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在BEMF补偿中,我们如何自己评估速度?
#评估工具 以上来自于谷歌翻译 以下为原文 In the BEMF compensation, How can we evaluate the speed by ourselves? #evaluation-tool |
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你好,
差异取决于保持电流为DC而其他电流为AC(正弦)的事实。 在第一种情况(DC)中,电流的rms值等于峰值,在第二种情况下,峰值是rms值的约1.4倍。 必须根据峰值评估KVAL值。 然而,保持电流通常应显着低于运行电流。请检查您是否确实需要保持高电流。它可能会导致散热问题。 问候 恩里科 以上来自于谷歌翻译 以下为原文 Hello, the difference depends on the fact that holding current is DC and the other currents are AC (sinusoidal). In the first case (DC) the rms value of the current is equal to the peak value, in the second one the peak is about 1.4 times the rms value. KVAL value must be evaluated according to the peak. However the holding current usually should be significantly lower tha the running current. Please, check if you really need to keep the holding current high. It could cause thermal issues. Regards Enrico |
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恩里科你好,
感谢您的回复。我对评估软件中INT_SPEED寄存器值的公式有另一个疑问。 根据L6470编程手册p43, Equation9)intersectSpeed [step / s] = INT_SPEED * 2 **( - 24)/ tick 另一方面,从AN4144应用笔记p15开始, 表3)INT_SPEED = intersectSpeed * 2 **(26)* tick 其中intersectSpeed = 4 * Rm /(2 * Pi * Lm)[step / s]虽然我认为这些是相同的公式, 指数2的指数不同。在评估软件中,似乎使用了表3的公式。 但是在dSPIN_Firmware_Library_STM32 C语言程序中,使用了Equation9。 例如,下面这个宏./*相交速度转换,范围0到3906步/秒* / #define IntSpd_Steps_to_Par(步骤)((uint16_t)(((步骤)* 4.194304)+0.5))我应该使用哪个公式? 谢谢 你好, 差异取决于保持电流为DC而其他电流为AC(正弦)的事实。 在第一种情况(DC)中,电流的rms值等于峰值,在第二种情况下,峰值是rms值的约1.4倍。 必须根据峰值评估KVAL值。 然而,保持电流通常应显着低于运行电流。请检查您是否确实需要保持高电流。它可能会导致散热问题。 问候 恩里科 以上来自于谷歌翻译 以下为原文 Hello Enrico, Thanks for reply.I have another question about the formula of INT_SPEED register value in the Evaluation Software. According to L6470 Programing manual p43, Equation9) intersectSpeed[step/s] = INT_SPEED * 2**(-24)/tick On the other hand, from AN4144 Application note p15, Table3) INT_SPEED = intersectSpeed * 2**(26) * tick where intersectSpeed = 4*Rm/(2*Pi*Lm) [step/s]Although I think that these are the same pupose formulas, the indices of the exponentiation of 2 differ. In the Evaluation Software, it seems that the formula of Table3 is used. But in the dSPIN_Firmware_Library_STM32 C language program, Equation9 is used. for example, this macro below./* Intersect Speed conversion, range 0 to 3906 steps/s */ #define IntSpd_Steps_to_Par(steps) ((uint16_t)(((steps)*4.194304)+0.5))Which formula should I use? Thanks Hello, the difference depends on the fact that holding current is DC and the other currents are AC (sinusoidal). In the first case (DC) the rms value of the current is equal to the peak value, in the second one the peak is about 1.4 times the rms value. KVAL value must be evaluated according to the peak. However the holding current usually should be significantly lower tha the running current. Please, check if you really need to keep the holding current high. It could cause thermal issues. Regards Enrico |
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正确的公式是表3中所示的公式,实际上在数据表的当前版本中,公式9(现在的公式10)已经修复。
因此,STM32库的公式是错误的。 感谢您的宝贵意见!它将在下一个版本中修复。 问候 恩里科 以上来自于谷歌翻译 以下为原文 The correct formula is the one indicated in Table 3, in fact in the current revision of the datasheet the Equation 9 (now Equation 10) had been fixed. As a consequence, the formula of the STM32 library is wrong. Thanks for your valuable feedback! It will be fixed in the next release. Regards Enrico |
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谢谢恩里科,
它非常澄清。 在这个场合,我在例子2中的AN4144应用笔记第17页中有一个令人担忧的部分 LM = 4MH; RM = 10欧姆 4 * Rm / Lm = 10000步/秒这是4 * Rm /(2 * Pi * Lm)= 1,592步/秒,不是吗? 而且,由于最高补偿率为0.04%, 如果开始Kval = 0,我认为转速为100% 补偿是2,500(= 100 / 0.04)步/秒。 (在应用笔记中,它被写为250步/秒。)无论如何,我现在使用L6470,其中五个设备是菊花链式的。 它工作得很好。谢谢你制作好装置! 以上来自于谷歌翻译 以下为原文 Thanks Enrico, It clarified very much. With the occasion, I have a worrisome part in AN4144 Application note p.17in Example2 Lm=4mH; Rm=10Ohm 4*Rm/Lm = 10000 step/sThis is 4*Rm/(2*Pi*Lm) = 1,592 step/s, isn't it? And more, since maximum compensation rate is 0.04%, if starting Kval=0, I think the rotation speed at 100% compensation is 2,500(=100/0.04) step/s. (In Application note, it is written as 250 step/s.)Any way, I am now using L6470 where five devices are daisy-chained. It works very well.Thank you for making nice device! |
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