The plate current in a triode is given by {I_ | vedclass

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Question

(A) The given equation for plate current is ${I_p} = 0.004 ({V_p} + 10{V_g})^{3/2} 	ext{ mA}$.Comparing this with the standard triode equation ${I_p}

Vedclass · Accepted Answer

$10, 16.7 	ext{ k}\Omega, 0.6 	ext{ m mho}$

Vedclass · Answer

(A) The given equation for plate current is ${I_p} = 0.004 ({V_p} + 10{V_g})^{3/2} 	ext{ mA}$.Comparing this with the standard triode equation ${I_p} = K ({V_p} + \mu {V_g})^{3/2}$,we find the amplification factor $\mu = 10$.To find the transconductance ${g_m}$,we differentiate ${I_p}$ with respect to ${V_g}$:${g_m} = \frac{\partial {I_p}}{\partial {V_g}} = 0.004 	imes \frac{3}{2} ({V_p} + 10{V_g})^{1/2} 	imes 10$.Substituting ${V_p} = 120 	ext{ V}$ and ${V_g} = -2 	ext{ V}$:${g_m} = 0.004 	imes 15 	imes (120 - 20)^{1/2} = 0.06 	imes (100)^{1/2} = 0.06 	imes 10 = 0.6 	ext{ m mho}$.Now,using the relation $\mu = {r_p} 	imes {g_m}$,we find the plate resistance ${r_p}$:${r_p} = \frac{\mu}{{g_m}} = \frac{10}{0.6 	imes 10^{-3} 	ext{ mho}} = \frac{10}{0.0006} \approx 16.67 	imes 10^3 \Omega = 16.67 	ext{ k}\Omega$.Thus,the parameters are $\mu = 10, {r_p} = 16.7 	ext{ k}\Omega, {g_m} = 0.6 	ext{ m mho}$.

$V_g$ (volt)	$0$	$-2$	$-4$	$-6$
$I_p$ (mA) for $V_p = 150$ $V$	$15$	$12.5$	$10$	$7.5$
$I_p$ (mA) for $V_p = 120$ $V$	$10$	$7.5$	$5$	$2.5$

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