A change of 0.8 ,mA in the anode current of a | vedclass

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Question

(A) The plate resistance $r_p$ is given by the ratio of change in anode potential to the change in anode current: $r_p = \frac{\Delta V_p}{\Delta i_p}

Vedclass · Accepted Answer

$6.25 \,V$

Vedclass · Answer

(A) The plate resistance $r_p$ is given by the ratio of change in anode potential to the change in anode current: $r_p = \frac{\Delta V_p}{\Delta i_p} = \frac{10 \,V}{0.8 	imes 10^{-3} \,A} = 12500 \, \Omega$.We know that the amplification factor $\mu = g_m 	imes r_p$,where $g_m$ is the transconductance.Also,$g_m = \frac{\Delta i_p}{\Delta V_g}$.Substituting $g_m$,we get $\mu = \frac{\Delta i_p}{\Delta V_g} 	imes r_p$.Rearranging for the change in grid voltage $\Delta V_g$: $\Delta V_g = \frac{\Delta i_p 	imes r_p}{\mu}$.Substituting the values: $\Delta V_g = \frac{4 	imes 10^{-3} \,A 	imes 12500 \, \Omega}{8} = \frac{50}{8} = 6.25 \,V$.

$A$ change of $0.8 \,mA$ in the anode current of a triode occurs when the anode potential is changed by $10 \,V$. If $\mu = 8$ for the triode,then what change in the grid voltage would be required to produce a change of $4 \,mA$ in the anode current?

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