Why are these mosfets passing large current, below turn on voltage ?

[MathWizard]

I was using LTspice to check some DC answers for a project and I never noticed this before, but 3 out of 3 mosfets I tried, all pass currents, even large currents, with their Vgs less than their turn on voltage.

So what DC model or equations are they using ?

I wanted to pass 80mA through a IRFZ44N, which has Vt=2.13V, but even with Vgs=2.12V, it's still maxing out and passing 304mA, which about max for 12.2V and a 40R resistor

An IRF510 did a lot better and only passed 2mA with it's Vgs a few dozen mV under it's Vth

[jpanhalt]

It appears you are simulating an IRLZ44, not an IRFZ44.  Note the lowest Vth can be 1V for the IRLZ44.

Datasheet snippet for IRLZ44 attached.

[Kleinstein]

The Vth prameter in the MOSFET model can be a bit different from the Vth used to characterize MOSFETs.
In the data-sheets Vth is given as the point where a small current, like 100 µA flows. The limit is set relatively arbitrary.
For the model chances are the Vth is more like the cross over from square root law for higher gate voltage to the exponential sub-threshold behavior. Especially for rather large FETs this may be at a not so small current, where one does not consider the FET as off.

[Siwastaja]

Also note Vgs_th is a poorly controlled parameter with a lot of unit-to-unit variation.

In switching application, you want to stay far from the uncertain region.

In linear application, you use feedback.

[MathWizard]

Ok I was just using LTSpice's Vt and Kp values, I know real-life values can vary a lot from the datasheet values.

Ok so something else to watch out for in sim's.

[JustMeHere]

There needs to be a high value resistor between gate and ground so the mosfet will properly turn off.  Either side of the power-gate resistor.  One side forms a voltage divider, but offers an layout advantage.  If the turn off resistor is high enough, the voltage divider can be ignored.