At 12:31 AM 3/18/2004 +0200, you wrote:
>Hi,
>
>Sorry to reply to my own mail, but I just had another thought.
>
>Won't using this algoritm change my 'duty cycle' from 1/4 to 1/32 since now
>I can only have one segment on at a time. Will the 80ma I can now supply
>that single segment make up for the change in duty cycle?

No, because the PIC can supply 25 * 8 = 200mA vs. 80mA. Thus your display
will be much dimmer.

>Also, could the algorithm be modified to the following:
>
>Segment Count, SC = 0
>
>1) Turn off all displays
>2) Turn on SC segment with 7445 driver.
>2) For each display (1 to 4), see if this segment is on or off
>3) Turn on those displays that have that segment on
>4) Increment SC
>5) If (sc > 7), se SC = 0
>6) Delay for fixed time
>
>That should change the duty cycle to 1/8, right?

Yes you can do this, and it is better. You must move the resistors to
the anodes from the segments (so there are fewer of them). But the
total display current is still limited to 80mA, so the net result is
that you've added a part which draws 50mA all by itself, and have a
display that is about 1/3 as bright, but save 4 port pins and 4 resistors
(or 1/0.5 network).

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

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