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C2000 28035 EPWM 波形的问题

Other Parts Discussed in Thread: CONTROLSUITE, TMS320F28027

我想得到一个EPWM1A、EPWM1B之间相差一个占空比相位的波形,为什么我初始化完成之后连波形都没有了,希望各位帮我找一下问题

初始化程序如下

void InitEPwm1Example()
{

EPwm1Regs.CMPA.half.CMPA = 2500;
EPwm1Regs.CMPB = 5000;
EPwm1Regs.TBPRD = 10000;

EPwm1Regs.TBPHS.half.TBPHS = 0x0000; // Phase is 0
EPwm1Regs.TBCTR = 0x0000; // Clear counter


// Setup counter mode
//EPwm1Regs.TBCTL.bit.CTRMODE = TB_COUNT_UPDOWN; // Count up
EPwm1Regs.TBCTL.bit.CTRMODE = TB_COUNT_UP;
EPwm1Regs.TBCTL.bit.HSPCLKDIV = TB_DIV1; // Clock ratio to SYSCLKOUT
EPwm1Regs.TBCTL.bit.CLKDIV = TB_DIV1;//0x0-1分频,TBCLK = SYSCLKOUT / (HSPCLKDIV × CLKDIV)

EPwm1Regs.TBCTL.bit.PHSEN = TB_DISABLE; // Master module
EPwm1Regs.TBCTL.bit.PRDLD = TB_SHADOW;
EPwm1Regs.TBCTL.bit.SYNCOSEL = TB_CTR_ZERO; // Sync down-stream module

// Setup shadowing
EPwm1Regs.CMPCTL.bit.SHDWAMODE = CC_SHADOW;
EPwm1Regs.CMPCTL.bit.SHDWBMODE = CC_SHADOW;
EPwm1Regs.CMPCTL.bit.LOADAMODE = CC_CTR_zero; // Load on Zero时间基准计数器等于周期
EPwm1Regs.CMPCTL.bit.LOADBMODE = CC_CTR_zero;

// Action Qualifier SubModule Registers
EPwm1Regs.AQCTLA.bit.ZRO = AQ_SET;//AQ_CLEAR;
EPwm1Regs.AQCTLA.bit.CAD = AQ_CLEAR;//AQ_SET达到CMPA的值变为低电平;
EPwm1Regs.AQCTLB.bit.CAU = AQ_SET;//达到CMPB的值变为低电平
EPwm1Regs.AQCTLB.bit.CBD = AQ_CLEAR;
//EPwm1Regs.AQCTLB.bit.PRD = AQ_CLEAR;

}


void InitEPwm2Example()
{

EPwm2Regs.CMPA.half.CMPA = 2500;
EPwm2Regs.CMPB = 5000;
EPwm2Regs.TBPRD = 10000;
// Setup TBCLK
EPwm2Regs.TBPHS.half.TBPHS = 0x0000; // Phase is 0
EPwm2Regs.TBCTR = 0x0000; // Clear counter


// Setup counter mode
EPwm2Regs.TBCTL.bit.CTRMODE = TB_COUNT_UP; // Count up
EPwm2Regs.TBCTL.bit.HSPCLKDIV = TB_DIV1; // Clock ratio to SYSCLKOUT
EPwm2Regs.TBCTL.bit.CLKDIV = TB_DIV1;//0x0-1分频,TBCLK = SYSCLKOUT / (HSPCLKDIV × CLKDIV)

EPwm2Regs.TBCTL.bit.PHSEN = TB_ENABLE; // Slave module
EPwm2Regs.TBCTL.bit.PRDLD = TB_SHADOW;
EPwm2Regs.TBCTL.bit.SYNCOSEL = TB_SYNC_IN;//TB_SYNC_DISABLE; // sync flow-through

// Setup shadowing
EPwm2Regs.CMPCTL.bit.SHDWAMODE = CC_SHADOW;
EPwm2Regs.CMPCTL.bit.SHDWBMODE = CC_SHADOW;
EPwm2Regs.CMPCTL.bit.LOADAMODE = CC_CTR_zero; // Load on Zero
EPwm2Regs.CMPCTL.bit.LOADBMODE = CC_CTR_zero;

