Peak inrush, the energy the switch has to absorb, and whether it stays inside its safe operating area. The energy is fixed at ½CV² whatever you do — only the time is yours.
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Charging a capacitor through a resistance dissipates exactly ½CV² in that resistance — the same energy that ends up stored in the capacitor, and completely independent of the resistance value. A 1000 µF bulk capacitor on a 12 V rail will always put 72 mJ into whatever charges it.
This is worth sitting with, because the instinct is to add series resistance to "reduce the inrush". It reduces the peak current, and it does nothing whatever to the energy. What it changes is the time, and therefore the power. Seventy-two millijoules in 50 µs is 1.4 kW; the same energy over 10 ms is 7 W. That is the entire principle behind soft-start.
E = ½CV² in the series element, always ·
Ipeak = V/R uncontrolled, or C·dV/dt with a rampτ = RC, 99% charged at 5τ
With nothing limiting it, the peak is V divided by the total series resistance — and that resistance is only the switch on-resistance and the wiring, typically tens of milliohms. Hundreds of amps is entirely normal for a moment. That is what welds relay contacts, nuisance-trips upstream protection, browns out a shared supply, and makes a connector arc on hot insertion.
Hot-swap designs rarely fail from the current rating. They fail because during the ramp the MOSFET has substantial voltage across it and current through it simultaneously, and the product can be hundreds of watts for milliseconds. That is a safe-operating-area question, not a continuous-rating question, and MOSFET SOA curves derate sharply for longer pulses.
Slowing the ramp reduces peak power but extends the time, and the SOA curve is not linear — which is why the slowest possible ramp is not automatically the safest. Check the actual SOA curve at your pulse width.
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