[SOLVED] 8051 Atmel 89C51 programming in C example

FenTrac said:

Attached below is a file with some ideas to look at. It is untested, but has comments on each line. I believe hexadecimal for 50,000 is C350 rather than 3C80. It's best to have a pull up resistor on PORT0 pin 0, to hold it high and have the switch pull it low when pressed. To set the pin 0 as an input, write a 1 to it. To test for even numbers, bit 0 is always a zero for even binary numbers.

View attachment 123561

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FenTrac said:

You can use the modulus operator in C represented by the % sign. It returns the remainder of division.

In the previous even number code, try just changing the line in the timer0 interupt handler:

Code:

if((a[i] & 0x01) == 0)  // even numbers all have bit 0 = 0

to:

Code:

if((a[i] % 3) == 0)  // the remainder if a[i] is a multiple of 3 = 0

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FenTrac said:

See if these changes to the interrupt routine help with the timing.

Code:

void timer0(void) interrupt 1 using 2 
{   
    TH0 = 0xC3;           // set timer0 to C350 = 50,000
    TL0 = 0x50;           // after each interrupt                    
    
    counter--;            // count the interrupt
    if(counter == 0)      // count 20 interrupts for 1 sec delay       
       {
       counter = 20;      // reset counter to 20
       if(started == 1)   // if timer is started
         {   
         while(i < 9)     // go through array "a" to find a multiple of 3  
           {
           if( (a[i] % 3) == 0) // multiple of 3 has remainder = 0                 
              {
              display = (a[i]);  // display a matching number 
              i++;          // increment i to the next number in "a"
              break;      // break out of the while loop after displaying a number
              } 
           i++;            // increment i to the next number in "a"
           }                 // end while i < 9     
         if(i == 9)       // if 9, all numbers have been tested
           {
           ET0 = 0;       // disable timer 0 interrupt 
           TR0 = 0;       // stop timer0 - all of "a" has been checked 
           started = 0;   // started = 0, stop until switch pressed again   
           }
         }  // end if started == 1    
       }  // end if counter == 0
}

Also noticed that timer0 is using 0x3C80 again, instead of 0xC350. Maybe that's what works best ?

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FenTrac said:

If it is too fast, increase the value of counter until the speed is what you want. It needs to be changed both in main, and in the interrupt routine where it is currently given a value of 20. (When testing on my AT89C51ED2 running at 60 Mhz at 2x I have to have counter = 200 to get the 1 second delay).
The interrupt routine posted above is designed to delay 1 second for each "multiple of three" number. In your example, this would take 3 seconds because 3 of the 9 numbers in "a" are multples of 3. The previous interrupt routine would have taken 9 seconds to check "a" since it was testing 1 number per second.

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FenTrac said:

The problem may be that TR0 needs to cleared to load TH0 and TL0.
The code change would be in the interrupt routine:

Code:

    TR0 = 0;                // stop counter 0 to reload
    TH0 = 0x3C;          // set timer0 to 3C80 to count up 50,000 before overflow
    TL0 = 0x80;           // after each interrupt
    TR0 = 1;                // start counter 0 after reload

I found in the Atmel C51 Hardare manual:
http://www.atmel.com/images/doc4316.pdf
"TH0 and TL0 can be read at any time but TRx bit must be cleared to preset their values, otherwise the behavior of the timer/counter is unpredictable."

Also on page 2-83 of the hardware manual, it shows the timer running at the system clock/6,
giving a 2 Mhz Timer Clock. That would account for your seeing 500 ms instead of 1 second. I think we were calculating 12 Mhz divided by 12 for a 1 Mhz Timer clock.

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