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I have a N-MOSFET as power switch like this:

enter image description here

I want to add a fuse on the power line but I have few capacitors on the across the load and I am worried that the inrush current might blow the fuse off. How can I limit the inrush current here? I saw that for P-channel one can add capacitor across drain and gate. I am not sure if this works for N-ch MOSFET as well.

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  • \$\begingroup\$ where is that fuse in your diagram? \$\endgroup\$ Dec 14 '18 at 12:43
  • \$\begingroup\$ whats wrong with the pmos version? \$\endgroup\$
    – Jogitech
    Dec 14 '18 at 12:45
  • \$\begingroup\$ Nothing wrong with pmos but my VDD will be On only if the FET is on. For pmos I need the VDD to be on all the time. \$\endgroup\$
    – zud
    Dec 14 '18 at 13:14
  • \$\begingroup\$ @MarcusMüller The fuse will be between the VDD line and the voltage source. \$\endgroup\$
    – zud
    Dec 14 '18 at 13:15
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    \$\begingroup\$ I am worried that the inrush current might blow the fuse off. Before actually solving this "problem" how are you sure it actually is a problem? Fuses are slow (even the fast ones). I would not try to "fix" this problem before I actually have confirmed that it is a problem. There are still many unknowns, like what capacitors (and what is their ESR), what is the supply (perhaps it has a limited output current: problem solved), and even: what fuse?? You're like the rookie engineer who sees problems everywhere due to lack of experience. \$\endgroup\$ Dec 14 '18 at 13:18
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With Nmos ist basically the same...cap between gate and source but an additional high side Driver with a high resistance on ist output and a diode in paralell to the resisor.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ why is the diode needed? \$\endgroup\$
    – Alex G
    Mar 22 at 20:23
  • \$\begingroup\$ @AlexG To allow for slow turn-on and fast turn-off, the diode is needed. It's also because if the power supply is removed, the capacitor may otherwise stay charged and damage the MOSFET gate (and with the gate driver being referenced to the source, the voltage in the capacitor is higher than the source, plus the MOSFET source pin will go to 0V once the main power supply is removed, so the gate is now at vSupply + vGate, i.e at 48v + 12v in this case, with the source side falling to 0v. \$\endgroup\$
    – JonasCz
    Jul 23 at 21:29
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In this case a cap between drain and gate would limit the rate at which Vds decreases (i.e., Vload increases). If limiting that voltage ramp will address your inrush current issue, that's the easy route.

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TL;DR A high gate resistor value will increase the time it takes to charge the mosfet gate, and so the time it takes for it to turn on.

Found this interesting document about this topic here

2.2. Switching Time and Drive Conditions

Bipolar transistors need a large base current to maintain low on-state voltage. In contrast, since power MOSFETs are voltage-controlled devices, they can be driven just by charging gate capacity, and are therefore a low in power consumption. Note, however, that power MOSFETs have a slightly large input capacitance Ciss. Thus, for high speed switching applications, it is necessary to quickly charge the input capacitance from a low-impedance signal source. Low-impedance drive is required to reduce turn-on time. However, the use of a high gate voltage results in the much charging of gate-source capacitance, resulting in an increase in td (off). Switching time can be controlled via gate resistance. If you want to change the turn-on and turn-off switching speeds separately, you can use diodes to change the gate resistance values for turn-on and turn-off.

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