flame rectification

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caosesvida

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I know this is a little off subject here, and I also see that there a a number of electronic savy gurus here. The question was is the electicity going through the flame AC or DC. My guess was that it is ac and the sensor picks up enough dc current to make the circuit work, there is a small amount of ac also but not enough to affect the circuit. here is a web page explaning the theory of operation for this. it is used on gas burners. thanks
http://www.heatincorporated.com/~technical_information/Troubleshooting/Flamerec.pdf
 
Re: flame rectification

Appears to me that you would have mostly DC with a smaller AC component. I wonder if this is cheaper or more reliable than the conventional heat sensor.
 
Re: flame rectification

Yes, the current would be mostly DC. They explain that when they talk about one plate being larger than the other. I agree with Rattus, there would be a small amount of AC current.

Two advantages I see over a conventional heat sensor:

1. Very fast response time. The sensor doesn't have to heat up or cool down.

2. Very fail safe. If a short were to occur across the sensor, the ignition control could detect it because the resulting current would be AC.

Steve
 
Re: flame rectification

I have thought about this some, and the "rectification" part is a mystery to me. Recitification requires a nonlinear element. One plate being larger than the other doesn't necessarly make it nonlinear.

As the voltage goes one way, electrons flow from the small plate to the large plate. As the voltage goes the other way, electrons flow from the large plate to the small plate. Seems to me like we would get the same resistance both times.

Taking a step back, I'm not even sure I understand the conduction process here. Do we have both electrons and positive ions flowing? Can an electron just jump off a wire into a gas? (I know they can jump from one solid wire to another, but into a gas?) I wonder if this is somewhat like a vaccuum tube where the eletctrons have to be emitted by heating one plate?

Does anyone else have a better insight into this?

Steve
 
Re: flame rectification

Steve,

Here is what I think:

When the voltage swings negative on one of the plates, electrons boil off to recombine with positive ions in the gas. That is, this plate functions as a cathode. The other plate functions as an anode. Then the process is reversed.

Here is the scenario:

Electrons from the ionized gas are collected by the anode. Positive ions flow toward the cathode where they recombine with the electrons boiling off the cathode.

Since the plates are of different sizes, the currents are unequal. This is not a good rectifier, but it seems to be adequate for sensing the presence of a flame.

I think too that the unequal plate size is a matter of physical space. Equal sized plates and AC could be used as well.
 
Re: flame rectification

Posted by Rattus:

electrons boil off to recombine with positive ions in the gas
Click to expand...
Thats what I was thinking also. But that is a lot different than conduction in a conductor. There is no "boiling off" required for a piece of wire to conduct.

And "boiling off electrons" requires the plate to be heated to a high temperature. That's where I suspect the real rectification occurs. The small plate is directly in the flame so it boils off a lot of electrons. The large plate acts as a large heat sink, so it doesn't get very hot. So it doesn't boil off a lot of electrons.

Also, I would suspect the two plates would be made of different materials. The small plate would be made of something that boils off electrons easily.

But, I'm still not sure that "boiling off electrons" are required for conduction through a ionized gas. A vaccuum tube is usually filled with inert gas, or it is truely a vaccuum. And you wouldn't have to boil off electrons to conduct through a liquid containing ions. So I still don't think we can say we know for sure how this works.

Steve
 
Re: flame rectification

Steve, if one plate is indeed hotter than the other, that might account for the difference in currents, or perhaps no current at all from the cooler plate. However, the anode does not have to be hot to collect electrons. The cathode must be hot to emit electrons.

The boiling off occurs in the vacuum tube, otherwise there would be no carriers to conduct the current. This also occurs in the filaments of a fluorescent tube where the electrons recombine with the ionized mercury vapor. There is no rectification because the filaments are identical as are their temperatures. The boiling off must occur on at least one plate (cathode) to "connect" so to speak to the ions in the flame.

I think I recall something called a mercury vapor rectifier; there were also thyratrons which worked something like an SCR and were also gas filled.

[ February 11, 2005, 10:41 AM: Message edited by: rattus ]
 
Re: flame rectification

Steve and Rattus

The PDF explains that it is the characteristics of the unit construction that rectify this AC. voltage which is being applied across the flame.

One plate is the unit burner which makes it a large plate and the other one is the sensor electrode which is usually a steel wire probe.

I use to work for a laundry outfit and troubleshoot boilers which used this type of flame sensor.

Quote from PDF

IF INSTEAD OF USING JUST WIRE YOU WERE TO PUT PLATES OF METAL ON THE ENDS OF THE
WIRES YOU WOULD INCREASE THE ABILITY TO CONDUCT THE CURRENT THROUGH THE FLAME. IF
YOU WERE TO GO ONE STEP FURTHER AND MAKE ONE PLATE VERY LARGE WHILE KEEPING THE
OTHER RELATIVELY SMALL, A UNIQUE CHARACTERISTIC WOULD OCCUR TO THE CURRENT AS IT
PASSES THROUGH THE FLAME. THE CURRENT WOULD TEND TO FLOW IN ONLY ONE DIRECTION AS
IT MADE ITS WAY THROUGH THE FLAME. THE RESULT WOULD BE THAT THE CURRENT WOULD NOW
BE DIRECT CURRENT, OR DC. WE CALL THIS EFFECT FLAME RECTIFICATION SINCE THE CURRENT IS
RECTIFIED FROM AC TO DC. THE FIGURE BELOW SHOWS THIS ARRANGEMENT:

The answer to your question is AC. is applied across and it is rectified. And the differential amp. has to see a current of so much of a micro amp. to satisfy the pilot circuit.

Ronald :)

[ February 14, 2005, 08:32 PM: Message edited by: ronaldrc ]
 
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