MOSFET Rds(on) – Common Questions Answered

For \$V_{GS}<V_{th}\$, there is weak-inversion current, which varies exponentially with \$V_{GS}\$, as given by

\$I_{D}\approx I_{D0}·e^\dfrac{V_{GS}-V_{th}}{n\frac{kT}{q}}\$

with

\$I_{D0}= I_{D}\$ when \$V_{GS}=V_{th}\$

\$k=\$ Boltzmann constant=\$1.3806488(13)·10^{−23} J·K^{-1}\$

\$T=\$ temperature in kelvins

\$q=\$ charge of a proton=\$1.602176565(35)·10^{−19}\$ C

\$n=\$ slope factor\$=1+\dfrac{C_D}{C_{ox}}\$

\$C_D=\$ capacitance of the depletion layer

\$C_{ox}=\$ capacitance of the oxide layer

You can use either experimental data, or a few points from graphs in the datasheet (like the one that Armandas suggests), to estimate \$I_{D0}\$ and \$n\$, and then use them to estimate \$I_{D}\$ for any \$V_{GS}\$ and \$T\$.

Reference: Modes of operation of a MOSFET.

Added: with my paragraph "You can use either..." I meant that you can do curve fitting to find the values for \$n\$ and \$I_{D0}\$ that best fit the data you have available, either from experiments (if you can do them), or from graphs from the datasheet (if there is any that is useful). In your case, Figure 2 (above), together with the equation above, might allow you extrapolate \$I_D\$ for lower \$V_{GS}\$ values. I'm not saying that you will end up with a high-quality estimate. I'm saying this is the best I could think of.

A supply of 10 volts across 1k means 10mA is the maximum current you can get thru the MOSFET. That is the problem with your experiment - no matter how much more you turned the juice up on the gate, the aiming point for drain current is 10mA and can not be exceded no matter how hard you try.

Of course Vds decreased - the mosfet was turning really hard-on and acting as a very small value resistor. This small resistance forms a potential divider with the 1k to make a very small voltage that is nowhere near 10V.