# Actuator resistance

Sorry, I’m not much good at electrical work, but I’m trying to understand why this actuator I purchased doesn’t use the rated power specs. I expect this question may have been asked before, but I can’t seem to phrase it correctly when looking for an answer.

It’s rated for 12VDC @ 400mA It uses 12V @ 2200mA on my bench power supply. If I limit the amps to 400mA, then it brings the voltage down to 2.2V.

So I suppose that means the resistance is 5 ohms when it should be 30 ohms. Why’s is the actual resistance so far off the rating?

Edit: I purchased the actuator on ebay, here. It didn't come with many technical specifications, sorry.

Apparently, it's a DC solenoid electromagnet, so I expect the design is comparable to an inductor, but I admittedly don't know much about inductors either, so...

• Can you show us the data sheet for the actuator? Dec 8, 2017 at 23:08
• Without telling us more about the actuator we can not really answer that. Sounds like stall current to me, but again, no data..no clue. Dec 9, 2017 at 0:13
• Are you sure the actuator is designed to operate on 12 Volts DC? An inductor (relay coil or similar) designed for AC use will draw a much higher than specified current on DC. Dec 9, 2017 at 2:37
• forums.adafruit.com/viewtopic.php?f=31&t=60298 Dec 9, 2017 at 5:22
• As mentioned in the linked adafruit forum the 400mA may be max holding current and the 12V max drive voltage at low duty cycle. The rating lable may also be totally wrong. Dec 11, 2017 at 22:33

Solenoids have a distinct set of characteristics that dictate that the amount of current required to pull them in is MUCH greater than that needed to hold them in. This is because of the magnetic circuit that closes the air-gap reduces, and in some models closes, when the actuator moves in.

As such, many solenoids are rated at an actuation voltage and a continuous holding current.

It would appear your solenoid requires 12V to get you the initial indicated pull force. This will result in a much larger current than the 400mA needed to hold the solenoid pulled back. Based on your figures, the coil is indeed ~5.4ohms at DC.

As such, you need to initially activate the solenoid with the full current for a brief period and then, assuming you want the solenoid to hold there, cut back on the current.

The current reduction can be accomplished by using two different drivers, one high and one low current.

A better alternative is to switch from the DC activation signal to a PWM controlled hold signal. The latter should be at a high enough frequency to maintain the lower constant current in the inductance of the coil without saturation or complete decay. Since the device data information is sparse at best, you may need to experiment a bit to determine the inductance when the actuator is pulled back. This value will be somewhat greater than when it is in the rest position.

The fly-back circuit should be designed to accommodate the holding current continuously and the peak-current sporadically.

The issue with this kind of circuit though is a failure in the control system or the MOSFET will result in full current flowing through the coil continuously which would cause the latter to overheat. As such, some additional protection is warranted.

Your linked ad shows the printed specs. It's for DC. We do not know what the printed 12VDC means. Add a series resistor for low enough current and measure how heavy mass the solenoid can lift It can be enough. If it isn't, you probably have got a fake product. I have more than once bought good looking stuff from webshops and got crap with fake well known logos.