MAX15162 high-side switch IC start-up behavior confusion

Question

We are using the MAX15162 "Circuit Breaker IC" (basically a high-side switch with current monitoring and start-up/MOSFET SOA protection), but don't understand exactly how its start-up process works.

In its datasheet it says that to keep the MOSFET in its SOA during start-up, it outputs a series of current pulses where the averaged startup current is 28 mA. This continues until either 1) the output is within 700 mV of the input (at which point it starts to enter normal operation mode) or 2) the start-up watchdog times out.

But if a load is resistive and draws more than 28 mA before Vout is equal to Vin, how is Vout ever going to reach Vin so that the IC enters normal mode?

Surely Vout would start at 0 V and start to climb. The 28 mA would start charging any capacitance on the output. But any resistive load would start drawing current too. Once Vout climbs such that 28 mA or more is going through the resistor, won't Vout stop climbing? Obviously I'm not understanding its behavior correctly.

Answer 1

As @Polynomial mentioned in a comment, we've come to the conclusion that this chip is designed to work in tandem with a separate switch downstream which actually connects the load. We came to this conclusion given the wording in the datasheet around the 28mA average start-up current/max. capacitive loading and a test we did. The test involved a +24V rail and 50Ω load, which would draw 24V/50Ω=0.5A. This was designed to be above the 28mA but below its max. current limit. This IC was incapable of switching on the 50Ω load. However, if we enabled this IC first, and then connected the load, it was able to work as specified.