# Saturation Current Equation

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#### aryajur

The saturation current equation is:

I = K (Vgs-Vth)²(1+λ Vds)

For a PMOS I simulate and plot the IV curves for Vgs = 0.65V and Vgs = 0.8V for a 0.25um process (Vth0 = 0.558; uo=100e-4; Cox = 6.06e-3 => uo Cox = 0.606e-4).
From the curve when I calculate K I get the following results:

Vgs = 0.65; K = 3.09e-4
Vgs = 0.8 ; K = 1.19e-4

I understand that mobility is dependant upon Vgs. It decreases for increasing Vgs, uo being the high field mobility. But how does K become more than 0.606e-4, I assume this means that mobility from simulation is > the low field mobility uo.
If anybody has an idea why I get such a high value of K then please reply. Thanks.

#### aryajur

Another confusng thing is K of a NMOS increases by increasing Vgs, while that of PMOS decreases with increasing Vgs.

#### Humungus

##### Full Member level 6
I think the problem comes from the fact that the equation you posted is only valid for deep strong inversion. If your Vth is 0.558 and you simulate at Vgs=0.65, the you are mostly between weak and moderate inversion. In this case the transistor is mostly governed by diffusion phenomena, which mean that the relationship among Id, Vgs and Vth is exponential, similar to that of the bipolar transistor. Seach for EKV model. Even at Vgs=0.8V, you still are at moderate inversion, so that neither WI nor SI approximations are valid. That is why you got weird values of K, which come from a very simple model.

### aryajur

Points: 2

#### aryajur

Thanks, I was thinking the same, but then the question came up, what is the criteria for deciding the Vth value for transistors. It is written upto 3 decimal places, so it must be something quite accurate. How is it determined from the curves (any curves) for the MOS?

#### ocarnu

##### Member level 5
Vth has a physical menaing: is the voltage at wich the charge in the chanell becomes 0 (beween accumulation and inversion). Try to read more about CMOS physics, the formulas for hand calculations have a very limited use.

#### Humungus

##### Full Member level 6
The given Vth in a model is just a fitting parameter. In general it is not the physical one as ocarnu pointed out. Vth given in model cards are extracted in several ways: linear interpolation of the simplest quadratic SI equation with the Vgs axis in a Id-Vgs courbe, second derivative of gm vs. Vgs, etc.

Anyway, all extracted values are, sey, within 15% of variation one of each other.

EKV model gives you an excellent aproximation in all regions of operation.

#### thcm

##### Member level 2
aryajur,this equation is only good for long channel device(L>2um),for DSM device,considering the effects
such as velocity satuaration and mobility degradation etc,device's IV value deviates greatly from the value computed by equation you listed.

however ,for hand calculation ,you can introduce an
term 1/(1+theta(vgs-vt)) to incorporates these effects.

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