"Bob Engelhardt" wrote in message
...

OK, here's the next thing about LEDs: they're current devices. I.e., you
design your circuit for the current through them, rather than the voltage
across them. And LED brightness is a matter of how much current you put
through them. For your lights, I would use the .027A (27 ma). The .067
is bogus: how can a bulb draw more than is going through the whole string?
Although the .067 might be a maximum current.

So, assuming a 6v DC supply and 3.1v dropped across the LED, there has to
be 2.9v dropped across the resistor. With .027A needed, a 2.9/.027 = 107
ohms. 100 ohms is a common value - use that. .027A & 2.9v gives .08
watts dissipated by the resistor - it will not heat up the candle stick.
And you can use the smallest resistor (1/8W).

Each LED will need a resistor.

BTW - it you want to use a separate wallwart for each, you can get them
for $2.59 each (shipped) he
http://www.ebay.com/itm/170898516293

If you want to carry this further, you can experiment with the brightness
by changing the 100 ohm resistor. Smaller resistor equals brighter LED.
But shorter life. Too small means zero life. If the .067A LED rating is
a maximum current, you couldn't use a resistor smaller than 2.9/.067 = 43
ohms. And at that, the life would probably be very short. IME

Have fun,
Bob

Sounds like a good plan, thanks, except I'm a little concerned about one
thing: actual wallwart voltage. Very often, they're marked as 6V, but
output will be a volt or two higher. In that case, I'm assuming I'd just
need to increase the resistance slightly to each bulb, correct? If 7V, then
3.9v dropped across the resistor so 3.9/.027= 144 ohms (I would use the
closest standard resistor, probably 150 ohms). So power would be a little
more, but I could still use a 1/4 watt resistor. Please correct me if I'm
wrong. Been years since I've done much with electronics calculations, but
they're coming back to memory somewhat.

Bud