How to power small solenoid with a 3.7V lithium battery pack

lithiumsolenoid

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.

schematic

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

Best Answer

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.