16F88 7 data bits

[rwskinner]

In the software serial routines, MELabs must be looking at the settings in order to determine how to send the data bits and set the parity right? Problem is it's a define and not a variable or it most likely could be changed on the fly.

I sure hate to bit bang this but I may have to.

Richard

[paul borgmeier]

>>I sure hate to bit bang this but I may have to.

you do not have to, PBP has done the work for you.

>>two adjacent pins connected together

another thought - no need for adjacent pins - use SEROUT for 8N1 and SEROUT2 for 7N1 ... using the same output port pin.

If PORTA.1 = 1 then ; check jumper
SEROUT PORTB.0, .... ; need 8N1
else
SEROUT2 PORTB.0, ... ; need 7N1
ENDIF

will that work?

Also, I do not see why you keep mentioning CCS??? We are taking PBP here

[rwskinner]

I'll give it a shot this morning. It "should" work.

Why do I keep mentioning CCS? I only mentioned it once in an effort to show it is possible to do it But upon closer examination, it also only handles it in software only.

>>>>two adjacent pins connected together
another thought - no need for adjacent pins - use SEROUT for 8N1 and SEROUT2 for 7N1 ... using the same output port pin.

Also, I do not see why you keep mentioning CCS??? We are taking PBP here

[paul borgmeier]

Hi Richard,

I hope this approach works for you.

Best of the Day,

[rwskinner]

Hi Richard,
I this approach works for you.
Best of the Day,

I ended up using SER2 for the 7,N,1 Stuff then
using HSER for the 8,N,1 Stuff.

Check the Dip switch, Configure TXSTA, RCSTA, SPBRG to the values needed or Zero if not needed. This worked great and allowed me to still use the modifiers like in SER2.

Thanks for the ideas.

Richard

[rwskinner]

Would it be possible to use the pic's hardware USART set at 9600,8,N,1 and receive and transmit data to the 7,N,1 device by simply massaging the bits some.

When I receive the bytes, I would shift them to the left once and store just the 7 bit value as a byte.

Before sending the bytes from the array, then make sure the 8th bit is a 1. Wouldn't the 7,N,1 system see this as 7 data bits and two stop bits?