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  • The plate current in a triode is given by <img alt="\mathrm{I_p=0.004\left(V_p+10 V_g\right)^{3 / 2} m A}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BI_p%3D0.004%5Cleft

The plate current in a triode is given by
\mathrm{I_p=0.004\left(V_p+10 V_g\right)^{3 / 2} m A}
Where \mathrm{I_{p},V_{p}} and \mathrm{V_{g}} are the values of plate current, plate voltage and grid voltage, respectively. What are the triode parameters \mathrm{\mu ,r_{p}} and \mathrm{g_{m}} for the operating point at \mathrm{V_{p}=120\ volt} and \mathrm{V_{g}=-2\ volt}

Option: 1

\mathrm{10,16.7\ \mathrm{k} \Omega, 0.6 \mathrm{~m}\ \mathrm{mho}}


Option: 2

\mathrm{15,16.7\ \mathrm{k} \Omega, 0.06 \mathrm{~m}\ \mathrm{mho}}


Option: 3

\mathrm{20,6\ \mathrm{k} \Omega, 16.7\ \mathrm{m\ mho}}


Option: 4

None of these


Answers (1)

\begin{aligned} &\mathrm{ I_p=0.004\left(V_p+10 V_g\right)^{3 / 2} \mathrm{~mA}} \\ &\mathrm{ \Rightarrow \frac{\Delta I_p}{\Delta V_g}=0.004\left[\frac{3}{2}\left(V_p+10 V_g\right)^{1 / 2} \times 10\right] \times 10^{-3}} \end{aligned}
\begin{aligned} &\mathrm{ \Rightarrow g_m=0.004 \times \frac{3}{2}(120+10 \times-2)^{1 / 2} \times 10 \times 10^{-3}} \\ &\mathrm{ \Rightarrow g_m=6 \times 10^{-4} \mathrm{mho}=0.6 \mathrm{~m} \mathrm{mho}} \end{aligned}
Comparing the given equation of \mathrm{I_{p}} with standard equation \mathrm{I_p=K\left(V_p+\mu V_g\right)^{3 / 2}} we get \mathrm{\mu =10}
\begin{aligned} &\mathrm{ \text { Also from } \mu=r_p \times g_m \Rightarrow r_p=\frac{\mu}{g_m}=\frac{10}{0.6 \times 10^{-3}}} \\ &\mathrm{ \Rightarrow r_p=16.67 \times 10^3 \Omega=16.67\ \mathrm{k} \Omega} \end{aligned}

Posted by

Kshitij

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