Ian Chapman wrote:
>
> It seemed so simple.  I wanted a yellow LED in my PIC16C84 design, so I
> connected it via a 3K3 resistor between RA4 and Vdd.  So far so good.
>
> Then I noticed something odd.  With the PIC spending most of its time in
> sleep mode, its quiescent supply current (with the LED off) varied from
> 30uA to 90uA in sympathy with the ambient light level - disconnection of
> the LED removed this variation.  A voltmeter (>1M impedance) showed RA4
> to sit at +3.6V relative to Vss when its open-drain output was inactive.
>
> It seems as if the photocurrent injected by the LED is entering RA4 and
> affecting the operation of the PIC (maybe via the Schmitt input buffer).
>
> Several questions spring to mind:
>
> 1.  What is causing the supply current variations?
> 2.  Is there any risk of misoperation of the PIC in this configuration?
> 3.  Would a non open-drain port pin be a better choice to drive an LED?
>
> Any thoughts?
>
> - Ian Chapman

Ian,

the inputs of CMOS circuits like the PIC don't like intermediate
input levels, because then you partially enable both of the input
amplifier transistors, making a weak rail-to-rail connection.
The more you come to the center voltage (2.5V) the more current
flows. The PIC datasheets show the input stage of the port pins,
and - as I think - also the current consumption under various
conditions.
If you would use a non open collector output, this effect should
not happen, because you short-circuit the LED
(plus-rail * LED * port-pin at Vdd).

Wolfram


--

+-----------------------------------------------+
! Wolfram Liebchen, liebchen@ipserv.ffo.fgan.de !
!        Forschungsinstitut fuer Optik          !
+-----------------------------------------------+