(a) Sketch the high-frequency equivalent circuit of a common-emitter amplifier that is fed with a voltage signal source $V_{\text {sig }}$ having a resistance $R_{\text {sig }}$ and that is loaded with a resistance $R_{L}$. In addition to the internal capacitances of the BJT, $C_{\pi}$ and $C_{\mu}$, there is a capacitance $C_{L}$ between the output node and ground.
(b) If the transistor is biased at $I_{C}=0.1 \mathrm{~mA}$ and has $\beta=100, V_{A}=40 \mathrm{~V}, f_{T}=500 \mathrm{MHz}$, and $C_{\mu}=$ $0.27 \mathrm{pF}$, find the values of $g_{m}, r_{\pi}, r_{o}$, and $C_{\pi}$.
(c) Use the method of open-circuit time-constants to obtain an estimate for the upper 3-dB frequency, $f_{H}$, for the case $R_{\text {sig }}=25 \mathrm{k} \Omega, R_{L}=25 \mathrm{k} \Omega$, and $C_{L}=0.5 \mathrm{pF}$. Also, find the midband gain.

Question

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The Answer is (a) The high-frequency equivalent circuit is shown in Fig. 10.2.1 below.
$(a) The high-frequency equivalent circuit is shown in Fig. 10.2.1 below. Figure 10.2.1 (b) \begin{aligned} I_{C} & =0.1 \mathrm{~mA}, \quad \beta=100, \quad V_{A}=40 \mathrm{~V} \ f_{T} & =500 \mathrm{MHz}, \quad C_{\mu}=0.27 \mathrm{pF} \ g_{m} & =\frac{I_{C}}{V_{T}}=\frac{0.1}{0.025}=4 \mathrm{~mA} / \mathrm{V} \ r_{\pi} & =\frac{\beta}{g_{m}}=\frac{100}{4}=25 \mathrm{k} \Omega \ r_{o} & =\frac{V_{A}}{I_{C}}=\frac{40}{0.1}=400 \mathrm{k} \Omega \ C_{\pi}+C_{\mu} & =\frac{g_{m}}{\omega_{T}}=\frac{4 imes 10^{-3}}{2 \pi imes 500 imes 10^{6}}=1.27 \mathrm{pF} \ \Rightarrow C_{\pi} & =1 \mathrm{pF} \end{aligned} (c) Capacitor C_{\pi} sees a resistance R_{\pi} \begin{aligned} R_{\pi} & =r_{\pi} \| R_{ ext {sig }} \ & =25 \| 25 \ & =12.5 \mathrm{k} \Omega \end{aligned} Capacitor C_{\mu} sees a resistance R_{\mu} , R_{\mu}=R_{L}^{\prime}+R_{ ext {sig }}^{\prime}\left(1+g_{m} R_{L}^{\prime} ight) where \begin{aligned} R_{L}^{\prime} & =R_{L}\left\|r_{o}=25 ight\| 400=23.5 \mathrm{k} \Omega \ R_{ ext {sig }}^{\prime} & =R_{ ext {sig }} \| r_{\pi} \ & =25 \| 25=12.5 \mathrm{k} \Omega \end{aligned} Thus, \begin{aligned} R_{\mu} & =23.5+12.5(1+4 imes 23.5) \ & =1211 \mathrm{k} \Omega \end{aligned} Capacitor C_{L} sees a resistance R_{L}^{\prime} , R_{L}^{\prime}=R_{L}\left\|r_{O}=25 ight\| 400=23.5 \mathrm{k} \Omega Now, \begin{aligned} au_{H}= & C_{\pi} R_{\pi}+C_{\mu} R_{\mu}+C_{L} R_{L}^{\prime} \ = & 1 imes 10^{-12} imes 12.5 imes 10^{3}+0.27 imes 10^{-12} \ & imes 1211 imes 10^{3}+0.5 imes 10^{-12} imes 23.5 imes 10^{3} \ = & 12.5 imes 10^{-9}+327 imes 10^{-9}+11.75 imes 10^{-9} \ = & 351.2 \mathrm{~ns} \ f_{H}= & \frac{1}{2 \pi au_{H}}=\frac{1}{2 \pi imes 351.2 imes 10^{-9}} \ = & 453 \mathrm{kHz} \ A_{M}= & \frac{V_{o}}{V_{ ext {sig }}}=\frac{r_{\pi}}{r_{\pi}+R_{ ext {sig }}} imes-g_{m} R_{L}^{\prime} \ = & \frac{25}{25+25} imes-4 imes 23.5 \ = & -47 \mathrm{~V} / \mathrm{V} \end{aligned}$

