Chapter 10.2, Problem 10.9E

To determine

The value of load resistance $R_L$ and the value of 3-dB high frequency $f_H$ of common emitter amplifier.

  $\begin{array}{l}{R}_L=\text{ 2}\text{.44 kΩ}\\ f_H=923.02\text{ MHz}\end{array}$

Given Information:

  $\begin{array}{l}{I}_E=1\text{ mA}\\ R_{sig}=5\mathrm{k}\Omega \\ R_L=5\text{ k}\Omega \\ \beta =100\\ V_A=100\text{ V}\\ C_{\mu }=1\text{ pF}\\ f_t=800\text{ MHz}\end{array}$

Calculation:

The equation for output resistance is given by,

  $r_o=\frac{V_A}{I_C}$

Plugging the values

  $\begin{array}{c}{r}_o=\frac{100}{1\text{ mA}}\\ =100\text{ kΩ}\end{array}$

The value of the mid-band gain is reduced to half and $R{\text{'}}_L$ is directly proportional to the mid-band gain. Therefore, the new value of $R{\text{'}}_L$ will be reduced to half.

Where,

  $\begin{array}{c}{R}_L{}^{\text{'}}=R_L\parallel r_o\\ =5\text{ kΩ}\parallel 100\text{ kΩ}\\ =\frac{5\times 100}{5+100}\\ =4.76\text{ kΩ}\end{array}$

And, mid-band gain $A_M=\frac{A_{M-old}}{2}$

Where, $A_M\alpha R_L{}^{\text{'}}$

Hence, $R_L{}^{\text{'}}=\frac{R^{\text{'}}{}_{L-old}}{2}$

Plugging the values

  $\begin{array}{c}{R}_L{}^{\text{'}}=\frac{4.76\text{ kΩ}}{2}\\ =2.38\text{ kΩ}\end{array}$

  $R_L{}^{\text{'}}=R_L\left|\right|r_o$

Plugging the values

  $\begin{array}{l}2.38\text{ kΩ}=R_L\left|\right|100\text{ kΩ}\\ 2.38\text{ kΩ}=\frac{100\times R_L}{100+R_L}\text{ kΩ}\\ R_L\text{= 2}\text{.44 kΩ}\end{array}$

The equation for resistance $r_{\pi }$ is given by,

  $r_{\pi }=\frac{\beta }{g_m}$

Plugging the values

  $\begin{array}{c}{r}_{\pi }=\frac{100}{0.04}\\ =2500\Omega \end{array}$

The equation for transconductance is given by,

  $g_m=\frac{I_C}{V_T}$

Plugging the values

  $\begin{array}{c}{g}_m=\frac{\text{1 mA}}{\text{25 mV}}\\ =0.04\text{ A/V}\end{array}$

The equation for 3 dB frequency is given by,

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

Where,

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

  $\begin{array}{c}{R}_{sig}{}^{\text{'}}=5\parallel 2.5\\ =1.67\text{k}\Omega \end{array}$

The value of $g_{m}R{\text{'}}_{L-old}$ is ,

  $\begin{array}{c}{g}_{m}R{\text{'}}_{L-old}=0.04\text{ A/V}\times 4.76\text{ kΩ}\\ =190.4\text{ V}/\text{V}\end{array}$

Hence, new values become half of the calculated old values,

  $g_{m}R{\text{'}}_L=\frac{g_{m}R{\text{'}}_{L-old}}{2}$

Plugging the values

  $\begin{array}{c}{g}_{m}R{\text{'}}_L=\frac{190.4\text{ V}/\text{V}}{2}\\ =95.25\text{ V}/\text{V}\end{array}$

Consider the expression for a capacitor $C_{\pi }$ as follow-

  $C_{\pi }+C_{\mu }=\frac{g_m}{2\pi f_t}$

Plugging the values

  $\begin{array}{c}{C}_{\pi }+1\text{ pF}=\frac{0.04}{2\pi \times 800\text{ MHz}}\\ C_{\pi }+1\times {10}^{-12}=\frac{0.04}{2\pi \times 800\times {10}^6}\\ C_{\pi }+1\times {10}^{-12}=7.95\times {10}^{-12}\\ C_{\pi }\approx 7\text{ pF}\end{array}$

The equation for input capacitance is given by,

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

Plugging the values

  $\begin{array}{c}{C}_{in}=7\text{ pF}+1\text{ pF}\left(1+95.25\text{ V}/\text{V}\right)\\ =103.25\text{ pF}\end{array}$

Now consider the expression for $f_H$ as follow-

  $\begin{array}{c}{f}_H=\frac{1}{2\pi \times 103.25\text{ pF}\times 1.67\text{k}\Omega }\\ =923.02\text{ MHz}\end{array}$