Question 11.18: Calculate the high 3 dB cut-off frequency fH of BJT Emitter ...

R_{B} = R_{1} || R_{2}= 75 × 10^{3} || 4.35 × 10^{3} = 3.66  k \Omega

We know that                f_{H}=\frac{1}{2\pi (R_{C_{b^{′}e}}C_{b^{′}e}+ R_{C_{b^{′}C}} C_{b^{′}C})}

where                              R_{C_{b^{′}e}}= r_{b^{′}e}\parallel \frac{R_{B}\parallel R_{s}+R_{L}\parallel R_{E} }{1+g_{m}(R_{L}\parallel R_{E} ) }

R_{C_{b^{′}C}}= (R_{B} || R_{s}) || [r_{b^{′}e} + (1 + \beta ) (R_{E} || R_{L})]

R_{C_{b^{′}C}} = (2.665 × 10^{3} || 200) || [1.45 × 10^{3} + (1 + 80) (330 || 55 × 10^{3})]

= 184.6 Ω

R_{C_{b^{′}e}}= 1.4\Omega \parallel \frac{(2.66 \times 10^{3}\parallel 200)+(5\times 10^{3}\parallel 330)}{1+57.14\times 10^{-3}\times (55\times 10^{3}\parallel 330)}

= 26.02 Ω

Hence, the 3 dB upper cut-off frequency is

f_{H}=\frac{1}{2\pi (R_{C_{b^{′}e}}C_{b^{′}e}+ R_{C_{b^{′}C}} C_{b^{′}C})}

=\frac{1}{2\pi (20.02\times 15\times 10^{-12} +184.6\times 1\times 10^{-12} )}

= 276.9 MHz