I think I can answer this one for you.  Tx is for PIC transmit, an output.
The pin should be set as an output, not an input.  It should not really
matter however as the pin directions for both Tx and Rx are overridden by
the PIC when you enable the USART.

And ... in your code below you are not setting any port direction.  You set
port direction in the TRIS register, not the PORT register.

Rgs
Ian.

-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Lucian
Sent: Sunday, 5 October 2003 10:54 pm
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [PIC]: 16F628 USART Problem


Hey guys,

I think I found the solution to my problem: it seems that the problem has
something to do with
the direction of the pins. I was setting them to 11 (input) as I remember
reading on the
datasheet. Now I've read on a program that the pins are set

        movlw b'00000100'       ; RB2(TX)=1 others are 0
        movwf PORTB

What I'd like to know : what Tx means ? Tx is for transmission from the
PIC's point of view, or from the (let's say) PC's point of view ? Or, in
other words, Tx is for transmitting a byte, or for receiving it from an
external device ? Cause when I set the ports like this, I receive (and print
on the LCD) some garbage, but not the transmitted data. (at least it
receives something :-) )
Can you make this clear for me ? :) Thank you.

Lucian


Lucian


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