Determination of constant of software serial (PIC16F)

[Dan]

Hi,All

This is a software serial's Technical data.

(Theorem of PIC)
1bit(Cycle) = (xtal/4)/baud

(In My Routine:Forward references)
1bit(Send)= 8n+16 (Cycle)
1bit(Recieve) = 8n+14 (Cycle)

n is a constant used in bwait2 routine.

(Example 1)
xtal=8000000
baud=9600

1bit(Cycle) = (8000000/4)/9600 = 208.33

208.33 = 8*n+15
n = (208.33-15)/8 = 24.16

(Example 2)
xtal=16000000
baud=9600

1bit(Cycle) = (16000000/4)/9600 = 416.66

416.66 = 8*n+15
n = (416.66-15)/8 = 50.20

(My Routine)
; ***********
;
; RS-232C Send 1byte
;
; Set W and Call (Return:W=0x09,Carry=1,Zero=1)
;
; ***********

    movwf   rxd
    movlw   .10
    movwf   cn
    bcf STATUS,C
            btfsc   STATUS,C    ;; 1
            bsf     GPIO,TX     ;; 1
            btfss   STATUS,C    ;; 1
            bcf     GPIO,TX         ;; 1
            call    bwait           ;; 2+(8*n+6) = 8*n+8
    rrf     rxd,f       ;; 1
            decfsz  cn,f            ;; 1
            goto    $-7     ;; 2 total:1+1+1+1+8*n+8+1+1+2 = 8*n+16 (Cycle)

bwait: call bwait2 ;2 total:2+(4n+2)+(4n+2) = 8n+6 (Cycle)
bwait2: movlw .256-n ;1
addlw 0x01 ;1
btfss STATUS,Z ;1
goto $-2 ;2
retlw 0x09 ;2 total:1+(1+1+2)n-1+2 = 4*n+2 (Cycle)

; ***********
;
; RS-232C Recieve 1byte (Return in W)
;
; ***********

            btfsc   GPIO,RX
            goto    $-1

    call    bwait2
            movwf   cn
            rrf     rxd,f       ;; 1
    call    bwait       ;; 2+(8*n+6) = 8*n+8
            btfss   GPIO,RX     ;; 1
            bcf     STATUS,C    ;; 1
    decfsz  cn,f        ;; 1
            goto    $-5     ;; 2 total:1+8*n+8+1+1+1+2 = 8*n+14 (Cycle)
            movf    rxd,w
            return

[Dan]

This is case AVR.

(Theorem of AVR)

1bit(Cycle) = xtal/baud

(In My Routine:Forward references)
1bit(Send)= 6n+23 (Cycle)
1bit(Recieve) = 6n+21 (Cycle)

n is a constant used in bwait2 routine.

(Example 1)
xtal=8000000
baud=9600

1bit(Cycle) = 8000000/9600 = 833.33
833.33 = 6*n+22

n = (833.33-22)/6 = 135.16

(Example 2)
xtal=9600000
baud=9600

1bit(Cycle) = 9600000/9600 = 1000
1000 = 6*n+22
n = (1000-22)/6 = 163

(My Routine)
; ***********
;
; RS-232C Send 1byte
;
; ***********

        LDI     cn,1+8+1
        COM     rxd
        RJMP    PC+5
        LSR     rxd           ;; 1
        brcs    PC+2          ;; 1
        sbi     PORTB,TX      ;; 2.or.1
        brcc    PC+2          ;; 1
        cbi     PORTB,TX      ;; 1.or.2
        rcall   bwait         ;; 3+6*n+11 = 6*n+14
        dec     cn            ;; 1
        brne    PC-7          ;; 2 Total:1+1+2+1+1+6*n+14+1+2 = 6*n+23 (Cycle)
        ret

; ***********
;
; RS-232C Recieve 1byte
;
; ***********

            SBIC    PINB,RX
            RJMP    PC-1
            LDI     rxd,0x80
            RCALL   bwait2
            RCALL   bwait      ;; 3+6*n+11 = 6*n+14
            SBIC    PINB,RX    ;; 3.or.2
            SEC                ;; 1
            ROR     rxd        ;; 1
            BRCC    PC-4       ;; 2 Total:6*n+14+3+1+1+2 = 6*n+21 (Cycle)

bwait: rcall bwait2 ;3+(3n+4)+(3n+4) = 6n+11
bwait2 ldi rs,n ;1
subi rs,1 ;1
brne PC-1 ;2/1
ret ;4 1+(1+2)n-1+4 = 3*n+4