Have any of our North American members tried to use tanning bed ballasts on 240V for lighting ? While browsing the Net, I found this website They have a 36 watt 220V choke, which I assume would work on 240V. If I decide to order, it would be to light a standard 36 watt (non-uv) lamp. My guess is that it would work, however I would like to know what you think.

PS : I am absolutely not a tanning bed enthusiast. I have never used them, nor do I intend to do so.

When using Euro 36w T8 lamp/ballast on 240v 60hz it may underdrive the lamp a tiny bit (nothing srious, probably not very noticeable) due to the 60hz, except that i think it'll work well

Ash: The rated voltage and frequency of the magnetic ballast must match the actual mains voltage and frequency.
The ability of the magnetic ballast to operate on variations in the mains voltage, is within a limit of 10%.
This isn't means that a 240V, 50hz ballast can be operated in the US 240V, 60hz dryer and A/C sockets, or in Israel and Europe, and an European/Israeli 230V 50hz ballast can be operate in the US dryer and A/C sockets or in the UK.
This isn't the case with electronic ballast.
http://www.eltam-eh.co.il/%D7%A9%D7%90%D7%9C%D7%95%D7%AA_%D7%95%D7%AA%D7%A9%D7%95%D7%91%D7%95%D7%AA.html

These ballasts are, however, rated for 60 Hz operation instead of 50Hz. If this project happens, they would be operated on 240V 60Hz, instead of 220V 60Hz.

In case of 60hz ballast on 60hz three won't be underdriving resulting from frequency. So we re just left with the overpowering due to the 240 vs 220. The overdriving will be quite small - i'd not expect any adverse effects on hte lamps or ballasts

Dor : The manufacturers rate the ballasts according to quite precise tolerances of lamps ratings. In fact, lamps and ballasts can still work quite well with way worse conditions, and nothing bad happens unless you go really too far

The result of 50 vs 60 hz, when using 50hz gear on 60hz, an inductive circuit will underpower a bit (about 5/6 if ignoring ballast losses), capacitive overpower a bit (by 6/5 if ignoring losses). Not too bad

Thanks for the info, Ash. It confirms what I thought. If ballast voltage tolerances are generally said to be +/- 10 % of rated voltage, a 220V ballast could go as high as 242 V. Even pushing it further than 242 V would probably not hurt it. A stuck starter would probably be worse than applying something of the order of 250 V (which would probably not happen) and these ballasts are probably quite good at surviving stuck starters.

When time permits (could be a long time), I might try to order some, although I probably should not wait too long before ordering...

Thanks again.

This is consistent wiht my experience. Have a soviet 9w PL fixture for 220v 50hz, been running it on 240v and higher (even 250 and higher) and nothing hapened to it or the lamp

What counts is only the part of the voltage droping on the ballast. The less voltage drop on the ballast in the 1st place, hte more the change will be pronounced - as the lamp arc voltage is about constant, so all the change is on the ballast only. Its going about like this :

Assume a 20w lamp (57v arc) on 220v ballast. Ballast gets 220-57 = 163v. at 240v itll be 240-57 = 183v. 183/163 = approx 1.12x overdrive
Assume a 40w lamp (103v arc) on 220v ballast. Ballast gets 220-103 = 117v. at 240v itll be 240-103 = 137v. 137/117 = approx 1.17x overdrive

Assume a 20w lamp (57v arc) on 220v ballast. Ballast gets 220-57 = 163v. at 240v itll be 240-57 = 183v. 183/163 = approx 1.12x overdrive
Assume a 40w lamp (103v arc) on 220v ballast. Ballast gets 220-103 = 117v. at 240v itll be 240-103 = 137v. 137/117 = approx 1.17x overdrive

You can't subtract voltages this way because the are not in phase. The voltage over the ballast is 90 degrees out of phase from the current (assuming it is an ideal inductor, ignoring its resistance), but the voltage over the lamp is approximately in phase with the current (the waveform is distorted, ignore that). Imagine the voltages represented by 3 vectors: the length of the first vector is proportional to lamp voltage, and the second vector represents ballast voltage. Adding the vectors together makes a right-angled triangle, since there is an angle of 90 degrees between them. Because of this, the total voltage (which is mains) is given by the Pythagorean theorem, so we get:

Vmains^2 = Vballast^2 + Vlamp^2, or

Vballast = sqrt(Vmains^2 - Vlamp^2)

So for your example for a 20 W lamp:

Vballast(220V) = 212.5 V
Vballast(240V) = 233.1 V

233.1/212.5 = 1.097

Also, power into a resistor is proportional to voltage squared, so the overdrive is 1.097^2 = 1.203x. For the 40 W it is 1.24x.

Power into the lamp is proportional to the voltage non squared due to the arc voltage being constant. Only the current changes and that is linear with the v drop on the ballast

I meant the power the ballast has to dissipate as heat.

Power into the lamp is proportional to the voltage non squared due to the arc voltage being constant. Only the current changes and that is linear with the v drop on the ballast

What matter with magnetic ballasts, is the ballast losses. And because the ballasts are more-less linear devices, their losses are proportional to square of current, so square of the voltage.
It is mainly the ballasts, what make the lamp circuit so vulnerable towards long time, but even slight mains overvoltages. Extra 10% of current mean only 10% lamp power increase, but ~20% ballast losses increase. And the extra 20% of ballast losses mean their temperature rise would be 20% higher.
In order to keep the mass reasonable, the ballast design have very little margin for any extra dissipated power: The designed temperature rise is usually about 70degC, already very close to the limit for usable life.
Therefore the long term overvoltage shall never exceed 5%, what mean about 6..7% current increase in higher arc voltage circuits (like 40/36W, lower arc voltage circuitss suffer less, as the ballast voltage is already high), what is about 12..15% power dissipation increase, so the temperature rise would become nearly 80degC. With 40degC ambient it mean 120degC in the winding...

With 10% voltage rise the current rise by about 12..15% (still not big deal for lamps), so ballast power dissipation increase by 25..30%, so temperature rise about 90..100degC. That mean the winding could reach easily more than 130degC, what could be survived only for shorter time.

Of course, the thermal budget strongly depend on fixture's thermal properties, the figures belong to more-less worst case scenarios...
But don't forget the winding is not the only thing affected by the elevated temperature. All the "crispy plastic" connector issues are strongly linked to the ballast operating temperature as well, so although the ballast itself may survive (they are made quite tough), the surrounding components could fail solely due to the elevated ballast temperature...