E Z Peaces wrote:
Don Klipstein wrote:
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 )

Wouldn't a diode reduce RMS voltage by 50%? If that's true, your chart
says current would be 68.5%. Power would be 34%. Your chart indicates
that the lumens of a 100W bulb would be 125/1750, or 7%, an 80%
reduction in lighting efficiency.

By formula, power consumption is proportional to voltage^1.6, and light
output is proportional to voltage^3.4. That would mean 9% of the light
for 33% of the power or a 73% reduction in lighting efficiency.

Disregarding forward voltage drop, wouldn't a half-wave rectifier reduce
RMS voltage by 50%?

Rats! It would reduce RMS voltage by 29.3%!

So, by formula, power would be 57% and lumens would be 31%, a 46%
reduction in lighting efficiency.