The plate resistance and the amplification factor of a triode are 10 kΩ and 20. The tube is operated at plate voltage 250 V and grid voltage – 7.5 V. The plate current is 140 mA. (a) To what value should the grid voltage be )b) To what value should the plate voltage be changed to take the plate current back to 10 mA?

Sol. r base p = 10 KΩ = 10 * 10^3 Ω μ = 20 V base p = 250 V V base g = –7.5 V I base p = 10 mA a) g base m = (δl base p/δV base g)V base p = constant ⇒ δV base g = δl base p/g base m = 15 * 10^-3 – 10 * 10^-3/μ/r base p = 5 * 10^-3/20/10 * 10^3 = 5/2 = 2.5 r’ base g = +2.5 – 7.5 = -5 V b) r base p = (δV base p/δl base p)V base g = constant ⇒ 10^4 = δV base p/(15 * 10^-3 – 10 * 10^-3) ⇒ δV base p = 10^4 * 5 * 10^-3 = 50 V V’ base p – V base p = 50 ⇒ V’ base p = -50 + V base p = 200 V