5
\$\begingroup\$

I have a 24VAC/60Hz solenoid. I used 12VDC just to try, and it worked good (well needless to say it worked as expected with 24VAC/60Hz).

What are the effects of a solenoid that's designed for AC running with DC? Applied force does not seem to be a problem in this case, so I think the main issue would be heat, as it is not running on the power source it was designed/optimized for. Technically speaking, what types of losses should be expected?

\$\endgroup\$
  • \$\begingroup\$ I don't think it will get any warmer... In fact, I'd say that it will be even more efficient. \$\endgroup\$ – Dzarda Jan 20 '14 at 8:01
  • \$\begingroup\$ @Dzarda well I have a small hint like that, but as it is optimized to AC this sounds a bit contradictory. I think that Eddy currents and saturation are the factors? \$\endgroup\$ – Diego C Nascimento Jan 20 '14 at 8:11
  • \$\begingroup\$ Precisely, you eliminate Eddy currents by passing through DC (please correct me), thus decrease magnetic flux-induced losses. On the other hand, the coil will become a simple resistor under DC, that could lead to more current being drawn... Well, you need to measure it anyway :) \$\endgroup\$ – Dzarda Jan 20 '14 at 8:21
  • \$\begingroup\$ @Dzarda thanks, I'm a bit unsure about Eddy Currents too, but I think that's correct. I don't think Joule heating would be a problem, as the coil is designed to 24VAC, running it in 12VCC should reduce it. But I don't know how the plunge acts, if it saturate at AC too. \$\endgroup\$ – Diego C Nascimento Jan 20 '14 at 8:29
  • \$\begingroup\$ If the DC current draw is similar to the AC current draw, you should be okay. As @Dzarda says, you need to measure it. \$\endgroup\$ – Spehro Pefhany Jan 20 '14 at 9:00
6
\$\begingroup\$

It's ampere-turns that produce the magnetic field to attract the moving part of a solenoid and operating at AC means the inductance of the coil comes into play and there is a limiting impedance restricting current. At DC the solenoid dc resistance may be very low and to get it to work correctly might mean operating it at quite low dc voltages compared to the ac operating voltage.

If you try running it at 24V dc it could easily burn-out so be careful. The dc voltage should be chosen so that the current is approximately the same as the RMS ac current at 24V AC.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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