# Have I worked out the correct AC power draw for this power supply?

I've purchased this power supply (480watt, 40amp, 220v AC to 12V DC).

From my (very limited) understanding, the maximum AC current drawn from my home circuit (assuming maximum load on the power supply) would be calculated as:

Have I got that right?

Secondly, if I expect a max load of 36Amps on the power supply, is the expected AC current draw +-1.9Amps?

The circuit breakers in my house are sitting at 20Amps, so I would like to understand how many of these 36Amp setups I can run on a single circuit.

Thanks

• You haven't allowed for the power supply efficiency. 480/220 = 2.2A, but if the power supply is 80% efficient, divide by 0.8 to get 2.7A. That's 7.5 on a 20A circuit.
– user16324
Commented Jan 22, 2018 at 22:15
• With efficiency figured in and some margin. You might get away with eight of these on a twenty amp breaker. And yes, you have the calculation correct. Commented Jan 22, 2018 at 22:16
• But if you try to turn them all on at the same time, the inrush current might trip the breaker. Commented Jan 22, 2018 at 23:08
• @BrianDrummond Thanks for info everyone, really appreciate it! Brian, I think your comment is the closest to answering the question. If you want to move this comment to an answer, I'll mark it as the accepted answer. Commented Jan 23, 2018 at 8:40
• @SimonB no. The mains electrical system is designed to cope with inrush current, since any motor, incandescent light, switching or capacitor-heavy power supply also does it. Go to a big box store and see hundreds of T8 or T5 fluorescents on the ceiling, they really pack out those circuits and they all have inrush, that 277V/20A circuit probably spikes at 80A, no big deal. This is why breakers have a thermal trip mode that applies to current 1-5x rating. Look at a breaker trip chart, it looks like that because of inrush. Commented Jan 23, 2018 at 16:08

you need to open it up and look at the fuse size (since there is data missing, like supply efficiency) and add that rating. IEC-60127 requires 30% headroom on the fuses of switch mode power supplies, so your maximum current draw can not be more than 70% of the fuse. NEC is 80% current draw allowed to the circuit distribution breaker. So if you add your fuse values its still within the NEC electrical wiring safety limits.

• Of course, I don't know the specific national requirements of neither you nor OPs home country, but I would guess that there are also some requirements to the electrical/electronical equipment that you plan to connect to the public grid (and also similar requirements from whoever holds your home insurance), and I would think that this random unit from banggood probably hasn't been tested to conform to those requirements anyway. I don't think opening the unit is the correct way of solving this issue, even though it probably is a safe way to ensure selectivity, at least. Commented Jan 23, 2018 at 8:18
• all switch mode power supplies have to conform to IEC-60127 no matter what certification authority has approved it (UL, CE). No matter of country of origin or use. If it isn't approved by anyone I would not use it. Commented Jan 23, 2018 at 17:12
• Yeah, well. I choose not to be so bastant about it. Different countries, different regulations. Also, several countries take the harmonized IEC standards and make their own names and tiny adjustments. India is one of them. (Also, CE is not a certification agency). But anyway - my main point is that a random SMPS that is found on a low cost Chinese page without any datasheet, is not likely tested to conform with any standard, only to the manufacturers own functional and/or technical requirements. People buy non-approved electronic equipment on eBay and AliExpress all day all year. Commented Jan 23, 2018 at 17:19
• I agree with you, though. I would also stay away. Mine and my family's safety is more important than saving  on a hobby project. Commented Jan 23, 2018 at 17:21

As part of that machine's UL listing, there should be a "nameplate" AC power draw (which it draws from the AC side, not what it produces). you rely on that nameplate number, which is either in amps or VA, when determining how many an AC circuit can handle.

VA is sister to "watts" -- simply put, watts are the part of the sinewave your load actually uses, and VA is the entire sinewave.

It is acceptable to measure the device's actual power draw at max load (assuming it is consistent) using a monitoring tool like a Kill-A-Watt, and use that number instead of nameplate.

Now if the load will be a contiuous load (>3 hours), you must do a derate - and that is 125%. Meaning if the machine nameplates at 800VA, you must derate it, counting it as if it is 1000VA. That applies against the 4400VA capacity of your 20A, 220V circuit.

Be careful not to double-derate, for instance I was looking at a 30A Tripp-Lite PDU that admonished users not to exceed 24A. That is the same derate.

That 125% is your design margin. It is the only margin you are required to have.

Inrush current is no big deal, the AC power system is already factored to cope with it, since it is also done by motors, incandescent lights, and almost any modern switching supply including electronic ballasts and LED drivers.

• A simpler way of putting that is that VA is what you get out, and Watts is what you put in. Commented Jan 23, 2018 at 2:22
• You and me both know that this random Chinese power supply unit has never been UL listed. Commented Jan 23, 2018 at 8:13
• @MrGerber then he'll need to get a Kill-A-Watt type measuring device and measure actual performance under load. Commented Jan 23, 2018 at 15:44