The plate current i_p in a triode valve is gi | vedclass

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(B) Given: $i_p = K(V_p + \mu V_g)^{3/2}$,$r_p = 10^4 \, \Omega$,$g_m = 5 	imes 10^{-3} \, 	ext{mho}$.First,calculate the amplification factor $\mu

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$-6 \, V, (1/30)^{3/2}$

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(B) Given: $i_p = K(V_p + \mu V_g)^{3/2}$,$r_p = 10^4 \, \Omega$,$g_m = 5 	imes 10^{-3} \, 	ext{mho}$.First,calculate the amplification factor $\mu = r_p 	imes g_m = 10^4 	imes 5 	imes 10^{-3} = 50$.For the grid cut-off voltage,$i_p = 0$,so $V_p + \mu V_g = 0$. Thus,$V_g = -V_p / \mu = -300 / 50 = -6 \, V$.Now,differentiate $i_p$ with respect to $V_g$ to find $g_m$: $g_m = \frac{\partial i_p}{\partial V_g} = K \cdot \frac{3}{2} (V_p + \mu V_g)^{1/2} \cdot \mu$.At the given operating point,$i_p = 8 \, mA$,so $8 = K(V_p + \mu V_g)^{3/2} \implies (V_p + \mu V_g)^{1/2} = (8/K)^{1/3}$.Substituting this into the $g_m$ equation: $5 	imes 10^{-3} = \frac{3}{2} K \mu (8/K)^{1/3} = \frac{3}{2} 	imes 50 	imes K^{2/3} 	imes 2 = 150 K^{2/3}$.Note: Since $i_p$ is in $mA$,$g_m$ is in $mA/V$. $5 	imes 10^{-3} \, 	ext{mho} = 5 \, mA/V$.$5 = 150 K^{2/3} \implies K^{2/3} = 5/150 = 1/30 \implies K = (1/30)^{3/2}$.

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