simple PWM design on PIC16F877A using ccs c compiler

[youngjkoh]

Hi, i'm new to the programming and I was trying to set up the a/d converter. I found this pwm example code from ccs examples, and I was expecting pinC7 as an analog input and the picC6 as a modulated output. However, when I checked the output on the oscilloscope it looked like it is not dependent on what input I put in C7! It looked like it goes high whenever Vcc is high... I don't understand.. Can anyone help me explaining it? Also, I was wondering how the timer works, and why we need it. Why is period=127? Thank you!

#if defined(__PCM__)
#include <16F877A.h>
#fuses HS,NOWDT,NOPROTECT,NOLVP
#use delay(clock=20000000)
#use rs232(baud=9600, xmit=PIN_C6, rcv=PIN_C7, BRGH1OK)  // Jumpers: 8 to 11, 7 to 12


void main() {
   char selection;
   byte value;


   printf("\r\nFrequency:\r\n");
   printf("    1) 19.5 khz\r\n");
   printf("    2) 4.9 khz\r\n");
   printf("    3) 1.2 khz\r\n");

   do {
     selection=getc();
   } while((selection<'1')||(selection>'3'));


   setup_ccp1(CCP_PWM);   // Configure CCP1 as a PWM

          //   The cycle time will be (1/clock)*4*t2div*(period+1)
	       //   In this program clock=20000000 and period=127 (below)
          //   For the three possible selections the cycle time is:
		    //     (1/20000000)*4*1*128 =  25.6 us or 39.062 khz
		    //     (1/20000000)*4*4*128 = 102.4 us or 9.765 khz
		    //     (1/20000000)*4*16*128= 409.6 us or 2.441 khz

   switch(selection) {
     case '1' : setup_timer_2(T2_DIV_BY_1, 127, 1);
                break;
     case '2' : setup_timer_2(T2_DIV_BY_4, 127, 1);
                break;
     case '3' : setup_timer_2(T2_DIV_BY_16, 127, 1);
                break;
   }



  setup_port_a(ALL_ANALOG);
  setup_adc(adc_clock_internal);
  set_adc_channel( 0 );
  printf("%c\r\n",selection);

  while( TRUE ) {
    value=read_adc();

    printf("%2X\r",value);

    set_pwm1_duty(value);          // This sets the time the pulse is
                                   // high each cycle.  We use the A/D
                                   // input to make a easy demo.
                                   // the high time will be:
                                   //  if value is LONG INT:
                                   //    value*(1/clock)*t2div
                                   //  if value is INT:
                                   //    value*4*(1/clock)*t2div
                                   // for example a value of 30 and t2div
                                   // of 1 the high time is 12us
                                   // WARNING:  A value too high or low will
                                   //           prevent the output from
                                   //           changing.
  }

}