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A basic question, how does the P-Mosfet reduce the inrush current.

schematic

simulate this circuit – Schematic created using CircuitLab Old image link

Can the P-Mosfet's drain and source orientation be swapped along with the diode and would that cause any changes to the protection.

Please help.

http://archive.eetasia.com/www.eetasia.com/ARTICLES/2001JUL/2001JUL03_AMD_POW_TAC.PDF%3FSOURCES%3DDOWNLOAD

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    \$\begingroup\$ eek..my eyes... \$\endgroup\$ – Trevor_G Sep 13 '17 at 15:47
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    \$\begingroup\$ @Trevor: The image really is that bad in the PDF! \$\endgroup\$ – Dave Tweed Sep 13 '17 at 16:11
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Surge is reduced by using a constant current source to charge the Gate capacitance of the P-FET. You can alter the amount of time taken in the linear region by altering the CC drive. If you drive the P-FET with a fast drive (high current therefore fast voltage change) you would get surge current.

No you can't turn the P-Fet around (swap Source and Drain) since you'd have simply a (body) diode which would conduct and drop the source voltage.

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  • \$\begingroup\$ Thanks for the answer, can we connect the gate directly to the ground instead of CC circuitry? Would that help in reducing the inrush current? \$\endgroup\$ – SIllyMan Sep 13 '17 at 16:15
  • \$\begingroup\$ No, that would defeat the surge protection. Read the note you reference .....this is slew rate control of a FET turn-on. \$\endgroup\$ – Jack Creasey Sep 13 '17 at 16:18
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As Trevor says, that's not a nice image to look at!

Quick answer to your question though: the FET slowly turns on, basically acting as a resistor that reduces in value. High resistance at the start reducing the high currents when you pretty much have a short to ground as the capacitor is empty. The FET turns on, reducing resistance as the voltage of the cap increases to match the source voltage.

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It should be obvious that if you swap drain and source, the body diode will no longer block the inrush current.

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