關於servo和timer

game-two

我想寫一個8051的c program,透過vb和電腦的rs232,
來同時控制6-9個servo,但8051的兩個timer,一個給rs232用了,而另一個則用來做約20ms一次的interrupt
那servo那兒的pwm應該怎麼寫?
在沒有外接pwm晶片的情況下怎樣可以在20ms的週期中輸出0.5ms-2.5ms的pwm給不同的servo?

我是一名新手,
這點一直都想不通......希望各位大大幫忙解困........


/* ser.c  */
#include <reg51.h>
sbit  wled  = P2^3;
sbit  spu0   = P2^4;
sbit  spu1   = P2^7;


sbit  wled2        = P2^6; //S

/* S3003 pulse time */
#define  DEA    3  /* servo backward  TC */
#define  DEB    19 /* servo forkward  TC */
#define  DEC    12 /* 10 11  12 13   14 15 servo center    TC */

#define  SERVOS 2

//int timercounter;
//int servocounts[SERVOS];
//int tmp;

int servo0_time;
int servo1_time;
int time,a,b,temp0,temp1,temp2;
unsigned char high,low;

mov_servo(void);
compare_SBUF(void);
/*------------------------------*/
delay(int d)
{
int i, j;
for(i=0; i<d; i++)
    for(j=0; j<50; j++)  ;
}
/*----------------------------*/
led_bl()
{
int i;
for(i=0; i<2; i++)
  {
   wled=0;  delay(50);
   wled=1;  delay(50);
  }
}
/*------------------------------------------------*/
del() /* 100 uS  0.1 mS delay */
{
char  i;
for(i=0; i<50; i++) ;               
}
/*---------------------------(s)*/




void t0isr(void) interrupt 1
{
compare_SBUF();
mov_servo();
TH0=high;
TL0=low;

}

void UART_int(void) interrupt 4

{
if (RI==1)
{
RI=0;
temp0=SBUF;
switch(SBUF)
{

case 0:        servo0_time=5;
                servo1_time=15;
                break;

case 1: servo0_time=15;
                servo1_time=5;
                break;

//case 2:

//case 3:

//case 4:

//case 5:

//case 6:

//case 7:

//case 8:

//case 9:

default: break;

}

temp1=servo0_time;
temp2=servo1_time;
}




}
/*-----------------------*/
mov_servo()
{

//timercounter++;
//  for (tmp=0; tmp < SERVOS; tmp++)
//  {
//      if (timercounter  < servocounts[tmp])
//          servocounts[tmp]=1;
//      else
//          servocounts[tmp]=0;
//  }


if (servo0_time > 0)
   {spu0=1;
        servo0_time--;}
         
else
   spu0=0;

if (servo1_time > 0)
   {spu1=1;
        servo1_time--;}
         
else
   spu1=0;


del();


}
/*------------------(S)*/

compare_SBUF()
{
if (SBUF==temp0)
{


servo0_time=temp1;
servo1_time=temp2;

}
}
/*-----------------(RS232)*/

init_UART()
{
  SCON=0x50;
  TMOD=0x21;
  TH1 =0xFD;
  TH0=high=(65536-20000)/256;
  TL0=low=(65536-20000)%256;


}


/*----------------(RS232)*/
main()
{

spu0=0;
spu1=0;
led_bl();
init_UART();
ET0=1;
ES=1;
EA=1;
TR1=1;
PS=1;


while(1);
}

/*----------------------*/

港豪奻辣炰邲

同意?上兄弟的意?~

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浙江老虎機上分器,上海老虎機上分器,江?老虎機上分器,水果機上分器,水果機?控器

dsprcz

Hi, you do not need to set all the 6 servos in phase simutaneously, you can make the 20ms for each servo by servo.
that means totally 120ms handling for each interrupt. using internal delay routine.

Or you can make some maths. to make delays for all 6 servos as well as the total periods.