A potential candidate is junction over temperature caused by dissipation during the capacitor charging.** As much energy ends up dissipated in the MOSFET as ends up stored in the capacitor, independent of how fast it turns on. **You have to look at the precharge duration and transient thermal impedance for that time scale to figure out if it's fine.

Not really. Only if MOSFET is placed into an ideal circuit. If for example ESR of capacitor is the same as MOSFET channel resistance and it turns on instantly, then MOSFET would dissipate only a half of total energy that gets stored in the capacitor, the other half would be dissipated by the capacitor itself. But then also comes resistance of wires and power source which will dissipate their share of the energy.

Obviously this was in reference to the ideal case, with only MOSFET Rdson as resistance.

However, the same capacitor energy is lost regardless of the MOSFET Rdson, switching time, source impedance, capacitor ESR, wiring resistance or anything else. The capacitor final energy is lost even with no resistance, as the source must deliver exactly twice that energy regardless of any resistances!!

Some analysis will show such, or a simple experiment with a pair of identical quality (film) capacitors.

Take one cap C1 and charge to V, other cap C2 has 0 volts, connect a DMM (use Hi Z State) across charged cap C1, disconnect charged cap C1 from charging source then short other cap C2 across charged cap C1 and note DMM reading, should be half the initial charged voltage as expected. Now both caps are charged to half the initial voltage, or V/2.

At start Cap energy = (C1Vi^2)2, where Vi is initial voltage.

Final Cap energy = (C1Vf^2)/2 + (C2Vf^2)/2), where Vf is final voltage on both caps.

Ratio Final Total Cap Energy to Initial Energy, or (Final EnergyC1 + Final EnergyC2)/(Initial EnergyC1), or [(C1Vf^2)/2 + (C2Vf^2)/2]/[(C1Vi^2)/2]

With C1 = C2, the ratio is 2(Vf/Vi)^2, or 1/2!!

So half the initial energy is lost in this process even though no resistance was included.

Just for fun we got a couple quality PolyPropylene caps.

C1 = 4.84807uF ESR 0.00503 ohms

C2 = 4.84523uF ESR 0.00521 ohms

Charged C1 to 10.0002 Volts, C2 was 0 volts

C2 Shorted to C1 and Voltage was 5.002 Volts on DMM Hi Z mode.

Did same test with a 10K resistor in series with C2 and got the same result, as with 1K and 100 ohms.

This shows the switching resistance doesn't matter wrt to the energy lost, sure it can be dissipated in different places and begs the question what about absolute zero switching resistance...where does the energy go, radiating magnetic fields

BTW this understanding led us to a solution back in ~80 to a CCD clock driver problem driving a bunch of high capacitance CCD clocks lines at high rates and over 15VPP. The usual solution back then was a massive clock driver which dissipated very high power, we came up with a driver based upon a technique we called "Reactive Clock" to exchange the CCD clock capacitive energy with the supply and saved over 95% of the driver power!! Published in EDN much later when we were allowed.

Edit: Forgot to mention, you can use unequal capacitors for C1 and C2, then:

Vf =Vi(C1/(C1 + C2))

Best,