Another glance on the MOS transistor equations

A few posts ago, I created the first part of an EKV-model discussion and we had a somewhat earlier post on MOS modeling. I haven’t compiled part two yet, but thought we could quickly revisit the Schichman-Hodges model and just look at the formulae again, and get a feeling for parameters and relationships. Also revisiting a unified expression for the current in “all” operating regions of the transistor.

Schichman-Hodges

For hand-calculations of MOS transistors we still, kind-of, use the Schichman-Hodges model from 1968 for the three regions (cut-off, linear, and saturation): $I_D = 0$ $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot \left( 2 \left( V_{gs} - V_{th} \right) - V_{ds}^2 \right)$ $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot \left( V_{gs} - V_{th} \right)^2$

where the parameters are effective carrier mobility, oxide capacitance per area, effective width and length of channel, gate-source voltage, drain-source voltage, and threshold voltage. Of course there should be large approximation signs, due to many reasons as discussed before. The transistor operates in different regions dependent on the voltages across it. See table further down below.

Also, the channel length modulation, lambda, is not included in the formula either.

Anyhow, I’ve always thought the formulas look a bit skewed and not that unified, arguably since they originally aimed at distinct operating regions. The drive and supply voltages were considerably higher than the threshold voltage and the chance/risk of the transistors sliding over between the different regions was very small.

There are suggestions on ways to glue the three regions together and use some indeces back and forth. We can for example introduce the channel length modulation for all three equations. The channel length modulation is (for hand calculations) assumed to be very small, and in the linear region, the inherent low drain-source voltage drop will further diminish its impact. $I_D = 0 \cdot \left(1 + \lambda \cdot \left( V_{ds} - V_{ds, sat} \right) \right)$ $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot \left( 2 \left( V_{gs} - V_{th} \right)\cdot V_{ds} - V_{ds}^2 \right) \cdot \left(1 + \lambda \cdot \left( V_{ds} - V_{ds, sat} \right) \right)$ $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot \left( V_{gs} - V_{th} \right)^2 \cdot \left(1 + \lambda \cdot \left( V_{ds} - V_{ds, sat} \right) \right)$

where I use the drain-source saturation voltage, which Sedra omitted. $V_{ds, sat} = V_{gs} - V_{th}$

Let us consider the linear region first and do some manipulations $\alpha = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L}$ $M = \left(1 + \lambda \cdot \left( V_{ds} - V_{ds, sat} \right)\right)$ $I_D = \alpha \cdot \left( 2 \left( V_{gs} - V_{th} \right)\cdot V_{ds} - V_{ds}^2 \right) \cdot M =$ $= \alpha \cdot \left(\left( V_{gs} - V_{th} \right)^2 - \left( V_{gs} - V_{th} \right)^2 + 2 \left( V_{gs} - V_{th} \right)\cdot V_{ds} - V_{ds}^2 \right) \cdot M =$ $= \alpha \cdot \left(\left( V_{gs} - V_{th} \right)^2 - \left( V_{gs} - V_{th} - V_{ds} \right)^2 \right) \cdot M =$ $= \alpha \cdot \left(\left( V_{gs} - V_{th} \right)^2 - \left( V_{gd} - V_{th} \right)^2 \right) \cdot M =$ $= \alpha \cdot \left(\left(\underbrace{ V_{gs} - V_{th}}_{V_{gs, eff}} \right)^2 - \left( \underbrace{ V_{gd} - V_{th} }_{V_{gd, eff}} \right)^2 \right) \cdot M =$ $= \alpha \cdot \left( V_{gs, eff}^2 - V_{gd, eff}^2 \right) \cdot M$

And actually, we also see that $V_{ds} - V_{ds, sat} = V_{ds} - (V_{gs} - V_{th}) = -V_{gd} + V_{th} = - V_{gd,eff}$

which compiles the equations as $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot 0 \cdot \left(1 - \lambda \cdot V_{gd, eff} \right)$ $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot \left( V_{gs, eff}^2 - V_{gd, eff}^2 \right) \cdot \left(1 - \lambda \cdot V_{gd, eff} \right)$ $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot V_{gs, eff}^2 \cdot \left(1 - \lambda \cdot V_{gd, eff} \right)$

and one could now introduce a couple of binary parameters such that all regions could be covered in one go. $I_D = \frac{\mu_0 \cdot C_{ox} } { 2 } \cdot \frac{W}{L} \cdot r_{on} \cdot \left( V_{gs, eff}^2 - r_{lin} \cdot V_{gd, eff}^2 \right) \cdot \left(1 - \lambda \cdot V_{gd, eff} \right)$

and the table compiles the choices. Now we have only two voltages to consider, and the different regions, and the physical parameters, of course.

Region $r_{on}$ $r_{lin}$ Conditions
Cut-off 0 $V_{gs,eff} < 0$
Linear 1 1 $V_{gs,eff} > 0, V_{gs,eff} > V_{ds,eff}$
Saturation 1 0 $V_{gs,eff} > 0, V_{gs,eff} < V_{ds,eff}$ Notably omitted

As always, the brasklapp, reservation is that the formulas above are far from accurate today, and we have not consider for example bulk effects and many, many more effects. However, the target was just to give another glance on the hand calculations and if there are ways to write them just a bit more intuitively…