无刷电感法启动算法.zip

对定位函数分析了一下，请高手斧正 //********************************************************************* //* This function detects the rotor position. It's input parameter is //* the previous rotor position. It returns the current rotor position //********************************************************************* /* !!!注意!!! !>HallPatt数组的值是8进制数字 !>steady = 1时, 不产生变化的相未进行电流比较，但x和z的移位还是执行了 steady 为0时，顺次通过对ACB三相加以正反向脉冲电流的比较结果组合生产相应的位置 steady 为1时，则 a> 先取得对应sense位置的下一预期位置与sense位置比较会产生变化的一相， b> 并对该相加以正反向脉冲电流取得比较结果 c> 根据比较结果来判断是当前位置是sense位置还是sense的下一预期位置 对steady为1时举例,调用zerostart_sensor(2, 1)的情况如下 a> y = ((HallPatt[2] & 7) ^ 2) & 7 = ((023 & 7) ^ 2) & 7 = 1 如果当前位置是位置2的下一预期位置(位置3)，B相的电流比较结果应该是1， b> 取得B相正反电流比较结果，有三种情况： 1> 正向明显大于反向 x = 1, z = 0，已经进入位置3 2> 正向明显小于反向 x = 0, z = 0，未进入位置3 3> 正向接近反向 x = 1 或 x = 0, z = 1，处于位置2和位置3的临界点 c> 判断， 如果是b>步的比较结果是情况3>则if(z & y) x = HallPatt[sense] & 7;得以执行，x被赋值3; 如果是b>步的比较结果是情况2>则if(!(x ^ (sense & y))) x = sense;得以执行，x被赋值2; 如果是b>步的比较结果是情况1>则else x = HallPatt[sense] & 7;得以执行，x被赋值3; 调用zerostart_sensor(4, 1)的情况如下 a> y = ((HallPatt[4] & 7) ^ 4) & 7 = ((046 & 7) ^ 4) & 7 = 2 如果当前位置是位置4的下一预期位置(位置6)C相的电流比较结果应该是1， b> 取得C相正反电流比较结果，有三种情况： 1> 正向明显大于反向 x = 2, z = 0，已经进入位置3 2> 正向明显小于反向 x = 0, z = 0，未进入位置3 3> 正向接近反向 x = 2 或 x = 0, z = 2，处于位置2和位置3的临界点 c> 判断， 如果是b>步的比较结果是情况3>则if(z & y) x = HallPatt[sense] & 7;得以执行，x被赋值6; 如果是b>步的比较结果是情况2>则if(!(x ^ (sense & y))) x = sense;得以执行，x被赋值4; 如果是b>步的比较结果是情况1>则else x = HallPatt[sense] & 7;得以执行，x被赋值6; */ unsigned char zerostart_sensor(unsigned char sense, bit steady) { unsigned char x; unsigned char y; unsigned char z; unsigned char resultfor,resultrev; x=0; //x is the rotor position z=0; //to indicate that there is no significant difference between 2 current peaks y = HallPatt[sense] & 7; //y is the expected commutation state y ^= sense; //y is now the bit that should change if the expected state is now the current state y &= 0x07; CC6_vLoadChannelShadowRegister_CC6_CHANNEL_3(0x20); CC6_vEnableShadowTransfer_CC6_TIMER_13(); //取得A相正反电流比较结果 if(!steady || (steady && y == 0x04)) { resultfor = currentcompare(0x16); //retrieve ADC conversion result of first current peak due to forward magnetic field in phase A resultrev = currentcompare(0x29); //retrieve ADC conversion result of second current peak due to reverse magnetic field in phase A if(resultfor > resultrev++) x++; if //if there is no significant difference between the 2 current peaks from phase A ( resultfor == resultrev || resultfor == (resultrev - 1) || resultfor == (resultrev - 2) ) { z++; }; } z<<=1; x<<=1; //取得C相正反电流比较结果 if(!steady || (steady && y == 0x02)) { resultfor=currentcompare(0x25); //retrieve ADC conversion result of first current peak due to forward magnetic field in phase C resultrev=currentcompare(0x1A); //retrieve ADC conversion result of second current peak due to reverse magnetic field in phase C if(resultfor>resultrev++)x++; if //if there is no significant difference between the 2 current peaks from phase C ( resultfor==resultrev || resultfor==(resultrev-1) || resultfor==(resultrev-2) ) { z++; } } z<<=1; x<<=1; //取得B相正反电流比较结果 if(!steady || (steady && y == 0x01)) { resultfor=currentcompare(0x19); //retrieve ADC conversion result of first current peak due to forward magnetic field in phase B resultrev=currentcompare(0x26); //retrieve ADC conversion result of second current peak due to reverse magnetic field in phase B if(resultfor>resultrev++)x++; if(resultfor==resultrev || resultfor==(resultrev-1) || resultfor==(resultrev-2))z++; //if there is no significant difference between the 2 current peaks from phase B } if(steady) { if(z & y) x = HallPatt[sense] & 7; //if the bit that is supposed to change(y) corresponds to the bit //where the current peaks are almost the same(z), then the expected state is now the current state else { if (!(x ^ (sense & y))) x = sense; else x = HallPatt[sense] & 7; } } return x; }