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This hopefully isn't too off topic for this forum, but here goes: after many days of tinkering I managed to get a circuit working that, in the end, energizes a solenoid to pull a lever. Sadly, even though the solenoid became energized and pulled its plunger in, it didn't have the strength to pull the lever. The solenoid I used says that it is rated to have a holding force of 1.1 lbs and an operating voltage of 12V.

I gave it 12V, and the solenoid snapped on and pulled its plunger in (it's a pull type of solenoid). But when tied the plunger to the lever, it seemed to lose all strength, not budging the lever at all...not even a wiggle. The same is true if I resisted the movement of the plunger with my hand--or just held it lightly in my hand. It seemed to lack any pull power when there was something there to actually pull. I then upped the voltage by stacking three 9V batteries in series, thinking it might give the solenoid some more muscle--but to no avail.

Is this related to its holding force? Should I buy a different solenoid that can hold more weight? Have others experienced this?

UPDATE

I measured the voltage of the batteries I was using to power the solenoid. It was very interesting. First off, they must have already been drained from powering the solenoid the last time because, although they're relatively new, their voltage from the get-go was reading about 14V instead of the 18V I would expect from putting two 9V in series. I connected them to the solenoid and watched the voltage drop down to less than 2V and continue dropping. So it would seem that at the current that the solenoid needs, the batteries can't sustain their voltage. Perhaps a 12V solenoid can't be powered with common batteries? Or am I thinking about this all wrong and I should expect to see the voltage of the batteries drop when connected to something that draws a current? Especially that much current? Is that normal operation for a battery? Or maybe I need to add a current-limiting resistor, while allowing enough current to go through to power the solenoid?

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  • \$\begingroup\$ I forgot to mention that I could also easily pull the plunger back out after it had been energized. There didn't seem to be much force keeping it in. And after I did, the plunger would slowly, almost reluctantly, pull it back in. It wasn't very snappy. \$\endgroup\$ – NickRamirez Mar 10 '14 at 18:00
  • \$\begingroup\$ What power supply did you use and did you measure the current and confirm the 12V wasn't in fact dropping to something a bit lower and therefore reducing its magnetic pull force. Is it a dc solenoid and do you have a link or data sheet for it? \$\endgroup\$ – Andy aka Mar 10 '14 at 18:07
  • \$\begingroup\$ Holding force will be the force it can apply once fully in : when there is no gap between plunger and pole piece. When there IS a gap (i.e. before pulling the plunger in) the force will be much lower. Reputable solenoids will have datasheets showing pull vs. distance at rated current. \$\endgroup\$ – Brian Drummond Mar 10 '14 at 18:13
  • \$\begingroup\$ I'm guessing that Andy has it right and your batteries are not able to hold up under the current draw that the solenoid wants. Measure your battery voltage while the solenoid is energized (or measure the solenoid current.) The pull force is proportional to the current, so if your voltage is dropping you won't be getting the current or the rated holding force. \$\endgroup\$ – John D Mar 10 '14 at 18:14
  • \$\begingroup\$ Also you may have underestimated the requirements. You can use a homemade dynamometer (a ruler and a low stiffness spring) to measure how much force the lever actually needs - it may be higher than what the solenoid can provide, hence the fact it doesn't budge. \$\endgroup\$ – Mister Mystère Mar 10 '14 at 18:17
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Force from a solenoid = \$(N\cdot I)^2 \dfrac{μ_0\cdot A}{(2 g^2)}\$

Where:

  • μ0 = 4π×10-7
  • F is the force in Newtons
  • N is the number of turns of the coil
  • I is the current in Amps in the coil
  • A is the area in length units squared (cross sectional area of the coil)
  • g is the length of the gap between the coil and the iron.

If you want to check your solenoids pull force and you know the number of turns and have a good estimate of the "gap" and the cross sectional area of the winding then use the formula.

However, it's notable that force is proportional to current-squared and if the 12V you used dropped down to (say) 6V under the load of the solenoid, the pull force would quarter because the current would have halved. It's also worth noting that the force is inversely proportional to the gap-squared - double the gap and the force quarters.

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  • \$\begingroup\$ There is a typo: "is proportional to [the current squared]" \$\endgroup\$ – Mister Mystère Mar 10 '14 at 18:16
  • \$\begingroup\$ I don't think the datasheet gives all of this information. But it's possible that the voltage dropped. But then again, I was giving it more than double the voltage it needed at one point. I tried putting a resistor in series, to better control the current (for the sake of efficiency), but then it didn't energize at all. I'm not totally clear on what amount of current a solenoid needs. Specs only ever mention voltage. \$\endgroup\$ – NickRamirez Mar 10 '14 at 19:00
  • \$\begingroup\$ @NickRamirez solenoids need current but if you look on the details linked on the page it tells you the dc resistance of the device and the power rating - from this you can reasonably infer current. Coil resistance is 36 ohms and this implies current should be 333mA. \$\endgroup\$ – Andy aka Mar 10 '14 at 19:10
  • \$\begingroup\$ An interesting thing to note is that for a constant voltage source, the quantity N*I in the force equation is a constant, because magnet wire has a resistance per unit length. Higher N means higher resistance and lower I. If you rewrite the equation in terms of voltage, you get F = V^2 mu0 / (8 pi g^2 O^2) where O is ohms/meter of the wire. The A and N terms disappear, meaning that for a given voltage and gap size, force is COMPLETELY determined by O. The number of turns simply serves to limit the current drain on the battery. And O is larger for smaller diameter wire. \$\endgroup\$ – Anachronist Jun 8 '17 at 20:00
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As you have noted, the 9v batteries you are using are drained with no load (14v for 2 equals 7v for a single 9v, which is ~1.15v for each of the alkaline cells inside a 9v battery). At load, it drops significantly. This is because two reasons. One, your solenoid has a large inrush and holding current. At 150% the typical voltage of 12v, there is larger current pull as well.

But Two, 9v batteries are NOT designed for large current draws. They are designed for long life at small currents (like smoke detectors). 9v batteries are typically mde up of 6 small packs of akaline cells in series. The typical capacity of a 9v is ~200 maH (This might be wrong, ill correct it later). As each cell is drained (simultaniously), the inernal resistance rises, the voltage lowers, and the current it can source does as well. A common solenoid will drain a 9v battery quickly. Other common consumer batteries have much higher capacity, and can withstand larger current drains for longer without too much voltage droop. AA batteries have current capacities of 2000maH, or 2 AH. C and D even more. 6v Lantern Batteries as well. Then there are the rechargeable batteries like 12v SLAs or LiPos, all that regularly power high drain motors for hours.

Think of it like this. Portable radios all use multiple D batteries in series. The only time you see a 9v in a radio is as a memory backup battery. Ditch the 9v Batteries, they are not useful for testing motors or solenoids beyond simply seeing if they are working.

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protected by Community Feb 18 at 14:29

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