I've got a small 1.5V 2W pull-type spring-loaded solenoid I'm trying to fire with a 3.7V (45mAh at 1C) lithium pack for about 0.5 sec. I am able to get it to work using a (1.5A max) voltage source at 3V, but the battery can't discharge the rated solenoid current (667mA at 3V) into the solenoid from the battery pack, it pretty much kills the pack instantly because it is trying to draw too much. I've tried charging a large electrolytic 470uF capacitor so that the current isn't drawn all at once from the battery, but it fires too weak I think because it can't supply enough current either.


Is there any way to do this given the circuit constraints?

  • \$\begingroup\$ On a somewhat different note: do you have the flyback catch diode for the solenoid? \$\endgroup\$ Sep 24 '15 at 22:59
  • \$\begingroup\$ That's a good idea, but that wouldn't be the problem in this situation, right? \$\endgroup\$
    – JBaczuk
    Sep 24 '15 at 23:01

Maybe consider using a buck regulator to convert the 3 volts to 1.5 volts. Given that they convert voltages with usually a power efficiency about 95%, the 1.333 amps required at 1.5 volts (2 watts) translates to just over 701mA on the 3V side.

However, given that you appear to say that the battery pack can supply only 667 mA, you might be screwed unless there is a guaranteed lower operating limit on the solenoid that is smaller than the numbers provided in the question.

If not either add a battery just for the solenoid circuit or make the basic battery bigger. I only have your data to go on.

  • \$\begingroup\$ So lowering the voltage would increase the current required. Lithium polymer battery packs this size are usually rated to supply up to 2C or 2x 45mA = 90mA. \$\endgroup\$
    – JBaczuk
    Sep 24 '15 at 23:19
  • \$\begingroup\$ "So lowering the voltage would increase the current required" - that does not compute - your solenoid is rated at 1.5 volts (according to your words) so not lowering the voltage isn't an option. \$\endgroup\$
    – Andy aka
    Sep 24 '15 at 23:26
  • \$\begingroup\$ I was referring to your suggestion of lowering the voltage from 3V to 1.5V using a buck converter. \$\endgroup\$
    – JBaczuk
    Sep 24 '15 at 23:27
  • \$\begingroup\$ Well I have no idea what "So lowering the voltage would increase the current required" means - have your tried researching a buck regulator. In your problem converting 3V to 1.5V means a current taken from the battery of 701mA can supply a current of 1.5 amps at the lower voltage. \$\endgroup\$
    – Andy aka
    Sep 24 '15 at 23:30
  • 1
    \$\begingroup\$ No, not a good idea - try charging a supercap instead then dumping its energy into the solenoid. \$\endgroup\$
    – Andy aka
    Sep 24 '15 at 23:31

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