On Mon, 25 Feb 2002 12:51:58 -0700, Tim Crist wrote:

>Okay,  I'll  try this again...
>
>To read from my EEPROM inside the 'F628 I put the desired EE address in =
w
>and call 'ReadEE' (see code snippet).
>
>I'll confirm the value in address 0x00 of the EE space equal to 0xAA, =
and
>address 0x01 equal to 0x55 by looking at the EEPROM memory window.  If I=
 run
>at 4Mhz and read the two addresses one after the other, 'ReadEE' sends =
back
>0xAA each time.  If I step through the program using the ICE2000, I'll =
get
>the correct values: 0xAA, and 0x55.  What gives?
>
>ReadEE
>        BANK1
>                movwf   EEADR                   ;Address to read
>                bsf     EECON1,RD               ;EE Read
>                movf    EEDATA,w                ;w =3D EEDATA
>        BANK0
>                return
>

I ran into something that might be similar in the 16C745's UART.

I wanted to send the same character 4 times as fast as I could. The
'745 runs at 24MHz. If I used the following sequence, some of the
bytes wouldn't transmit.

    movlw  'A'
    movwf  TXREG          ; write it
    btfss      PIR1, TXIF    ; TXREG empty?
    goto      $-1                 ; no, wait until it is
    movwf   TXREG         ; else write it again
    btfss      PIR1, RCIF
    goto      $-1
    movwf  TXREG
    btfss      PIR1, TXIF
    goto      $-1
    movwf  TXREG

I could never get all four bytes sent. Usually two or sometimes three
came across. I ended up having to insert a NOP between each "movwf
TXREG" instruction and the "btfss PIR1, TXIF" following it.=20

I suspect that there's a race condition between the PIR1 status bit
and the TXREG. Since you'd normally be changing w before writing TXREG
again anyway, you can burn that cycle by reloading w before you test
the PIR1 bit.

Placing a NOP between the bsf and the register read may cure your
problem. (Don't forget, the bsf does its write in the last Q cycle of
an instruction. Even at 4MHz, that's not much time before the read
occurs.)


Regards, Bob

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