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周羽姿

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[问答]

如何防止L298发热?

您好,我在运行电机L298N IC时正在使用L298NBut。电机规格如下: - 电压= 12V,失速电流= 50A,额定电流= 5A。能够
电机规格如下: - 电压= 12V,失速电流= 50A,额定电流= 5A。我可以使用Sense电阻来限流吗?如果是,那我该如何计算Rs的值?

#L298

以上来自于谷歌翻译


以下为原文




Hello, I am using L298NBut while running the motor L298N IC. Motor specification are as follows:- Voltage=12V, Stall current=50A, Rated current=5A. Can
Motor specification are as follows:- Voltage=12V, Stall current=50A, Rated current=5A. Can I use Sense resistor for current limiting? If yes then how do I calculate the value of Rs?


#l298

回帖(1)

刘璐

2019-1-17 14:00:45
我认为L298的功率级太小,无法驱动这个电机。每个输出的最大电流为2 A,并行模式允许您达到4 A.我建议您查看L6206,也许它更符合您的需求。
在任何情况下,没有人强迫您使用电机达到其最大额定值。
您可以使用分流电阻+ L6506等PWM电流控制器来实现电流限制。
电流限制器值根据以下公式设置:
Ilimit = Vref / Rsense
非常重要注意:L6235嵌入了两个独立的电流控制器。你必须只使用其中一个。
实际上,在并行模式下,L298必须被视为单个全桥。

以上来自于谷歌翻译


以下为原文




I think the power stage of the L298 is too small to drive this motor. The maximum current for each output is 2 A and parallel mode allows you to reach up to 4 A. I suggest you to have a look to the L6206, maybe it fits better your needs.
In any case, no one forces you to use the motor to its maximum rating.
You can implement the current limiting using a shunt resistior + a PWM current controller like the L6506.
The current limiter value is set according this formula:
Ilimit = Vref/Rsense
VERY IMPORTANT NOTE: the L6235 embeds two independent current controller. You must use ONLY ONE OF THEM.
In fact, in parallel mode, the L298 must be considered as a single full-bridge.
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