When we are analyzing MOSFET within a circuit (e.g. as an amplifier) with drain or source resistor we must consider the load line of transistor, like so: enter image description here

When used in triode mode a.k.a. ohmic mode there is said that drain current increases linearly with drain-source voltage and when used in active region there is often said that drain current changes almost nothing with drain-source voltage.

  • How is this possible if increasing drain current decreases drain-source voltage and vice versa when considering load line?

The load line in your graph shows what current flows (Id) over a varying Vgs.

Your text describing triode mode and saturation concerns the behavior of Id over varying Vds (not Vgs) while keeping Vgs constant.

Here the load line crosses the horizontal (well almost) part of the blue curves, that is the saturation region.

The triode mode region is the part on the left where the blue lines are nearly vertical.

  • \$\begingroup\$ So varying Vds, while Vgs is constant we consider the output characteristic without the load line? \$\endgroup\$ – Keno Jun 27 '17 at 19:36
  • 1
    \$\begingroup\$ You could make a plot of that yes, but notice how steep the triode part on the left side is. It is not very useful. And load lines are used for amplifiers where the input is varied (Vgs) and the output voltage (Vds) varies more. That is the amplification you get. Keeping Vgs constant will not show amplification so plotting a load line is quite useless. \$\endgroup\$ – Bimpelrekkie Jun 27 '17 at 19:37

Imagine your mosfet is a rheostat controlling a load. As the resistance lowers there is more drain current and a natural lowering of drain source voltage. Ultimately your rheostat (aka mosfet) will conduct nearly 5 amps when shorted out hence there is 12 volts across your load implying your load is a little over 2 ohms resistance. When your pot (aka mosfet) is not conducting current there is the full 12 volts across the it and zero volts across the load.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.