On Jun 5, 9:53 am, "Dave Plowman (News)" wrote:
In article ,
Derek Geldard wrote:





IKEA have *New*-ly started offering a light module with 3 very bright
white LED's in a single plastic capsule "Grosby" £2.99.
They are very bright and eminently suitable for repacing 10W halogen
capsules to illuminate double/single glass fronted display cabinets
which fails often and offer severe difficulty in replacement.
However they are powered by 3 AAA batteries.
I've measured the current consumption, they draw 56 mA from 3 AAA
cells in series. So I rekon they'll last one day or thereabouts.
So what is needed is a means of efficiently powering them from the
mains, that avoids the high power consumption and failure of the old
10W halogens.
I'm considering such tricks as powering them from 3 AAA rechargables
and charging them via a timeswitch or an electronic substitute so
that they charge only (say) 10% of the day (which surely should be
enough) to save energy. Given the heat losses in "Wall Wart" type
transformers.
Any better ideas from the panel ?

LEDs aren't concerned about voltage as such but the current passing
through them. So you could use any DC supply with a suitable series
connected resistor or better one resistor per actual LED.

There's a formulae to calculate the resistor value for a single LED. It
is:-

VS - VF
R = ----------
I

Where R is the required resistance value in ohms
VS is the supply voltage
VF the forward voltage drop of the LED
I the current required in amps

So if we take 12 volts, a current of 20mA (about yours per LED), and a

Assuming they are connected in parallel.

forward voltage drop of 2 volts which is an average, we get a figure of
500 ohms per LED. Use the nearest available resistor value as it's not
critical.

However, white LEDs often have a VF rather higher than others and it could

s/often/always/

be 4 volts. Or anything from approx 1 to 5 volts. ;-)

Show me a white LED datasheet that specifies a Vf anywhere near 1V ;-)

MBQ