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I've got a SMPS which I'll be supplying with 230Vac. It outputs 24Vdc. Maximum power output is 600W. The efficiency is 89%.

I know that on the DC side, I'll be drawing 240W, or 10A. How can I work out the AC power draw?

Also, what cable size should I be looking at if I wanted to draw the full 600W from the PSU? On the input side, I would have thought 2.5mm^2 is ok, but surely not for the DC side with over 20A.

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With 240 W out, the input power will be 240 W plus whatever the supply uses/wastes internally. You say the supply is 89% efficient at full load. How efficient is it at the 40% load of 240 W?

To be pessimistic, let's assume 80% efficiency at 240 W. Then do the math. (240 W)/80% = 300 W. To get the current, divide that by the voltage: (300 W)/(230 V) = 1.3 A.

Cable size is a function of the maximum current it must be able to sustain. Get a wire chart and look it up.

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  • \$\begingroup\$ So would a cable that carries 10A @ 10kV would be the same that carries 10A @ 1V? (Obviously insulation not considered, just the copper). \$\endgroup\$ – CircularRecursion Sep 3 '18 at 0:29
  • \$\begingroup\$ Yes, assuming the I*R heat losses are not a factor. Thicker insulation will change the temperature, and the temperature time constant. Thus system stabilization will be longer. \$\endgroup\$ – analogsystemsrf Sep 3 '18 at 5:32

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