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JEE Main & Advanced Physics Semiconducting Devices Question Bank Critical Thinking

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    The plate current ip in a triode valve is given \[{{i}_{p}}=K{{({{V}_{p}}+\mu \,{{V}_{g}})}^{3/2}}\] where ip is in milliampere and Vp and Vg are in volt. If rp = 104 ohm, and \[{{g}_{m}}=5\times {{10}^{-3}}mho,\] then for \[{{i}_{p}}=8\,mA\] and \[{{V}_{p}}=300\,volt,\] what is the value of K and grid cut off voltage                                    [Roorkee 1992]

    A)            ? 6V, (30)3/2                          

    B)            \[-\,6V,\,{{(1/30)}^{3/2}}\]

    C)            + 6V, (30)3/2                          

    D)            + 6V, (1/30)3/2

    Correct Answer: B

    Solution :

                       \[\mu ={{r}_{p}}\,{{g}_{m}}=50\] From \[{{i}_{p}}=KV_{p}^{^{3/2}}\] Þ \[\frac{\Delta {{V}_{p}}}{\Delta {{i}_{p}}}={{r}_{p}}=\frac{2i_{p}^{-1/3}}{3{{K}^{2/3}}}\] Þ \[{{g}_{m}}=\frac{\mu }{{{r}_{p}}}=\frac{3\mu {{K}^{2/3}}i_{p}^{1/3}}{2}\]\[=\frac{3}{2}\mu {{K}^{2/3}}\left[ {{K}^{1/3}}{{({{V}_{p}}+\mu {{V}_{g}})}^{1/2}} \right]\] \[=\frac{3}{2}\mu K{{({{V}_{p}}+\mu {{V}_{g}})}^{1/2}}\]= 75 K (ip/K)1/3 Because ip was in mA, gm is substituted as 5 m℧ Þ \[5=75{{k}^{2/3}}i_{p}^{1/3}\]\[=75\,{{k}^{2/3}}{{(8)}^{1/3}}\]Þ \[k={{\left( \frac{1}{30} \right)}^{3/2}}\] Cut off grid voltage \[{{V}_{G}}=-\frac{{{V}_{p}}}{\mu }=-\frac{300}{50}=-6V\]


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