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I want to power four 5 V devices via a single 12 VDC 8 A (96 watt) supply using a buck converter for each device.

I know I could try and redo my circuit but this means disconnecting and rewiring just to test, so I wonder if someone with far more knowledge can do the math and say "Yes" or "No."

The devices are connected via buck converters 5 V, 2.5 A maximum.

  • Raspberry Pi Zero 2W requires 12.5 watts
  • Raspberry Pi Zero 2W requires 12.5 watts
  • Raspberry Pi Zero 2W requires 12.5 watts
  • Raspberry Pi 4B requires 15 watts

Total 52.5 watts.

Is this 12 V supply good enough after conversion to power everything without any power dropouts?

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  • \$\begingroup\$ Peter_Brown_USA - Hi, Just an FYI - some (most?) of the cheap DC-DC converters sold on marketplaces like Ebay, Amazon, AliExpress etc. have exaggerated ratings (as well as often having counterfeit ICs, high output noise etc. and are lacking a trustworthy datasheet & post-sales support). You didn't ask for places to get buck converters (which is good, as such questions are off-topic here), but in case you were thinking of buying from such places, I wouldn't. Personally (unless I'm designing the converter) I always buy from reputable suppliers, even though it costs $. \$\endgroup\$
    – SamGibson
    Nov 16, 2023 at 17:43

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That will work just fine. The buck converters will need to be at least 52.5/96 = 54.7% efficient, which is such a low requirement that practically any buck converter on the planet can do it.

The Pi 4B will of course need a buck converter that can provide 3A, not just 2.5A.

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  • \$\begingroup\$ So all this conversion works with watts (I am a noob at electronics :-) ) taking in the efficiencies into the equations \$\endgroup\$ Nov 16, 2023 at 14:19
  • \$\begingroup\$ @Peter_Brown_USA Exactly! Switching converters are generally designed to lose as little power (Watts) as possible in the conversion, so the available output power is generally a large fraction of the input power (i.e. 90% or more for a buck). As a result, if you convert to a lower voltage, you will be able to draw a higher current from the output than is consumed at the input. \$\endgroup\$ Nov 16, 2023 at 16:43

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