Mplab c18. Контроллер pic18f452 . Реализация software UART.
Непойму как рассчитать DelayRXHalfBitUART, DelayTXBitUART ... и т.д. для кварца 5mhz и скорости 2400 baud.
вот нашёл примерчик .... но всё равно непонятно ... плз. нужна помощь
#include <p18cxxx.h>
#include <stdlib.h>
#include <timers.h>
#include <delays.h>
#include <usart.h>
#include <sw_uart.h>
void PutROMStringUART(static const rom char *);
void PutRAMStringUART(const ram char *data);
void str2ram(static char *dest, static char rom *src);
void main (void)
{
char RAMmsg[32];
OpenUART();
while (1)
{
str2ram(RAMmsg,"Hello World! [RAM]");
WriteUART('A');
WriteUART('B');
WriteUART('C');
PutROMStringUART("Hello World! [ROM]");
PutRAMStringUART(RAMmsg);
Delay10KTCYx(200);
}
}
void str2ram(static char *dest, static char rom *src)
{
while ((*dest++ = *src++) != '\0')
;
}
void PutROMStringUART(static const rom char *data)
{
do
{ // Transmit a byte
WriteUART(*data);
} while( *data++ );
}
void PutRAMStringUART(const ram char *data)
{
do
{ // Transmit a byte
WriteUART(*data);
} while( *data++ );
}
void DelayTXBitUART(void) {
// delay for: ((((2*FOSC/(4*baud))+1) / 2) - 12 cycles
// = ((((2*16000000 / (4*4800)) + 1) / 2) - 12
// 821 cycles @16mhz @4800 baud
_asm
movlw 255 // 255 * 3 = 750 cycles
delayloop:
decfsz WREG, 1, 0 // 2 cycles on skip, one otherwise
bra delayloop // 2 cycles
nop // to make the last iteration 3 cycles instead of just 2
movlw 23 // 23 * 3 = 69 cycles (750 + 69 = 819 cycles)
delayloop:
decfsz WREG, 1, 0 // 2 cycles on skip, one otherwise
bra delayloop // 2 cycles
nop // to make the last iteration 3 cycles instead of just 2
_endasm
}
void DelayRXHalfBitUART(void) {
// delay for: ((((2*FOSC/(8*baud))+1) / 2) - 9 cycles
// = ((((2*16000000 / (8*4800)) + 1) / 2) - 9
// 408 cycles @16mhz @4800 baud
_asm
movlw 136 // 136 * 3 = 408 cycles
delayloop:
decfsz WREG, 1, 0 // 2 cycles on skip, one otherwise
bra delayloop // 2 cycles
nop // to make the last iteration 3 cycles instead of just 2
_endasm
}
void DelayRXBitUART(void) {
// delay for: ((((2*FOSC/(4*baud))+1) / 2) - 14 cycles
// = ((((2*16000000 / (4*4800)) + 1) / 2) - 14
// 819 cycles @16mhz @4800 baud
_asm
movlw 255 // 255 * 3 = 750 cycles
delayloop:
decfsz WREG, 1, 0 // 2 cycles on skip, one otherwise
bra delayloop // 2 cycles
nop // to make the last iteration 3 cycles instead of just 2
movlw 23 // 23 * 3 = 69 cycles (750 + 69 = 819 cycles)
delayloop:
decfsz WREG, 1, 0 // 2 cycles on skip, one otherwise
bra delayloop // 2 cycles
nop // to make the last iteration 3 cycles instead of just 2
_endasm
}
на си 
спасибо.