L
Lostgallifreyan
- Jan 1, 1970
- 0
I'm not sure if have it figured out by now, but here is a simple way
to figure it.
Consider the led's red or green has a forward drop of about 2 volt,
and its equivalent resistance at 20 ma will be 100 ohms. Now take 2)
100 ohm resistors to make a voltage divider ( + 5 volt to resistor a
with resistor b in series then connecting to ground). now with the
back to back led connected to the center of your divider being 2.5
volt and the other end of the led to the Pic output, A high or a low
will give roughly 2.5 volt difference on the led. You may or may not
even need a resistor to drop the extra .5 volt but if you do, you end
up with 3 resistors and the led instead of the extra opamp's. I would
still consider at least a transistor buffer on the Pic outputs to
drive the leds. JTT
Do you really think that 24 transistors are easier to handle than an op-amp
IC to make the 2.5V rail?