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High speed RS232 for PIC

First flight-shot

Let’s make our transmitter routine, assuming that the data byte is passed in W, inserting delay n instead of nop delays.

movwf data ; put data to rotation buffer
rrf data, W ; prepare data for output:
bcf port, 0 ; 0: sending start bit
xorwf data, F ; 1: '1' in DATA means toggle port
movlw 1 ; 2: load port-toggling mask
skpnc ; 3: first bit is already in C
xorwf port, F ; 4: send bit 0
delay 3 ; 5, 6, 7
btfsc data, 0 ; 8: test data
xorwf port, F ; 9: send bit 1
delay 2 ; 10, 11
btfsc data, 1 ; 12: test data
xorwf port, F ; 13: send bit 2
delay 2 ; 14, 15
btfsc data, 2 ; 16: test data
xorwf port, F ; 17: send bit 3
delay 3 ; 18, 19, 20
btfsc data, 3 ; 21: test data
xorwf port, F ; 22: send bit 4
delay 2 ; 23, 24
btfsc data, 4 ; 25: test data
xorwf port, F ; 26: send bit 5
delay 2 ; 27, 28
btfsc data, 5 ; 29: test data
xorwf port, F ; 30: send bit 6
delay 3 ; 31, 32, 33
btfsc data, 6 ; 34: test data
xorwf port, F ; 35: send bit  7
delay 3 ; 36, 37, 38
bsf port, 0 ; 39: sending stop bit
return

As we now see, this routine needs delays for 2 and 3 instruction cycles. If we use simple nop to make delays, it would take additional 20 words of program memory making routine 43 words long. Let us consider alternatives to save a word or two.

movwf data ; put data to rotation buffer
rrf data, W ; prepare data for output:
bcf port, 0 ; 0: sending start bit
xorwf data, F ; 1: '1' in DATA means toggle port
movlw 1 ; 2: load port-toggling mask
skpnc ; 3: first bit is already in C
xorwf port, F ; 4: send bit 0
goto $+1
nop
btfsc data, 0 ; 8: test data
xorwf port, F ; 9: send bit 1
goto $+1
...    
btfsc data, 6 ; 34: test data
xorwf port, F ; 35: send bit  7
goto $+1
nop
bsf port, 0 ; 39: sending stop bit
return ; Routine length: 35 words

Sighting the limits

What is the highest speed one may achieve with this approach? Marginal timing error 20% is reached when it is equal to a half of the instruction cycle, which gives 2.5 IC per bit or 400000 baud. Generalizing, we may say that BAUD RATE = CLOCK/10 is the estimated limit. What is the lowest speed that requires bit aligning? The problems begin when cumulative timing error reaches 20%. Averaging bit error to 0.5 IC, we get equality 9/2 IC = bit*0.2 that gives us 22.5 cycles per bit, or 44K baud @ 4 MHz and 178K baud @ 10 MHz. Of course, it depends on rate/clock ratio and, for some particular cases, there may be no need in bit alignment. Table 2 shows timing errors1 with equal and aligned loops for two most used clock frequencies and five highest baud rates. For comparison purposes, it also lists lowest bit errors achieved with hardware UART2.

 

8 data bits + 1 start bit

1 The following formulas were used to calculate errors:
 Figure

2 UART errors calculated using SPBRG Calc [2]

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