I
ian field
Guest
"Bob Monsen" <rcmonsen@gmail.com> wrote in message
news:KK%1j.24352$JD.21058@newssvr21.news.prodigy.net...
for lower voltages - one of the UK TV makers used the "wattless dropper" for
the 300mA heater chain, the capacitor was 4.3uF which by dividing the 300mA
down to the required current and then dividing the capacitance by the same
amount gives the value required.
news:KK%1j.24352$JD.21058@newssvr21.news.prodigy.net...
There's a simple formula for driving filaments that more or less holds true"ian field" <dai.ode@ntlworld.com> wrote in message
news:IeX1j.31839$dN2.17700@newsfe6-gui.ntli.net...
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
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
doesn't care about this, so it may be a better choice.
As others have pointed out, the best and cheapest choice may be a glue gun
and a neon indicator bulb which is rated for 240VAC.
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.