2
\$\begingroup\$

I want to build a battery-powered device that actuates a 24 V solenoid when a button is pressed. How could I power it from a 18650-sized lithium-ion cell? It should work down to about 3 V, when it's nearly flat. Or I could put two cells in series, if necessary.

In my case, the the solenoid seems to be rated at 900 mA, since its coil resistance is 26.6 Ω measured with a multimeter. Only a short pulse is required; there's no need for a hold current.

I'd like to keep the device small, cheap, and robust. If I got a little boost regulator module rated for 30 V / 1 A output, is there any chance it would work from 3 V (or 6 V)? An input current of at least 7.2 A (or 3.6 A) seems like a lot, for both the regulator and the battery.

I could slowly charge capacitors to 24 V (or even 30 V). But I've already experimented with it using a 20 V supply (which IS enough for the solenoid). At that voltage, 6000 µF wasn't enough to actuate the solenoid, but 7500 µF worked. The capacitor bank was annoyingly large though.

My favorite idea is this: build up a current of 900 mA (or more) in a coil and let that discharge into the solenoid. Could this work? It would have to be a big enough coil to power the solenoid long enough for it to actuate. Or would there be sparks or something because the current in the solenoid coil needs time to ramp up from 0? I have included a schematic below (ignore the coil values). A logic circuit would close both switches first, and very soon after that, open SW2.

current pulse generator idea

\$\endgroup\$
1
  • \$\begingroup\$ Few solenoids require that sort of power. If it's not HUGE it is probably AC rated and the DC resistance is not what sets current. Try it on 24V with various series resistors or a variable supply and find the real DC turn on voltage. Datasheet? \$\endgroup\$
    – Russell McMahon
    Commented Aug 31, 2014 at 12:32

3 Answers 3

3
\$\begingroup\$

You want to power a 24V 900ma solenoid (21.6W) from a battery rated at 3.7V 3.4AH (12.58WH)? You could build a charge pump to make 24V from 3..3.7V, but (a) your battery certainly won't last very long between charges, and (b) you should expect significant losses from the charge pump, so your battery should be expected to last an even shorter time between charges.

Your thought of storing energy in a coil would consume a lot of extra energy, too, and as @JohnD pointed out would give you unpredictably high voltage spikes.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ I agree the capacitor is the way to go. I think we use energy conservation to guesstimate the recharge time. He needs 7500uF at 24 V ~about 2 Joules... then pick a battery current. I wonder if you could make a Cockroft-Walton type thing, but replace the diodes with fets driving them alternately? \$\endgroup\$ Commented Aug 31, 2014 at 16:15
  • \$\begingroup\$ Note that I just want to quickly flick the solenoid, so a short pulse of power is enough. I've edited the question to mention this. I'd expect to get enough juice for hundreds of pulses from that battery.And in my tests with the capacitor bank, they didn't get fully emptied but still had a few volts in them after a manual button push. With a short pulse I'd expect them to retain at least 12 volts, so that helps. \$\endgroup\$
    – Rennex
    Commented Aug 31, 2014 at 20:28
  • \$\begingroup\$ @Rennex... OK, I feel a lot better about that - with very brief pulses, your battery should last a lot longer than if run continuous. A charge pump would be a good way to go - you'd be drawing power similarly to the way a good photoflash would draw power (they often pump up from <10V to >4KV). \$\endgroup\$ Commented Aug 31, 2014 at 20:35
2
\$\begingroup\$

This page has a great example and schematic for charging up a capacitor to trigger a 24v solenoid: http://rayshobby.net/minty-water-valve-controller

It looks like he used a 3.6v battery and a 2200uF capacitor.

The schematic from his post:
enter image description here

And out of curiosity, what did you end up doing?

\$\endgroup\$
1
\$\begingroup\$

You can't change the current in in inductor instantaneously, so your idea will result in the voltage on the solenoid flying to very high levels, likely high enough to cause insulation breakdown and/or arcing somewhere. (Probably the switch, and would likely damage the contacts.)

The boost converter idea would work if you could get a battery that could support the large input current.

Otherwise the capacitor bank, or a redesigned solenoid would seem to be the best alternatives.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.