Lighting-Gallery.net
Lanterns/Fixtures => Modern => Topic started by: slipperypete on December 20, 2011, 12:07:07 AM
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Hi everyone, I am new to the club here so please forgive me if I sound stupid ??? I heard it through the grapevine that you can make a 50 Watt mercury vapor fixture using a 40 Watt single lamp pre heat ballast. That beingsaid I have a rapid start single lamp 40 Watt ballast and I was wondering if that ballast could be used for a 50 Watt mercury vapor fixtur. And I apologize if this sounds stupid, but where would I find a 50 Watt mercury vapor bulb? All we got here are 80 100 160 175 and 250 Watt bulbs. ThanxThanx
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With discharges the voltage is steered by the lamp (load), the current by the ballast. So ballast have to be designed so, it force the correct current trough the discharge with voltage drop of the lamp, but should keep the current, if the arc voltage change.
For the lamp is important, what current it is designed for versus what current is delivered by the ballast.
And as the ballast is not "perfect" current source, it's current (so internal losses) sightly depend on the load voltage. Therefore for the ballast stress is important, whether the used lamp have about the same voltage drop the ballast is designed for.
The F40T12 ballast is designed to deliver 0.43A into about 100V arc load, while the 50W MV is designed to be fed by 0.6A and have voltage drop about 90V.
So when combining these two:
As a discharge ballast load, 90V vs 100V are so close, so they could be treated as equivalent, so (after lamp warmup) the ballast would not see any difference between F40T12 and MV50W.
The MV is a high pressure lamp, so it behave differently when cold and hot. When cold, nearly all of the mercury fill is liquid, so in the arctube is quite low pressure of argon gas. As it is low pressure, the arc voltage is low (about 20..40V), so most ballasts deliver a bit higher current, what mean power delivered to the lamp is rather small (less then 1/2). As the arctube heat up, the mercury evaporate, rising the pressure inside the arctube, so the arc voltage, so the power delivered from the ballast. With MV it all settle, when all of the Hg is evaporated, at this point the arc have about it's final voltage drop (even when further heating up, there is no more material to add to the arctube). Then the temperature settle, only te brightness increase may be observed in this stage.
The 50W MV would be fed by only 0.43A, what mean it will run at about 36W after warmup, what should be OK for the lamp.
But because of such low current, there is some risk with some ballasts, then the current would be too low, so the lamp would not heat up to vaporize it's mercury dose, at least it would take longer. If this problem is about to occur, the arctube would need very long time to warm up (1/2 hour or so) or it will not warm up at all. But I think, at least indoors the 0.43A would just work.
But do not use HF electronic ballast! The high frequency would cause resonances inside the arctube, what would overstress it, so it may fail soon.
So what to do in your case:
- Either find ballast model, what is known to work well with 50W MV
- Or try the combination you want to use and check the arc voltage, how the arctube warmup.
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I find a 600 MA primay 120 current to 24 volt transforer works well as a ballast
and you wire as a auto tranformer configuration or bucking the tranformer what it called.
40 watt preheat ballast works ok but under drive it a little.
but for rapid start i thought you need a cathode connected for it to work. otherwise it just glow dimly.
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I find a 600 MA primay 120 current to 24 volt transforer works well as a ballast
and you wire as a auto tranformer configuration or bucking the tranformer what it called.
Sorry, I don't understand what you mean...
for rapid start i thought you need a cathode connected for it to work. otherwise it just glow dimly.
Any RS OCV is enough to strike the MV. Of course, if the ballast connection need complete filament circuits to operate, you have to "mimic" the filament presence by 10 Ohm/2W resistors.
But with RS may be another problem: Many of them underdrive the fluorescent lamps they are rated for (quite common is ~0.27A instead of 0.43A; here I mean real lamp current, not the mains input current printed on the label), so for the MV the power would be really too low. But if it is two lamp ballast and run both lamps in phase, you may connect both outputs in parallel to get the reasonable current for the 50WMV.
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@ medved what i mean using a tranformer is making a home made auto transfomer ballast.
since it need 90 volt operating. well what you do is you take a transformer just hooking like you do on the primamy on 120 volts. then you take the secondary side and one side on hot line so ether it will buck the voltage or boost it by 24 volts ac. as you can get close to the spec of 96 volts on buck. or you can go the other way and get 144 volts on boost out of the tranformer.
