Re: SOLAR LIGHT BATTERY ELIMINATOR QUESTIONS

MrYooneek@xxxxxxxxx wrote:
Hi, I am planning to use a power supply instead of the batteries in my
solar landscape lights because of a shade problem and the uniqueness of
the lights. With the two re-chargeable batteries installed I am
measuring 3vdc @ 500 micro amps. I am wondering if this measurement is
correct.

Was that with the light operating? It sounds like standby current draw. Or are you measuring the battery voltage and reading the current rating off the battery (which would likely be 500 milli-amp, not micro-amp).

My experience is that most leds require 10 milliamps. Did they
find a way to operate them so efficiently? There are two low intensity
leds in each lamp and one flickers. I figure at 500 microamps, apiece,
hooking 11 of these lamps in parallel I would need a power supply
capable of an output of 3 volts at 55 microamps?

How do you figure this? 500 x 11 = 5500uA (or 5.5mA). Assuming the 500uA is right.....

Since the ultra low
current I figure a plug in xfmr like a phone charger would work and I
would have to use a resistor to limit the current since most of these
are in the 100 ma and up range? What about a battery impedance problem,
by leaving the batteries out and hooking the power suppy directly to the
battery terminals? Will this be a problem? Many thanks in advance.
Dale

You'd want a resistor in for sure - measure the current draw when a light is on and apply Ohm's Law to figure out a resistor for the number of lights you want to hook up. Assuming the LED's are really 10mA each and there are two each light, that gives you 220mA total. If you use a 9VDC plug-pack/wallwart you'd want a resistor about 15 Ohms. There's probably an intensity control built into the lights so you may want to play with just one light to start with in case you fry it.

Cheers.

Ken
.

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