Q. The plate current in a triode is given by $I_{p} = 0.004\ \left( V_{p} + 10V_{g} \right)^{3/2}mA$ Where $I_{p},\ V_{p}$ and $V_{g}$ are the values of plate current, plate voltage and grid voltage, respectively. What are the triode parameters $\mu,\ r_{p}$ and $g_{m}$ for the operating point at $V_{p} = 120\ volt$ and $V_{g} = - 2\ volt$
Explanation
(a) $I_{p} = 0.004\ \left( V_{p} + 10V_{g} \right)^{3/2}mA$
$\Rightarrow \frac{\mathrm{\Delta}I_{p}}{\mathrm{\Delta}V_{g}} = 0.004\left\lbrack \frac{3}{2}\left( V_{p} + 10V_{g} \right)^{1/2} \times 10 \right\rbrack \times 10^{- 3}$
$\Rightarrow g_{m} = 0.004 \times \frac{3}{2}(120 + 10 \times - 2)^{1/2} \times 10 \times 10^{- 3}$
$\Rightarrow g_{m} = 6 \times 10^{- 4}mho = 0.6\ m\ mho$
Comparing the given equation of $I_{p}$ with standard equation $I_{p} = K\left( V_{p} + \mu V_{g} \right)^{3/2}$ we get $\mu = 10$
Also from $\mu = r_{p} \times g_{m} \Rightarrow r_{p} = \frac{\mu}{g_{m}} = \frac{10}{0.6 \times 10^{- 3}}$
$\Rightarrow r_{p} = 16.67 \times 10^{3}\Omega = 16.67k\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.