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@ medved what i mean using a tranformer is making a home made auto transfomer ballast.
since it need 90 volt operating. well what you do is you take a transformer just hooking like you do on the primamy on 120 volts. then you take the secondary side and one side on hot line so ether it will buck the voltage or boost it by 24 volts ac. as you can get close to the spec of 96 volts on buck. or you can go the other way and get 144 volts on boost out of the tranformer.
I see. But that may work with an incandescent or similar resistive load, but it would not work with any discharge.
The MV would most likely not ignite, or if it will so, the current would be very high ans fry both the lamp and the transformer.
It is, because the voltage is dictated by the discharge, so in order to get stable power, you need a source, what adopt the actual output voltage towards the actual arc need so, the current is the required one.
With magnetic ballasts this may be done transforming the voltage high enough (at least 200V for MV, usually about 220..280V) and put an impedance in series with the lamp in order to regulate the power.
As you don't want to dissipate any extra power, the impedance should be either inductance or capacitance.
As the discharge is highly nonlinear load, the impedance should have high impedance on higher frequencies (mains harmonics), in order to smoothen out the shape of the current, so an inductive component should be present.
So you need a transformer and a reactive component.
These two functions do not have to be physically separated components, you may integrate them into one autotransformer unit by designing the magnetic circuit so, the magnetic flux generated by the primary may go around the secondary, if that is loaded. But that require special core consruction - between primary and secondary windings have to be a magnetic shunt.
Such integrated magnetic device is then called HX (auto)transformer, it is used for HID's, as well as for fluorescents (for fluorescents extra windings for cathode heating are added to the primary coil, or the device integrate two secondaries to serve two lamp circuits - each having it's own magnetic shunts to the primary).
That is, what differ ordinary transformers from ballast HX autotransformers: Ordinary transformers are designed t have as little as possible, as you need the output voltage to be fixed and independent on load current. But on the other hand ballast HX transformers are designed to have their output "weak", so they rather maintain constant current into loads with varying voltage drop.
Sometimes a capacitor is used as the main ballasting impedance (some HPF fluorescent and HID CWA). With fluorescents the reason is, then the transformer part does not have to feature the magnetic shunt, while the lead phase lamp circuit is compensated by the main inductance of the transformer part (by simple gap in the transformer magnetic circuit). The main advantage is, then the primary does not have to handle the full apparent power as with the HX (OpenCircuitVolts x lamp current = about 110VA for the single F40T12), but only the real power (so about 50W for single F40T12, include ballast losses).
The CWA use magnetic shunt between primary and secondary, but this is made with reduced cross-section, so it saturate when the current reach predefined value. The leakage inductance (formed by the shunt) then form series resonant circuit with the capacitor (remember the CWA circuit), where the total impedance is the capacitor reactance minus the reactance of the leakage inductance, what is for low currents (where the shunt does not saturate, so the inductance is high) very low total impedance, the lamp current would increase. But as the lamp current increase beyond predetermined level, the shunt start to saturate, so the inductance drop. But as the inductance drop, the resulting impedance of the series LC increase, so the current does not rise further.
This arrangement ensure, then the lamp current remain constant, regardless of the lamp and mains voltages. As the lamp voltage is always the same (on MV it is pretty constant over life and drive power), such ballast compensate out mains fluctuations, it deliver the "constant wattage" into the constant voltage load.
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But do not use HF electronic ballast! The high frequency would cause resonances inside the arctube, what would overstress it, so it may fail soon.
Sorry for exhuming such an old topic, but this is what I'm just pondering over. I've got several Tesla 50W MBF lamps. I do like their daylight colour but not their flicker - especially the pulsating intervals, caused probably by load management (http://en.wikipedia.org/wiki/Load_management) signals present in our power system.
I thought the HF ballast might solve my problem. As far as I understand resonances, they appear at particular frequencies (and their harmonics) but HF ballast frequencies vary (according to the manufacturer) in some range so I don't necessarily need to have a problem like this, do I?
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Or the ballast frequency lock itself on the lamp resonance, so "makes sure" it stay at the worst frequency...
The flicker could be to a big extend eliminated by series capacitor, so e.g. F58T8 ballast in series with 4.7uF/450VAC capacitor (adjust the capacitor to have exactly 0.6A after the lamp warms up). This forms a ballast behaving very close to US CWA , so able to supress the mains variation and all subharmonic components.
Definitely using a high frequency ballast is really asking for the resonance troubles...
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OK, thanks - the capacitor looks like a better idea.
Didn't know that a plain short arc-tube could shift the HF ballast frequency that way...
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The selfoscillating (and some other as well) ballasts shift their frequency according to the load voltage.
