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  • The plate current in a triode valve is given <img alt="\mathrm{i_p=K\left(V_p+\mu V_g\right)^{3 /

The plate current \mathrm{i_p} in a triode valve is given \mathrm{i_p=K\left(V_p+\mu V_g\right)^{3 / 2}} where \mathrm{i_p} is in milliampere and \mathrm{V_p \: and \: V_g} are in volt. If \mathrm{r_p=10^4 \mathrm{ohm},\: and\: g_m=5 \times 10^{-3} \mathrm{mho}}, then for \mathrm{i_p=8 \mathrm{~mA} \: and\: V_p=300 volt} , what is the value of \mathrm{K} and grid cut off voltage
 

Option: 1

-6 \mathrm{~V},(30)^{3 / 2}
 


Option: 2

-6 \mathrm{~V},(1 / 30)^{3 / 2}
 


Option: 3

+6 \mathrm{~V},(1 / 30)^{3 / 2}
 


Option: 4

+6 \mathrm{~V},(1 / 30)^{3 / 2}


Answers (1)

best_answer

\mathrm{ \mu=r_p g_m=50 }

\mathrm{ From \: i_p=K V_p^{3 / 2} \Rightarrow \frac{\Delta V_p}{\Delta i_p}=r_p=\frac{2 i_p^{-1 / 3}}{3 K^{2 / 3}} }

\mathrm{ \Rightarrow g_m=\frac{\mu}{r_p}=\frac{3 \mu K^{2 / 3} / i_p^{1 / 3}}{2}=\frac{3}{2} \mu K^{2 / 3}\left[K^{1 / 3}\left(V_p+\mu V_g\right)^{1 / 2}\right] }

\mathrm{ =\frac{3}{2} \mu K\left(V_p+\mu V_g\right)^{1 / 2}=75 K\left(i_p / K\right)^{1 / 3} }

\mathrm{ Because \: i_p \text{was in }m A, g_m \text{is substituted as} \: 5 \mathrm{mv} }

\mathrm{ \Rightarrow 5=75 K^{2 / 3} i_p^{1 / 3}=75 K^{2 / 3}(8)^{1 / 3} \Rightarrow K=\left(\frac{1}{30}\right)^{3 / 2} }

Cut off grid voltage \mathrm{ V_G=-\frac{V_P}{\mu}=-\frac{300}{50}=-6 \mathrm{~V} }

Hence option 2 is correct.

Posted by

mansi

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