For a 6F6 triode tube, the plate current, plate voltage, and grid voltage are related by i_p = 41(V_p + 7 V_g)^ 1.41 , where V_p and V_g are in volts and i_p is in microamperes. If the tube is operated at V_p = 250 V and V_g = -20 V , calculate the mutual conductance.

Step-by-step solution

The mutual conductance g_m of a triode is defined as the rate of change of plate current with respect to grid voltage, keeping the plate voltage constant. g_m = ( i_p V_g )_ V_p Given the relation for the plate current: i_p = 41(V_p + 7 V_g)^ 1.41 Differentiating i_p with respect to V_g : g_m = 41 1.41 (V_p + 7 V_g)^ 1.41 - 1 7 g_m = 404.67 (V_p + 7 V_g)^ 0.41 Substituting V_p = 250 V and V_g = -20 V : g_m = 404.67 (250 + 7(-20))^ 0.41 g_m = 404.67 (250 - 140)^ 0.41 g_m = 404.67 (110)^ 0.41 g_m 404.67 6.87 g_m 2780