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.

Explanation

Given : `V_p = 250V, V_2 =-20V `
` (a) I_p = 41 (V_p + 7 V_g)^1.41`
` = 41 (250-140)^1.41`
` = 41 xx (110^(1.41))`
` = 30984 mu A = 30mA. `
` (b) I_p = 41 (V_p + 7V_0)^(1.41) `
Differentiating now, we get
` dip = 41 xx 1.41 xx `
`(V_p + 7V_g)^(0.41) xx (dV_p + 7dV_g) `
` Now, r_p = dV_p/di_p [V_g rarr Constant ] `
` or dV_p/ di_p = ((1 xx (10^6))/(41 xx 1.41 xx (110^0.41)))`
` = (10^6) xx 2.51 xx (10^-3) `
` = 2.5 xx (10^3) Omega = 2.5 K Omega `
(c) From above,
` dI_2 = 41 xx 1.41 xx 6.87 xx 7dV_g `
` g_m = (dI_p/dV_g) `
` = 41 xx 41 xx 6.87 xx 7 mu mho `
` = 2780 mu mho `
` = 2.78 milli mho `
(d) Amplification factor
`mu = r_p xx g_m `
` = 2.5 xx (10^2) xx 2.78 xx (10^-3) `
` = 6.95 ~~ 7 .`