// Action Qualifier SubModule Registers
EPwm2Regs.AQCTLA.bit.ZRO = AQ_SET;//AQ_CLEAR;
EPwm2Regs.AQCTLA.bit.CAD = AQ_CLEAR;//AQ_SET;
EPwm2Regs.AQCTLB.bit.CAU = AQ_SET;
EPwm2Regs.AQCTLB.bit.CBD = AQ_CLEAR;
}

  • 有没使能PWM的时钟

    还有如下:


    // For this example, only initialize the ePWM
    EALLOW;
    SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 0;
    EDIS;

    InitEPwm1Example();
    InitEPwm2Example();
    InitEPwm3Example();

    EALLOW;
    SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 1;
    EDIS;

    建议用CONTROLsuite的例程去修改。

    ERIC

  • 初始化程序确实没有是能时钟,但是在主程序中有这一部分啊,这样不能吗???

    void main(void)
    {
    // Step 1. Initialize System Control:
    // PLL, WatchDog, enable Peripheral Clocks
    // This example function is found in the DSP2803x_SysCtrl.c file.
    InitSysCtrl();

    // Step 2. Initalize GPIO:
    // This example function is found in the DSP2803x_Gpio.c file and
    // illustrates how to set the GPIO to it's default state.
    // InitGpio(); // Skipped for this example

    // For this case just init GPIO pins for ePWM1, ePWM2, and TZ pins
    InitEPwm1Gpio();
    InitEPwm2Gpio();
    InitTzGpio();

    // Step 3. Clear all interrupts and initialize PIE vector table:
    // Disable CPU interrupts
    DINT;

    // Initialize the PIE control registers to their default state.
    // The default state is all PIE interrupts disabled and flags
    // are cleared.
    // This function is found in the DSP2803x_PieCtrl.c file.
    InitPieCtrl();

    // Disable CPU interrupts and clear all CPU interrupt flags:
    IER = 0x0000;
    IFR = 0x0000;

    InitPieVectTable();

    // Interrupts that are used in this example are re-mapped to
    // ISR functions found within this file.
    EALLOW; // This is needed to write to EALLOW protected registers
    PieVectTable.EPWM1_TZINT = &epwm1_tzint_isr;
    PieVectTable.EPWM2_TZINT = &epwm2_tzint_isr;
    EDIS; // This is needed to disable write to EALLOW protected registers

    // Step 4. Initialize all the Device Peripherals:
    // This function is found in DSP2803x_InitPeripherals.c
    // InitPeripherals(); // Not required for this example


    EALLOW;
    SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 0;
    EDIS;

    InitEPwm1Example();
    InitEPwm2Example();

    EALLOW;
    SysCtrlRegs.PCLKCR0.bit.TBCLKSYNC = 1;
    EDIS;

    // Step 5. User specific code, enable interrupts
    // Initalize counters:
    EPwm1TZIntCount = 0;
    EPwm2TZIntCount = 0;

    // Enable CPU INT3 which is connected to EPWM1-3 INT:
    IER |= M_INT2;

    // Enable EPWM INTn in the PIE: Group 2 interrupt 1-3
    PieCtrlRegs.PIEIER2.bit.INTx1 = 1;
    PieCtrlRegs.PIEIER2.bit.INTx2 = 1;

    // Enable global Interrupts and higher priority real-time debug events:
    EINT; // Enable Global interrupt INTM
    ERTM; // Enable Global realtime interrupt DBGM

    // Step 6. IDLE loop. Just sit and loop forever (optional):
    for(;;)
    {
    asm(" NOP");
    }

    }

  • 没看你的程序,你看一下wpwm1a没有波形,但是epwm1b是不是高电平?如果是,初始化没问题

  • 是高电平,但是为什么没有波形

  • 那就说明已经配置好了,但是你没有给他比较值,让它输出PWM。。。

    for(;;)
    {
    EPwm1Regs.CMPA.half.CMPA = 500(这个数根据你的计数器数到的最高点来算的,假如你计数器值1000,给cmp500就有50%的PWM输出,给300就有60的%波形输出。); 
    EPwm1Regs.CMPB = 500; 
    }

    你在配置好pwm后,加上这句,就有pwm波形了。。。没时间细说,你可以查下CMP寄存器

  • 你百度一下PPT ‘TMS320F28027课件_第七讲_ePWM’,这个比表详细

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