I am getting several PSUs (LRS-350-24) from Mean Well. Here's the specs:


1) The input current is stated to be 3.4A/230VAC at 88%. 350W output at 88% efficiency would mean 398W input/230V = 1.73A input which is 50% of the rated input current @ 3.4A/230VAC.

So should i take it as 1.73A or 3.4A? Because up to 3 PSUs will be tied to a switch rated at 10A, 240V. 3.4A would have exceeded the limit.

2) The stated inrush current is 60A/230VAC. Does this change if say, i am only going to use 50-60% of the PSU's output?

Reason is because the circuit for this PSU is tied with other lighting with 10A type C breaker which may trigger due to the inrush since the minimum is 5x.



1 Answer 1


The input to this supply is rectified and the rectifier output likely charges a capacitor bank. This means the supply will draw current only during part of the sine wave input, during the period when the rectified voltage is higher than the capacitor voltage (biasing the diodes on), so peak input current is higher because all of the input current happens in this short period (Tc in the diagram below). I would go with the higher number when choosing a switch rating. Since there is no current during the remainder of the sine wave, the input power is instantaneously zero during this time, and the average power input (average of instantaneous voltage x instantaneous current) is likely what they use to calculate efficiency.

This same capacitor bank is the culprit for your inrush current, so it is pretty much independent of the load during start-up.

Good luck! enter image description here

  • \$\begingroup\$ Is the 3.4A peak input current independent of the load too? Because the PSU will only be at 50% load max. \$\endgroup\$
    – HA E
    Jun 5, 2019 at 14:02
  • \$\begingroup\$ I think you can use the input current proportional to the load. \$\endgroup\$ Jun 5, 2019 at 18:22

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