Question

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

Answer 1

Since $i_p=k{(V_g+\frac{V_p}{\mu })}^{\frac{3}{2}}$
Differentiating the equation
$d{i}_p=K\frac{3}{2}{(V_g+\frac{V_p}{\mu })}^{\frac{1}{2}}d{V}_g\Rightarrow \frac{d{i}_p}{d{V}_g}=\frac{3}{2}K{(V_g+\frac{V_p}{\mu })}^{\frac{1}{2}}{g}_m=\frac{3}{2}{(V_g+\frac{V_p}{\mu })}^{\frac{1}{2}}$
From (1)
$i_p={\left[\frac{3}{2}K{(V_g+\frac{V_p}{\mu })}^{\frac{1}{2}}\right]}^3\times K^{\prime }{i}_p=K^{\prime }(g_m{)}^3\Rightarrow g_m\alpha 3\sqrt{i_p}$