Electric Solenoid Current Draw

For a 24VDC solenoid coil, 1.17 Amps looks correct. When current is flowing, the rated 28W will be dissipated as heat in the coil winding. If you have several of these in a small enclosed area, the heat build up may be excessive. Think about adequate cooling when designing/implementing your project to maximize solenoid lifetime and reliability. Good luck!

Thank you mjb1962853 and old salt. Very much appreciated you sharing your wisdom and experience.

roy hammy-

You don't state whether you are dealing with AC or DC electrical. If it is DC your calculation is correct. If you are dealing with AC the calculations are more complicated than the simplicity of Ohms Law. In AC you are dealing with several different factors but you would get an answer close to the resistance of the DC calculations. This is called "reactance". In almost all instances the calculation's for simple AC circuits can be made with Ohms law, I=E/R, to provide the data you are trying to obtain. When Inductance and Capacitance are introduced into the calculations, that's when things get much more complicated.

This may be an oversimplification to some persons but for most purposes it is near to the exact answer. Adding inductors, capacitors, phase angle, etc. would make it seem very complicated and the original question get lost in all the calculations.

Good Luck, Old Salt

My error I am working with 24 volt DC

Would there be an inrush current, even with DC? I don't know, just asking. I can't think of an obvious reason, but there might be some effect as the armature moves towards the electromagnet.

In any case I would suggest for design OP uses a safety factor at least 2 on the calculated 1.17 amp.

Yes, there would be an inrush unless there is some external circuitry to reduce or eliminate it. A safety factor would certainly be a prudent item to include within the calculations/specifications.

Good Luck, Old Salt

Not usually on DC - the resistive component dominates. I have often seen the opposite - the build up of dc current lags due to the coil inducatnce with an R/L time constant.

With AC when the armature (or solenoid) is open the current is higher until the armature pulls in changing (shortening) the magnetic path length.

Perhaps you might want to go with a little bit of a larger unit. I would use 2 amps per coil. I have seen delivered voltage levels collapse a little bit (then recover) on cheaper DC power supplies. The collapsing voltage can cause problems in response time of some of the solenoids.

Cost difference would be minimal and being safer is better then being sorry.

You are correct and this is the plan as we are looking for longevity even though this project is the prototype. The power supply will be of higher capacity.

You can buy a digital multimeter for a modest banknote.

Measure the resistance (ohms Ω) of the solenoid.

The current taken (cold) will be (supply voltage)/ohms. The wattage would be supply voltage x current.

The copper winding will heat-up when on continuously, and increase in resistance by up to 25%.

Sometimes one gets AC solenoids used on DC. Usually, they need an additional resistance automatically switched-in after pull-in, to give holding current without overheating the solenoid coil. A lot of DC motor contactor solenoids used to have "economy resistors" switched-in for continuous hold. Not likely in your case - the solenoid resistance would be much lower than you would expect for the wattage and supply voltage [watts/supply voltage].

First even remotely helpful post when reading from the top down!!!

You beat me to it!!

That is exactly what he needs to do.

He could probably quite easily add the economy resistors (start with a value the same as the cold DC coil resistance) without much of a problem......if we knew more about the circuit!! That would reduce coil heating.....

Some coils will simply pull in with the resistor always in circuit, just a bit slower....testing needed....

The inrush current will depend on how much metal is inside the solenoid when power is applied. That may also be a function of how much stroke you have. If the stroke is small, you probably have a good bit of metal already in the coil. You might want to get a sample and connect it up with an Oscilloscope and some very small (shunt) resistors to observe the current.

You might consider looking for a suitable varistor since they are sometimes used as a surge protection device. It may slightly affect your timing but for a PLC application that should not be a big deal. If not already stated, you will need a good reversed bias diode to protect the PLC output since turning off the solenoid will create an inductive spike. Just like on a relay, the reverse biased diode takes care of the collapsing magnetic field when the circuit is opened.

1. For 28VA burden, at 24V, whether AC or DC, the current has to 1.2A.

2. You are talking of High Pressure Solenoid Valve. If the construction is that solenoid operates a pilot which in turn operates a the main valve then 28VA continuous burden calculated in correct.

3. If it is a direct operated solenoid with long travel or travel to open a valve with high pressure difference across its seat, then this may be a two coils in one solenoid. One high resistance and another low resistance coil, both in series, the low resistance provided for initial pull, the high resistance initially shorted by built in contact, which opens once the plunger in pulled in, the current reduces to value sufficient to hold the solenoid opened. In this case, you can know about such construction by connecting DC amp meter in series and noting down the steady state current. If this is the case then continuous burden on DC supply is 24 x measured current (which should be less than 28W).

4. Since you are also talking of use of PLC in the system to protect electronics as one of the contributor suggested, installed a Snubber Diode or Free Wheeling Diode to absorb surge during switching. Also select all auxiliary relays with built in free wheeling diodes.

The Rule of Thumb is to use 3X rated current for solenoid application. AC will sometimes actually be higher and DC usually lower but the power supply should be great enough, the wire large enough, that the solenoid does not get a voltage drop ahead of the coil. The worst thing on a solenoid is to half pull in and just set there. They will burn up. Most of the energy is expended to pull the solenoid in, and it needs very little to keep it activated. Be sure your supply, wiring, fusing, etc meets the 3X and you won't go wrong. If you need to lower that, you can, but you will need to make specific calculations.

Be sure to put in the reverse diodes across the coils as mentioned earlier or you will have failures on the energizing switches.

True for AC - not DC coils

"True for AC - not DC coils"

It is always better to use the 3X rule for DC also because of voltage drop in the line. While the current is not there, DC is much more vulnerable to voltage drop on solenoids and sometimes will "hang" and not operate completely. You really need to be sure full operating voltage is at the solenoid when it is first operated. Once it operates, it is not as important. Also DC is somewhat more sensitive to line and coil inductance to be able to get the current started flowing. Impedance is important in DC circuits also and is sometimes ignored. Most of these relays I have worked with are high speed machine processes and believe me, you must not ignore impedance and use only resistance in those cases.

I design the electrical part of large factory machines that typically bristle with many 100's of solenoids, both hydraulic and pneumatic types.

We exclusively use 24VDC for the power to these solenoids.

For a 30W hydraulic solenoid, we design for 2A per solenoid , but we know well that this is well more than what is actually drawn, the overhead that is allowed for in this figure is always way more than is ever needed.

I don't think it necessary to go to 3x rated current, this seem a little excessive to me.

....and to me!!

An excellent answer. GA.

A diode or similar snubber across the coil would of course kill dangerous voltage surges, but will add considerably to the drop off time due to freewheeling action of the reverse diode. If keeping this time short is important, one should use a varistor instead.

A diode with a resistor in series approximately equal to the coil resistance will provide a faster drop out and still limit the terminal volts to 1 pn.