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TMS320F280049C: 关于0%和100%占空比

Tobby Guo
Prodigy 70 points
Part Number: TMS320F280049C


你好：

使用UPDOWN模式，部分配置如下：

EPwm1Regs.CMPCTL.bit.LOADAMODE = CC_CTR_ZERO_PRD;//CC_CTR_ZERO_PRD;
EPwm1Regs.CMPCTL.bit.LOADBMODE = CC_CTR_ZERO_PRD;//CC_CTR_ZERO_PRD;
EPwm1Regs.CMPCTL.bit.SHDWAMODE = CC_SHADOW;
EPwm1Regs.CMPCTL.bit.SHDWBMODE = CC_SHADOW;

EPwm1Regs.AQCTLA.bit.CAD = AQ_SET; 
EPwm1Regs.AQCTLA.bit.CAU = AQ_CLEAR; 
EPwm1Regs.AQCTLB.bit.CBD = AQ_CLEAR;
EPwm1Regs.AQCTLB.bit.CBU = AQ_SET;

实际需要0%和100%占空比，最初的代码使用

EPwm1Regs.CMPA.bit.CMPA = 1;
EPwm1Regs.CMPB.bit.CMPB = f_ePWM_Timer_TBPRD - 1;

EPwm2Regs.CMPA.bit.CMPA = 1;
EPwm2Regs.CMPB.bit.CMPB = f_ePWM_Timer_TBPRD - 1;

实测在0%和100%占空比有很小的尖刺

查找280049的手册

认为CMPA/CMPB=TBPRD应该是可以的，然后继续查找

对这个的理解就是CMPA/B只能使用1到TBPRD-1的值，而且这种很小的尖刺还无法避免，但是这个尖刺是无法接受的。

然后使用0和TBPRD的值进行测试，这个尖刺消失，查找论坛，针对type4 EPWM，没有找到明确的结论说CMPA/B是否可以使用0和TBPRD的值，能不能帮忙确认一下，多谢？

如果不能使用0和TBPRD，我需要0%和100%占空比，有没有推荐的方案，多谢

  • 0
    TI__Guru** 101655 points

    你可以看一下这个E2E帖子，针对type-4类型的PWM的：e2e.ti.com/.../tms320f28075-type-4-epwm-to-generate-full-range-0-100-duty-cycle-waveform-w-o-isr-intervention

    另外，还有一种产生0~100%占空比的方法是通过设置AQ动作模块来产生，即直接将输出拉高或者拉低。

  • 0 回复 Green Deng
    Prodigy 70 points

    使用AQ动作来产生动作，代码如下：

    if(st_SPWM2R3S.f_A > f_DeadCntDuty) //A相
    {
    EPwm1Regs.AQCTLA.bit.CAD = AQ_SET;
    EPwm1Regs.AQCTLA.bit.CAU = AQ_CLEAR;
    EPwm1Regs.AQCTLB.bit.CBD = AQ_CLEAR;
    EPwm1Regs.AQCTLB.bit.CBU = AQ_SET;

    EPwm2Regs.AQCSFRC.all = 0x06;

    EPwm1Regs.CMPA.bit.CMPA = (f_ePWM_Timer_TBPRD * st_SPWM2R3S.f_A - DeadTime); //有效占空比
    EPwm1Regs.CMPB.bit.CMPB = (f_ePWM_Timer_TBPRD * st_SPWM2R3S.f_A + DeadTime);

    EPwm2Regs.CMPA.bit.CMPA = f_ePWM_Timer_TBPRD - 1;
    EPwm2Regs.CMPB.bit.CMPB = f_ePWM_Timer_TBPRD - 1;
    }
    else if(st_SPWM2R3S.f_A < -f_DeadCntDuty)
    {
    EPwm1Regs.AQCSFRC.all = 0x09;

    EPwm2Regs.AQCTLA.bit.CAD = AQ_SET;
    EPwm2Regs.AQCTLA.bit.CAU = AQ_CLEAR;
    EPwm2Regs.AQCTLB.bit.CBD = AQ_CLEAR;
    EPwm2Regs.AQCTLB.bit.CBU = AQ_SET;

    EPwm1Regs.CMPA.bit.CMPA = 1;
    EPwm1Regs.CMPB.bit.CMPB = 1;

    EPwm2Regs.CMPA.bit.CMPA = (f_ePWM_Timer_TBPRD * (1.0f + st_SPWM2R3S.f_A) - DeadTime) ;
    EPwm2Regs.CMPB.bit.CMPB = (f_ePWM_Timer_TBPRD * (1.0f + st_SPWM2R3S.f_A) + DeadTime);

    }
    else
    {
    // EPwm1Regs.CMPA.bit.CMPA = 1;
    // EPwm1Regs.CMPB.bit.CMPB = f_ePWM_Timer_TBPRD - 1;
    //
    // EPwm2Regs.CMPA.bit.CMPA = 1;
    // EPwm2Regs.CMPB.bit.CMPB = f_ePWM_Timer_TBPRD - 1;

    EPwm1Regs.AQCSFRC.all = 0x05;
    EPwm2Regs.AQCSFRC.all = 0x05;
    }

    实测波形如下

  • 0 回复 Tobby Guo
    Prodigy 70 points

    从上到下依次位1A,2A,1B,2B

    从上面的波形可以看出，这些毛刺是互补存在的，不是干扰，是实际端口波形

    将上述互补尖刺展开，ns级别的信号

    能不能给点建议，多谢

  • 0 回复 Tobby Guo
    Prodigy 70 points

    关于这个尖刺该如何消除，能不能给点建议，多谢

  • +1 回复 Tobby Guo
    TI__Guru** 101655 points

    你好，这个问题为你在E2E上咨询了一下，你留意一下工程师的回复：https://e2e.ti.com/support/microcontrollers/c2000-microcontrollers-group/c2000/f/c2000-microcontrollers-forum/1007139/tms320f280049c-use-aq-module-to-generate-pwm-wave-with-0-100-duty-cycle 

  • 0 回复 Green Deng
    Prodigy 70 points

    正在跟踪中，多谢