Plus the arc voltage gets altered by the resonance.
That means the frequency will respond to the arc changes in the lamp and the arc will respond to the frequency, nice feedback system.
So unless the ballast is specially designed to prevent those resonances (the fluorescent ballast isn't for sure), the Murphy's laws "take care" it will combine in the worst way possible...
With the capacitor you may adjust the current above or below the choke rating, but going above means overheating it. But as the F58 is a 0.65A rated and the 50W MV 0.6A rated, the overheating would not be any problem here.
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35W MH ballast works perfectly with the 50W MV lamps. use a magnetic ballast though. lamp current is right at .42A and 50W
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Alights, lamp current or line current? If the lamp current is 0.42A, that's worse than an F40T12 ballast - you might as well use one of those.
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With LAMP current of 0.42A the LAMP power won't be more than about 35W. However the ballast INPUT power could well be about the 50W...
But from an MH 35W ballast feeding a 50W MV I would expect the lamp current nearly 0.5A, so lamp power around the 40W mark.
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will check the input current in 2 weeks when I'm back in town. I want to say it was .65A or something
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It doesn't really matter; the input current won't give us a good estimation of lamp power. The best metric we have without lamp current is the 50 watt input power, which translates roughly to 40 watt lamp power.
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Lamp current was at .42A like stated above. .
to find wattage I did .42×120V=50.4W
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Oh, sorry. I asked if you meant lamp or line current and you didn't say which, so I assumed the latter.
To get lamp power you have to multiply the lamp current by the arc voltage (about 100V for a 50 watt MV lamp), not the line voltage. Then you have to multiply the result by ~0.85, which is the inherent power factor of a discharge lamp. So we're talking lamp power of about 35 watts.
Strange though, that's rather low; no better than an F40T12 ballast. As Medved said, a 35 watt MH ballast should push around 0.5A. Do you know who made the ballast?
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ahh makes sense! I will remember that for future reference
Thankyou Trent
bummer the ballast under powers the lamp.it's an advance ballast kit
however the 50W MH ballast seems to overpower it, and I've been using it with 75/80w MV
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Yeah, the 50 watt ballasts are best suited to 75 watt lamps.
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I will repeat what was already said, but that error is far too common here:
For assessing the real power delivered to a lamp you have always study the lamp circuit (between the lamp and ballast), the mains current does not tell anything.
I know here are many statements (e.g. about 2xF40 lamps) like "That ballast draws 0.8A from the mains, so it is full power ballast" or so, but such conclusion is completely wrong: The fact the ballast consume 0.8A does not automatically mean it feed the lamps by the full 40W. The only thing it means it loads the mains by 0.42A, but it could still feed the lamps just by 25W.
OK, it does mean the total lamp power plus losses can not be more than 100W, but that information is not what was supposed to be judged...
Only with special cases the lamp current is equal to the mains current: Either not compensated series reactor ballast, or (way more rare case) just by a coincidence the mains and lamp current happen to match. The series reactors are quite common in the 230V area, but not in the uncompensated form.
And as the F40 arc voltage (~105V) is very close to the US mains voltage (120V), the high power factor (~0.95) F40 ballasts with the typical ~85% efficiency happen to have the lamp current not that far from the mains input current, but that belong to the coincidence case - if another ballast would be just normal power factor (~0.6) and feed the lamp by just 25W (common with the shop lites), the mains current would be similar 0.4..0.45A per lamp. And the conclusion "0.45A mains current => lamp fed by 40W" would be completely wrong.
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I know here are many statements (e.g. about 2xF40 lamps) like "That ballast draws 0.8A from the mains, so it is full power ballast" or so, but such conclusion is completely wrong: The fact the ballast consume 0.8A does not automatically mean it feed the lamps by the full 40W.
I've seen cases of such statements here, but the other way around. Many members deem the Advance Mark III Energy Saver ballasts to be reduced power because they are 0.73A (and probably because they have "Energy Saver" in the name), but I have confirmed through my own measurements of lamp current that they are full power.
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I think I just recently seen a few 35 watt metal halide ballasts on E-bay, if I see them again, I'll post the links.
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The 50 watt mercury vapor ballast I have drives the lamps at approx. 60 watts :-\ 0.61A @ 99 volts
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That's exactly perfect. 99V by 0.61A is about 51 watts. You have to factor in the lamp's power factor, which throughout my testing I've found to be around 0.85.
The 50W MV lamp is rated for 0.60A.
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Oh yeah lol, I forgot about that :-[