Hi Tomas,

The power dissipated by the PIC in this case is NOT (Vdd-Vss)*Current.
It is (Vdd-Vpin)*Current. So, if you are supplying the PIC with 5V,
and the voltage under a 25mA load on the pin is 3V, then the
contribution to the total PIC power dissipation from this pin is
(5-3)*0.025=3D50mW. This is all being dissipated by a tiny FET driver
inside the IC.

At 3V supply, the situation would be something like the following: 1V
on the pin at a 25mA load, giving (3-1)*.025=3D50mW still. The voltage
drop between the PIC's Vdd pin and the output pin comes from the
current (25mA) times the on resistance of the FET driver for that pin.

Another problem which Mark alluded to is that the voltage on the
output pin may drop considerably (as in the above example, 2V drop) at
such high currents. This is because you are beginning to take the FET
driver out of the ohmic region into the constant current region.

I suspect that if you shorted one of the output pins to ground and
drove it high, it would probably source about 100mA. This is the
reason why some people don't feel the need to use current limiting
resistors with LEDs on PIC pins - it will often work out that you will
get something like 20mA out of the pin when you force it to the Vf of
a typical LED. The problem with this is that PIC output pin resistance
varies from IC to IC, and with temperature, and the same is true of
the Vf of an LED, so that perhaps 10% of the products produced this
way will have out of spec currents which will damage either the LED or
the PIC output driver.

You are extremely ambitious, which is good to a point. You remind me
of myself at about 16 years old. That's about the time when I joined
the PICLIST (I'm 28 now) and I have learned a LOT. Please try to be
patient and learn from the good people here - even those who can be
gruff at times. If you really do take your product to production in
the millions, you will learn many things along the way, one of them
being that you WILL see almost every failure mode imaginable. Stuff
that you would never see doing hobby pic projects (because it has a
probability of 0.01%) will show up in some of your production units.
You therefore must rely on a good deal of planning and testing, and
re-design using info from failed units, to perfect your design.
Anything which is slopped together and just "works" will probably work
in 80% of production units, but there WILL also be that extra 20%
fallout or more due to something you didn't see in your single
prototype.

Sean


On Sun, Jul 13, 2008 at 9:49 PM, Tom=E1s =D3 h=C9ilidhe <toe@lavabit.com> w=
rote:
>
> Take the following simple circuit:
>
> PIC pin set high  -->  resistor -->  LED  -->  ground
>
> Now here's what I'm curious about:
>
> The datasheet cites the maximum current for an I/O pin as 25 mA.
>
> What I think is a little strange about this is that you can power the
> PIC at different voltages; I mean if the PIC's powered by 5 V, then 25
> mA equates to a power dissipation of 125 mW. However if you run the PIC
> at 3 V, then 25 mA means a power dissipation of only 75 mW.
>
> If you're allowed source 25 mA when running at 5 V, then should you not
> be allowed to source 41 mA when running at 3 V? (41 =3D 25 * 5 / 3)
>
> Just a thought.
>
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