Question

A plate current of 10 mA is obtained when 60 volts are applied across a diode tube. Assuming the Langmuir-Child relation $i_p\infty V_p^{3/2}$ to hold, find the dynamic resistance r_p in this operating condition.

Answer 1

According to Lamgmuir-Child Law,
the relation between plate current (i_p) and the plate voltage (V_p) is given by
$i_{\mathrm{p}}=C{V_{\mathrm{p}}}^{3/2}...\left(1\right)$

Differentiating equation (1) with respect V_p, we get:
$\frac{d{i}_{\mathrm{p}}}{d{V}_{\mathrm{p}}}=\frac{3}{2}C{V_{\mathrm{p}}}^{1/2}...\left(2\right)$

Dividing (2) and (1), we get:
$$\begin{gathered} \frac{1}{i_p}\frac{d{i}_p}{d{v}_p}=\frac{3/2C{V_p}^{1/2}}{C{V_p}^{3/2}} \\ \Rightarrow \frac{1}{i_{\mathrm{p}}}.\frac{d{i}_{\mathrm{p}}}{d{v}_{\mathrm{p}}}=\frac{3}{2V_{\mathrm{p}}} \end{gathered}$$

The dynamic resistance is given by:
$$\begin{gathered} \frac{d{v}_{\mathrm{p}}}{d{i}_{\mathrm{p}}}=\frac{2V_{\mathrm{p}}}{3i_{\mathrm{p}}} \\ {r}_{\mathrm{p}}=\frac{2V_{\mathrm{p}}}{3i_{\mathrm{p}}} \\ {r}_{\mathrm{p}}=\frac{2\times 60}{3\times 10\times {10}^{-3}} \\ {r}_{\mathrm{p}}=4\times {10}^3=4\mathrm{k\Omega } \end{gathered}$$