Chapter 7, Problem 7.32P

Consider the circuit shown in Figure P7.32. The transistor parameters are $\beta =120$ , $V_{BE}\text{(on)=0}\text{.7V}$ , and $V_A=\infty$ . (a) Find $R_C$ such that $V_{CEQ}=2.2\text{V}$ . (b) Determine the midband gain. (c) Derive the expression for the corner frequencies associated with $C_C$ and $C_E$ . (d) Determine $C_C$ and $C_E$ such that the corner frequency associated with $C_E$ is $f_C=50\text{Hz}$ and the corner frequency associated with $C_C$ is $f_C=50\text{Hz}$ .

Figure P7.32

To determine

The value of $R_c$

Answer to Problem 7.32P

The value of collector resistor, $R_c$ is $7.65\text{kΩ}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

From figure, the value of emitter current is,

  $I_{EQ}=0.2\text{mA}$

The value of base current is,

  $I_{BQ}=\frac{I_{EQ}}{1+\beta }$

Substitute $0.2\text{mA}$ for $I_{EQ},$ and 120 for $\beta$ in the equation

  $\begin{array}{cc}{I}_{BQ}& =\frac{0.2\times {10}^{-3}}{1+120}\\ & =\frac{0.2\times {10}^{-3}}{121}\\ & =1.65\mu \text{A}\end{array}$

The value of collector current is,

  $I_{cQ}=\left(\frac{\beta }{1+\beta }\right)I_{EQ}$

Substitute $0.2\text{mA}$ for $I_{EQ}$ and 120 for $\beta$ in the equation.

  $\begin{array}{l}{I}_{CQ}=\left(\frac{120}{1+120}\right)\left(0.2\times {10}^{-3}\right)\\ =\left(\frac{120}{121}\right)\left(0.2\times {10}^{-3}\right)\\ =\left(\frac{120}{121}\right)\left(0.2\times {10}^{-3}\right)=0.1983\text{mA}\end{array}$

The value of emitter voltage is,

  $I_{BQ}{R}_i+V_{BE}(\text{on})+V_E=0$

  $V_E=-\left[I_{BQ}{R}_i+V_{BE}(\text{on})\right]$

Substitute $1.65\mu \text{A}$ for $I_{BQ},0.7\text{V}$ for $V_{BE}(\text{on})$ and $10\text{kΩ}$ for $R_i$ in the equation.

  $V_E=-\left[\left(1.65\times {10}^{-6}\right)\left(10\times {10}^3\right)+0.7\right]$

  $=-[0.0165+0.7]$

  $=-0.7165\text{V}$

The value of collector voltage is,

  $V_{CEQ}=V_C-V_E$

  $V_c=V_{CEQ}+V_E$

Substitute $2.2\text{V}$ for $V_{\text{CEQ}}$ and $-0.7165\text{V}$ for $V_E$ in the equation.

  $V_{CEQ}=V_C-V_E$

  $V_C=2.2-0.7165$

  $V_C=1.4835\text{V}$

The value of collector resistor is,

  $R_C=\frac{V^{+}-V_C}{I_{CQ}}$

Substitute 3 V for $V^{+},1.4835\text{V}$ for $V_c$ and $0.1983\text{mA}$ for $I_{ce}$ in the equation

  $\begin{array}{cc}{R}_c& =\frac{3-1.4835}{0.1983\times {10}^{-3}}\\ & =\frac{1.5165}{0.1983\times {10}^{-3}}\\ & =7.65\text{kΩ}\end{array}$

To determine

The mid band gain

Answer to Problem 7.32P

The value of mid-band gain $A$ , is $-25.8$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Draw the small signal equivalent circuit of figure to derive expression for mid band gain.

  

Apply voltage division rule to calculate the voltage across the base input resistor associated withthe input portion of the circuit.

  $V_{\pi }=\left(\frac{r_{\pi }}{r_{\pi }+R_i}\right)V_i$

Apply Kirchhoff's current law at output node in Figure 1

  $\frac{V_0}{R_L}+\frac{V_o}{R_C}+g_m{V}_{\pi }=0$

  $V_o\left(\frac{1}{R_L}+\frac{1}{R_C}\right)=-g_m{V}_{\pi }$

  $V_o\left(\frac{1}{R_C\parallel R_L}\right)=-g_m{V}_{\pi }$

  $V_o=-g_m\left(R_C\parallel R_L\right)V_{\pi }$

Substitute $\left(\frac{r_{\pi }}{r_{\pi }+R_i}\right)V_i$ for $V_{\pi }$ in the equation.

