If {R_p} = 7 text{ k}Omega and {g_m} = 2.5 te | vedclass

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(A) The amplification factor $\mu$ is given by the product of plate resistance ${R_p}$ and transconductance ${g_m}$.$\mu = {R_p} 	imes {g_m} = 7 	im

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$-2.86$

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(A) The amplification factor $\mu$ is given by the product of plate resistance ${R_p}$ and transconductance ${g_m}$.$\mu = {R_p} 	imes {g_m} = 7 	imes 10^3 \Omega 	imes 2.5 	imes 10^{-3} 	ext{ mho} = 17.5$.Also, the amplification factor is defined as $\mu = - \frac{\Delta {V_p}}{\Delta {V_g}}$ for a constant plate current.Given $\Delta {V_p} = 50 	ext{ V}$, we have $17.5 = - \frac{50}{\Delta {V_g}}$.Therefore, $\Delta {V_g} = - \frac{50}{17.5} \approx - 2.86 	ext{ V}$.

If ${R_p} = 7 \text{ k}\Omega$ and ${g_m} = 2.5 \text{ millimho}$, then on increasing the plate voltage by $50 \text{ V}$, how much must the grid voltage be changed so that the plate current remains the same? (in $V$)

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