There's a simple formula for driving filaments that more or less holds true for lower voltages - one of the UK TV makers used the "wattless dropper" for the 300mA heater chain, the capacitor was 4.3uF which by dividing the 300mA down to the required current and then dividing the capacitance by the same amount gives the value required.
Use a transformer. Or a 240V lamp.
Using a resistor will result in a 20-fold increase in power consumption. Also, if the lamp fails open (which it will eventually), it will have to withstand the entire 240V.
That never bothered the radio manufacturers who made AC/DC sets with series heater chains and the dial lamps in series with the heater chains.
????? The rectifier tubes were designed with a tap for a #47 lamp. They didn't use a string of resistors simply to light the pilot lamp.
Hmm... That's a misunderstanding. As long as the lamp is good, the resistor has to withstand almost the entire 240Vac. To be precise, 240-12=228Vac. It also has to dissipate the 12W mentioned by me and others. Once the lamp fails open (as incandescents usually do) the resistor feels neither voltage nor current.
petrus bitbyter
The current neon bulb style is NE-2. I can buy these from Dick Smith for around 45 c, and a 270k resistor for 6 c . I can probably borrow a glue gun from my father. I think thats the simplest and cheapest way to do it.
A resistor would be a dreadful choice for a whole host of reasons, not least the fact that you'll create 20W of waste heat for every 1W of lamp rating.
Graham
I meant that the (12V) lamp will have to withstand 240V, as the resistor won't be dropping any of it if the lamp isn't drawing any current.
This may not be a problem, but there's no easy way to tell. Lamps aren't designed to be used this way, so the only voltage figure likely to be given is the one which produces the desired power. There isn't likely to be a separate insulation breakdown voltage quoted for the (unlikely) case where an open filament results in a much higher voltage.
Read my post properly. You need a resistor of more than twice that power.
Rob.
As I stated in my previous post, I'm sticking with neon lamps. The existing resistor has testing fine, so now I just have to pay a whole
44 cents for an NE-2 lamp. Cheap and easy enough.
If the lamp filament opens there will be no current flowing. The resistor is in series with the lamp, so no current means no voltage drop across the resistor.
Yes, there will be the the full 240V at the lamp socket when the lamp opens or it is removed. This could be a shock hazard depending upon the design of the lamp socket.
Lamps are designed to run at a specified voltage and current. No more, no less. How one achieves the correct operating conditions is down to practicality, safety, and common sense.
In Europe it was common practice to save the cost and weight of a transformer by having all the tubes with equal current rating instead of equal voltage so they could be wired all in series, if a panel lamp was required this would also be added to the series chain with the same current rating as all the tube heaters. These were generally known as AC/DC sets and were easy to spot when first switched on, as the bulb filament had much lower thermal inertia than the tube heaters it would light overbright for a split second as the heaters got up to temperature - so lamp failure was far more common than heater failure and the blown lamp would have the full
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