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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.

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Solution

## 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: $d{i}_{\mathrm{p}}=K\frac{3}{2}{\left({V}_{\mathrm{g}}+\frac{{V}_{\mathrm{p}}}{\mu }\right)}^{1/2}d{V}_{\mathrm{g}}\phantom{\rule{0ex}{0ex}}⇒{g}_{\mathrm{m}}\mathit{}\mathit{=}\mathit{}\frac{d{i}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}=\frac{3}{2}K{\left({V}_{\mathrm{g}}+\frac{{V}_{\mathrm{p}}}{\mu }\right)}^{1/2}$ From (1), plate current can be written in terms of transconductance as: ${i}_{\mathrm{p}}\mathit{}={\left[\frac{3}{2}K{\left({V}_{\mathrm{g}}+\frac{{V}_{\mathrm{p}}}{\mathrm{\mu }}\right)}^{1/2}\right]}^{3}×K\mathit{\text{'}}\phantom{\rule{0ex}{0ex}}\text{H}\mathrm{ere},{K}^{\mathit{\text{'}}}\mathrm{is}\mathrm{a}\mathrm{constant}=\left(\frac{2}{3}{\right)}^{3}×\frac{1}{{K}^{2}}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{p}}\mathit{}=\mathit{}K\mathit{\text{'}}\left({g}_{\mathrm{m}}{\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒{g}_{\mathrm{m}}\propto \sqrt[3]{{i}_{\mathrm{p}}}$

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