0
\$\begingroup\$

Quick background:

I'm working on a very wide input range for a battery powered application (capability for two different types of chemistries). In the original design, a low side NFET was used as reverse polarity protection. This works, of course, at higher battery levels but when it starts getting lower, Vgs is just simply not high enough to sustain the in-rush currents from the devices being powered.

So I was able to find a low Iq charge pump from LT and was using it somewhere else in the circuit but I thought.. "why not use it to drive the gate of that low side FET as well!" See what I'm talking about below.

Reverse polarity low side NFET

So the big question is, any gotchas here that I'm not thinking of? I'm going to have it prototyped up shortly but I figured it would benefit the community if I share my concerns here!

Thanks y'all

\$\endgroup\$
  • \$\begingroup\$ do you have any specs for Vmin max on each port that are relevant? as well RgsCiss =T issues during transition with high current \$\endgroup\$ – Sunnyskyguy EE75 Sep 23 '18 at 21:12
  • 1
    \$\begingroup\$ I'd say give it a try. I think it will work. I wonder though, if you quickly connect the circuit in forward, and then reverse polarity if the mosfet will still be on due to the very large 10M resistor. \$\endgroup\$ – Drew Sep 23 '18 at 21:46
  • \$\begingroup\$ Good point @drew \$\endgroup\$ – jaredwolff Sep 24 '18 at 0:55
  • \$\begingroup\$ @TonyEErocketscientist Vmin/max for the Schottky diodes? Left side is either 0 or 5V. Battery+ ranges from about 7.4V down to 1.4V (or less due to high current transients) The Ciss is nearly 600pf so that may not play so nicely with the 10M. I'm just trying to strike a good compromise between static current draw (/battery life) and functionality. \$\endgroup\$ – jaredwolff Sep 24 '18 at 1:04
  • \$\begingroup\$ Heat dissipation is always a factor when bridge two low ESR voltages (battery & supply) at different potentials. I^2ESR \$\endgroup\$ – Sunnyskyguy EE75 Sep 24 '18 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.