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[Lcdproc] Using multiple display


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  • From: eric.ross.c AT gmail.com (Eric Ross)
  • Subject: [Lcdproc] Using multiple display
  • Date: Tue Aug 22 18:47:01 2006

Joris Robijn wrote:
> On 22 Aug 2006 at 13:18, Eric Ross wrote:
>
>> Hello there
>>
>> I'm planning to use 2 HD44780 compatible displays to show some info, but I
>> cant really figure out if I can use lcdproc to control two unrelated
>> displays (each one displaying different data).
>
> Yes you can, but then you should start two instances of LCDd. And you
> need two parallel ports...

Well, I was thinking about somehow using one parallel port and 4 bits to
control one display and next 4 bits to control the second one... nevermind :P

> Or alternatively write a client that displays the data in such a way that
> each display shows exactly what it should.
>
> The data that is visible is placed by a client program on what is called
> a screen. A screen can span multiple vertically stacked HD44780 displays.

So one possible solution is connect the two display to the same parallel
port,
and make the client do the magic of doing the separation of the data, even
when
in reality the displays are stacked together. I'm right here ?

I've looking at the man page reading about the protocol, and it appears to be
possible to write such thing

> The situation for one or multiple displays on one LPT port is documented
> online.

Yep, I'm reading the user guide for the 0.4.x version, but I dont understand
the wiring using the 4-bit interface:

printer port LCD
D0 (2) D4 (11)
D1 (3) D5 (12)
D2 (4) D6 (13)
D3 (5) D7 (14)
D4 (6) RS (4)
D5 (7) RW (5) (LCD3 - 6) (optional - pull all LCD RW low)
D6 (8) EN (6)
D7 (9) EN2 (LCD2 - 6) (optional)

So the first display must be wired like this:

printer port LCD
D0 (2) D4 (11)
D1 (3) D5 (12)
D2 (4) D6 (13)
D3 (5) D7 (14)
D4 (6) RS (4)
D6 (8) EN (6)

but the second display, this way?:

printer port LCD
D0 (2) D4 (11)
D1 (3) D5 (12)
D2 (4) D6 (13)
D3 (5) D7 (14)
D4 (6) RS (4)
D6 (8) EN (6)
D7 (9) EN2 (LCD2 - 6) (optional)

is this correct ?

Thanks in advance.
Eric




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