Question

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.

Answer 1

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

\[ i_p    =   K'( g_m  )^3 \]

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