I am looking for a MOSFET that is capable of dealing of a max pulsing voltage of 60A(between drain and source) and came across these 2 MOSFETS

https://www.onsemi.com/pub/Collateral/NDPL070N10B-D.PDF https://pdf1.alldatasheet.com/datasheet-pdf/view/257362/SUNTAC/IRFZ44N.html

both are capable of dealing with the Amperage and voltage I require

but the only thing I don't understand is that the Vgs is +,-20V . Excuse me if I am incorrect but that would mean I would have to provide +20V at the gate to turn the MOSFET on (enhancement N type)? and for it to turn off -20V or just down to 0V.

Is this correct? Thankyou

  • 2
    \$\begingroup\$ No, those are abs max ratings, beyond which you destroy the device. Stay well clear of them. Check what voltage Rds(on) is specified at : usually 10V. That and 0V are your target On-Off voltages (though 12 or 15V for ON won't hurt either) \$\endgroup\$
    – user16324
    Jun 20, 2020 at 18:30
  • \$\begingroup\$ Ok good because I was thinking that was a little excessive.. thankyou \$\endgroup\$ Jun 20, 2020 at 18:34

1 Answer 1


As @Brian Drummond said, you are looking at the absolute max specs. Both parts say that you'll need 4V (says it can work at 2V, but you don't want to operate at the limit). Apply more than 4V to turn it on, less than 2V to turn it off. If you have it, 10V will work perfectly but you don't have to go that high if you don't have it available.

Your Rds(on) tells you voltage drop you'll see across the FET and how much heat will be generated when it is on. I think of it like a resistor that is added to my power path. 60A at 12.8mohms isn't a huge voltage drop (60*0.0128=0.768V) but parts will heat up if it sustained. As the part heats up, that resistance will increase too.

  • \$\begingroup\$ This is just wrong. It mistakes the threshold voltage Vgs(th) with the voltage required for full switching. At 4V you may get as little as 1 mA through the FET (see the OnSemi datasheet for measurement conditions for Gate Threshold voltage.) \$\endgroup\$
    – user16324
    Jun 21, 2020 at 15:27
  • \$\begingroup\$ I'll give you that 4V is the minimum for full switching, but I think my point is still the same. You don't need the full 10V. According to the graphs in the datasheet, if you want to pull 60A, then you only need 5.5V. \$\endgroup\$
    – PiMaker0
    Jun 21, 2020 at 16:09
  • \$\begingroup\$ Again no (wrt the OnSemi NDPL070 part). The ID/VDS curves show(at 60A) for Vgs=6V, Vds=1.5V at 60A, for 90W dissipation. You don't want to operate it there. And remember those curves aer typical, not guaranteed. Are you looking at teh other datasheet? \$\endgroup\$
    – user16324
    Jun 21, 2020 at 16:39
  • \$\begingroup\$ First off, thanks for taking the time to look at this with me. I was looking at the ID/VGS curve of the NDPL070 part. The way I read it, for Vds of 10V, you only need a Vgs of 5.5V to allow 60A Id. The point I was trying to make is that Vgs is on a curve and doesn't require a particular voltage to be considered "on." To your point, the higher the voltage, the more current can be allowed through. \$\endgroup\$
    – PiMaker0
    Jun 21, 2020 at 16:55
  • 1
    \$\begingroup\$ You are understanding the curve, but the problem is that a Vds of 10V doesn't count as even close to "fully switched on" even if that was guaranteed rather than typical. With a 12V supply it would only leave 2V across the load, with a 600W device dissipation (at 60A). It's practically still an amplifier in that region, not a switch. To get guaranteed Rds(on) 12.8 mOhm or less, use 10V. \$\endgroup\$
    – user16324
    Jun 21, 2020 at 18:02

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