Question

The plate current in a triode can be written as
$i_p=k{\left(v_g+\frac{V_p}{\mathrm{\mu }}\right)}^{3/2}$
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
$i_{\mathrm{p}}=K{\left(V_g+\frac{V_{\mathrm{p}}}{\mu }\right)}^{3/2}...\left(1\right)$
Differentiating the equation w.r.t $V_{\mathrm{G}}$ ,we get:
$$\begin{gathered} d{i}_{\mathrm{p}}=K\frac{3}{2}{\left(V_{\mathrm{g}}+\frac{V_{\mathrm{p}}}{\mu }\right)}^{1/2}d{V}_{\mathrm{g}} \\ \Rightarrow g_{\mathrm{m}}\mathit{}\mathit{=}\mathit{}\frac{d{i}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}=\frac{3}{2}K{\left(V_{\mathrm{g}}+\frac{V_{\mathrm{p}}}{\mu }\right)}^{1/2} \end{gathered}$$
From (1), plate current can be written in terms of transconductance as:
$$\begin{gathered} i_{\mathrm{p}}\mathit{}={\left[\frac{3}{2}K{\left(V_{\mathrm{g}}+\frac{V_{\mathrm{p}}}{\mathrm{\mu }}\right)}^{1/2}\right]}^3\times K\mathit{\text{'}} \\ \text{H}\mathrm{ere},K^{\mathit{\text{'}}}\mathrm{is}\mathrm{a}\mathrm{constant}=(\frac{2}{3}{)}^3\times \frac{1}{K^2} \\ {i}_{\mathrm{p}}\mathit{}=\mathit{}K\mathit{\text{'}}(g_{\mathrm{m}}{)}^3 \\ \Rightarrow g_{\mathrm{m}}\propto \sqrt[3]{i_{\mathrm{p}}} \end{gathered}$$