The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as
$i_{p} = 41 (V_{p} + 7 V_{g})^{1.41}$
Where $V_{p}$ 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.

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Solution

Given: $V_{p} = 250 V; V_{2} = - 20 V I_{p} = 41 (V_{p} + 7 V_{g})^{1.41} = 41 (250 - 140)^{1.41} = 41 \times (110)^{1.41} = 30984 μ A = 30 m A (b) I_{p} = 41 (V_{p} + 7 V_{0})^{1.41}$
Differentiating now, we get
$= d i p = 41 \times 1.41 \times (V_{p} + 7 V_{4})^{0.41} \times (d V_{p} + 7 d V_{g}) N o w, r_{p} = \frac{d V_{p}}{d i_{p}} o r \frac{d V_{p}}{d i_{p}} = \frac{1 \times 10^{6}}{41 \times 1.41 \times 110^{0.41}} = 10^{6} \times 2.51 \times 10^{- 3} = 2.5 \times 10^{8} Ω = 2.5 K Ω .$
(c) From Above,
$d I_{2} = 41 \times 1.41 \times 6.87 \times 7 d v_{g} g_{m} = \frac{d I_{p}}{d V_{g}} = 41 \times 41 \times 6.87 \times 7 μ m h o = 2780 μ m h o = 2.78 milli mho$
(d) Amplification factor
$μ = r_{p} \times g_{m} = 2.5 \times 10^{2} \times 2.78 \times 10^{- 3} = 6.95 \approx 7$

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