# Trying to simulate the linear region of a MOSFET

I wanted to simulate the linear region, so I tried this: simulate this circuit – Schematic created using CircuitLab

I expected the triangle wave to be curved, but it didn't have those features. Am I doing something wrong or am I expecting the wrong results?

How does one usually simulate the linear region in SPICE?

Or do I have to make a lumped model of an NMOS for it to display the linear region?

EDIT: simulate this circuit

I suppose this would be the more appropriate setup. The triangle $V_{ds}$ no longer creates a triangular $V_{ov}$. Making everything clearer to think about. Still, as in the previous schematic, the question remains, I was expecting curves, instead of straight lines.

EDIT:

Has anybody tied breadboarding the fist circuit and measured it with an oscilloscope? I just realized that it might not actually have significant curves. As the $V_{in}$ varied in phase with $V_{ds}$, that might compensate for the the expected upside down parabola of a varying $V_{ds}$, making it a lot less curved.

• If you're connecting a voltage source across the FET, the FET isn't doing anything useful. Normally you would connect your input signal in series with the bias generator V_in (6v) and measure the source and drain voltages. Or perhaps I really don't understand what you're trying to do. – user_1818839 Dec 6 '15 at 16:03
• Spice uses lumped models (for FETs and for pretty much everything except transmission lines) so that part of your question makes even less sense. – Fizz Dec 6 '15 at 21:19
• Also, you'll have to clarify what you mean by linear region before someone can answer. – Fizz Dec 6 '15 at 22:19
• @RespawnedFluff OK, I meant the reverse of a lumped model. A mass collection of different components, meant to simulate a single one. – kozner Dec 7 '15 at 5:45

## 1 Answer

The SPICE model will work correctly. As drawn, your FET is in the linear region, and if you superimpose a small triangle on the input, you'll get a small triangle at the output because it modulates the RDSON of the device.

• But would it displace curves instead of straight lines? This would all be much easier if I only an an oscilloscope... – kozner Dec 7 '15 at 5:44
• In your circuit, the FET has 4k in the drain, and 2k in the source. With 6 V on the gate pin, the VDS will be small (10's of mV), and VD ~= VS = 2 V. This gives the FET VGS=4 V, and an RDSON proportional to 1/(VGS-VT). The 2k & 4k make an approximately constant drain current, and so the VDS will be proportional to the RDSON of the device. – jp314 Dec 7 '15 at 5:53
• ^Sorry, my meant to type "display" – kozner Dec 7 '15 at 6:06
• I didn't get all that you were saying, but why is $R_{dson}$ proportional to to $1/(V_{ov})$, instead of $V_{ds}$?? – kozner Dec 7 '15 at 6:11
• in a FET, with VGS > VT, and VDS 'small', the RDSON is proportional to 1/(VGS-VT). You can derive that from the Shichman-Hodges model – jp314 Dec 8 '15 at 3:41