# 12 V Lamp On 240 V

Discussion in 'Electronic Basics' started by Dave.H, Nov 24, 2007.

1. ### Dave.HGuest

I need to operate a 12 volt AC/DC filament indicator lamp on 230/250
volt mains. I'm guessing a resistor would do the trick. If so, what
value?

2. ### Greg NeillGuest

What's the current requirement for the lamp?
(alternatively, what's its wattage rating?)

3. ### Dave.HGuest

Website doesn't say. It's a 12 V LES style lamp. I can probably get a
5 watt wirewound resistor for it.

4. ### Dave.HGuest

I looked on the page for the replacement lamps. 50 mA

5. ### Robert WilsonGuest

I make that to be 4560W. Be advised though that the resistor will

The calcs, like an LED.

R = (240-12)/0.05 gives 4560R.

Power rating is P = 0.05^2*4560.

I would round it up to a 4k7 resistor.

May your luck be in. Don't blame me if you blow yourself up.

Rob.

6. ### Greg NeillGuest

So worst case is you'd want to drop 250 - 12 = 238 Volts
at 50 mA across the resistor. Right away you can see
that it'll have to dissipate 238V*50mA = 12 Watts.
That's a lot of heat to shed just to support an indicator
lamp.

The resistance would be 238V/50mA = 4700 Ohms near enough.

Maybe you should look at finding a 240V indicator lamp?

7. ### Dave.HGuest

Thanks. Found a 4.7k 5 watt, wirewound, ceramic cased resistor on the
Dick Smith website.

8. ### Dave.HGuest

I have a neon indicator lamp installed at the moment, it's just about
dead, and there's no way to hold it in place. I don't have a hot glue
gun.

9. ### petrus bitbyterGuest

Well, according to Ohms law:

R=(240-12)/50=4,56k so take 4k7.
Wattage (240-12)*50=11400mW so you will need at least a 12W resistor.
The thing also will become pretty hot. Generally speaking, this is not the
best way to use a 12V lamp.

A capacitor of about 68uF might do a better job. No need to say it should be
able to handle at least 250Vac.

Electronics may do even better but are more complicated. Not the thing I'd

BTW. Why do you need this extraordinary application? Did you try to find
other solutions?

petrus biybyter

10. ### Dave.HGuest

Perhaps an LED instead? A regular red, 20 mA one will do.

11. ### petrus bitbyterGuest

You can buy one. Small types are dead cheap overhere. Guess the resistor you
need may be more expensive. Especially as it will frie your equipment.
Otherwise, you can use silicone kit or construction kit. They're available
in small tubes.

Neon indicator lamps are for sale in many styles. A lot of them with build
in resistors. Whats the current lamps type? Has it a thread or a bayonet
fitting? Has it a resistor build in? I should not be surprised if the type
is available if only you know where to look for it.

petrus bitbyter

12. ### Michael A. TerrellGuest

Why don't you use a 12.6 volt "Filament" transformer?

--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

13. ### Phil AllisonGuest

"Dave.H"

** Only if you intend to kill yourself or the next person who tries to
change that bulb.

What an extremely stupid question.

........ Phil

14. ### ian fieldGuest

A LED lends itself to the "wattless dropper" trick, at the mains frequency
the impedance of a capacitor is much higher than the load so it acts similar
to a constant current source. The capacitor must be rated for mains - at
least 400V, preferably 600V, and you need a resistor in series, about 1k2 to
limit current surge and the LED must have an inverse parallel diode to stop
its PIV being exceeded, try a 0.047uF to start with and work up to the
required brightness with successive higher preferred values.

15. ### BaronGuest

Are you aware that "Omron" do a 240v/12v panel lamp. Its basically a
low voltage transformer and lamp in a single hole fitting. Not a cheap
as putting a new neon lamp in though. You can buy those for a few
pence.

16. ### Meat PlowGuest

Use a transformer.

heh

18. ### ian fieldGuest

Well he was raised by dingoes!

19. ### Roger DewhurstGuest

Why not rely on the impedance of a capacitor? A good textbook will give
the the formula relating impedance to frequency and capacitance.

R

20. ### Bob MonsenGuest

He can do this with his 12V incandescent. Use a 0.68uF cap and a 10 ohm 5W
resistor in series with the 240V and the bulb. As you pointed out, the cap
must be rated for 400VAC, which means it won't be cheap.

If he uses an LED, he'll need another diode in parallel, backwards, and
probably a zener in parallel to prevent overvoltage. The incandescent