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Many load switches implement a fix or controlled slew-rate of its output current in order to prevent inrush currents.

I was wondering what is the step response of this kind of switches after the slew-rate time. To be exact, if I use such a switch to charge 200uF capacitors and my circuit draw some energy from those capacitors (not fully discharged, just a little dip), will the switch re-charge the capacitors instantly or will they be charged smoothly due to slew-rate time?

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    \$\begingroup\$ It cannot be instant but limited by I, low ESR of cap, RdsOn of switch and energy available (Power*time) and soak time at voltage max. This can easily vary from seconds to minutes. \$\endgroup\$ Sep 4 '17 at 17:23
  • \$\begingroup\$ So, the slew-rate doesn't matter and the maximum allowed current would be flow throught the load switch. \$\endgroup\$
    – MrBit
    Sep 4 '17 at 17:29
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    \$\begingroup\$ fastest slew rate is limited by cap specs , ESR*C=T and dv/dt =Ic/C \$\endgroup\$ Sep 4 '17 at 17:33
  • \$\begingroup\$ So, once the load switch is enabled and Vout reaches its maximum value, the response would be same as it would with a conventional high side switch (P-channel mosfet). \$\endgroup\$
    – MrBit
    Sep 4 '17 at 17:45
  • \$\begingroup\$ essential same but with some memory effects. \$\endgroup\$ Sep 4 '17 at 17:55
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My reading of the Vishay SiP32508 high-side switch is that if the voltage on both ends of the IC doesn't droop below 1 - 1.1 V, then the slew control circuitry won't engage, and the IC will behave as a 44-mOhm Rds(on) transistor switch.

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