Expression for gds as a function of VGS in weak inversion?

Hi! I need a way to show analytically that the output conductance of a MOS decreases as the VGS decreases in WEAK INVERSION. Classical books provide expressions only for STRONG INVERSION, like

gds = IDS/Va = lambda*IDS

Does anybody know is such an expression exists FOR WEAK INVERSION?. I need something of the form:

gds = gds(VGS), in weak inversion.

Thanks, Jorge.

In gds=IDS/Va the IDS is an operating point current which is dependent from VGS by an exponential function (page 36, **broken link removed**). Va (Early-voltage) is a constant, process dependent.

Thanks for the reply frankrose. I'm afraid I need something more physical (it's for a journal paper!). Ideally, a general expression that would hold both in strong an weak inversion. So far the best I have found is the following expression from Tsividis' book "Operation and Modelling of the MOS transistor" (eq. 8.2.48 in the 2nd edition / eq. 7.2.58 in the 3rd edition):

gds = (W/L)uCox*[Vgb-VFB-PSIsL-gamma*sqrt(PSIsL)]

(PSIsL is the surface potential at the drain) which I guess can be rewritten in terms of Vgs as:

gds = (W/L)uCox*[Vgs-Vsb-VFB-PSIsL-gamma*sqrt(PSIsL)]

However, I'm not sure if this expression is valid all the way from strong to weak inversion (that would be great) and in linear and saturation regions. Can someone please confirm this?

Note that I am not interested in the modelling of the Ids dependence on Vds (channel length modulation, DIBL, etc): I just need an expression to show that, when Vgs diminishes, the drain-to-source conductance diminishes too!

Thanks, Jorge.

If you look on the charge based models like EKV or ACM, you can seen they have nothing to do with gds for saturated transistor.
So, it does not matter in which inversion level mosfet is working, the ideal transistor has gds→0, Vds » Vdssat. It is because the normalized channel charge (and then current) is defined as a subtraction of so called forward (calculated in respect to source potential) and reverse one (calculated in respect to drain potential). In Saturation reverse part is going to 0, resulting with flat output characteristic.

As I know, the finite output resistance of mosfet is modeled by adding such things like: channel length modulation, drain induced barrier lowering and hot carrier injection.
And all of these effects are putted more or less independent to inversion region.
However, You can find that each of mentioned effects is dependent to overdrive voltage and/or affecting threshold voltage, so results with a Vgs influence on the gds.

Hi Dominik, thanks for your reply. I see your point that gds is normally due to CLM, DIBL, etc, but I get the feeling that we are missing something important here: my intuition tells me that, as the inversion level is reduced (by lowering VGS), the channel should become less and less conductive, right? For instance, in the limit case of VGS=0 the thing is totally OFF and there is no channel under the gate, and you essentially end up with an infinite resistance between source and drain. This channel resistance as a function of VGS is what I'm trying to write analytically. Does this make sense to you (or maybe I'm totally off here?)?

//edit - I see there is a problem with latex…

My understanding of your idea is as follow:
1. the channel conductance is defined as:
g_ds = I_D(V_GS,V_DS)/V_DS
And it is rather clear that with shorted source and gate, the channel diminish and gds→0 (for enhancement mode fets).
However it is also clearly stated in a posts above.
For example, including only first order channel modulation (as a constant) we get:
g_ds=Λ_m·I_D(V_GS)
and the drain current is this factor related to gate-to-source voltage. More accurate modeling replaces constant factors by appropriate functions of biasing, i.e. threshold voltage as a function of drain-to-source voltage, etc.

In long transistors (L≥1µm) the accurate results are achieved by using channel length modulation effect and estimating gds as (the full equation consists of a material constants which you can find in Enz, Vittoz book)
g_ds ~ I_D(V_GS)/L · [1/(V_DS - (V_GS-V_Th)/n)]^½

Concluding, the strongest V_GS therm is included in the relation for drain current, which has exponential relation with Vgs in weak inversion. It is a reason of exploding resistance there.