# A triode has plate characteristics in the form of parallel lines in the region of our interest. At a grid voltage of – I volt the anode current – I (In milli amperes ) is given in terms of plate voltage V  ( in volts)  by the algebraic relation.I = 0.125 V – 7.5For grid voltage of – 3 volts, the current at anode voltage of 300 volts is 5 milliampere. Determine the plate resistance (rp), transconductance (gm) and the amplification factor (μ) for the triode.

Kevin Nash
askIITians Faculty 332 Points
10 years ago
Hello Student,
I = 0.125 V – 7.5
⇒ dl = 0.125 dV or dV / dl = 1 / 0.125 = 8
We know that plate resistance, rp = dV / dl = 8mΩ
The transconductance , gm = [dl / dVg] v = constt
At Vg = - 1 volt , V = 300 volt, the plate current
I = [0.125 x 300 – 7.5] mA = 30 mA
Also it is given that V = - 3V, V = 300 V and I = 5mA
∴ gm = [30 – 5 / - 1 – (- 3)] = 25 / 2 x 10 -3 = 12.5 x 10-3 s
The characteristics are given in the form of parallel lines.
Amplification factor
= rp gm = 8 x 103 x 12. 5 x 10-3 = 100
Thanks
Kevin Nash