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I have a cheap PSU of this sort, used for my 3d printer. It has a 110V input, and a 12V output with 20A maximum current. Unfortunately, it doesn't have detailed specs/data available beyond that.

Now, I would like to place a simple ON-OFF switch (like this) at a convenient physical position, and in series with the AC power line.

Of course, it's important that the switch I choose (for example a 3A-rated one) is safe enough for the input current of the PSU. By conservation of energy, can I assume that the input doesn't draw more than (12 * 20 / 110) = 2 to 3 A? Or will it be significantly more than that, since I'm underestimating losses in the form of heat?

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    \$\begingroup\$ You are correct, and the type of switch you show would be fine just as long as it is mains (120 V) rated. I'd suggest you use a 10A 120 V switch which are very common. \$\endgroup\$ Commented Dec 17, 2017 at 16:37
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    \$\begingroup\$ You're right that the nominal current will about 2 A. However many Mains power supplies do have a large inrush current when they're powered up. Then large capacitors need to charge and this results in a short peak current. So 2 A will be too little and might work at first but probably will not last long. I would opt for a 5 A or more switch, that should be robust enough. \$\endgroup\$ Commented Dec 17, 2017 at 16:38
  • \$\begingroup\$ See this I think it could help you: magna-power.com/learn/kb/… \$\endgroup\$
    – gustaudio
    Commented Apr 16 at 19:17

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Cheap PSU may have not power factor corrector. In such case it draws non-sinusoidal current with RMS value which is larger than P/Uin (power per input voltage). Then, input power is larger then output (because of non-unity efficiency). So, even nominal input current is estimated at about 3 A. Taking into account already mentioned large inrush cuurrents, it's a good idea to use 10 A rated switch for long reliable operation. But 3A rated switch may also work for a while :)

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Input power supply units must have current protection. Typically a fuse. The fuse burns out during a malfunction. When the current is much higher than the current of normal operation.

The rated current of the ON-OFF switch must exceed the rated current of the protective element in the power circle.

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If you want to allow for the power supply being less than 100% efficient, add an efficiency factor to your calculation.

If we assume an efficiency of 80%, then it comes out to (12 X 20 / 110) / 0.8 = 2.7.

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