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My project requires a solenoid with a stroke of 15mm and the travel speed must be 15mm in 2ms. What can be done to get higher speed of travel in the armature?

Donald Wehmeyer
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  • If you have a look on the math behind the movement of the armature, you will see that not only the force exerted by the coil but also the mass of the armature plays a role: If you are able to reduce the mass of the armature, do so. – 0x6d64 Jan 26 '12 at 11:17

2 Answers2

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You can drive the solenoid with a higher voltage.

As long as your duty-cycle is low enough that you don't have heating issues, it should work fine.

It's worth noting that 15 mm in 2 ms is pretty fast. That translates to 7.5 m/s, or 27 kph.

How big is your solenoid? Can you tell us more about the solenoids you are looking at?

Connor Wolf
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  • Note: The speed you calculated is the average speed. If you assume a velocity of 0 at t=0 and a constant acceleration, the final velocity has to be twice that value. – 0x6d64 Jan 26 '12 at 11:01
  • To add to what @ox6d64 said, consider the energy requirement. Let's say the total mass moved is 30 g. With a final velocity of 15 m/s, that comes out to 3.4 J. That plus any inefficiency has to be dumped onto the coil in a short time. That's 82 V on a 1 mF capacitor, for example. – Olin Lathrop Jan 26 '12 at 12:58
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What @Fakename and Olin said. Plus ...

Minimise mass to be accelerated.

Consider finding some other way of adding energy (may not be viable).

Reduce time constant: Time constant = L/R
If you can INCREASE R time constant goes down.
But as t goes down so does current, so you need more V.
So eg if a solenoid had a 100 ohm coil and was made to run from 12V the end current would be i=V/R = 12/100 = 120 mA.

BUT if you added a 900 ohm resistor in series and ran it from 120 V the end current would be the same BUT the time constant would be 10% of previous.

BUT 120V x 120 mA = ~14 Watt. Not terrible but not desirable. SO instead running it from a constant current source achieves much of the same result.

Energy: As Olin notes, there will be an energy calculation that show what you have to put in at absolute minimum. Noy=te that in his example mF is (correctly) miliFarad = 1,000 uF.

Russell McMahon
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