24 Feb
2010
24 Feb
'10
8:30 a.m.
Dave McGuire wrote:
On Feb 23, 2010, at 9:14 PM, Brent Hilpert wrote:
I just looked up my old notes and tried it again:
- two LEDs in reverse parallel, or LED and diode
- C = 0.22 uF, 200V
- R ~= 680 to 1K, 1/4W
This targets around 10mA at 120VAC.
~0.2W consumption. Far more efficient than the straight R solution which
consumes around 1.2W, and hence 2W resistor.
Being pedantic, the power company might not like it as it adds reactive
(out-of-phase, power factor) current to the grid. On the other hand, they might
like it as it is capacitive reactance which helps counteract all the inductive
reactance from all those little unloaded wall-warts and transformers people
plug in.
I think I've seen such a circuit used in manufactured equipment too, but I
can't remember what.
I have on occasion run single LEDs from line voltage by paralleling
the LED
with a diode, then putting the pair in series with a C and R. C is
calculated
to provide most of the voltage drop via capacitive reactance, a
small R is
still present to limit inrush current. The diode limits the reverse
voltage and
permits bidirectional current flow so the cap sees AC.
Power consumption and heat dissipation are reduced compared to a
solution using
only R, and a power resistor is not needed.
I haven't tested the circuit over many on/off cycles though, to
find out
whether perhaps the peak in-rush current reduces LED life.
Wow...that's pretty impressive work! Elegant in its simplicity.