Question

我正在使用MSP430F5438A和MSP-FET430UIFdebugger。我运行一个简单的代码来测量ADC采样所需的时间。代码很简单，如下所示。问题是当我调试相同的代码时，当我观察TA1R寄存器，它给出0X2D（即十进制45），因此转换时间为45 /（8MHZ），这给出了5.6微秒。但是不使用调试器，当我在LCD中输出相同的TA1R值时，它给出了十进制的175是22微秒。为什么它不同？我在采样完成后立即停止计时器然后在LCD中显示值。这是代码..（我只使用SCLK发布了code.im的逻辑）< / p>

static bool flag=0;    
static uint8 temp=0;

main()

{
ADC12CTL0 = ADC12SHT02 + ADC12ON; // Sampling time, ADC12 on
ADC12CTL1 = ADC12SHP; // Use sampling timer
ADC12IE |=BIT0; // Enable interrupt
ADC12CTL0 |= ADC12ENC;

P6SEL |= BIT0; // P6.0 ADC option select

updateLcd();

TA1CCTL0 = CCIE; 

TA1CCR0 = 65535;

TA1CTL = TASSEL_2 + MC_2 + TACLR;//continous mode,SMCLK

 ADC12CTL0|= ADC12SC;
while(flag==0);
TA1CTL=MC_0;//Halt the timer
updateLcd();
}


#pragma vector = ADC12_VECTOR
__interrupt void ADC12_ISR(void)

 {
switch(__even_in_range(ADC12IV,34))
{
case 0: break; // Vector 0: No interrupt
case 2: break; // Vector 2: ADC overflow
case 4: break; // Vector 4: ADC timing overflow
case 6: // Vector 6: ADC12IFG0
temp=ADC12MEM0;

flag=1;

 break;
case 8: break; // Vector 8: ADC12IFG1
case 10: break; // Vector 10: ADC12IFG2
case 12:break; // Vector 12: ADC12IFG3
case 14: break; // Vector 14: ADC12IFG4
case 16: break; // Vector 16: ADC12IFG5
case 18: break; // Vector 18: ADC12IFG6
case 20: break; // Vector 20: ADC12IFG7
case 22: break; // Vector 22: ADC12IFG8
case 24: break; // Vector 24: ADC12IFG9
  case 26: break; // Vector 26: ADC12IFG10
 case 28: break; // Vector 28: ADC12IFG11
case 30: break; // Vector 30: ADC12IFG12
case 32: break; // Vector 32: ADC12IFG13
case 34: break; // Vector 34: ADC12IFG14
default: break; 
}
}

static void updateLcd(void) {


lcdBufferClear(0);
lcdBufferPrintString(0, "Timer Test", 0, eLcdPage0);
lcdBufferSetHLine(0, 0, LCD_COLS-1, 7);
lcdBufferPrintString(0, "Timer value:", 0, eLcdPage3);
lcdBufferPrintInt(0,temp ,70, eLcdPage4);
lcdBufferPrintInt(0,TA1R,30,eLcdPage5);
lcdSendBuffer(0);
}

MSP430UIF调试器显示不同的定时器值

0 个答案: