Question 11.11: A certain BJT transistor has rπ = 2 kΩ and β = 100 at 1 MHz ...

We know that               f_{T}=\beta f_{\beta }=\beta _{f} f

or                                     f_{T} = 5 × 20 × 10^{6} = 100  MHz

f_{\beta }=\frac{f_{T}}{\beta } =\frac{100\times 10^{6} }{100}=1  MHz

We know that                f_{\beta }=\frac{1}{2\pi C_{\pi }r_{\pi } }

1 × 10^{6}=\frac{1}{2\pi C_{\pi }\times 2\times 10^{3} }

Hence,                            C_{\pi }=\frac{1}{2\pi \times 2\times 10^{3} \times 1\times 10^{6} }=80   pF