The plate current in a triode can be written as \[i_p = k \left( V_g + \frac{V_p}{\mu} \right)^{3/2}\] Show that the mutual conductance is proportional to the cube root of the plate current.

#### Solution

Given:- The plate current varies with plate and grid voltage as

\[i_p = K \left( V_g + \frac{V_p}{\mu} \right)^{3/2}........... (1)\]

Differentiating the equation w.r.t `V_G,` we get:-

\[d i_p = K\frac{3}{2} \left( V_g + \frac{V_p}{\mu} \right)^{1/2} d V_g \]

\[ \Rightarrow g_m = \frac{d i_p}{d V_g} = \frac{3}{2}K \left( V_g + \frac{V_p}{\mu} \right)^{1/2}\]

From (1), plate current can be written in terms of transconductance as:-

\[i_p = \left[ \frac{3}{2}K \left( V_g + \frac{V_p}{\mu} \right)^{1/2} \right]^3 \times K'\]

Here, \[K'\text{ is a constant } = \left(\frac{2}{3} \right)^3 \times \frac{1}{K^2}\]

\[ i_p = K'( g_m )^3 \]

\[ \Rightarrow g_m \propto \sqrt[3]{i_p}\]