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我有一个直流电机,有刷(100V,20A,240N/m)和一个MOSFET H桥,由一个DSPICC33 FJ128 PWM驱动。我能够正确地移动引擎并给它最大的扭矩(240n/m @ 20AMP)。现在我有一个问题,我必须避免与发动机停止(I= 0AMP)他反应。一种可能的外力,与此作用力相反。是否有一种威廉希尔官方网站
来施加扭矩,使发动机保持卡住(停止)并且难以转动?无刷电机是可能的,但在那些没有刷子?
以上来自于百度翻译 以下为原文 Hi to all. I have a dc motor with brushes (100V 20A 240N / m) and a MOSFET H-bridge driven by a dspicc33fj128 PWM. I was able to move the engine correctly and give it the maximum torque (240N / m @ 20Amp). Now I have the problem that I have to avoid that with the engine stopped (I = 0Amp) he reacts to a possible external force immediately, opposing this force. Is there a technique to apply a torque that keeps the engine stuck (stopped) and hard to turn? In brushless motors is it possible, but in those without brushes? |
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3个回答
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对于具有永磁激励的刷电机,“堵塞”,短路电线!(例如,通过驱动两个半桥到GND)电机的“硬”位置将取决于电机的结构。(不适用于定子电场励磁的电机,转子绕组与定子串联。)这样的电动机在没有电流的情况下具有零转矩。
以上来自于百度翻译 以下为原文 For brushed motors with permandnt magnet excitation to "block", short the wires! (e.g. by driving both half-bridges to GND) How "hard" the motor will hold the position somewhat depends on the motor's construction. (Won't work for motors where the stator field is excited electrically, the rotor windings being in series with the stator. Such motors have zero torque without current.) |
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TIYULAM,我想你需要一个带有PI控制器位置反馈的电机编码器(这是一个大直流电机!)T Yorky
以上来自于百度翻译 以下为原文 thieulam, I would say you need an encoder on the motor with positional feedback for a PI controller. (That's a big DC motor!) T Yorky |
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你没有说你是如何驾驶你的H桥。如果您使用0%占空比为停止,100%为全速与方向逻辑信号。它将短路马达的引线来制动它。大多数伺服驱动器使用“锁定反相驱动器”,您可以谷歌它。在50%占空比与锁定反相驱动电机将停止和完全制动。小于50%的电动机将在一个方向上移动,大于相反方向的50%。
以上来自于百度翻译 以下为原文 You did not say how you are driving your H-Bridge. If you are using 0% duty cycle as stop and 100% as full speed with a direction logic signal. It will take shorting the motor leads to brake it. Most servo drives use "Locked Anti-Phase Drive", you can Google it. At 50% duty cycle with Locked Anti-Phase Drive the motor will be stopped and fully braked. Less than 50% duty the motor will move in one direction, greater than 50% in the opposite direction. |
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