> best rgds,
> Davis

In addition to Bob Ammerman's reply, to put the resistance in
a formula

R = (E-Vf) x 1000/I

Where R=ohms, E=supply, Vf=LED forward voltage, I=LED
current in mA

To use Bob's example

R = (12-2.1) x 1000/25 = 396. You should get good brightness
with most LEDs using more than 390, say 1k8 (=5.5mA). Driving
a red/yellow/green with a PIC pin would be fine in most cases
with 1k

This applies to DC supply. There are times, say when using a
regulator that you don't want to load any more than necessary,
when you'd rather put the LED on the AC from the transformer.
What you do is put a small diode (like a 1N914) in reverse
parallel to the LED and use half the resistance that you would
have done had it been a DC voltage

You can also put an LED acrross the mains, but I think it's
frowned on, by adding a 220nF X2 rated capacitor in series
(R and the reverse parallel diode are there too) to limit the
current

Another usage is with pulsing drive, such as moving signs and
infra-red transmitters. In these cases you calculate the average
amperage. This is related to the duty cycle of the wave (as per
AC). For example, an infra-red transmitter LED may have just
1 or 2 ohms in series, because it is being driven with pulses
that can be measured in microseconds at a fairly low duty cycle.
IOW, roughly, if the duty cycle is 1:99 (the LED is on for only 1%
of the time), the resistor is reduced accordingly. During the 'on'
time, the LED may be passing several amps, but it is for such
short duration that it doesn't cause destructive heating. This
information should be included in the LED's datasheet

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics