In article , 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%?
The answer is no, an "ideal diode" (with zero voltage drop) reduces RMS
voltage by about 29.3%, to original voltage times the square root of 1/2.
RMS means "square root of mean square".
This means that supply voltage being blocked half the time has the
average square or any other average-of-function-of input voltage being
halved.
Square root of mean square of 50% "duty cycle" is about .707.
======================
As for another way to analyze this: Adding an "ideal diode" in series
with an "ideally resistive load" powered by AC means that power
consumption is halved.
Since Ohm's Law says that power cosumption by an ideal linear resistor
is proportional to square of voltage, the "representative voltage" as a
result of adding an ideal diode to halve the duty cycle is not halved but
reduced by 29.3%.
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.
Which is true if RMS voltage is halved. Adding a diode to block half of
the AC cycle ideally does not halve RMS voltage, but ideally reduces RMS
voltage by a factor of square root of 2.
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.
By one somewhat-published "one-size-fits-all" sort of formula, this is
true, assuming effective/RMS voltage is halved.
Disregarding forward voltage drop, wouldn't a half-wave rectifier reduce
RMS voltage by 50%?
No, the "mean square" of input voltage is halved. Half the squares of
input voltage are unchanged, and half the squares of input voltage are
changed to zero.
Square root of mean square of input voltages becomes 70.7% of original.
70.7% of original effective voltage applied to an "ideal resistor"
(resistance not varying with input voltage or temperature) cuts
its power consumption and dissipation (heat output) by half.
70.7% of original effective voltage applied to a typical incandescent
lightbulb means power consumption about 59% rather than 50% because the
filament's resistance varies directly with its temperature.
(And the lower filament temperature here roughly halves the already-low
percentage of its 'outgo' being in the form of visible light.)
Please feel free to followup or to e-mail me ) if I have
been insufficient so far in explanation here.
- Don Klipstein )