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?=20

Also, could the algorithm be modified to the following:

Segment Count, SC =3D 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 =3D 0
6) Delay for fixed time

That should change the duty cycle to 1/8, right?

  _____ =20

Andr=E9 Miller =09


-----Original Message-----
From: pic microcontroller discussion list =
[mailto:PICLIST@MITVMA.MIT.EDU] On
Behalf Of Andre Miller
Sent: 18 March 2004 12:06 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [PIC:] Multiplexing seven segment displays, correct =
resistor
values?

=20
Hi Byron,

Thanks, I guess I should have gone and read that datasheet. I just =
assumed
you meant a BCD to 7 Seg display driver. I understand now what you =
meant.

I'll add the driver chip and implement the algorithm you suggested and =
come
back if I have more questions :-).

> Your algorithm is slightly more complex now. Also to prevent bleed=20
> some
rearranging is in order:
>=20
>  0) Turn off all displays. Prevents bleed.
>  1) Set the correct segment for display X by incrementing segment=20
> count
(SC) on
>     7445 DCBA input. I'd normally use a shadow register for this but =
I'm
>     paranoid.
>  2) Check to see if the segment is supposed to be on.
>  3) If it's supposed to be on, then turn on the display
>  4) If SC > 7 then SC =3D 0 and X =3D X+1
>  4) If (X > 4), set X =3D 1 [You can also do this by shifing a 0 =
through
>     the 4 bits assigned to the BC557.
>=20
> Clock speed should be fine. Note that you should delay for each=20
> segment
whether or not it is turned on or off.


  _____ =20

Andr=E9 Miller =09

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