Posted by admin | Posted in Electronics | Posted on 16-03-2010
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Running a small 4W fluorescent lamp from a resistor?
I’ve heard that this is possible, but resistors aren’t used for most fluorescent lights as they consume almost as much power as the lamp it’s self. I live in the uk where the line voltage is about 240V and I was thinking of using a 10 Watt wirewound resistor but i’m not sure how many ohms it needs to be to run a small 4w fluorescent lamp. The rest of the fluorescent lamp would be the same as if i was using a normal ballast and i’l be using a starter too.
The resistors would be comming from Maplin Electronics, there range of 10w wirewound resistors is here: http://www.maplin.co.uk/module.aspx?moduleno=2181
The following analysis is only approximate since the current through the lamp does not vary linearly with voltage when it is running, but the results should be in the ballpark.
The answer will depend heavily upon the average voltage across the lamp when it is running. For a small lamp, that might be 30V. This means that your resistor would need to drop 240-30=210V. For a 4W lamp, the current is 4W/30V=0.133A. The power that the resistor would need to dissipate would be P=I*V=0.133A*210V=28W. A rule of thumb would be to double that, say 50-60W resistor. The resistance would be R=E/I=210V/0.133A=1580ohm
So the resistor would dissipate 28W/4W= 7 times the power that the lamp dissipates. That is why it is preferred to use a reactance (inductor or capacitor) rather than a resistor to limit the current.
Here is a design that uses capacitors plus some other elements:
http://www.giangrandi.ch/electronics/neon/neontube.html
Starting can be an issue in general. The lamp will start easier if you place a grounded (earthed) metallic surface near the lamp. Then the electric field strength will be stronger from the electrode to the earthed surface (being closer) than it will be to the far electrode, making it easier to ionize the gas near the electrode. Once any of it is ionized, it is easier for it to propogate down the tube with the metallic surface nearby. Once the entire length of the tube is ionized, the metallic surface is no longer needed since the gas will stay ionized long enough to remained sustained as current changes direction each half cycle. Note, this effect depends upon one side of the supply being earthed, which is usually the case.
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