Question 17.PE.NAQ.1: For the cascaded amplifier shown in the following figure, wh......

The overall upper cut-off frequency can be determined by determining the upper cut-off frequency for each stage. The upper cut-off frequency for the third stage amplifier (f_{H3}) is given by

f_{\mathrm{H3}}={\frac{1}{2\pi R_{3}^{\prime}\left[C_{\mathrm{e}}+C_{\mathrm{c}}(1+g_{\mathrm{m}}R_{\mathrm{C3}})\right]}}

where

R_{3}^{\prime}=\frac{(R_{\mathrm{C2}}|R_{\mathrm{B5}}|R_{\mathrm{B6}}+r_{\mathrm{bb′3}})\times r_{\mathrm{b^{\prime}e3}}}{(R_{\mathrm{C2}}\left|R_{\mathrm{B5}}\right|\left|R_{\mathrm{B6}}+h_{\mathrm{ie3}}\right)}={\frac{(1000|\vert2000\vert|200+200)\times800}{(100\vert|2000\vert\vert200+1000)}}=245.34 \ \Omega

f_{\mathrm{H3}}={\frac{1}{2\times3.1414\times245.34\ [40\times10^{-12}+5\times10^{-12}(1+60\times10^{-3}\times1\times10^{3})]}}=1.88~\mathrm{MHz}

The upper cut-off frequency for the second stage (f_{H2}) is given by

f_{\mathrm{H2}}=\frac{1}{2\pi\,R_{2}^{\,\prime}\left[C_{\mathrm{e}}+C_{\mathrm{c}}(1+g_{\mathrm{m}}R_{\mathrm{L2}}^{\,\prime})\right]}

The effective value of load resistance for the second stage (R_{L2′} ) is given by

R_{L2^\prime}=R_{C2}||R_{B5}||R_{B6}||h_{ie3}=1000||2000||200||1000= 133.33 \,\Omega

The value of R_{2^{\prime}} is given by

R_{2^\prime}=\frac{\left(R_{C1}||R_{B3}||R_{B4}+r_{bb^{\prime}2}\right)\times r_{b{\prime}e} }{\left(R_{C1}||R_{B3}||R_{B4}+h_{ie2}\right) }=\frac{\left(1000||2000||200+200\right)\times 800 }{\left(1000||2000||200+1000\right)}=245.34\,\Omega

Therefore,

f_{H2}=\frac{1}{2\times 3.1414\times 245.34[40\times 10^{-12}+5\times 10^{-12}]\left(1+60\times 10^{-3}\times 133.33\right)}=7.632  MHz

The upper cut-off frequency for the first stage (f_{H1}) is given by

f_{\mathrm{H} 1}=\frac{1}{2 \pi R_{1^{\prime}}\left[C_{\mathrm{e}}+C_{\mathrm{c}}\left(1+g_{\mathrm{m}} R_{\mathrm{L} 1^{\prime}}\right)\right]}

The effective load resistance for the first stage (R_{L1^′}) is given by

R_{\mathrm{L} 1}^{\prime}=R_{\mathrm{C} 1}\left\|R_{\mathrm{B} 3}\right\| R_{\mathrm{B} 4}\left\|h_{\mathrm{ie} 2}=1000\right\| 2000\|200\| 1000=133.33 \Omega

The value of R_{1^′} is given by

R_{1}^{\prime}=\frac{\left(R_{\mathrm{s}}\left\|R_{\mathrm{B} 1}\right\| R_{\mathrm{B} 2}+\tau_{\mathrm{bb}^{\prime} 1}\right) \times r_{\mathrm{b}^{\prime} \mathrm{e} 1}}{\left(R_{\mathrm{s}}\left\|R_{\mathrm{B} 1}\right\| R_{\mathrm{B} 2}+h_{\mathrm{ie} 1}\right)}=\frac{(500\|2000\| 200+200) \times 800}{(500\|2000\| 200+1000)}=235.292 \Omega

f_{\mathrm{H} 1}=\frac{1}{2 \times 3.1414 \times 235.292\left[40 \times 10^{-12}+5 \times 10^{-12}\left(1+60 \times 10^{-3} \times 133.33\right)\right]}=7.967 \mathrm{MHz}

The overall upper cut-off frequency is limited by the cut-off frequency of the first stage as it is around four times less than the cut-off frequencies of the other two stages. The overall upper cut-off frequency is approximately 1.88 MHz, that is, 1880 kHz.

Ans. (1880)