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rwskinner
Joined: 08 Dec 2006
Posts: 125
Location: Texas
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| Baud Rates and modes |
 Posted: Fri Feb 15, 2008 7:03 am |
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I have a project that requires 9600,7,N,1 (Modbus ASCII Slave).
From what I understand the Pic USART will not handle 7 data bits, so does this mean when (7,N,1) is chosen that the program will use a software USART and simply bit bang it and interrupts can't be used? |
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SET
Joined: 15 Nov 2005
Posts: 161
Location: Glasgow, UK
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| Posted: Fri Feb 15, 2008 8:25 am |
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| If you use 8 bits, then on receive your Ok - you'll simply have an extra bit (least significant if I remember right?), so you can right shift the result by one place. For transmit, left shift and make the LSB a 1, the MODBUS kit should then handle that Ok. |
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rwskinner
Joined: 08 Dec 2006
Posts: 125
Location: Texas
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| Posted: Fri Feb 15, 2008 5:05 pm |
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| SET wrote: | | If you use 8 bits, then on receive your Ok - you'll simply have an extra bit (least significant if I remember right?), so you can right shift the result by one place. For transmit, left shift and make the LSB a 1, the MODBUS kit should then handle that Ok. |
This was my exact, original idea until I was told the bits would still be off since there is no parity bit. I'm not real familiar so please jump in and correct me.
If I shift the data, then the data itself will be contain in 7 bits. We agree.
But if I'm sending it in a 8,N,1 format aren't I still sending 1 too many bits to the 7,N,1 system, or will the extra bit just be interpreted as an extra stop and be ignored.
7,N, 1
1 Start
7 Data
1 Stop
(9 bit times)
8,N,1
1 Start
8 data (even though the last bit is zero)
1 Stop
(10 bit times)
Just curious, and willing to learn. |
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Ttelmah
Guest
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| Posted: Sat Feb 16, 2008 5:27 am |
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What you have to do, is make sure that the last bit of your 8bit data, is always a '1'. This then gives the effect of an extra 'stop' bit (which is unimportant, since 'stops' can be of indefinate length in RS232).
So:
| Code: |
void send8(int8 chr) {
chr != 0x80;
putc(chr);
}
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Obviously, if using streams, use 'fputc;' to the required stream, and then either just call this with the individual characters, or use this as the 'target' function for printf.
Best Wishes |
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Neutone
Joined: 08 Sep 2003
Posts: 839
Location: Houston
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| Posted: Sat Feb 16, 2008 4:33 pm |
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| Ttelmah wrote: | What you have to do, is make sure that the last bit of your 8bit data, is always a '1'. This then gives the effect of an extra 'stop' bit (which is unimportant, since 'stops' can be of indefinate length in RS232).
So:
| Code: |
void send8(int8 chr) {
chr != 0x80;
putc(chr);
}
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Obviously, if using streams, use 'fputc;' to the required stream, and then either just call this with the individual characters, or use this as the 'target' function for printf.
Best Wishes |
That would in effect be 7,N,2 or 7,N,1 with an extra bit time between each byte. MODBUS specification regarding time between bytes in a packet allows for this. It would be different for data going in the other direction. You cant receive 7,N,1 data with a USART set for 8,N,1. You could however receive data sent as 7,N,2 using the USART set as 8,N,1 with no problems. I would expect almost every implementation of MODBUS ASCII to allow 8,N,1, the PI–MBUS–300 specification requires it.
The format for each byte in ASCII mode is:
Coding System: Hexadecimal, ASCII characters 0–9, A–F
One hexadecimal character contained in each
ASCII character of the message
Bits per Byte: 1 start bit
7 data bits, least significant bit sent first
1 bit for even/odd parity; no bit for no parity
1 stop bit if parity is used; 2 bits if no parity
Error Check Field: Longitudinal Redundancy Check (LRC)
The important part is 2 stop bits if no parity. |
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