C2000 28035 EPWM 波形的问题-TI中文支持网

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;
}

Eric Ma：

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

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

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

}

void InitEPwm2Example()
{

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

zehao pan：

回复 Eric Ma:

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"); }

}

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

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

}

void InitEPwm2Example()
{

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

user4880999：

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

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

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

}

void InitEPwm2Example()
{

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

zehao pan：

回复 user4880999:

是高电平，但是为什么没有波形

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

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

}

void InitEPwm2Example()
{

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

user4880999：

回复 zehao pan:

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

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

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

}

void InitEPwm2Example()
{

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

user4880999：

回复 user4880999:

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

C2000 28035 EPWM 波形的问题

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