In article , Tony wrote:
Don Klipstein which is me wrote:

I snip to what I said about diodes to edit for space

The diode types do indeed extend filament life greatly - but I consider
them energy wasters. They only reduce power consumption by about 40-41%
typically, while typically reducing light output by generally at least
70%. Energy efficiency is roughly halved. To restore original light
output, you would need to nearly double the amount of electricity used -
costing much more than what is usually saved by saving lightbulbs.

I'm sure they reduce power by less then 40%, I thought it was closer to
30%. I did a test in my shop some time ago but don't know where I wrote
down the numbers.

A silicon diode combined with 120 volts AC means RMS voltage
delivered to the load very close to 84 volts, 70% of full voltage.

I did some testing in this area, which I publish in:

http://members.misty.com/don/incchart.html

For the "USA-usual" 120V 100W "A19" lightbulb with rated life
expectancy of 750 hours and light output 1710-1750 lumens (1670-1710
for "Soft White", at 84 volts:

RMS current is 82.5% of "full". .825 times .7 is .5775, or close
enough to 58 watts. Add to this power consumption of the diode, which
in this specific case is about .45 watt - total is about 58.5 watts,
58.5% of "full power".

Though I do report 28.4% of "full" light output.

120V incandescents with vibration resistant filaments and/or of lower
wattage will have wattage decreased a little less and lumens decreased a
little more than is the case of the example above by use of a diode, as
long as design watts and design amps don't get so low that an economy of
scale makes a vaccum fill gets better than a gas fill. (Vaccum in 120V
incandescents is used in one extreme example of a 60 watt one, most
tubular 40 and 25 watt ones, and most of lower wattages with
multi-supported filaments such as C-7 filament style.)

In the above web article of mine, I do state that a specific 25 watt
120V gas-filled incandescent at 84 volts has current consumption 83.5% of
"full", meaning power consumption about 58.5% of "full". Add another half
percent for the diode, and this data point becomes 59%.
I have yet to determine light output here, but I would guesstimate about
24-25% of "full".

- Don Klipstein )