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I have a power supply (original Raspberry Pi power supply, datasheet) with 5.1 V which delivers up to 3 A. In case of a short-circuit or if more than ~ 3.4 A is drawn, the power supply will change to a short-circuit state.

This behaviour is used for a simple reverse voltage protection. A Schottky diode is placed anti-parallel and therefore will cause a short-circuit in case of reversed polarity. It will draw all the current and cause a high voltage drop below 0.3 V and so will protect the circuit.

The question though is, how many amperes do I need for the Schottky diode considering that the power supply will switch to a short-circuit state in that case? I consider that the diode will only get the ~ 3.4 A for milliseconds. Will I really need a diode with >= 3 A or isn't a diode with 1 A also be enough?

The Schottky diode I use is this one (SB 140 DIO, 40 V, 1 A), see datasheet. I also have one with 5 A, but I doubt if such a huge diode is really required.

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    \$\begingroup\$ I consider that the diode will only get the ~ 3.4 A for milliseconds <-- It's not a foldback power supply so, unless the diode short is manually removed, the current will be present indefinitely \$\endgroup\$
    – Andy aka
    Commented Mar 14, 2023 at 16:23
  • \$\begingroup\$ But how then can the short-circuit protection be effective? I assumed that once the limit is exceeded, it oppresses the current for a time period X. Once this has passed and the limit is again exceeded for a time period of Y, it is again oppressed. So I would assume that the diode most of the time isn't able to draw significant current. \$\endgroup\$ Commented Mar 14, 2023 at 16:31
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    \$\begingroup\$ The USB-C power supply is almost certainly a switching type, so will likely try to drive 3.4A into the load for some milliseconds, fail and reset, wait some number of milliseconds, then try again. The off-wait could be as short as 20ms, depending on how aggressively it re-tries. Could "see" this if an oscilloscope were connected for a one-second test. \$\endgroup\$
    – rdtsc
    Commented Mar 14, 2023 at 17:38
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    \$\begingroup\$ Forward voltage of only 0.3V under such conditions seems very optimistic. The diode datasheet says it will typically be around 0.4V. Now if you hot-plug the power supply into the diode it might see a 50A+ spike, not 3.4A (due to output capacitance in the power supply) and the voltage could get to more like -0.75V which would likely be destructive to the Pi. \$\endgroup\$ Commented Mar 14, 2023 at 17:52
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    \$\begingroup\$ It's possible it could retry for 4ms every 5000ms, or some other very low duty cycle. Really need an oscilloscope to actually see what it's doing for accurate analysis. \$\endgroup\$
    – rdtsc
    Commented Mar 14, 2023 at 19:41

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The conservative (i.e. worst case) assumption is that a power supply that's shorted becomes a current source: it will provide 3A continuously as long as the output voltage is below the nominal.

Switching power supplies may well revert to a pulsed mode of operation when their output is short-circuited, but you can't depend on that if you want to be robust (otherwise why have the diode at all).

So, the diode needs to be sized for continuous 3A operation.

Instead, I'd suggest using an ideal diode IC on the input. These days they are cheap and plentiful. Then you won't be shorting the power supply. And if you use four ideal diodes, you can have a full bridge, and won't even care about what the supply polarity is :)

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  • \$\begingroup\$ Thank you for the hint with the ideal diode! But I doubt if I have enough space for it. Do you think this is enough? i.prvy.eu/f3bhzl4.png \$\endgroup\$ Commented Mar 14, 2023 at 18:29

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