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I need to power my Pi 3A+ via the pins (not the onboard micro USB) so I can put the power connector anywhere I want in my custom box. I have a separate micro USB female connector connected via two wires to the first 5V pin and one of the ground pins on the Pi. Using the same power adapter that works very well with the onboard connector, I often get under-voltages, and so I worry about regulating the voltage.

I've read hats-master and it says that it's fine to do things that way, as long as I "implement a duplicate power safety diode", and it also provides this diagram:

backpowering-diagram

Seems to me that this diagram shows not one but two diodes, plus a polyfuse. I could solder these on the 5V input (between my own micro USB female connector and Pi's pins) but I don't understand what kind of diodes I need, and what polyfuse.

I would appreciate some simple answers telling me what components I should buy, and if it's fine to just solder them as per the diagram above.

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    I thought diodes caused a voltage drop - how is that supposed to work? – CoderMike May 28 at 9:21
  • The only way powering via the 5V and ground pins could be worse is if you are using higher resistance wires to the pins. – joan May 28 at 9:26
  • @joan the wires are very short (4-5 cm), really thin and have low resistance. They're just a strand off a 40-pin ribbon. I think the resistance is negligible in this case. – Alex Deva May 28 at 9:44
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If you already get under-voltages without any diodes, it means your USB socket, wires and connectors that go on the GND/5V pins have a high resistance which results in a significant voltage drop.

If you have a voltmeter, you could actually measure how much voltage arrives on your USB socket, and how much of that makes it to the Pi board. Otherwise, get wires with copper core which is at least half as thick as a GPIO pin. If you only have thin wires, it's possible to put several of them in parallel to get the right total thickness, but it will be a mess compared to a single thick wire.

If your wires are really short, it's less of a problem, but the connections between the connectors and the pins (and the connectors and the copper inside the jumper wires) remain just as significant regardless of the length. Soldering works wonders compared to low-quality connectors.

Adding any kind of protection will only make things worse, especially a regular diode. Inside a custom enclosure there's very little risk that two power supplies will ever be connected to the Pi at the same time, so it's not worth the effort.

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Last time I looked at the documentation it specified an "ideal diode" - which is NOT an actual diode.

enter image description here

There are NO polyfuses or diodes on the Pi4.

The wires in a ribbon cable are woefully inadequate! Use REAL wires, as short as possible.

If you actually read the HATS Master it says "Raspberry Pi Model A+, B+, Raspberry Pi 2B and 3B have an 'ideal' reverse current blocking diode (ZVD) circuit on their 5V input. The 5V GPIO header pins connect to the 5V net after the micro-USB input, polyfuse and input 'ideal' diode made up of the PFET and matched PNP transistors. The ideal diode stops any appreciable current flowing back out of the 5V micro USB should the 5V net on the board be at a higher voltage than the 5V micro USB input." There is NO MENTION of actual diodes!

A physical diode would cause a 0.7V drop - causing a 5V supply to produce 4.3V - well below the 4.65 trigger voltage.

You could do this, but if the connection to the Header pins is the ONLY power source there is no need.

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    Obviously there aren't, hence the instructions to add them :) The source of the diagram is in the link. It's an official Raspberry Pi repo on GitHub and it talks about powering a Pi via the pins. Attributes like "woefully inadequate" and "real" mean very little to me. After all, a ribbon cable is used all over the industry to pass 5V between devices, and I'm using even shorter lengths (I'm cutting them in halves). What's a "real" wire, and where's the border to woeful inadequacy? – Alex Deva May 28 at 9:56
  • Also, a regular diode will dissipate 1.5 Watt at 2A: simply put, it will get hot. – Dmitry Grigoryev Jun 29 at 12:57

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