用28335的定时器触发adc采样，两个中断都能进去，但是采样值不对，求大神们帮忙看看！！！

likai hu

Other Parts Discussed in Thread: CONTROLSUITE

采集的是一个正弦波，通过数组Voltage1来观察数据和波形。但是采样值基本无变化，这是为什么呢？下面附上程序：

主函数：

void main(void)
{
	 EALLOW;
	      SysCtrlRegs.HISPCP.all = ADC_MODCLK;	// HSPCLK = SYSCLKOUT/2*ADC_MODCLK
	      EDIS;
	int i;
	InitSysCtrl();//初始化系统函数
	IER = 0x0000;//禁止CPU中断
    IFR = 0x0000;//清除CPU中断标志
    InitPieCtrl();//初始化PIE控制寄存器
    InitPieVectTable();//初始化PIE中断向量表
	InitCpuTimers();
	   InitEPwm1Gpio();
	   InitEPwm2Gpio();
	   InitEPwm3Gpio();
   DINT;
   InitAdc();//ADC初始化
      EALLOW;
      SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 0;//禁止PWM时钟
      EDIS;
      InitEPwm1Example();
      InitEPwm2Example();
      InitEPwm3Example();
          ConfigCpuTimer(&CpuTimer0,30,1000);
          StartCpuTimer0();//启动定时器0
          PieCtrlRegs.PIECTRL.bit.ENPIE = 1;//使能PIE模块的总中断
            EALLOW;
            SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 1;//使能PWM时钟
            EDIS;

          EALLOW;  // This is needed to write to EALLOW protected register
           PieVectTable.TINT0 = &TINT0_ISR;//使能Timer0中断
           PieVectTable.ADCINT = &ADCINT_ISR;//使能ADC中断

          EDIS;    // This is needed to disable write to EALLOW protected registers
          PieCtrlRegs.PIEIER1.bit.INTx7 = 1; //使能PIE中断的CPU定时器0
          PieCtrlRegs.PIEIER1.bit.INTx6 = 1; //使能PIE中断中的ADC中断
          IER |= (M_INT1); // Enable CPU Interrupt 1
          EINT;   //使能全局中断
          ERTM;   //使能实时中断
for (i=0; i<BUF_SIZE; i++)
      {
        Voltage1[i] = 0;
        Voltage2[i] = 0;
        Voltage3[i] = 0;
      }

定时器中断函数：

Uint16 q=0;
interrupt void  TINT0_ISR(void)      // CPU-Timer 0
{
  // Insert ISR Code here
	    q++;//判断是否进入中断
		AdcRegs.ADCTRL2.bit.SOC_SEQ1 = 1;//软件触发启动SEQ1
	    PieCtrlRegs.PIEACK.all = PIEACK_GROUP1;

ADC中断函数：

Uint16 qa=0;
interrupt void ADCINT_ISR(void) // ADC
{
// Insert ISR Code here
qa++;//判断是否进入中断
// To receive more interrupts from this PIE group, acknowledge this interrupt
// PieCtrlRegs.PIEACK.all = PIEACK_GROUP1;

// Next two lines for debug only to halt the processor here
// Remove after inserting ISR Code
Uint16 a[10],b[10],c[10];
Uint16 i,j,r,s,t;
Voltage1[ConversionCount1++] = AdcRegs.ADCRESULT0 >>4;
Voltage2[ConversionCount2++] = AdcRegs.ADCRESULT1 >>4;
Voltage3[ConversionCount3++] = AdcRegs.ADCRESULT2 >>4;
if(ConversionCount1>(511))
ConversionCount1=0;
DELAY_US(100);

AdcRegs.ADCTRL2.bit.RST_SEQ1 = 1; // Reset SEQ1
AdcRegs.ADCST.bit.INT_SEQ1_CLR = 1; // Clear INT SEQ1 bit
PieCtrlRegs.PIEACK.all = PIEACK_GROUP1; // Acknowledge interrupt to PIE

ADC初始化函数：

extern void DSP28x_usDelay(Uint32 Count);

	    EALLOW;
		SysCtrlRegs.PCLKCR0.bit.ADCENCLK = 1;//使能ADC时钟
		ADC_cal();
		EDIS;
AdcRegs.ADCTRL1.bit.RESET = 1;    //Reset ADC module
		    asm(" RPT #20||NOP");
    AdcRegs.ADCTRL3.all = 0x00E0;  // 给ADC内部上电
    DELAY_US(ADC_usDELAY);         // 在ADC转换前要延时一段时间
AdcRegs.ADCTRL1.bit.ACQ_PS = 0xff;  // Sequential mode: Sample rate   = 1/[(2+ACQ_PS)*ADC clock in ns]
    	                              //                     = 1/(3*40ns) =8.3MHz (for 150 MHz SYSCLKOUT)
    	      					    //                     = 1/(3*80ns) =4.17MHz (for 100 MHz SYSCLKOUT)
    	      					    // If Simultaneous mode enabled: Sample rate = 1/[(3+ACQ_PS)*ADC clock in ns]
    	         AdcRegs.ADCTRL3.bit.ADCCLKPS = 0;//不分频
    	         AdcRegs.ADCTRL1.bit.SEQ_CASC = 1;        // 1  Cascaded mode
    	         AdcRegs.ADCTRL1.bit.CONT_RUN = 1;       // Setup continuous run
    	         AdcRegs.ADCTRL1.bit.SEQ_OVRD = 1;       // Enable Sequencer override feature
    	         AdcRegs.ADCMAXCONV.bit.MAX_CONV1 = 0x2;  // convert and store in 8 results registers
    	         AdcRegs.ADCCHSELSEQ1.bit.CONV00 = 0x0; // Setup ADCINA3 as 1st SEQ1 conv.
    	         AdcRegs.ADCCHSELSEQ1.bit.CONV01 = 0x1;
    	         AdcRegs.ADCCHSELSEQ1.bit.CONV02 = 0x2;// Setup ADCINA2 as 2nd SEQ1 conv.
    	         AdcRegs.ADCTRL2.bit.INT_ENA_SEQ1 = 1;  // 向CPU发出中断申请

这个问题已经困扰我好几天了，实在是没检查出来是什么问题，希望看到的大神停下来帮忙指点一下，感激不尽！！！

8 年多前