测1.616V 电压,结果大部分集中在1.6333-1.6068V之间。还有少量数据偏差较大,问:为什么会有这么大的偏差,问题来至哪里?如何改善?
#include <msp430F5529.h>
#define Num_of_Results 8
vola
tile unsigned int results[Num_of_Results];
// Needs to be global in this
// example. Otherwise, the
// compiler removes it because it
// is not used for anything.
int main(void)
{
WDTCTL = WDTPW+WDTHOLD; // Stop watchdog timer
P6SEL |= 0x01; // Enable A/D channel A0
ADC12CTL0 = ADC12ON+ADC12SHT0_8+ADC12MSC; // Turn on ADC12, set sampling time
// set multiple sample conversion
ADC12CTL1 = ADC12SHP+ADC12CONSEQ_2; // Use sampling timer, set mode
ADC12IE = 0x01; // Enable ADC12IFG.0
ADC12CTL0 |= ADC12ENC; // Enable conversions
while(1)
{
ADC12CTL0 |= ADC12SC; // Start convn - software trigger
__bis_SR_register(LPM4_bits + GIE); // Enter LPM4, Enable interrupts
__no_operation(); // For debugger
}
}
#if defined(__TI_COMPILER_VERSION__) || defined(__IAR_SYSTEMS_ICC__)
#pragma vector=ADC12_VECTOR
__interrupt void ADC12ISR (void)
#elif defined(__GNUC__)
void __attribute__ ((interrupt(ADC12_VECTOR))) ADC12ISR (void)
#else
#error Compiler not supported!
#endif
{
static unsigned char index = 0;
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
results[index] = ADC12MEM0; // Move results
index++; // Increment results index, modulo; Set Breakpoint1 here
if (index == 8)
{
index = 0;
}
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;
}
}
去掉三个最大值和三个最小值后,求平均
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