Can anyone suggest the best (simplest) technique for determining the impredance of a 20W fluourescent lamp at a specific frequency?
Can it be calculated as an approximation or does it need to be measured in situ?
Thank you,
Henry
I have -no- idea. Though, wouldn't it look like a bidirectional diode, Vf ~
80V, with chunky reverse recovery time?Tim
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The lamp itself ? Considering the underlying physical processes, I wouldn't be surprised if the current was dependent on the voltage in a nonlinear way. And then there is possibly a pressure and temperature dependence.
Rene
-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com & commercial newsgroups - http://www.talkto.net
Being a gas-filled tube, isn't it going to be two state with hysteresis? You certainly will need some kind of current limiting.
...Jim Thompson
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Being gas discharge, it also likely has a negative resistance characteristic.
Jeff
Fluorescent impedance, of course, is negative.
That is why a ballast or its equivalent is required.
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** With or without the ballast choke ?Make a huge difference.
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To a first approximation, the impedance of a fluorescent lamp tube is zero, once the discharge has been struck.
If you design on this basis - aiming to regulate the current through the tube more or less indpendent of the voltage across it - you won't go far wrong.
Traditional fluorescent lamp drivers used a big inductor to limit the AC current through the tube. Modern systems use smaller inductors driven by high-frequency switches to create the same effect in a smaller, cheaper, package that generates less waste heat.
-- Bill Sloman, Nijmegen
I guess not. Once the charges are separated, after a certain amount of charge has passed through, there won't flow any current anymore, or only against a rather high voltage. This repeats every cycle.
Rene
-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com & commercial newsgroups - http://www.talkto.net
These are confusing answers to a difficult question, the Z of flo lamps. Let me add to the confusion. It depends on the frequency of the driving waveform. At low frequencies,
10KHz, the lamp continues arcing thru zero crossings, thus needs no high voltage at leading edge which improves luminous efficacy (Lumens/Watt). The current and voltage waveform can be purely sinusoidal. So lets just talk of normal solid state drive at 50Khz. The voltage and current are in phase so the arcing lamp looks purely resistive. If the lamp current is allowed to increase the voltage will drop and this is why we say it has a negative resistance and must be driven by a source that has a larger positive resistance than the lamp's negative resistance such that the sum of the two resistors is positive. If you only want to know the Z of a flo lamp at one V*I operating point at high frequency then it is just a resistor. To change operating points you must add a positive |Z| to stabilize and control your operating point. Possible positive Zs are capacity, inductance or current control. Voltage control cannot be used because there are lamp operating points with different currents but the same voltage. Get a flo lamp V*I curve and you will understand. Capacity and inductance have positive or negating phase relationships but still have positive |Z| because as their current icreases their voltage also increases. Makes sense to me. Harry
Sorry Rene - this sisn't quite right. As far as I know, fluorescent lamps work as "glow discharge lamps" where positve ion bombardment knocks secondary electrons off the cathode, so there is no question of "the charges being separated".
Fluorescent lamp starters work by heating up electron-emitting filaments at both ends of the lamp. Once the dischage has been established, positve ion bombardment keeps the surface of the electrodes hot, presumably for long enough to restart the discharge after the AC current has passed through zero, close to the point where the AC voltage is at a peak (positive or negative).
Sam's F-lamp FAQ gives more detail. Sam has poached some my posts from here for his other optical FAQ's and seems to build pretty good expositions. See
-- Bill Sloman, Nijmegen
Hello Bill,
In Europe many of them work that way, causing that dreaded start-up delay. In the US most appear to be apply a high enough voltage for an instant start which then collapses to a pre-defined value after start. Even the really old ones in our garage do that. Click - ON.
Regards, Joerg
I was thinking DC and slow AC, my fault. The light isn't emitted in the cathode but in the gaz itself. The fluorecent lamp emit in the ultraviolet, and the coat makes visible light from it by fluorescence. I'm positive that charged ions from the gaz collide with each other and thus produce the ultraviolet. It cannot be the cathode, it would have to be far too hot, above 6000 Kelvin. IMO the glowing cathode is a normal cathode as from the tubes, to lower the exit work for the electrons. If there wasn't the hot cathode, it would have to be a field emitter as in the electron gun of an electron microscope. With a sufficiently small radius and a sufficiently high voltage, the electrons come out too.
Rene
-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com & commercial newsgroups - http://www.talkto.net
When I was workiing on electron microscopes (1982-91 at Cambridge Instruments in Cambridge, U.K.) the regular electron source was a hair-pin shaped tungsten filament that was heated to the point where electrons boiled off.Ttungsten boiled off too, so they lasted about a week
Where we wanted a stable, brighter source, we used a lathanum boride single crystal mounted on a tungsten hair-pin, which gave good - if rather directional - electron emission at slightly lower temperatures. They lasted about six months.
There were some field emmision sources on the market. Cold field emitters are too unstable to be much good - the individual atoms on the tip move around unpredictably.
"Hot" field emissionsources were nicer - you ran them hot enough to get useful field emission at lower fields than you need for a cold field emission source, so the emitting tip can have a larger radiius of curvature, and there are enough atoms in the emitting tip that they could move around without changing the emitted current to and detectable extent.
None of this has much to do with fluorescent tubes, where - during sustained operation - the bulk of the electrons are generated by positive ion bombardment of the cathode. The electrons moving through the gas ionise enough gas atoms to keep up the supply of positive ions, which eventually migrate to the cathode.
At least in Eorpean fluorescent tubes, the two filaments inside the tube glow during start - implying that they are heated hot enough to produce some electron mission,
-- Bill Sloman, Nijmegen
Measure it - current through tube, voltage across tube versus time. Work out a model from that.
A silly guess would be a switch providing a breakdown voltage of around 80 volts and a series resistor of about 50 Ohms. I would still measure it.
I think puting it in the simplest way possible is to think of the tube itself as current dependant resistor with a negative slope, so at zero current the resistance is open circuit to voltages less than the striking voltage, at 1 amp or so it may be a few ohms.
There is a response time of the ionized gas so if the current falls to zero the tube remains 'struck' for some short time, and also the thermal response time of the electrodes has an effect (mainly on striking voltage)
I did some work on improving the power factor of electronic ballast, the simplest way was to omit the resorvoir capacitor, this made for an odd tube current/voltage waveform :- a 40khz switching voltage modulated at 4 times line frequency.
Colin =^.^=
Do you mean of the lamp itself? Its a highly nonlinear V vs I function. The folks over on sci.engr.lighting might be able to provide a model for you.
BTW, you misspelled 'impudence'. ;-)
-- Paul Hovnanian mailto:Paul@Hovnanian.com ------------------------------------------------------------------ Don\'t get even -- get odd! :¬þ
This will get you started:
At low frequency it is a completely different characteristic.
I prefer 'imprudence', personally. ;-)
Actually, like with diac's and other trigger diodes, it is discontinuous. It abruptly changes from a high resistance, moderate voltage to a low to negative resistance low voltage. The transition is called plasma inception point, and is hideously pressure and temperature dependant. As Rene intimated.
-- JosephKK Gegen dummheit kampfen die Gotter Selbst, vergebens. --Schiller
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