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

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## Given: Plate voltage, VP = 250 V Grid voltage, VG = $-$20 V (a) As given in the question, plate current varies as, ${i}_{\mathrm{p}}=41\left({V}_{\mathrm{p}}+7{V}_{\mathrm{g}}{\right)}^{1.41}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{p}}=41\left(250-140{\right)}^{1.41}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{p}}=41×\left(110{\right)}^{1.41}\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{p}}=30984.71\mathrm{\mu A}=30.98\mathrm{mA}\phantom{\rule{0ex}{0ex}}$ (b) ${i}_{\mathrm{p}}=41\left({V}_{\mathrm{p}}+7{V}_{\mathrm{G}}{\right)}^{1.41}$ Differentiating this equation, we get: $d{i}_{\mathrm{P}}\mathit{}=41×1.41×\left({V}_{\mathrm{p}}+7{V}_{\mathrm{g}}{\right)}^{0.41}×\left(d{V}_{\mathrm{p}}+7d{V}_{\mathrm{g}}\right)\left(1\right)\phantom{\rule{0ex}{0ex}}\text{P}\mathrm{late}\mathrm{resistance}\mathrm{is}\mathrm{defined}\mathrm{as}:\phantom{\rule{0ex}{0ex}}{r}_{\mathrm{p}}={\left(\frac{d{V}_{\mathrm{p}}}{d{i}_{\mathrm{p}}}\right)}_{{\mathrm{V}}_{\mathrm{g}}=\mathrm{Constant}}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{equation}\left(1\right),\phantom{\rule{0ex}{0ex}}\frac{d{V}_{p}}{d{i}_{p}}=\frac{1×{10}^{6}}{41×1.41×{110}^{0.41}}\phantom{\rule{0ex}{0ex}}\frac{d{V}_{p}}{d{i}_{p}}={10}^{6}×2.51×{10}^{-3}\phantom{\rule{0ex}{0ex}}\frac{d{V}_{p}}{d{i}_{p}}=2.5×{10}^{3}\mathrm{\Omega }=2.5\mathrm{K\Omega }$ (c) From above: $\mathrm{As}d{I}_{{}_{\mathrm{P}}}=41×1.41×\left(250+7×\left(-20\right){\right)}^{0.41}×7d{V}_{g,}\phantom{\rule{0ex}{0ex}}{g}_{\mathrm{m}}=\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}{\right)}_{{V}_{\mathrm{P}}=\mathrm{Constant}}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{equation}\left(1\right),\phantom{\rule{0ex}{0ex}}\frac{1}{7}{\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{{V}_{\mathrm{P}}=\mathrm{Constant}}=41×1.41×\left(250+7×\left(-20\right){\right)}^{0.41}\phantom{\rule{0ex}{0ex}}{\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{{V}_{\mathrm{P}}=\mathrm{Constant}}=41×1.41×\left(110{\right)}^{0.41}×7\mathrm{mho}\phantom{\rule{0ex}{0ex}}{\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{{V}_{\mathrm{P}}=\mathrm{Constant}}=41×1.41×6.87×7\mathrm{mho}\phantom{\rule{0ex}{0ex}}{\left(\frac{d{I}_{\mathrm{p}}}{d{V}_{\mathrm{g}}}\right)}_{{V}_{\mathrm{P}}=\mathrm{Constant}}=2.78\mathrm{milli}\mathrm{mho}$ (d) Amplification factor, $\mu \mathit{}={r}_{\mathrm{p}}×{g}_{\mathrm{m}}\phantom{\rule{0ex}{0ex}}=2.5×{10}^{3}×2.78×{10}^{-3}\phantom{\rule{0ex}{0ex}}=6.95\approx 7$  Suggest Corrections  0      Similar questions  Related Videos   PN Junction
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