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I made a silly beginner's mistake: I purchased some surplus rotary solenoids from electronic goldmine without paying close attention to their power requirements ("Oh, they are so cheap!"), so I assumed they were 12v and reasonably powerful. There are no specs - just the catalog blurb (below).

Neither assumption is true, as it turns out. My benchtop power supply only goes to 30 volts and at that power, the solenoids have almost no force.

Before I try to find a 40 volt power supply, I'd like to know: will force increase by any significant amount with the increased voltage?

Of course a 40 volt power supply will probably cost more than just sourcing 12 volt solenoids but I thought I would ask.

enter image description here

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  • \$\begingroup\$ you could start by measuring the resistance of solenoid coil or measuring current that goes into it with voltage set to some fixed value, this will give you an estimate, how much power you are putting into it. And of course - with increasing voltage you are putting more power into it. \$\endgroup\$
    – miceuz
    Commented Oct 10, 2012 at 17:48
  • \$\begingroup\$ @miceuz - my bench supply indicates that the solenoid is drawing .12A @ 28V and .13A @ 30.7V, so it is not drawing much current. \$\endgroup\$
    – spring
    Commented Oct 10, 2012 at 18:01

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For devices like this, force is proportional to the square of the voltage, minus nonlinearities and losses. So the force you'll get at 40 VDC is (40*40)/(30*30) or about 76% more than 30 VDC, assuming you're already in the zone that overcomes starting friction etc. There's also a return spring -- that has a static force to overcome, so the equation may become: (40*40*kL - kS)/(30*30*kL - kS) kL is the "force of inductor based on squared voltage constant" and kS is "force of spring constant." You can see from this that the higher the kS, the more relative force you'll actually get out at 40 volts than at 30 volts.

That being said, check if your benchtop power supply can be bridged; mine does 30V single, but bridged it can go to 60V.

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  • \$\begingroup\$ Thanks. I just have a hobbyist bench supply from Edmund Scientific (5v, 12v, variable < 30v) so can't be bridged. Would you know offhand if there are any consumer type electronics that I could scavenge a @40v power supply from? I don't recall ever encountering anything like that. \$\endgroup\$
    – spring
    Commented Oct 10, 2012 at 20:56
  • \$\begingroup\$ Most AC-supply audio amplifiers (think home stereo) above ~100W RMS have bipolar 36V supplies in them. \$\endgroup\$
    – HikeOnPast
    Commented Oct 11, 2012 at 3:39
  • \$\begingroup\$ I've got one of these supplies ($200) and it can be bridged with itself to create 60V/5A (or 30V/10A): goo.gl/DWrgj \$\endgroup\$
    – Jon Watte
    Commented Oct 11, 2012 at 5:25
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Power delivered is equal to:

\$P=V^2/R\$

or

\$P=I * R\$

Now you are delivering

\$28 * 0.12 = 3.36 W\$ of power

and coil resistance is

\$28 / 0.12 = 233 Ohm\$

at 40V you'll be delivering

\$40^2/233 = 6.87W\$

So the answer is yes - you will definately increase the force by increasing voltage.

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