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I am learning electronics and I just got stumped. I was just looking at a flame sensor and it appears to have only has one connector.

I thought sensors in general need two wires to form a complete circuit. How can current travel from the flame to the rod to create a complete circuit? If you wanted to measure the voltage or current of the sensor on an open flame, how can this be done with a multimeter? What do you reference it against? I thought the metal collar was a grounding point, but this is used to bolt the sensor to a furnace. Is the furnace acting as a reference point?

enter image description here

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  • \$\begingroup\$ That sensor has got 2 terminals \$\endgroup\$ – JonRB Jan 9 '15 at 18:19
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The Flame rod in your picture is not a thermocouple.

The other connection in the circuit is the flame being in contact with a metal surface that acts as a ground.

Most every flame rod system I have worked on puts and AC voltage on the flame rod. Depending on the system this voltage may be between 80 and 300 volts.

Without a flame present in the system you can only verify an AC voltage from the Flame Rod to ground.

When a flame contacts the flame rod part of the flame MUST be in contact with a metal surface - such as a nozzle where the gas and air combine.

A DC electrical current is established that flows from the flame rod through the flame and to ground. The DC current can be measured with a micro amp meter in series.

I believe that the general principal is that the AC voltage ionizes carbon particles in the flame which conduct the current. Gas and air mixtures affect the flame rod signal.

IF the flame pulls away from the nozzle, such as a higher firing rate, there will be intermittent flame detection as the ground path is usually lost.

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    \$\begingroup\$ Great answer! I still don't quite understand the AC vs DC part. Essentially, you're saying that the flame is conductive and current through the flame is responsible for continuity with the device. \$\endgroup\$ – user148298 Jan 12 '15 at 17:25
  • \$\begingroup\$ So theoretically, you should get continuity if you put both ends of the multimeter leads in the same flame without touching? \$\endgroup\$ – user148298 Jan 12 '15 at 17:28
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    \$\begingroup\$ Think of it this way - the flame is a diode. If you put a diode in series with an AC voltage you would get a DC recitified Voltage and current out. \$\endgroup\$ – Tinkerer Jan 12 '15 at 20:45
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    \$\begingroup\$ Your continuity suggestion is a good idea but you would probably read a really high resistance. A better experiment would be to put a flame rod in a flame and attach one multimeter leads to it and the other to ground on a diode test scale. You may have to swap lead polarities as off the top of my head I don't know if the flame rod would be the anode or cathode. I've never done such a test but it would be interesting to see the result. \$\endgroup\$ – Tinkerer Jan 12 '15 at 20:55
  • \$\begingroup\$ Aren't flames naturally ionized? youtube.com/watch?v=a7_8Gc_Llr8&feature=youtu.be&hd=1 \$\endgroup\$ – Phil Frost Feb 13 '17 at 16:21
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The flame sensor in your photo uses the electrode in the flame vs. the frame of the furnace (which includes the root of the flame) and is connected to the mounting lug on the flame sensor (that's why there is a ceramic insulator in there).

The flame has nonlinear (like a diode) characteristics that allow it to be very reliably distinguished between a proper gas flame and an open, a short or a resistance due to soot or moisture between the electrode and ground. This is vitally important, because if the gas valve is left on with no flame, the gas will eventually find an ignition source, and Ka BOOM!

Don't fool with this stuff lightly, there is significant redundancy and other considerations required to make a safe, approved and legally acceptable ignition controller.

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I'm not a plumber, but I think the answer is that the flame carries the current and that the other terminal is the (grounded) nozzle of the flame jet. (Flame is a plasma and a conductor.. pretty cool.)

Edit: Poking around on the web, this might not be a thermocouple but a flame rectification sensor.

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I suggest that it is actually the hot sensor rod that does the rectifying. A hot object sheds electrons more easily than a cooler one. This was/is used to great effect in vacuum tubes/valves, Cathode-Ray TV tubes and even fluorescent lamps (though both ends of those are hot).

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The ionization that occurs in the flame envelope causes the continuity. The larger the ground area, the better and more consistent the signal will be back to your Fireeye relay. Most common misunderstanding in this process is the fact that the ground is not part of the flame rod assembly. The flame envelope has to consume the space between the rod and the ground area. Simple touching flame to the tip of the rod generally won't complete the circuit. Hope this helps

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The flame sensor when heated up acts like a cathode of the vacuum tube. It emits electrons which flowing to the ground through the flame, like in the vacuum diode.

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