Figure 10.2.1
(b)

\begin{aligned}I_{C} & =0.1 \mathrm{~mA}, \quad \beta=100, \quad V_{A}=40 \mathrm{~V} \f_{T} & =500 \mathrm{MHz}, \quad C_{\mu}=0.27 \mathrm{pF} \g_{m} & =\frac{I_{C}}{V_{T}}=\frac{0.1}{0.025}=4 \mathrm{~mA} / \mathrm{V} \r_{\pi} & =\frac{\beta}{g_{m}}=\frac{100}{4}=25 \mathrm{k} \Omega \r_{o} & =\frac{V_{A}}{I_{C}}=\frac{40}{0.1}=400 \mathrm{k} \Omega \C_{\pi}+C_{\mu} & =\frac{g_{m}}{\omega_{T}}=\frac{4 imes 10^{-3}}{2 \pi imes 500 imes 10^{6}}=1.27 \mathrm{pF} \\Rightarrow C_{\pi} & =1 \mathrm{pF}\end{aligned}

(c) Capacitor

C_{\pi}

sees a resistance

R_{\pi}

\begin{aligned}R_{\pi} & =r_{\pi} \| R_{ ext {sig }} \& =25 \| 25 \& =12.5 \mathrm{k} \Omega\end{aligned}

Capacitor

C_{\mu}

sees a resistance

R_{\mu}

,

R_{\mu}=R_{L}^{\prime}+R_{ ext {sig }}^{\prime}\left(1+g_{m} R_{L}^{\prime} ight)

where

\begin{aligned}R_{L}^{\prime} & =R_{L}\left\|r_{o}=25 ight\| 400=23.5 \mathrm{k} \Omega \R_{ ext {sig }}^{\prime} & =R_{ ext {sig }} \| r_{\pi} \& =25 \| 25=12.5 \mathrm{k} \Omega\end{aligned}

Thus,

\begin{aligned}R_{\mu} & =23.5+12.5(1+4 imes 23.5) \& =1211 \mathrm{k} \Omega\end{aligned}

Capacitor

C_{L}

sees a resistance

R_{L}^{\prime}

,

R_{L}^{\prime}=R_{L}\left\|r_{O}=25 ight\| 400=23.5 \mathrm{k} \Omega

Now,

\begin{aligned} au_{H}= & C_{\pi} R_{\pi}+C_{\mu} R_{\mu}+C_{L} R_{L}^{\prime} \= & 1 imes 10^{-12} imes 12.5 imes 10^{3}+0.27 imes 10^{-12} \& imes 1211 imes 10^{3}+0.5 imes 10^{-12} imes 23.5 imes 10^{3} \= & 12.5 imes 10^{-9}+327 imes 10^{-9}+11.75 imes 10^{-9} \= & 351.2 \mathrm{~ns} \f_{H}= & \frac{1}{2 \pi au_{H}}=\frac{1}{2 \pi imes 351.2 imes 10^{-9}} \= & 453 \mathrm{kHz} \A_{M}= & \frac{V_{o}}{V_{ ext {sig }}}=\frac{r_{\pi}}{r_{\pi}+R_{ ext {sig }}} imes-g_{m} R_{L}^{\prime} \= & \frac{25}{25+25} imes-4 imes 23.5 \= & -47 \mathrm{~V} / \mathrm{V}\end{aligned}

(A) Sketch the High-Frequency Equivalent Circuit of a Common-Emitter Amplifier Vsig V_{\text {sig }}Vsig ​

(A) Sketch the High-Frequency Equivalent Circuit of a Common-Emitter Amplifier $V_{\text {sig }}$