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 \[{{V}_{G}}=-\frac{{{V}_{p}}}{\mu }=-\frac{300}{50}=-6V\]You need to login to perform this action.
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