> I have an LED that needs 1.6Volts to light up. and I have a 9 volt
battery.
> now I see every calulation to go from Volts, watts, resistance, amps,
etc..
> What I have not seen is what the end result is. It I have a 9 volt battery
> and a 40ohm resistor then I end up with a amp of 0.225.

You know that the LED will use far less than 225mA ? Probably it will
be bright enough at 5mA using a resistor of 9000/5 = 1400 ohms, the
LED specs will give you a more exct answer

Also, a load of 225mA will fry a PIC output. Max is 25mA but you'd
want to keep it under that. Maybe 20mA for prolonged use

> Ok now is it better to have a PICs chip connected to the ground pin
> of an LED or is it best to have the PIC control the voltage to the
> device? will either solution result in a voltage leak or drain?

Microchip claim 25mA source or sink for a PIC pin (although I'm
sure someone here mentioned that the PIC can actually sink very
slightly more than 25mA, but for practical purposes it was not
that important)

Assuming then that sink/source are both 25mA, it wouldn't appear
to matter whether you have the LED as high side (PIC grounding
the LED) or low side (PIC supplying the current). If you're using the
PIC as a current source (LED is between PIC and ground) then
setting the PIC pin low will put both sides of the LED at 0V and you
should have no leakage. Similarly if the LED is on the high side.
When the PIC pin is at 5V, then both sides of the LED are at 5V.

However, that's assuming that the specs in the DC section of the
manual are understated. A "low" is given as 0.7V max and a "high"
as Vdd-0.6V min. I've just measured an F628 with a driven LED
and the "high" is Vdd-0.03, which implies pretty low leakage

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