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Topic: 5V to 24V to switch a solenoid (Read 11819 times) previous topic - next topic


So i am trying to use an arduino micro to switch a 24V solenoid, i am using a LM2557 DC-DC step up converter to boost the output of the arduino. (see link)
everything is fine until i hook up the solenoid, when i hook it up the out put from the arduino drops to 2.5V and the output across the step up converter holds at 4V. this could be because the solenoid draws 3.0 W (.125 amps) which the arduino does not put out. also i'm sure amps are lost as i step up the voltage.
i'm wondering if there is a way to boost the output current of the arduino? or there is there another way of making this work?
or should i try suing a relay or some other collection of components?

help, thanks


im not an expert, but id think youd be better using the arduino to activate a relay, and run the 24v through that from a separate supply. is this for a sprinkler/pump valve? most 5v relays can handle the voltage for a sprinkler valve with issue. on the coil of the relay, you will want a protection diode for when the EM field breaks down.  you can get solid state relays if you want to avoid the constant click/clack of a relay.

you might need a pullup/darlington type transistor approach depending on the relay and the load you need to switch.  ive built fluid level managers (water level in a decorative pond) that used 24vac sprinkler solenoids in the valves, but not using arduino (yet) so i know an itsy bit.

id really recommend using the arduino to control a relay or optocoupler because you are much less likely to damage your arduino if some part of it goes wrong. dont forget the protection diode!



As pyrotuc mentioned, the idea would be to have the Arduino control the power rather than provide it.  That can be done via a relay (which in turn might be controlled by a smaller transistor) or a MOSFET like in this tutorial page.  Be sure to choose your components such that the MOSFET etc are rated at over 24V in your case.

Cheers ! Geoff
"There is no problem so bad you can't make it worse" - retired astronaut Chris Hadfield


Just to amplify the point with inductive loads. Capacitors like to keep volts constant and inductors like to keep current constant. When you switch off the current going through an inductor the volts will go very high until the FET or bipolar transistor goes short and probably dies. If you connect a diode from the junction of the inductor and the transistor to the supply such that the diode is normally off, when you switch off the junction will be clamped at the voltage of the forward biased diode above the supply.


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Your answer may already be here: https://forum.arduino.cc/index.php?topic=384198.0

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