选择哪种电机驱动器IC用于驱动12V 2.5A电磁阀

你好，
 
 
 通常，IC的电流额定值（rms或DC）应大于目标输出电流，在您的情况下为2.5 A.
 如果要主动控制输出电流，则需要带集成电流控制的IC，如L6207 / 27。否则，您可以使用开环方法通过PWM调制应用适当的电压来控制螺线管电流（Vout，平均值= VS x DutyCycle，Iout，mean = Vout，mean / Rsolenoid）。在这种情况下，您可以使用L6206 / 26将过流阈值设置为略高于电磁阀的额定电流（这样可以防止电气过应力）。
 
 您还应该考虑IC的功耗，以避免发热。
 
 有关IC的应用说明，请参阅更多详细信息。
 
 Regadrs
 恩里科



以下为原文





Hello,


In general the current rating (rms or DC) of the IC should be greater than your target output current, in your case 2.5 A.
If you want to actively control the output current you need an IC with integrated current control such as the L6207/27. Otherwise you can control the solenoid current using an open loop method applying the proper voltage through a PWM modulatio (Vout,mean = VS x DutyCycle, Iout,mean = Vout,mean/Rsolenoid). In this case you can use the L6206/26 setting the overcurrent threshold just above the current rating of the solenoid (this way it is protected from electrical overstress).

You should also consider the power dissipation of the IC in order to avoid thermal issued.

More details are available on the application notes of the ICs.

Regadrs
Enrico

你好。我几天前将L6206 IC连接到MicroController，发出PWM信号。到L6206的输入引脚。首先，我只使用了第一座桥，并保持了2个桥臂打开（浮动）。威廉希尔官方网站 图是数据手册中给出的威廉希尔官方网站 图。我用这个设置吹了2个L6206。原因不明。
 
 
然后我将所有未使用的引脚（第二个桥）连接到地并检查输出。有效。
 
但我仍然不确定是什么让前两个IC无法正常工作。通过不工作，我的意思是当我将它连接到威廉希尔官方网站 时，他们的+ ve和-ve点变短了。
 
在所有这些内部，微控制器也变热，实际上不应该发生。如果由mcu提供的驱动器为0到3.3v，可以采取哪些措施来保护mcu免受驱动器IC上的高电压影响。
 
有关L6206 IC工作技术的任何建议都会受到欢迎。
 
最好的祝福
kASHYAP GADA
gada.kashyap（@）的Gmail（。）的COM



以下为原文





Hello. I Had few days back, connected the L6206 IC With a MicroController giving out PWM signals. to the input pins of the L6206. Firstly I was using just the first bridge and had kept the 2 bridge pins open(floating). The circuit diagram is the one given in the data sheet. I blew 2 L6206 with this setup. The reason is unknown.


Then I connected all the unused pins (2nd bridge) to ground  and checked the output. It worked.  

But I am still not sure what made the first two ICs non working. By non working i mean that when i connect it into the circuit their +ve and -ve points come shorted.  

Also within all this the micro controller also getting hot, which shouldnt actually happen. If the drive provided by the mcu is 0 to 3.3v what can be done to protect the mcu from the high voltage on the driver IC.  

Any Suggestions on my working techniques with L6206 ICs would be welcomed.

Best Regards  
kASHYAP GADA
gada.kashyap(@)gmail(.)com