Question 17.PE.MCQ.23: For the transistor discussed in Question 21, what is the val......

(d)f_{\mathrm{T}}=h_{\mathrm{fe}}f_{\beta}

=2{2}{4}\times358.63\times10^{3}=80.333\ \mathrm{MHz}

For  f=f_{\mathrm{T}},

A_{\mathrm{i}}=-\frac{224}{1+\left\{(80.333\times10^{6})/(358.63\times10^{3})\right\}}

=-{\frac{224}{1+224j}}

\left|A_{\mathrm{i}}\right|={\frac{224}{\sqrt{1^{2}+\;224^{2}}}}=1

Therefore,

\angle A_{\mathrm{i}}=90^{\circ}-\tan^{-1}\left(224\right)=90^{\circ}-89.744^{\circ}=0.256^{\circ}