timer A for microseconds.

felipe paulo

...  Enter in interrup, add a value in queue type FIFO, exits, remove and shows this value...

If the above takes more than 130 microseconds, then the next interrupts will be requested before the previous one is handled. Interrupt will pile up or mess-up. Stack and FIFO might overflow.

How fast is the CPU running? How much can it do within 130 microsecond?

CPU running 1MHz.

in interrupting he gets a sample of ADC,  add in FIFO and verify if the loop was run 128 times

the timer are as configured:

  CCTL0 = CCIE;
  CCR0 = 130;
  TACTL = TASSEL_2 + MC_1;   

felipe paulo

... in interrupting he gets a sample of ADC,  add in FIFO and verify if the loop was run 128 times ...

That probably will take more then 130 MCLK.

Try set MCLK to 8 MHz and keep SMCLK at 1 MHz

  BCSCTL1 = CALBC1_8MHZ;
  DCOCTL = CALDCO_8MHZ;
  BCSCTL2 = DIVS_3;

I set up like this?

is almost solved the problem. Put 130 microseconds in timer e then exit, it worked. But now i trying do in the timer, a ADC converter, but he don´t want leave the interrupt.

  BCSCTL1 = CALBC1_8MHZ; //MCLK = 8MHZ
  DCOCTL = CALDCO_8MHZ;
  BCSCTL3 = DIVS_3;   //SMCLK = 8MHZ/8=1MHZ
 
  ADC10CTL1 = INCH_7 + CONSEQ_1;            // A2/A1/A0, single sequence
  ADC10CTL0 = ADC10SHT_2 + MSC + ADC10ON + ADC10IE;
  ADC10DTC1 = 0x03;                         // 3 conversions
  ADC10AE0 |= 0xC0;                         // P1.7,6 ADC10 option select
 
    InitLCD();  //Inica o LCD
    LCDClear(); //Limpa a tela
  __enable_interrupt();  //Habilita interrupções

  TACCTL0 = CCIE;                           // TACCR0 interrupt enabled
  TACCR0 = 16;            //Ciclos de SMCLK
  TACTL = TASSEL_2   MC_1   ID_3;                  // SMCLK/8 = 125Khz, upmode
  LPM0;   //Entra em MODO low power mode 0
  LCDSetPosition(0,0); //Define posição
  printDecimal(points); //Exibe a variável pontos
}

// Timer A0 interrupt service routine
#pragma vector=TIMER0_A0_VECTOR
__interrupt void Timer_A (void)
{

    ADC10CTL0 &= ~ENC;
    while (ADC10CTL1 & BUSY);               // Wait if ADC10 core is active
    ADC10SA = (unsigned int)&conv_ADC10;    // Inicializa o buffer de dados
    ADC10CTL0 |= ENC   ADC10SC;             // Inicia a amostragem/conversão
 
}
 
// ADC10 interrupt service routine
#pragma vector=ADC10_VECTOR
__interrupt void ADC10_ISR(void)
{    
  while (ADC10CTL1 & BUSY);
  points++; //Increment points for me know how many times the timer was executed
            //if was executed 150 times, he leave of interrupt, get the values and show in LCD.
  if(points == 150){
  LPM0_EXIT;
  }
}


but does not work

if I put the TIMER with ACLK instead of SMCLK could be even faster?
Why Are the SMCLK 1MHZ he does not do what I gotta do this time with this range.

felipe paulo

    while (ADC10CTL1 & BUSY);               // Wait if ADC10 core is active

This kind of statement ha sno place inside an ISR. It will execute for an indeterminate length. And since an ISR cannot interrupt anotehr ISR, this may lock-up the whole system.

ISRs are fast-in and should be fast-out too. No waiting inside. The ISR is called when something happens, to immediately react on the event. They are definitely not meant to wait until something happens.

Independent of that, ENC does not start the conversion. ADC10SC starts it. (SC for Start Conversion). ENC is just enabling or disabling the encoder (e.g. to emergency-stop a sequence)

Finally, ADC10_VECTOR is called whenever a conversion is done. So if it is called, you know for sure that a conversion was done, and a result is ready. So you should copy the content here and not in the timer ISR.

IN the tiemr ISR, there are two cases: either the sequence is ready when teh timer expires. Then it is superfluous to wait for the ADC: it's done and you only need to set ADC10SC again to start the next sequence. Or it isn't done. In this case your timer is too fast. Always. I tmakes no sense to wait for the ADC then, as you'll have to always wait. Since the ADC requires always the same tiemr for a sequence. You must either speed up the ADC by selecting a different SHTx or slow down the timer.
Which one happens should be obvious after a little elementary school math.