I understand the specs stamped on almost all of the DC power adapters I've seen, but I have one with some extra values that confuse me. Specifically, the adapter says:

INPUT:  100-240VAC, 50/60Hz
   max. 700mA, 50VA

I'm in the U.S., so the input is 120VAC at 60Hz. I'm using this to intermittently power a 12VDC solenoid that draws 540mA.

What I don't understand is the line in the middle. I know 50VA indicates apparent power, which is an upper bound on real power. That seems plausible given that the real power at the output could be 48 Watts.

But I don't understand how that's consistent with the 700mA limit. If that that's an RMS current, then it could never reach 700mA at any voltage in the given range without exceeding the 50VA. So I guess the 700mA is the maximum instantaneous current draw. Is that right?

Does that "max" line contain useful information to someone (like me) who's trying to ensure the adapter can power a 12VDC solenoid that draws 540mA?

  • 1
    \$\begingroup\$ Yes, it's the peak surge current on power up. The question has been asked dozens of times here. \$\endgroup\$
    – Transistor
    Oct 26, 2021 at 15:49
  • 1
    \$\begingroup\$ Does that "max" line contain useful information to someone (like me) who's trying to ensure the adapter can power a 12VDC solenoid that draws 540mA? No, that max. 700 mA applies only to the AC input. As mentioned already, it is a peak current and generally only flows very briefly when you connect the adapter to mains voltage. So the mains voltage connection needs to be able to support 700 mA for a short time (that is usually no issue but it is mentioned anyway). The 12VDC, 2A is all that you need to be concerned about regarding the relay. \$\endgroup\$ Oct 26, 2021 at 15:52
  • 1
    \$\begingroup\$ "the real power at the output could be 48 Watts." No, unless you want the magic smoke to escape. It shouldn't be more than 12 V x 2 A = 24 W. \$\endgroup\$
    – Graham Nye
    Oct 26, 2021 at 17:30
  • \$\begingroup\$ @Transistor: If there's a canonical version of this question, I'd support closing this as a duplicate. But I haven't found one. \$\endgroup\$ Oct 27, 2021 at 16:39
  • \$\begingroup\$ I agree. Nothing good showed up in the "Related" questions list (on the right hand side of the page). I considered looking for one but there are so many questions related to power supplies that I wasn't sure I could find an appropriate one. We need to come up with a title for the canonical version. Yours isn't bad. \$\endgroup\$
    – Transistor
    Oct 27, 2021 at 16:44

2 Answers 2


700 mA is the max. input current. At 100 Vac low line voltage, that's 70 W of power at the input to the transformer. But there is a further restriction down to 50 VA, probably caused by the transformer design. You don't say whether this is a switching or linear supply, but the VA rating indicates to me that it probably is linear, with a standard, grunt 60 Hz power transformer.

Also, the "real power at the output" is rated for 24 W max., not 48.

Does that "max" line contain useful information to someone (like me) who's trying to ensure the adapter can power a 12VDC solenoid that draws 540mA?

No. What matters to you are the output specs. Note that if this is a linear supply, with a simple diode/bridge and filter cap on the secondary, the unloaded output voltage might be considerably higher than 12 V. This gets into the quality of the steel in the transformer core and other aspects of transformer design.

If it is a switching supply, then it probably has a minimum output current spec needed for stable regulation. This rarely is provided in the specs on the case, but a safe guess is 10% of the max. rated current. In your case, your load is over twice this value (200 mA), so you're good.


Switching supplies have wide voltage input range. The supply you have is a "worldwide" model that has to cope with the following extremes:

  1. Japan, with nominal line voltage of 100VAC, and at low line condition of -10%, so at the absolute minimum of 90VAC.

  2. UK, with nominal line voltage of 240VAC, and at high line condition of +5%, so at the absolute maximum of 252VAC. Some application areas require an even higher margin like +7% or even +10%.

Since it is required to operate indefinitely at those two extremes, there will be a further safety factor added so the absolute min-max voltage range is slightly larger in practice.

At 90VAC, 700mA, the supply will draw no more than 63VA. This is a condition with highest primary current, and thus highest losses from the primary rectification, primary bulk capacitor ripple heating, primary switch conduction loss, and primary winding I^2R loss. The real power consumed will be around 30-35W most likely, with 24W (full rating) taken from the output, and 5-10W dissipated. Modern, less cost-cutting designs will dissipate on the lower end of this, but lowest line condition is often the worst for efficiency.


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.