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李路明

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[问答]

L6470与我的电机不兼容

嗨,

我用步进电机探索EVAL6470H。
根据电机电气参数,INT_SPEED应为2000步/ s左右。问题是L6470对此参数的最大限制为976步/秒。
因此,如果我希望我的应用程序以6000步/秒的速度工作,那么在加速到那一点时我无法获得稳定的电流(但我可以''欺骗''参数,这样我就可以获得6000步/秒的所需电流)。
为什么要将INT_SPEED参数限制得那么低,你有一个解决方案吗?或者我的电机不能使用这个运动控制?

谢谢。
#L6470

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Hi,

I'm exploring the EVAL6470H with my stepper motor.
according to the motor electrical parameters, the INT_SPEED should be around 2000 steps/s. The problem is that the L6470 has a maximal limitation of 976 steps/s for this parameter.
So if I want my application to work at 6000 steps/s, I cannot get a steady current while accelerating to that point (I can, though, ''trick'' the parameters so I would get the wanted current at 6000 steps/s).
Why do you limit the INT_SPEED parameter so low, and do you have a workaround solution for me, or is it impossible to use this motion control with my motor?

Thank you.  
#l6470

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刘璐

2019-6-14 09:39:46
在这种情况下,您可以将INT_SPEED参数设置为最大值。
 
如果您需要有关电压模式的更多详细信息,请参阅具体的应用说明http://www.st.com/st-web-ui/static/active/en/resource/technical/document/application_note/DM00061093.pdf

 
问候
恩里科

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In this case you can set the INT_SPEED parameter to the maximum value.

If you need more details to the voltage mode, you can consult the specific application note http://www.st.com/st-web-ui/static/active/en/resource/technical/document/application_note/DM00061093.pdf
.

Regards
Enrico
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刘璐

2019-6-14 10:01:08
我懂了。
 
您可以发布应用和电机的参数(目标电流,电源电压,Rm,Lm,ke等)吗?
 
恩里科

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以下为原文





I see.

Can you post the parameters of your application and motor (target current, supply voltage, Rm, Lm, ke, etc)?

Enrico
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李路明

2019-6-14 10:07:16
当然。
 
目标电流 - 1A
电源电压 - 24V
Rm - 6.2欧姆
Lm - 1.88 mH
ke - 0.009 V / Hz
 
这给出了大约2100步/秒的INT_SPEED,但我限制在~1000步/秒。
 
例如,如果我想以6000步/秒的速度运行,我可以计算一个变通方法的最终斜率,它将以该速度给出所需的电流(通过交叉图形),并获得高达1000步/秒的稳定电流,但是从1000到6000步/秒我得到一个过流,因为最后的斜率开始得太早。
 
 
谢谢。

以上来自于谷歌翻译


以下为原文





Sure.

target current - 1A
supply voltage - 24V
Rm - 6.2 Ohm
Lm - 1.88 mH
ke - 0.009 V/Hz

This gives an INT_SPEED of ~2100 step/s, but I'm limited to ~1000 step/s.

If I want to run at 6000 step/s for example, I can calculate a workaround final slope that will give the wanted current at that speed (by intersecting the graphs), and I get a steady current up to 1000 step/s, but from 1000 to 6000 step/s I get an overcurrent since the final slope starts too early.


Thank you.
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刘璐

2019-6-14 10:22:20
好,
 
 
我做了一些计算。
首先,使用“理想”补偿曲线,器件在6000步/秒速度之前达到占空比的100%(约为4600步/秒)。高于此速度的测量不会进一步增加施加到电动机的电压正弦波。
 
试试这个配置,让我知道它是如何工作的:
kval_acc = kval_dec = kval_run = 0x42
st_slp = 0x06
fn_slp_acc = fn_slp_dec = 0x0b
int_spd = 0x3fff
 
恩里科

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以下为原文





Ok,


I made some calculations.
First of all also using the ''ideal'' compensation curve the device reaches the 100% of duty cycle before the 6000 step/s speed (at about 4600 step/s). That meas that above this speed the voltage sinewave applied to the motor is not further increased.

Try this configuration and let me know how it work:
kval_acc = kval_dec = kval_run = 0x42
st_slp = 0x06
fn_slp_acc = fn_slp_dec = 0x0b
int_spd = 0x3fff

Enrico
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