The plate current $i_{p}$ in a triode valve is given $i_{p} = K\left( V_{p} + \mu V_{g} ight)^{3/2}$ where $i_{p}$ is in milliampere and $V_{p}$ and $V_{g}$ are in $volt$. If $r_{p} = 10^{4}\ ohm,$ and $g_{m} = 5 imes 10^{- 3}mho$, then for $i_{p} = 8\ mA$ and $V_{p} = 300\ volt$, what is the value of $K$ and grid cut off voltage

Question

The plate current $i_{p}$ in a triode valve is given $i_{p} = K\left( V_{p} + \mu V_{g} ight)^{3/2}$ where $i_{p}$ is in milliampere and $V_{p}$ and $V_{g}$ are in $volt$. If $r_{p} = 10^{4}\ ohm,$ and $g_{m} = 5 	imes 10^{- 3}mho$, then for $i_{p} = 8\ mA$ and $V_{p} = 300\ volt$, what is the value of $K$ and grid cut off voltage

Dr. Abubacker · Accepted Answer

(b) $\mu = r_{p}g_{m} = 50$
From $i_{p} = KV_{p}^{3/2} \Rightarrow \frac{\mathrm{\Delta}V_{p}}{\mathrm{\Delta}i_{p}} = r_{p} = \frac{2i_{p}^{- 1/3}}{3K^{2/3}}$
$\Rightarrow g_{m} = \frac{\mu}{r_{p}} = \frac{3\mu K^{2/3}/i_{p}^{1/3}}{2} = \frac{3}{2}\mu K^{2/3}\left\lbrack K^{1/3}\left( V_{p} + \mu V_{g} \right)^{1/2} \right\rbrack$
$= \frac{3}{2}\mu K\left( V_{p} + \mu V_{g} \right)^{1/2} = 75\ K\ \left( i_{p}/K \right)^{1/3}$
Because $i_{p}$ was in $mA,\ g_{m}$ is substituted as $5\ m℧$
$\Rightarrow 5 = 75K^{2/3}i_{p}^{1/3} = 75\ K^{2/3}(8)^{1/3} \Rightarrow K = \left( \frac{1}{30} \right)^{3/2}$
Cut off grid voltage $V_{G} = - \frac{V_{P}}{\mu} = - \frac{300}{50} = - 6V$

Explanation

Reviewed by Dr. Abubacker

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