This might be a silly question but if I'm using a 12V power supply rated to provide 5A does that mean that if I was to regulate the voltage down to 5V I could theoretically draw 12A at 5V?

If Watt = Amp X Voltage, then the supply should be apply to provide 60W in any configuration of amps and Volts(provided the voltage is under 12V)

Is this correct?

  • 2
    \$\begingroup\$ Yes, theoretically. Practically there will be loss in the conversion. Note, that linear regulators won't do the trick, you need a switching one. \$\endgroup\$
    – Eugene Sh.
    Jul 30, 2018 at 20:08
  • 1
    \$\begingroup\$ Normally No, unless the spec says so. \$\endgroup\$ Jul 30, 2018 at 20:16
  • \$\begingroup\$ Look into the data sheet. Most power supplies can't do this, but for those who can it will be mentioned. \$\endgroup\$
    – starblue
    Jul 30, 2018 at 20:17
  • \$\begingroup\$ Seems like we have different understanding of the question.. Are you talking about adjustable power supply alone, or a power supply and some kind of circuit to regulate it's output? \$\endgroup\$
    – Eugene Sh.
    Jul 30, 2018 at 20:19
  • \$\begingroup\$ Not likely. Some components of the power supply may be limited by current, not power. This depends a lot on the design of the power supply. \$\endgroup\$ Jul 30, 2018 at 20:33

3 Answers 3


No regulator is perfect, and you will not get 100% efficiency. In general, it is Power in - inefficiency = power out, when using a switching buck or step down regulator. Average efficiency is 80 to 90%, so you could get 10.8 amps at 5V from a 60W supply input.

  • \$\begingroup\$ 12A must use a very low RDSon mosfet. \$\endgroup\$
    – Jason Han
    Jul 31, 2018 at 1:15

Almost certainly not.

For linear supplies, the lower the output voltage the greater the power dissipated by the regulator.

For switching supplies, inductor/transformer saturation is usually the limiting factor, and this is largely independent of output voltage.

  • \$\begingroup\$ I guess I framed by question incorrectly. I'm not looking for a precise 1:1 conversion. I was planning on using 2XLM2596 switching regulators which can output 3A each at 5V. I wanted to ensure that a 12V 5A supply could supply 6A at 5V. \$\endgroup\$
    – shekit
    Jul 30, 2018 at 21:22
  • \$\begingroup\$ Yes a 60w supply could with the correct regulator provide a 30W output. Those two seem appropriate but you are better off getting a 7 amp or higher regular instead of 2 small ones. If you need 6 amps, get a slightly bigger regulator as a margin of safety. \$\endgroup\$
    – Passerby
    Jul 30, 2018 at 23:05
  • \$\begingroup\$ @Passerby another reason for using 2 regulators was to reduce noise in the circuit. I have a Raspberry Pi and an amplifier connected to speakers which is picking up a lot of noise. It does reduce if I use an LC filter but only really goes away if I power the pi and amplifier separately and share ground between the two. I was thinking I could use one regulator for the amplifier and one for the Pi and this will hopefully help reduce the noise. \$\endgroup\$
    – shekit
    Jul 31, 2018 at 0:46
  • \$\begingroup\$ That's a legit use case. \$\endgroup\$
    – Passerby
    Jul 31, 2018 at 1:06

No. This means the power supply is designed to produce 12V constantly, up to a maximum of 5A. It will not output other voltages.

If you connected it to a 0.4167 ohm resistor (which would draw 12A at 5V, or 28.8A at 12V) it may (depending on the quality) turn itself off, output only 0.1 volts, make a strange noise, go bang and/or catch fire.

Yes. You can add a second conveter, after this one, to convert 12V to 5V and you will be able to get 11-ish A out of that one. The output of your main power supply would be the input of your second power supply. Or you could get one that outputs 5V to start with.


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