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在进行eqep测位置时,qep的 position counter register中的计数要么是0要么是最大值QPOSMAX,不论电机的转速是多少,position counter register中的计数始终在0和设置的最大值QPOSMAX之间跳动,不知道为什么,请大家帮忙分析一下,谢谢。下面是我的eqep的初始化:
EQep1Regs.QDECCTL.bit.QSRC=00; // QEP quadrature count mode
EQep1Regs.QEPCTL.bit.FREE_SOFT=2;
EQep1Regs.QEPCTL.bit.PCRM=00; // PCRM=00 mode - QPOSCNT reset on index event
EQep1Regs.QEPCTL.bit.UTE=1; // Unit Timeout Enable
EQep1Regs.QEPCTL.bit.QCLM=1; // Latch on unit time out
EQep1Regs.QPOSMAX=0xffffffff;
EQep1Regs.QEPCTL.bit.QPEN=1; // QEP enable
EQep1Regs.QCAPCTL.bit.UPPS=5; // 1/32 for unit position
EQep1Regs.QCAPCTL.bit.CCPS=7; // 1/128 for CAP clock
EQep1Regs.QCAPCTL.bit.CEN=1; // QEP Capture Enable
EQep1Regs.QPOSCTL.bit.PCE = 1;
// EQep1Regs.QEINT.bit.QDC = 1;
EQep1Regs.QEINT.bit.UTO = 1;
条件:采用增量式编码器(只有ABZ信号,没有UVW);基于28335EQEP的速度和位置控制
1: EQEP配置:
EQep1Regs.QUPRD=Period_SD;
EQep1Regs.QDECCTL.bit.QSRC=00; // QEP quadrature count mode
EQep1Regs.QEPCTL.bit.FREE_SOFT=2;
EQep1Regs.QEPCTL.bit.PCRM=00; // PCRM=00 mode - QPOSCNT reset on index event
EQep1Regs.QEPCTL.bit.UTE=1; // Unit Timeout Enable
EQep1Regs.QEPCTL.bit.QCLM=1; // Latch on QPOSCNT :0, unit time out :1
EQep1Regs.QPOSMAX=0xffff;//9999;
EQep1Regs.QEPCTL.bit.QPEN=1; // QEP enable
EQep1Regs.QCAPCTL.bit.UPPS=0; // 1/1 for unit position
EQep1Regs.QCAPCTL.bit.CCPS=0; // 1/1 for CAP clock
EQep1Regs.QCAPCTL.bit.CEN=1; // QEP Capture Enable
EQep1Regs.QPOSCNT=0;
EQep1Regs.QCTMR=0;
EQep1Regs.QPOSLAT=0;
EQep1Regs.QCTMRLAT=0;
EQep1Regs.QCPRD=0xffff;
2:速度采样:无论高低速都采用M法
if(EQep1Regs.QFLG.bit.UTO==1)
{
X=EQep1Regs.QPOSLAT;
。。。。。。
T = EQep1Regs.QCTMRLAT;
}
EQep1Regs.QCLR.bit.UTO=1;
供参考