DSP28335 eCAN跟usbcan通讯问题

2019-03-26 15:14发布

硬件:DSP28335的eCAN-B与vp230连接,USBcan采用的是周立功的usbcan-i与上位机连接。

出现的问题是上位机上的监控软件没法接受到DSP的eCAN模块发过来的数据。


下面是代码:
void main(void)
{


   struct ECAN_REGS ECanbShadow;

   InitSysCtrl();//开启系统时钟和外设时钟

   InitECanGpio();//这里用到的是ecan-b的GPIO12和GPIO13
   DINT;
   InitPieCtrl();
   IER = 0x0000;
   IFR = 0x0000;



   InitPieVectTable();



    MessageReceivedCount = 0;

    EALLOW;
    ECanbShadow.CANTIOC.all = ECanbRegs.CANTIOC.all;
    ECanbShadow.CANTIOC.bit.TXFUNC = 1;
    ECanbRegs.CANTIOC.all = ECanbShadow.CANTIOC.all;

    ECanbShadow.CANRIOC.all = ECanbRegs.CANRIOC.all;
    ECanbShadow.CANRIOC.bit.RXFUNC = 1;
    ECanbRegs.CANRIOC.all = ECanbShadow.CANRIOC.all;
    EDIS;


    ECanbRegs.CANME.all = 0;



    ECanbMboxes.MBOX0.MSGID.all = 0x01C40000;//发送 标准帧
    ECanbMboxes.MBOX1.MSGID.all = 0x18040000;//接受


    ECanbRegs.CANMD.all=0x00000002;//1->RX 0->tx



    ECanbRegs.CANME.all = 0x00000003;



    ECanbMboxes.MBOX0.MSGCTRL.bit.DLC = 8;
    ECanbMboxes.MBOX1.MSGCTRL.bit.DLC = 8;


    ECanbMboxes.MBOX0.MSGCTRL.bit.RTR = 0;
    ECanbMboxes.MBOX1.MSGCTRL.bit.RTR = 0;



    ECanbMboxes.MBOX0.MDL.all = 0x9555AAA0;
    ECanbMboxes.MBOX0.MDH.all = 0x89ABCDEF;



    EALLOW;
    ECanbRegs.CANMIM.all = 0xFFFFFFFF;
    ECanbRegs.CANMIL.all = 0;


    ECanbShadow.CANMC.all = ECanbRegs.CANMC.all;
    ECanbShadow.CANMC.bit.CCR = 1;
    ECanbRegs.CANMC.all = ECanbShadow.CANMC.all;
    EDIS;



    // Wait for CCE bit to be set..
    do
    {
      ECanbShadow.CANES.all = ECanbRegs.CANES.all;
    } while(ECanbShadow.CANES.bit.CCE != 1 );

    // Configure the eCAN timing
    EALLOW;
    ECanbShadow.CANBTC.all = ECanbRegs.CANBTC.all;


    ECanbShadow.CANBTC.bit.BRPREG = 9;    // (BRPREG + 1) = 1
    ECanbShadow.CANBTC.bit.TSEG2REG = 1 ; // to the CAN module
    ECanbShadow.CANBTC.bit.TSEG1REG = 6;  // Bit time = 2us,则为500kbps
    ECanbShadow.CANBTC.bit.SAM = 1;
    ECanbRegs.CANBTC.all = ECanbShadow.CANBTC.all;

    ECanbShadow.CANMC.all = ECanbRegs.CANMC.all;
    ECanbShadow.CANMC.bit.CCR = 0;
    ECanbRegs.CANMC.all = ECanbShadow.CANMC.all;
    EDIS;



    do
    {
      ECanbShadow.CANES.all = ECanbRegs.CANES.all;
    } while(ECanbShadow.CANES.bit.CCE != 0 );

    // Configure the eCAN for self test mode
    // Enable the enhanced features of the eCAN.
    EALLOW;
    ECanbShadow.CANMC.all = ECanbRegs.CANMC.all;
    ECanbShadow.CANMC.bit.ABO = 1;
    ECanbShadow.CANMC.bit.STM = 0;    // Configure CAN for self-test mode
    ECanbShadow.CANMC.bit.SCB = 1;    // eCAN mode (reqd to access 32 mailboxes)
    ECanbRegs.CANMC.all = ECanbShadow.CANMC.all;
    EDIS;


    // Begin transmitting
    for(;;)                                
    {

       ECanbRegs.CANTRS.all = 0x00000001;  // Set TRS for all transmit mailboxes,MX0开始发送
       while(ECanbRegs.CANTA.all != 0x00000001 ) {}  // Wait for all TAn bits to be set..发送完成后对应的位被置1
       ECanbRegs.CANTA.all = 0x0000FFFF;   // Clear all TAn
       MessageReceivedCount++;

       //Read from Receive mailboxes and begin checking for data */

    }
}


我的想法是邮箱0设为发送邮箱,邮箱1设为接受。系统频率为100Mhz,can波特率为500k.程序运行后一直等待在while(ECanbRegs.CANTA.all != 0x00000001 ) {}这一步,也就是说没有发送完成等待置位 此帖出自小平头技术问答
友情提示: 此问题已得到解决,问题已经关闭,关闭后问题禁止继续编辑,回答。
12条回答
play_uyy
2019-03-27 18:01
Laspide 发表于 2014-5-29 08:49
ECanbShadow.CANBTC.bit.BRPREG = 9;    // (BRPREG + 1) = 1
    ECanbShadow.CANBTC.bit.TSEG2REG = 1 ; ...

The following bit timing rules must be fulfilled when determining the bit segment values:
· TSEG1(min)>=TSEG2
· IPT <= TSEG1 <=16 TQ
· IPT <= TSEG2 <=8 TQ
· IPT = 3/BRP (the resulting IPT has to be rounded up to the next integer value)
· 1 TQ <= SJW min[4 TQ, TSEG2] (SJW = Synchronization jump width)
· To utilize three-time sampling mode, BRP >= 5 has to be selected
TSEG2 是可以选1的,100M对应的就是500Kbps.这个我用示波器测过的

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