Chapter 10, Problem 10.30P

To determine

The $r_{\pi },A_M\text{and }{f}_H$

  $\begin{array}{l}{r}_{\pi }=5\text{ k}\Omega \\ A_M=-29.02\text{ V/V}\\ f_H=1007.3\text{ KHz}\end{array}$

Given Information:

  $\begin{array}{l}{R}_{sig}=10\text{ k}\Omega \\ R_{B1}=68\text{ k}\Omega \\ R_{B2}=27\text{ k}\Omega \\ R_E=2.2\text{ k}\Omega \\ R_C=4.7\text{ k}\Omega \\ I_C=0.8\text{ mA}\\ C_{\mu }=0.5\text{ pF}\\ f_T=1\text{ GHz}\\ R_L=10\text{ k}\Omega \\ \beta =160\end{array}$

  

Fig: Given circuit

Calculation:

Write the equation for trans-conductance $g_m$ in terms of collector current $I_C$ and temperature equivalent voltage $V_T$ is given by

  $g_m=\frac{I_C}{V_T}$

Where

  $V_T=25\text{ mV}$

Plugging the values

  $\begin{array}{l}{g}_m=\frac{\text{0}\text{.8 mA}}{25\text{ mV}}\\ g_m=0.032\text{ A/V}\end{array}$

The expression for frequency $f_T$ in terms of capacitance and trans-conductance is given by,

  $f_T=\frac{g_m}{2\pi \left(C_{\pi }+C_{\mu }\right)}$

Plugging the values

  $\begin{array}{l}1\text{ GHz}=\frac{0.032\text{ A/V}}{2\pi \left(C_{\pi }+0.5\text{ pF}\right)}\\ \left(C_{\pi }+0.5\text{ pF}\right)=\frac{0.032\text{ A/V}}{2\pi \left(1\text{ GHz}\right)}\\ C_{\pi }=\frac{0.032\text{ A/V}}{2\pi \times 1\times {10}^9}-0.5\text{ pF}\end{array}$

  $C_{\pi }=4.59\text{ pF}$

Write an equation for the effective load resistance of the amplifier

  $R_L^{\text{'}}=R_C\parallel R_L$

Plugging given values.

  $\begin{array}{l}{R}_L^{\text{'}}=4.7\text{ k}\Omega \parallel 10\text{ k}\Omega \\ R_L^{\text{'}}=\frac{\text{4}\text{.7 k}\Omega \times 10\text{ k}\Omega }{\text{4}\text{.7 k}\Omega +10\text{ k}\Omega }\\ R_L^{\text{'}}=3.19\text{ k}\Omega \end{array}$

The expression for total input capacitance is given by,

  $C_{in}=C_{\pi }+C_{\mu }\left(1+g_m{R}_L^{\text{'}}\right)$

Plugging given values.

  $\begin{array}{l}{C}_{in}=4.59\text{ pF}+0.5\text{ pF }\left(1+0.032\text{ A/V}\times 3.19\text{ k}\Omega \right)\\ C_{in}=4.59\text{ pF}+0.5\text{ pF }\left(103.08\right)\\ C_{in}=56.13\text{ pF}\end{array}$

Write an equation for the equivalent base resistance of the amplifier

  $R_B=R_{B1}\parallel R_{B2}$

Plugging given values.

  $\begin{array}{l}{R}_B=68\text{ k}\Omega \parallel 27\text{ k}\Omega \\ R_B=\frac{\text{68 k}\Omega \times 27\text{ k}\Omega }{\text{68 k}\Omega +27\text{ k}\Omega }\\ R_B=19.32\text{ k}\Omega \end{array}$

Write the equation for resistance $r_{\pi }$ in terms of trans-conductance $g_m$ as follows:

  $r_{\pi }=\frac{\beta }{g_m}$

Plugging the values

  $\begin{array}{l}{r}_{\pi }=\frac{160}{\text{0}\text{.032 mA/V}}\\ r_{\pi }=5\text{ k}\Omega \end{array}$

Write an equation for the effective signal resistance of the amplifier

  $R_{sig}^{\text{'}}=r_{\pi }\parallel R_B\parallel R_{sig}$

Plugging given values.

  $\begin{array}{l}{R}_{sig}^{\text{'}}=5\text{ k}\Omega \parallel 19.32\text{ k}\Omega \parallel 10\text{ k}\Omega \\ R_{sig}^{\text{'}}=5\text{ k}\Omega \parallel \frac{\text{19}\text{.32 k}\Omega \times 10\text{ k}\Omega }{\text{19}\text{.32 k}\Omega +10\text{ k}\Omega }\\ R_{sig}^{\text{'}}=5\text{ k}\Omega \parallel 6.58\text{ k}\Omega \\ R_{sig}^{\text{'}}=\frac{\text{5 k}\Omega \times 6.58\text{ k}\Omega }{\text{5 k}\Omega +6.58\text{ k}\Omega }\\ R_{sig}^{\text{'}}=2.84\text{}\text{k}\Omega \end{array}$

Write the equation for 3-dB frequency as follows:

  $f_H=\frac{1}{2\pi C_{in}{R}_{sig}^{\text{'}}}$

Plugging given values.

  $\begin{array}{l}{f}_H=\frac{1}{2\pi \times 55.63\text{ pF}\times 2.84\text{k}\Omega }\\ f_H=\frac{1}{2\pi \left(55.63\times {10}^{-12}\text{ F}\right)\left(2.84\times {10}^3\Omega \right)}\\ f_H=1007.3\text{ KHz}\end{array}$

Write the equation for the corresponding gain $A_M$ of the amplifier given as follows:

  $A_M=-\left(\frac{R_B}{R_B+R_{sig}}\right)\left(\frac{r_{\pi }}{r_{\pi }+\left(R_B^{}\parallel R_{sig}^{}\right)}\right)g_m{R}_L^{\text{'}}$

Plugging given values.

  $\begin{array}{l}{A}_M=-\left(\frac{\text{19}\text{.32 k}\Omega }{\text{19}\text{.32 k}\Omega +10\text{ k}\Omega }\right)\left(\frac{5\text{ k}\Omega }{5\text{ k}\Omega +\left(19.32\text{k}\Omega \parallel 10\text{k}\Omega \right)}\right)\left(\text{0}\text{.032 A/V}\right)\left(\text{3}\text{.19 k}\Omega \right)\\ A_M=-\left(\frac{\text{19}\text{.32 k}\Omega }{\text{29}\text{.32 k}\Omega }\right)\left(\frac{5\text{ k}\Omega }{5\text{ k}\Omega +6.589\text{ k}\Omega }\right)\left(\text{0}\text{.032 A/V}\right)\left(\text{3}\text{.19 k}\Omega \right)\\ A_M=-29.02\text{ V/V}\end{array}$