thin "wires"- one
soldered/welded to the outer of a small short tube at each end.
The other wire goes through the small short tubes. At the actuator end,
it slided freely in the tube but is terminated immediately after -
looking as if it just has a small ring soldered to the end. At the bulb
end, it just goes into the tube - presumably connected to the bulb in
Now, clearly, if the thin wires were rigid bars, it would be obvious how
it works. Heat the bulb bar, it expands relative to the other bar and
exerts force on the actuator. But these are just very thin, coiled,
wires. I really cant see how any force is transferred to the actuator.
I've tried heating the bulb - but there is no visible movement at the
other end. Which proves nothing, as the thing was replaced - presumably
because it didn't work..
Anyone know how these work?
A picture would help a lot, but it sounds like it might be
a flame failure thermocouple. The end is heated in a the
pilot flame and it generates just enough current to hold a
gas solenoid valve open. If the pilot light goes out, the
gas solenoid valve closes, cutting off the gas.
LOL, thanks Andrew. I think that you have the answer.
I hadn't thought that they might actually be wires and that it could be
electrical rather than mechanical.
They actually can make a solenoid that operates off the current
generated by heating a thermocouple? How many mV do they give out?
Or does it have an amplifer and separate power souce?
Presumably the guy tested its output and discarded it as below spec.
So,how do they fail? Thermocouples are pretty reliable, IME.
I did measure one once -- IIRC, it was about 20mV.
I don't have one here to check now.
After 5-10 years of a pilot flame playing on it, they fall to bits.
Often the end gets eaten away, which should be obvious if it's not
symetrical any more. OTOH, you might find it still works fine. They
cost so little that engineer might simply replace one to see if it
was the cause of a problem.
I don't doubt you, but I am still trying to get my head around how 20mV
can produce enough mechanical force to move anything, eg a solenoid. If
it was compressed air, and not gas,it could maybe have moved a needle
into a tiny bleed hole, that then allowed the build up of static air
pressure to move the valve. But this would leak gas all the time that
the needle wasn't blocking the hole.
The two metal strips are only in contact when the flame heats them
therefore the valve is only open and the gas flows only when the metal
strips are heated. If the flame goes out (or the power fails) the metal
strips seperate, the valve closes, the gas is cut off and the flame is
The current doesn't move anything. It provides only the holding
current for the solenoid valve against a spring. The valve is
initially opened by pressing a button and holding it in until
the thermocouple warms up (a few seconds) and can hold it open.
If the flame goes out, the thermocouple must cool and release
the solenoid within 30 or 60 seconds (I forget which), shutting
off the gas.
sense - the energy comes from the person and not the thermo-generated
power. With no air gap, I suppose not much power is needed to hold
against a little spring.
Out here in the sticks there is no mains gas and I am not familiar with
gas fires! Last one I had, you turned a little brass tap and stuck a
match in the front.
Andrew has given you the correct answer in this case,
but in general don't "diss" low voltage/high current scenarios.
The clasic one I am thinking of is those Weller soldering guns.
The voltage across the bit is almost immeasurable but the
current induced into what is effectively a shorted-turn is plenty
to heat the bit.
It is rather easier to imagine that something plugged into a mains
supply can do useful work..
In this case, a small metal cylinder has its tip heated in a small gas
flame and this produces enough power to hold a solenoid against a
spring. And the spring has to be strong enough to *always* pull the
armature away in the case of the power failing - as it is a safety device.
So not only producing power, but enough power to hold against a
reasonably strong spring.
And doing this with 20mV. Even if it produced an amp, that is just 20mW.
That's less than an ant having a bad ant day..
Ah I see what you are saying, but the return spring for the button you press
much stronger, and separate from, the spring that closes the pilot valve.
I pulled a faulty ITT gas valve apart before disposing of it a few months
and I would estimate the weight of a couple of pound coins would be enough
to cancel the spring force. The pilot valve itself was a (synthetic)? rubber
membrane that required inlt a few mm travel.
I wasn't actually thinking of the spring for the button return.
I still don't understand how these valves work and don't have one to
look at (or even gas out here in the sticks).
IIUC, you turn the knob and push. This push lifts a hammer against a
spring and then releases it, striking a piezo element and causing a
spark at the pilot jet to light it. The push must first open the pilot
light valve, pushing a plate in the valve against the armature of an
elctromagnet and supplying gas to the pilot light. This plate is
presumably at the centre of the rubber membrane you mentioned.
The pilot lights, heating a thermopile and generating electricty which
is fed to the electromagnet. That holds the plate against the armature,
keeping the pilot light valve open and it stays open until the gas
supply is interrupted or the pilor light is blown out and the thermopile
The operator doesn't know all this - but merely keeps the button pushed
for ten or so seconds and then releases it - at which point the pilot
should stay lit.
The operator then turns the valve to a "heat" setting.
Somehow the plate must be interlocked with the main heater gas supply
valve, so that only if the plate is against the armature, will the main
heating gas supply go to the burner, where it will be lit by the pilot
flame. If the thermopile cools and the plate is released, somehow it has
to then stop the main heating gas supply. Presumably by moving
something, using energy stored in a spring. That spring is presumably
compressed by the operator pushing the button - but has to be held back
by the electromagnet - and when released has enough energy stored within
it to close the main gas valve, and then some for a safety factor.
That is my quandry - the thermocouple/thermopile has to produce enough
power to not only hold the pilot light valve open, but to resist the
spring energy that has to be stored so that the main heater gas valve is
closed if the thermocouple cools.
Unless, of course, I am totally misunderstanding what is going on...
Try looking at this page and the associated hyperlinks to thermocouple etc.
I know it is for a propane installation but the principle is the same. On
the two diagrams the gas enters the valve from the left and exits from the
lower port on the right.
So the valve actually interrupts the total gas supply, and not just the
pilot supply! Presumably that means:
The gas goes through the ESV to the inlet of the rotary gas valve that
controls the amount of gas going to the burners. In the "light"
position, the burner supply is zero, becuase of this rotary valve. Once
the pilot is lit, the knob is turned to regulate the amount of gas going
to the burners.
If the pilot light is out, even though the rotary gas valve would allow
gas to the burners, the ESV has the supply interrupted.
There must be a physical stop to prevent the knob being pushed in far
enough to mechanically close the ESV, at any point other than the
The penny has finally dropped, perhaps.
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