I'm looking at this datasheet, which has this graph on page 3:
The FET has an \$R_{DS(on)}\$ of \$14 m\Omega\$; how can \$V_{DS}\$ be > 10 V at 100 A? Shouldn't that be 1.4 V? And if it was 10 V, wouldn't the FET evaporate at 100 A = 1000 W?
mosfetspecifications
I'm looking at this datasheet, which has this graph on page 3:
The FET has an \$R_{DS(on)}\$ of \$14 m\Omega\$; how can \$V_{DS}\$ be > 10 V at 100 A? Shouldn't that be 1.4 V? And if it was 10 V, wouldn't the FET evaporate at 100 A = 1000 W?
Best Answer
The 14 mOhm you mention is valid when Vgs >= 10V (in other words, the FET is fully turned on). The graph you show illustrates the FETs behaviour for Vgs in the range 2..8V, the range in which the FET is between fully off and fully on. In this operational mode the Isd is determined mostly by the Vgs (the FET behaves as a voltage-controlled constant current source). For this mode to work there must be a minimal Vds, otherwise that voltage would be the limiting factor for the Isd. Hence the Vds >= 10V requirement.