The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as i base p = 41(V base p + 7 V base g)^1.41 Where V base p and V base g are in volts and I base p in microamperes. The tube is operated at V base p = 250 V, V base g = - 20 V. Calculate (a) the tube current, (b) the plate resistance, (c) the mutual conductance and (d) the amplification factor.

Sol. V base p = 250 V, V base g = –20 V a) ip = 41(V base p + 7V base g)^1.41 ⇒ 41(250 – 140)^1.41 = 41 * (110)^1.41 = 30984 μA = 30 mA b) I base p = 41(V base p + 7V base g)^1.41 Differentiating, dip = 41 * 1.41 * (V base p + 7V base g)^0.41 * (dV base p + 7dV base g) now r base p = dV base p/di base p V base g = constant. Or dV base p/di base p = 1 * 10^6/41 * 1.41 *110^0.41 = 10^6 * 2.51 * 10^-3 ⇒ 2.5 * 10^3 Ω = 2.5 KΩ c) From above, dl base p = 41 * 1.41 * 6.87 * 7 d V base g g base m = dl base p/dV base g = 41 * 1.41 * 6.87 * 7 μ mho = 2780 μ mho = 2.78 milli mho. d) Amplification factor μ = r base p * g base m = 2.5 * 10^3 * 2.78 * 10^-3 = 6.95 = 7