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com2 ,problems with bit stop

 
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marcelo_bahia



Joined: 15 May 2009
Location: argentine

Posted: 22 June 2009, 16:54 PM    Post subject: com2 ,problems with bit stop

I use the following configuration in Z24-P to transmit,com2 only tx pin11 8-p-2
(8 data bits, even parity, 2 stop bits at 115000.)
my problem is that no appear 2 stop bits y and alters the output bit
will have some problem with the progracion, which may be the error?
the z24 has a max232 connected, the images are a departure max232,view photo 8-p-1 (1 bit stop) and 8-p-2 (2 bit stop).

Code:
Const txPin as Byte = 11
Const channel as Byte = 2

Dim oq(1 to 40) as Byte

Dim b as Byte
Dim i as Byte
Dim Bstop as Byte

Sub Main()
 Bstop=2
 b=75  'letter K in ascii
   
  ' initialize the queues
  Call OpenQueue(oq)

  ' define the characteristics of the serial channel ,( channel 2,no Rx,only tx,8bitdata and 2bit parity even,bit stop)
  Call DefineCom(2, 0,txPin, &H38,Bstop)
 
  ' open the serial channel
  Call OpenCom(2, 115600, 0, oq)
 
 do
  'output port serie com2 micro 
  Call PutQueue(oq,b,1)
  Call PutQueue(oq,b,1)
 loop


[admin edit: added code tags]



Schematic Z24 max232.doc
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com_2.bas
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 Filename:  com_2.bas
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set 8-p-2.jpg
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from Max232 tx ,Z24 set com2 tx
8bit Parity even 2 bit stop

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set 8-p-1.jpg
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from Max232,Z24 set com2 tx
8bit parity even 1 bit stop

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dkinzer
Site Admin


Joined: 03 Sep 2005
Location: Portland, OR

Posted: 22 June 2009, 17:54 PM    Post subject: Re: com2 ,problems with bit stop

marcelo_bahia wrote:
my problem is that no appear 2 stop bits
You have discovered a previously undetected problem that seems to have been introduced around v2.5.0 of the VM. You can work around the issue by specifying a larger value for the stopBits parameter to DefineCom(), for example:
Code:
Call DefineCom(2, 0, 0, &H38, 3)
This works because the maximum number of stop bits on a hardware UART channel is two and, as the documentation states, any value above 1 will select two stop bits.
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marcelo_bahia



Joined: 15 May 2009
Location: argentine

Posted: 23 June 2009, 11:58 AM    Post subject:

results

1)changing Call DefineCom(2, 0,0, &H38,3) ,for 2 stop bits as shown in photo 3 bit.jpg

2)changing Call DefineCom(2, 0,0, &H38,2) ,for 1 stop bits as shown in photo 2 bit.jpg

thank you Don



2 bit.jpg
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3 bit.jpg
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