Analyzing \$g_m\$ vs \$V_{in}\$ for a common source config

Question

I had this doubt while going through the book "Design of Analog CMOS Integrated Circuits" by B. Razavi, if we are to plot \$g_m\$ vs \$V_{in}\$ for the above circuit, then we have \$g_m\$ =0 till \$V_{in}\$ crosses the threshold voltage(\$V_t\$). Now, the MOSFET enters the Saturation region directly, and we have \$g_m=k(V_{in}-V_{t})\$ that gives \$g_m\$ to be directly proportional to \$V_{in}\$. My main problem arises is when the MOSFET leaves the saturation region and enters the triode region. We'll have \$g_m=k(V_{ds})\$ which has no term of \$V_{in}\$ in it. How do I go on analysing the circuit in the triode region? What will the final relation on graph look like?

The book provides as the solution, I didn't quite get the solution provided in the book (on page 49). Any insights would help.

Answer 1

To find \$g_m\$ you need to take derivative of \$I_D = f(V_{IN}) \$

In saturation when \$V_{IN} > V_T\$, we have the following: $$g_m = K_P(V_{IN} - V_T)$$

Assuming that \$I_D = \frac{K_P}{2}(V_{IN} - V_T)^2\$

The MOS will enter the triode region when \$V_{DS} < V_{GS} - V_{T}\$:

The drain current in the triode region is equal to:
$$I_D = K_P\: \left( V_{GS} - V_T - \frac{V_{DS}}{2} \right)V_{DS} = K_P\: \left( V_{IN} - V_T - \frac{V_{DD} - I_D R_D}{2} \right)(V_{DD}-I_D R_D) $$

And the \$g_m = \frac{dI_D}{dV_{IN}}\$ in triode region is equal to:

$$g_m = \frac{1 - \sqrt{1 + K_P R_D ((2 + K_P R_D (V_{IN} - V_T) - 2 V_{DD} ) (V_{IN} - V_T))} + K_P R_D (V_{IN} - V_T)}{R_D \sqrt{1 + K_P R_D ((2 + K_P R_D (V_{IN} - V_T) - 2 V_{DD} ) (V_{IN} - V_T))}} $$

And if I plot in LTspice assuming:

\$V_{DD} = 10V , R_D = 1k\Omega, K_P = 0.001, V_T = 1V\$

I get this:

If we plot \$g_m\$ in the triode region for \$V_{IN}\geqslant \frac{\sqrt{1 + 2V_{DD} R_D K_P} \:\: - \: 1}{K_P R_D} + V_T = 4.58258V\$ (for this Vin MOS enters triode region). I get exactly the same plot as in the LTspice.