hi, not sure if this is really the right section on here, but has anyone out there successfully dimmed a standard fluorescent tube? Even more specifically, has anyone dimmed recessed "u" shape 9/18/36w fluorescent fittings? Reason i ask is i was in a lecture theatre at a university last year and the guy dimmed the recessed fluorescent fittings in the ceiling, for a slide show, with no ill effects. i didn't think it was feasible to dim fluorescents.
dimming fluorescents?
- Thread starter Steve
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Found this, look at AW10u
http://www.letsautomate.com/10030.cfm
Dont know if it will be any use to you.
http://www.letsautomate.com/10030.cfm
Dont know if it will be any use to you.
im not trying to do it myself, but i have been reading up on the subject, it seems people have been dimming "ordinary" fluorescent lights for a while with no ill effects, in fact prolonging tube life, i was just wondering if anyone here had tried it! and on what sort of dimmer and ballast?
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B
Big_Spark
I've done it thousands of times...literally, but you need to have the right control gear in the fitting and the right dimmer is used for the type of load your attaching to it.
You have to use the appropriate control gear that supplies the tube heaters with a separate circuit.
The normal circuit puts the heaters in series with the tube current, so if you try dimming with normal control gear, you starve the heaters & electrons can't leave the cathodes. Separatly heated, there's no problem.
The normal circuit puts the heaters in series with the tube current, so if you try dimming with normal control gear, you starve the heaters & electrons can't leave the cathodes. Separatly heated, there's no problem.
That unit is merely a switching device for home automation, it doesn't dim, advice above is correct, you need special dimmable ballasts, you will also need to match tubes correctly to the correct ballast this usually means same manufacturer and also tubes will need to be what's called burnt in, this means that the tubes need to be on at full level continously for about 3 days otherwise premature failure will occur (usually the heaters go).Hotwire said:
B
Big_Spark
I think you need to stop talking as if the heater is a tube is a seperate element when they are not..
A fluorescent lamp consists of a glass tube, filled with argon gas, and dosed with a small drop of mercury. At each end there is a cathode, which are normally made of tungsten, because of its low evaporation rate. The glass tube is coated on the inside with a layer of phosphors.
When a fluorescent lamp is turned on, the current heats the cathodes at each end of the lamp, causing them to emit electrons. A stream of electrons flows through the gas from one electrode to the other. These electrons bump into the mercury atoms and excite them. As the mercury atoms move from the excited state back to the unexcited state, they give off ultraviolet photons. These photons hit the phosphor coating the inside of the fluorescent tube, and this phosphor creates visible light. Reducing the current across the cathode will reduce the electron flow from it, and thus the amount of ultrviolet photons produced, and this in turn will reduce the amount of visble light emitted by the lamp.
In summary:
There is a stream of electrons flowing between the electrodes at both ends of the fluorescent bulb.
The electrons interact with mercury vapor atoms floating inside the tube
The mercury atoms become excited, and then when they return to an unexcited state they release photons of light in the ultraviolet region of the spectrum
These ultraviolet photons collide with the phosphor coating the inside of the bulb and the phosphor creates visible light.
The amount of electrons flowing in the tube id directly proportional to the current flowing across the cathode at any one time.
A fluorescent lamp consists of a glass tube, filled with argon gas, and dosed with a small drop of mercury. At each end there is a cathode, which are normally made of tungsten, because of its low evaporation rate. The glass tube is coated on the inside with a layer of phosphors.
When a fluorescent lamp is turned on, the current heats the cathodes at each end of the lamp, causing them to emit electrons. A stream of electrons flows through the gas from one electrode to the other. These electrons bump into the mercury atoms and excite them. As the mercury atoms move from the excited state back to the unexcited state, they give off ultraviolet photons. These photons hit the phosphor coating the inside of the fluorescent tube, and this phosphor creates visible light. Reducing the current across the cathode will reduce the electron flow from it, and thus the amount of ultrviolet photons produced, and this in turn will reduce the amount of visble light emitted by the lamp.
In summary:
There is a stream of electrons flowing between the electrodes at both ends of the fluorescent bulb.
The electrons interact with mercury vapor atoms floating inside the tube
The mercury atoms become excited, and then when they return to an unexcited state they release photons of light in the ultraviolet region of the spectrum
These ultraviolet photons collide with the phosphor coating the inside of the bulb and the phosphor creates visible light.
The amount of electrons flowing in the tube id directly proportional to the current flowing across the cathode at any one time.
You've summarised how a cold cathode lamp works, Big_spark.
Normal fluorescent lamps have hot cathodes. Electrons boil off the hot cathode until the cloud of electrons creates so high a negative charge that no more boil off. The voltage across the lamp pulls some electrons from the cloud to the other cathode.
That's why you cant dim a fluorescent tume with a normal ballast - the heaters don't provide the cloud of electrons when they arn't hot enough.
Normal fluorescent lamps have hot cathodes. Electrons boil off the hot cathode until the cloud of electrons creates so high a negative charge that no more boil off. The voltage across the lamp pulls some electrons from the cloud to the other cathode.
That's why you cant dim a fluorescent tume with a normal ballast - the heaters don't provide the cloud of electrons when they arn't hot enough.
i was under the impression that after striking the heaters were switched out although they obviously get some current from the fact the electrons leave partway along the filaments rather than right at the end that is connected to the power supply
plugwash said:
Partly switched out.
On startup, the whole of the available power runs through the heaters to heat 'em up quick. Starter switch shorts out the tube so the arc can't strike.
Starter switch opens, causes a voltage surge because of the choke, strikes the arc and the tube current passes through the heaters, then through the arc.
The arc positions itself where the electrons boil off the heaters. It won't strike on the unheated cathode because there arn't any electrons there. The arc current keeps the heaters hot.
Cold cathode works differently. The voltage is higher and the electrons flowing in the arc strike the cathode and cause other electrons to jump out. Less efficient than heated cathodes.
For those doubting thomas's:
http://members.misty.com/don/f-dim.html
http://members.misty.com/don/f-dim.html
but they are! the heaters are both in series with the starter switch contact .Big_Spark said:
You are talking about neons(cold cathode) where there is only one contact each end of the tube (cathode and anode, polarity determined after burn in)usually a wire , not two pins as in flourescent tubes, heaters are across the pins at either end. It's the glowing of the heaters you see upon starting.
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