At 10:37 4/18/2000, Chris Smith wrote:
>Wiring your own is fairly easy.  Each LED should get around 1.5 to 2.2v
>(check on package)  if you wire them in series (positive to negative) in a
>daisy chain, all you need to do is to divide the voltage available by the
>voltage of each LED to get the number of LEDs in each string.  For most
>LEDs, 7 would be the correct number (2v times 7= 14v)

Ok, yes, you can wire them in series.  Shoulda thought of that.  But you
*absolutely* need to have a current-limiting series resistor, because these
things basically act like switches, on or off.  If you don't use the
resistor you'll either see darkness or learn what burning epoxy smells like...

I'd use a maximum of 5 reds in series, myself, on a 12v automotive
circuit.  Assume the red LEDs are 1.8v each, totalling 9.  The formula is
Rlimit [ohms] = (Vsrc - Vled) [volts] / I [amps].  If you use milliamps,
the answer is in kilohms.

So for 20 ma at 14.3 volts, you need 260 ohm, quarter watt. (Worst case)
At 13.5v you get 17 ma.
At 12.5 v you're down to 13 ma.

If you use six @ 1.8v, total 10.8v.  For the same conditions, resistor
would be 175 ohms, quarter watt.  That would give you:
14.3v >> 20 ma
13.5v >> 15 ma
12.5v >> 10 ma

For seven @ 1.8v, total 12.6v.  Resistor 85 ohms, quarter watt.
14.3v >> 20 ma.
13.5v >> 11 ma.
12.5v >> no light.

So you see that the bigger your series string is, the more light variation
you will get from different system voltages (but smaller power dissipation
by the resistor).  For contrast:

One LED @ 1.8v.  Resistor 625 ohms, half watt.
14.3v >> 20 ma.
13.5v >> 19 ma.
12.5v >> 17 ma.

david

David Beierl - Providence, RI
  http://pws.prserv.net/synergy/Vanagon/
'84 Westy "Dutiful Passage"
'85 GL "Poor Relation"