Electronic – Can you use less then the rated current on a High Side Smart Switch such as BTS555

The mechanical design of the switch affects both the voltage and current ratings. The higher the voltage, the greater the chance of arcing as the switch opens. The arcing can reduce the switch reliability over time by inducing oxidation of the surface of the switch contacts and hence increasing the resistance of the contacts. In addition, the voltage rating is provided to offer an insulation guarantee. The switch must be safe to operate at the rated voltage and pose no risk of shock to the user.

Too high a current results in a voltage drop across the switch reducing its functionality (an ideal switch should have 0 resistance and hence 0 voltage drop across it).

In your load circuit, it appears you'd like to use one switch per resistor. Since the resistors are in parallel, the maximum current a switch will see will be 35A/2 = 17.5A when you turn on a second parallel resistor. The 20A switch should suffice. when a third switch is turned on, the current in each switch reduces to 35A/3 = 11.67A.

To summarize, for your application, you need only consider the current rating of the switch. If your application is mass-market, to minimize cost you would perhaps want a lower voltage-rated switch which will be less expensive.

I recently received 4 of these :-) !!!
I haven't tried them yet.
IPB180P04P4L-02 :-) :-) :-) :-)
IPB180P04P4L-02 datasheet here P Channel MOSFET, 40V, 180A,
1.8 milliOhm typical at 100A (!) with Vgs = 10V at 25 C.
2.4 milliOhm max same conditions.
At Vgs = -10V Rdson is about ruler straight through 180- Amps.
It will be higher at higher temperatures. Pricing here $6.38/1 at E14 / Farnell and less in volume
$2.75/1 Mouser - 6000 in stock.
$1.31/1 Avnet !!!, - no stock.

Drive resistors are non critical for occasional on/off. With 1K you'll get maybe 10's of uS switch time worst case but minimal loss overall.

BUT Vgs_max matters. For the IPB180P04P4L-02 it's 16V max, so a 10V clamp (zener) woul;d be in order if using more than 10V drive. (You usually get to exceed Vgsmax only once per MOSFET lifetime.

Check data sheet for your FETs.

Your loss calculation has gang aglae.

4.2 mOhms for Vgs = 10 V ... If my calculation is right, the power loss would be about 0.25 W for a load of 60 Amps.

Power = I^2 x R = 3600 x 0.0042 =~ 15 Watts.

For the IPB180P04P4L-02 at 60A at say 2 mR that's ~= 7 Watts.
My package is the TO263-7-3.
3 n+ 1 cut off + 1 in TO263 pkg with bent pins for SMD.
Tjc = 1 K/W = 7 K rise at 7 Watts. You want some sort of semi real heatsink for this at say 10 Watts to be safe.
At 5 C/W (easily enough achieved that's about 60C rise for 10W.
Lower C/W heatsink = cooler.