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I have a project where I am trying to control the position of a solenoid valve. The valve operates at 12V, 0-2A (0 = close, 2A = open). The system is open-loop and needs a fast response circuit.

I plan to control the circuit using an Arduino PWM "Analog" output.

The first thought that came to mind was an LED driver, since their primary purpose is maintaining accurate, constant current. I already have a high power LED driver IC that has either analog 0-2V or PWM dimming.

Is there any reason why this would not work, or is there a better alternative? I guess my main uncertainty is that an LED is a diode, and a solenoid is a giant inductor, so I'm not sure if this demands a different drive methodology.

Thanks :)

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    \$\begingroup\$ What do you mean by "control the position"? Are you trying to do something other than open/closed without any feedback? Does the solenoid datasheet suggest that this is possible, in any reliable, repeatable way? \$\endgroup\$ – Elliot Alderson May 11 at 15:03
  • \$\begingroup\$ Is the delay introduced by the LED driver acceptable? PWM --> DIM implies that at least a few PWM cycles are integrated to produce the 0.0 - 2.0V equivalence. \$\endgroup\$ – rdtsc May 11 at 15:16
  • \$\begingroup\$ Also once you build up current in the solenoid you need somewhere for that current to go at turn-off. An anti-parallel diode will work, but it won't be fast. A TVS rated for sufficient energy can allow faster turn off, and there are other schemes as well. \$\endgroup\$ – John D May 11 at 15:27
  • \$\begingroup\$ @ElliotAlderson The valve is designed with a certain spring force such that a given current will move it to a certain position with a relatively linear relationship. \$\endgroup\$ – kobra20 May 11 at 16:33
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    \$\begingroup\$ @kobra20 You should use a bi-directional TVS that won't break down with the normal max applied voltage, but will clamp the back EMF at a voltage that is lower than your driver's max voltage rating. It also has to handle the 1/2LI^2 energy of the inductor at whatever rate you're turning on and off. \$\endgroup\$ – John D May 11 at 16:56

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