Question

The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as
i_p = 41 (V_p + 7 V_g)^1.41
Here, V_p and V_g are in volts and i_p in microamperes.
The tube is operated at V_p = 250 V, V_g = −20 V. Calculate (a) the tube current, (b) the plate resistance, (c) the mutual conductance and (d) the amplification factor.

Answer 1

Given:
Plate voltage, V_P = 250 V
Grid voltage, V_G = $-$20 V

(a) As given in the question, plate current varies as,
$$\begin{gathered} i_{\mathrm{p}}=41(V_{\mathrm{p}}+7V_{\mathrm{g}}{)}^{1.41} \\ {i}_{\mathrm{p}}=41(250-140{)}^{1.41} \\ {i}_{\mathrm{p}}=41\times (110{)}^{1.41} \\ {i}_{\mathrm{p}}=30984.71\mathrm{\mu A}=30.98\mathrm{mA} \end{gathered}$$

(b) $i_{\mathrm{p}}=41(V_{\mathrm{p}}+7V_{\mathrm{G}}{)}^{1.41}$
Differentiating this equation, we get:
$$\begin{gathered} d{i}_{\mathrm{P}}\mathit{}=41\times 1.41\times (V_{\mathrm{p}}+7V_{\mathrm{g}}{)}^{0.41}\times (d{V}_{\mathrm{p}}+7d{V}_{\mathrm{g}})(1) \\ \text{P}\mathrm{late}\mathrm{resistance}\mathrm{is}\mathrm{defined}\mathrm{as}: \\ {r}_{\mathrm{p}}={\left(\frac{d{V}_{\mathrm{p}}}{d{i}_{\mathrm{p}}}\right)}_{{\mathrm{V}}_{\mathrm{g}}=\mathrm{Constant}} \\ \mathrm{From}\mathrm{equation}(1), \\ \frac{d{V}_p}{d{i}_p}=\frac{1\times {10}^6}{41\times 1.41\times {110}^{0.41}} \\ \frac{d{V}_p}{d{i}_p}={10}^6\times 2.51\times {10}^{-3} \\ \frac{d{V}_p}{d{i}_p}=2.5\times {10}^3\mathrm{\Omega }=2.5\mathrm{K\Omega } \end{gathered}$$

(c) From above:
$$\begin{gathered} \mathrm{As}d{I}_{{}_{\mathrm{P}}}=41\times 1.41\times (250+7\times (-20){)}^{0.41}\times 7d{V}_{g,} \\ {g}_{\mathrm{m}}=(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}{)}_{V_{\mathrm{P}}=\mathrm{Constant}} \\ \mathrm{From}\mathrm{equation}\left(1\right), \\ \frac{1}{7}{\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{V_{\mathrm{P}}=\mathrm{Constant}}=41\times 1.41\times (250+7\times (-20){)}^{0.41} \\ {\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{V_{\mathrm{P}}=\mathrm{Constant}}=41\times 1.41\times (110{)}^{0.41}\times 7\mathrm{mho} \\ {\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{V_{\mathrm{P}}=\mathrm{Constant}}=41\times 1.41\times 6.87\times 7\mathrm{mho} \\ {\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{V_{\mathrm{P}}=\mathrm{Constant}}=2.78\mathrm{milli}\mathrm{mho} \end{gathered}$$
(d) Amplification factor,
$$\begin{gathered} \mu \mathit{}=r_{\mathrm{p}}\times g_{\mathrm{m}} \\ =2.5\times {10}^3\times 2.78\times {10}^{-3} \\ =6.95\approx 7 \end{gathered}$$