Common drain JFET output resistance "problem"

The output resistance of small signal model for common drain JFET amplifier equals

$$\frac{1}{g_\text{m}} || R_{\text{ds(on)}} \parallel R_\text{S}$$ and since $$R_{\text{ds(on)}} =\frac{1}{g_\text{m}}$$ we can also write this formula as $\frac{1}{g_m} \parallel \frac{1}{g_m} \parallel R_\text{S}$ right?

Which could be derived to

$$R_{\text{OUT}} = \frac{1}{g_\text{m}}/2 \parallel R_\text{S}$$ right?

Answer

$g_m$ is a transistor transconductance. In saturation region we viewing the transistor (FET) as a voltage controlled current source.

Vin is a input voltage and the output is a current, hence $g_m=\frac{Io}{V_{in}}$

Hence, for the FET $g_m$ is equal to $gm = \frac{dI_d}{dV_{gs}}$ (slope of the Id = f(Vgs) function)

In saturation region the Drain terminal behaviour just like an current source controlled via $V_{gs}$ voltage. And this is why you see the voltage controlled $(V_{gs})$ current source $I_d = g_mV_{gs}$ in the small signal equivalent circuit. Look at the answer given by KingDuken.

But this "drain" current source is not ideal. For the ideal current source the output current (drain current $I_D$) does not depend upon the voltage across it ($V_{ds}$). But in the real transistor $V_{ds}$ voltage due to channel length modulation will have small effect on the drain current.

And to "model" this effect (to represent channel length modulation on the small-signal equivalent circuit), we add a resistor $r_o$ parallel to the drain current source.

As you can see $r_o \approx \frac{1}{\lambda I_D}$ represent variation of $I_D$ with $V_{DS}$.

And $R_{ds(on)}$ is a FET resistance in the triode region when FET is full-on and $V_{ds}$ is very low $V_{ds}<<(V_{gs} - V_{th})$.

We can estimate the lambda value if we solve this set of equations:

$$I_{d1}=K(V_{gs} - V_{th})^2 (1+\lambda V_{ds1})$$
$$I_{d2}=K(V_{gs} - V_{th})^2 (1+\lambda V_{ds2})$$

$$I_{d1} - I_{d2} = K \lambda (V_{ds1} - V_{ds2}) (V_{gs} - V_{th})^2$$

Use the above to calculate $\lambda$

$$\lambda=\frac{I_{d1} - I_{d2}}{K(V_{ds1} - V_{ds2}) (V_{gs} - V_{th})^2}$$

Or this one

$$\lambda=\frac{I_{d2} - I_{d1}}{I_{d1}V_{ds2} - I_{d2}V_{ds1}}$$

Additional we can find $K$ factor

$$K = \frac{I_{d1}V_{ds2} - I_{d2}V_{ds1}}{(V_{ds2} - V_{ds2})(V_{gs} - V_{th})^2}$$

But we never do this type of calculation when designing circuit using a discrete FET's.