TI中文支持网我的芯片是TMS320F28035，我测试了C2000Ware_3_02_00_00这个库里面的cla_adc这个例程，然后发现那个adc触发是pwm周期内多次触发，而不是pwm一次，adc触发cla一次。我把代码贴出来
#include "DSP28x_Project.h" // Device Headerfile and Examples Include File
#include "CLAShared.h"
#include <string.h>
#include <stdint.h>
//
// Prototype statements
//
__interrupt void cla1_isr2(void);
//
// Globals
//
#pragma DATA_SECTION(ConversionCount, "Cla1ToCpuMsgRAM");
#pragma DATA_SECTION(VoltageCLA, "Cla1ToCpuMsgRAM");
Uint16 ConversionCount;
Uint16 LoopCount;
Uint16 VoltageCLA[NUM_DATA_POINTS];
extern Uint16 Cla1funcsLoadStart;
extern Uint16 Cla1funcsLoadSize;
extern Uint16 Cla1funcsRunStart;
void Gpio_select(void)
{
EALLOW;
GpioCtrlRegs.GPADIR.bit.GPIO12 = 0x01; // All outputs
GpioCtrlRegs.GPADIR.bit.GPIO24 = 0x01;
GpioCtrlRegs.GPBDIR.bit.GPIO34 = 0x01;
GpioDataRegs.GPADAT.bit.GPIO12 =0x01;
GpioDataRegs.GPADAT.bit.GPIO24 =0x01;
GpioDataRegs.GPBDAT.bit.GPIO34 =0x01;
EDIS;
}
void main(void)
{
Uint16 adc_tmp,adc_val;
InitSysCtrl();
DINT;
InitPieCtrl();
IER = 0x0000;
IFR = 0x0000;
InitPieVectTable();
EALLOW; // This is needed to write to EALLOW protected register
PieVectTable.CLA1_INT2 = &cla1_isr2;
EDIS; // This is needed to disable write to EALLOW protected registers
InitAdc(); // For this example, init the ADC
AdcOffsetSelfCal();
PieCtrlRegs.PIEIER11.bit.INTx2 = 1; // Enable INT 11.2 in PIE (CLA Task2)
IER |= M_INT11; // Enable CPU Interrupt 11
EINT; // Enable Global interrupt INTM
ERTM; // Enable Global realtime interrupt DBGM
memcpy((uint16_t *)&Cla1funcsRunStart,(uint16_t *)&Cla1funcsLoadStart,
(unsigned long)&Cla1funcsLoadSize);
EALLOW;
Cla1Regs.MVECT2 = (Uint16)((Uint32)&Cla1Task2 – (Uint32)&Cla1Prog_Start);
Cla1Regs.MVECT8 = (Uint16)((Uint32)&Cla1Task8 – (Uint32)&Cla1Prog_Start);
Cla1Regs.MPISRCSEL1.bit.PERINT2SEL = CLA_INT2_ADCINT2; Cla1Regs.MMEMCFG.bit.PROGE = 1; // Map CLA program memory to the CLA
Cla1Regs.MCTL.bit.IACKE = 1; // Enable IACK to start tasks via software
Cla1Regs.MIER.all = (M_INT8 | M_INT2); // Enable Task 8 and Task 2
Cla1ForceTask8andWait(); // Force CLA task 8.
// This will initialize ConversionCount to zero
AdcRegs.ADCCTL1.bit.INTPULSEPOS = 0;
AdcRegs.INTSEL1N2.bit.INT2E = 1; //Enable ADCINT2
AdcRegs.INTSEL1N2.bit.INT2CONT = 0; //Disable ADCINT2 Continuous mode
AdcRegs.INTSEL1N2.bit.INT2SEL = 1; //setup EOC1 to trigger ADCINT2 to fire
AdcRegs.ADCSOC1CTL.bit.CHSEL = 2; //set SOC1 channel select to ADCINA2
AdcRegs.ADCSOC1CTL.bit.TRIGSEL = 5; //set SOC1 start trigger on EPWM1A
AdcRegs.ADCSOC1CTL.bit.ACQPS = 6; EDIS;
InitEPwm1Gpio();
EALLOW;
SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 0;
EDIS;
EPwm1Regs.AQCTLA.bit.ZRO = AQ_SET; // PWM toggle high/low
EPwm1Regs.AQCTLA.bit.CAU = AQ_CLEAR;
EPwm1Regs.ETSEL.bit.SOCAEN = 1; // Enable SOC on A group
EPwm1Regs.ETSEL.bit.SOCASEL = 4; // Select SOC from from CPMA on upcount
EPwm1Regs.ETPS.bit.SOCAPRD = 1; // Generate pulse on 1st event
EPwm1Regs.CMPA.half.CMPA = 0x2EE; // Set compare A value
EPwm1Regs.TBPRD = 0x5DC;
EPwm1Regs.TBCTL.bit.CTRMODE = 0; // count up and start
EALLOW;
SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 1;
EDIS;
Gpio_select();
for(;;)
{
adc_val=ConversionCount;
adc_tmp=VoltageCLA[0];
LoopCount++;
}
}
__interrupt void cla1_isr2()
{
GpioDataRegs.GPATOGGLE.bit.GPIO24=1;
AdcRegs.ADCINTFLGCLR.bit.ADCINT2 = 1; PieCtrlRegs.PIEACK.all = 0xFFFF;
}
这个代码我就是把pwm1a的信号输出，还有就是在__interrupt void cla1_isr2()这个中断里面翻转了一下IO,然后抓取了pwm的波形跟__interrupt void cla1_isr2()中断的波形，发现pwm输出是20khz,而中断的频率到了250khz，跟我的理解不一样啊，我的理解应该是pwm一个周期应该产生一次adc触发，cla收到adc触发后，在Cla1Task2这个任务里转换一次adc,然后产生__interrupt void cla1_isr2()中断告知cpu,不知道我的理解是否正确。我把波形贴出来，大神帮我看看，为什么实际产生的波形跟我的理解有误差呢？
我的问题：
1、为什么那个__interrupt void cla1_isr2()中断不是pwm1a在一个周期内只产生1次？
Green Deng：
1、我不确定你测的GPIO波形是否正确？正常的GPIO波形应该是0~3.3V的，为什么你的波形幅值这么奇怪。。。
2、例程里面采的是ADCINA2通道，把结果存到了ADCRESULT1的，这点没有问题啊
user6392471：
回复 Green Deng:
GPIO是正常的，那个波形是由于有led后跟一个限流电阻，我量的led端电压
adcresult1问题我找到资料了，但是那个cal中断就不知道是为什么了
