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 amplification factor.

Step-by-step solution

The plate current of a triode is generally expressed as i_p = K(V_p + V_g)^n , where is the amplification factor. Comparing the given equation i_p = 41(V_p + 7 V_g)^ 1.41 with the standard equation, we directly get = 7 . Alternatively, the amplification factor is defined as = - ( V_p V_g )_ i_p . For a constant i_p , differentiating the given equation yields: 0 = 41 1.41 (V_p + 7 V_g)^ 0.41 (dV_p + 7 dV_g) dV_p + 7 dV_g = 0 = - dV_p dV_g = 7