Voltage control for actuator?

Hello all,

I have a electromechanical shutter for a microscope which I would like to control with an Arduino. To open the shutter takes a ~20 ms pulse of +70VDC across a 12-ohm coil, and to keep the shutter open requires +7VDC. The shutter closes by itself after current is removed.

I'm sure I can handle the software and timing side of things, but I don't know the correct way to go about controlling the current to the actuator coil. Is there some sort of switchable voltage regulator I can control from some digital output pins of the Arduino?

Thanks for any and all help!

That's a lot of power you need. :o

To keep it open you need 7 / 12 = 0.58 A To open it you need 70 / 12 = 5.8 A That means you need 406 Watts to open it and 4 Watts to keep it open. It's not the sort of thing you will find off the shelf.

In order to reduce the power supply requirements down to 70V at 0.6A I would consider a circuit that charges up a capacitor to 70V through a resistor that would allow 0.58 amps to flow normally. That puts it at about 108R. I haven't worked out the capacitor value yet. I'll work on it if you are interested.

Then all you need to do is switch it on. Luckily you should be able to do this with a high voltage FET.

Thanks, Mike -- I appreciate the advice. I think the reason for the large pulse is to get the shutter to open really fast...

Looking at the power requirements of the controller that the company sells (which simultaneously has too many and too few features, and costs more than the shutter itself), I suspect they too are charging a capacitor to open the shutter.

The shutter spec is here: http://www.uniblitz.com/data/resources_filelibrary/cs35_2-15-08.pdf and the controller: http://www.uniblitz.com/data/resources_filelibrary/vcm-d1_user_manual_2_0.pdf

On the controller spec, I notice that there are two leads to the actuator coil (top of page 20), and one ground. I'll have to call tech to see if these are just wired in parallel because the wires in the connector cable are too thin otherwise, or whether the leads operate separate coils...

Anyhow, what would the time constant to fill the capacitor on such an RC circuit be? Less than 100 ms would be great.

Finally, based on my questions, you can probably guess that I know some basic electronics (and can learn), but never had formal EE courses (though I'm dandy on the software side). Would you rate a setup with a such a capacitor as "probably too dangerous for a newb"?

Sorry for additional noise:

I called tech support for more information about the two leads to the coils -- it turns out it's just a +/-/ground setup, but poorly labeled in the diagram. Also, they provide some suggested circuit diagrams for the shutters: http://www.uniblitz.com/data/resources_filelibrary/circuit__descriptions.pdf

Their basic approach is indeed the capacitor-driven pulse.

I think the circuit for my shutter (labeled "typical VS35 drive") is more complex in this case than it would be with the arduino + power FET because their example has some transistor logic for switching based on an input pulse. Whereas I'd just use the arduino to turn the FET on and off, right?

Also, the circuit suggestion has a timer to deliver a pulse from 70V to open, then 7V to hold for the first few seconds, and then 5V for the remainder of the hold time (the higher hold voltage is used at first to reduce mechanical bounce). However, I'm told that this is optional -- holding at +5 or +7V should be OK for the model I have. (Though would there be an easy way to have the arduino switch the circuit to a lower voltage? Just another digital output pin to a transistor controlling an additional resistor?)

Basically, at this point, my main question is how do-able you think such a set-up would be for a newbie... Is it on the order of "fun learning exercise" or closer to "easy way to hurt yourself and damage an expensive shutter"?

Thanks for all the help!

I have just done the maths on the capacitor value and it works out to be a rather big beast. I get a value of just over 1800 uF. That is from the amount of energy you have in a 5.8A 20mS pulse has to be supplied in a capacitor with 63 volts across it using the Q=VC formula. I have had a look at those circuits and surprisingly one of them works in exactly the opposite to my idea. A capacitor is used to initially short out the limiting resister until it charges and then current is limited by the resistor across it. That way also has the advantage of automatically discharging the capacitor when the shutter is off. While this is good it puts up the requirement for the capacity of the power supply.

You can see from the commercial unit that this is a substantial piece of kit and given the power levels a home made soloution is going to be a bit bigger.

how do-able you think such a set-up would be for a newbie.

It is hard to say I can only go off my experience when I was starting out. After building a transistor radio and playing about with some amplifiers I made a Cathode Ray tube circuit. This produced a spot on a CRT tube and then I experimented with magnetic deflection with it. This required a me to make a mains power supply and voltage ladder to generate 1200V for the high tension part of the circuit. I didn't have any trouble with that. I was 14 at the time and it was the mid 60's when people were not as paranoid / concerned about safety as they are today. So I would say it is eminently doable for a beginner but not for an idiot. I always say that ignorance is a prerequisite for learning but shouldn't be elevated to an art form. The main problem is that high voltage components, transformers and capacitors are not as readily available as they were in the 60's when valve amplifiers were still common place. Maybe you could make things easer by using a relay to trigger the shutter rather than a FET, in that way you would keep the high voltage away from your Arduino. I would have thought there would be little chance of damage to your shutter. The only way you could damage it would be to keep it open too long with too much current through it so it burnt out but the series resistor should stop that.

Anyway best of luck and if you want me to review any schematics before you decide to make it then I will be happy to do so.

Thanks for the advice... that is some capacitance that would be needed.

I'm not great at understanding circuits (no real intuition). I follow how the "limited drive circuit" to which you referred (the one that works the opposite of your initial thinking) operates, and I basically understand how the circuit you described to me would work. I'm not totally clear on how the "typical shutter drive" from the schematics works, though. Initially I thought it was like the one you first proposed, but it appears that the capacitance values are far too low (that circuit is for a smaller shutter that takes a +65VDC pulse and has a slightly lighter load across the coil, but not much). My guess is that that circuit is just switching on and then immediately off a +65VDC source, and smoothing that initial pulse out with the capacitor/resistor. Might that be at all correct?

I'll have to look around and see what sort of power supplies are available, etc., before I decide one way or another, but it's good to hear that I should be able to make something like this work...