  $V_o=-g_m\left(R_C\parallel R_L\right)\left(\frac{r_{\pi }}{r_n+R_i}\right)V_i$

  $\frac{V_o}{V_i}=-g_m\left(R_C\parallel R_L\right)\left(\frac{r_{\pi }}{r_{\pi }+R_i}\right)$

  $A_v=-g_m\left(R_C\parallel R_L\right)\left(\frac{r_{\pi }}{r_{\pi }+R_i}\right)$

The value of base input resistance.

  $r_{\pi }=\frac{\beta V_T}{I_{cQ}}$

Substitute $0.1983\text{mA}$ for $I_{cQ},26\text{mA}$ for $V_T$ and 120 for $\beta$ in the equation

  $\begin{array}{cc}\begin{array}{l}{r}_{\pi }=\frac{(120)\left(26\times {10}^{-3}\right)}{0.1983\times {10}^{-3}}\\ =15.73\text{kΩ}\end{array}& \end{array}$

The value of transconductance is,

  $g_m=\frac{I_{CO}}{V_T}$

Substitute $0.1983\text{mA}$ for $I_{cQ}$ and $26\text{mA}$ for $V_r$ in the equation.

  $\begin{array}{cc}{g}_m& =\frac{0.1983\times {10}^{-3}}{26\times {10}^{-3}}\\ & =7.627\text{mA}/\text{V}\end{array}$

The value of mid-band gain is,

  $A_v=-g_m\left(R_c\parallel R_L\right)\left(\frac{r_s}{r_x+R_i}\right)$

Substitute $7.627\text{mA}/\text{V}$ for $g_m,7.65\text{kΩ}$ for $R_C,20\text{kΩ}$ for $R_L,15.73\text{kΩ}$ for $r_{\pi }$ and $10\text{kΩ}$

for $R_i$ in the equation.

  $A_v=-\left(7.627\times {10}^{-3}\right)\left[\left(7.65\times {10}^3\right)\parallel \left(20\times {10}^3\right)\right]\left(\frac{15.73\times {10}^3}{\left(15.73\times {10}^3\right)+\left(10\times {10}^3\right)}\right)$

  $=-\left(7.627\times {10}^{-3}\right)\left[\frac{\left(7.65\times {10}^3\right)\times \left(20\times {10}^3\right)}{\left(7.65\times {10}^3\right)+\left(20\times {10}^3\right)}\right]\left(\frac{15.73\times {10}^3}{\left(15.73\times {10}^3\right)+\left(10\times {10}^3\right)}\right)$

  $\begin{array}{l}=-\left(7.627\times {10}^{-3}\right)\left(3.383\times {10}^3\right)\hfill \\ =-25.8\hfill \end{array}$

To determine

To derive: The expression for the corner frequencies associated with $C_C$ and $C_E$ .

Answer to Problem 7.32P

The expression for the corner frequency associated with $C_C$ is

  $f_C=\frac{1}{2\pi \left(R_c+R_L\right)C_c}$

The expression for the corner frequency associated with $C_E$ is.

  $f_E=\frac{1+\beta }{2\pi \left(r_{\pi }+R_i\right)C_E}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Derive the expression for the comer frequency associated with $C_C$

  ${\tau }_C=\left(R_C+R_L\right)C_C$

  $\frac{1}{2\pi f_c}=\left(R_c+R_L\right)C_c$

  $f_C=\frac{1}{2\pi \left(R_C+R_L\right)C_C}$

Thus, the expression for the corner frequency associated with $C_C$ is

  $f_C=\frac{1}{2\pi \left(R_c+R_L\right)C_c}$

Derive the expression for the comer frequency associated with $C_E$

  ${\tau }_E=\left(\frac{r_{\pi }+R_i}{1+\beta }\right)C_E$

  $\frac{1}{2\pi f_E}=\left(\frac{r_{\pi }+R_i}{1+\beta }\right)C_R$

  $f_E=\frac{1+\beta }{2\pi \left(r_{\pi }+R_i\right)C_E}$

Thus, the expression for the corner frequency associated with $C_E$ is.

  $f_E=\frac{1+\beta }{2\pi \left(r_{\pi }+R_i\right)C_E}$

To determine

The value of $C_C$ and $C_E$

Answer to Problem 7.32P

The value of capacitance, $C_C$ when $f_c=50\text{Hz}$ is $0.115\mu \text{F}$

The value of capacitance, $C_E$ when $f_E=10\text{Hz}$ is $74.8\mu \text{F}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Determine the value of capacitance, $C_C$ when $f_c=50\text{Hz}$

  $\begin{array}{cc}{f}_c& =\frac{1}{2\pi \left(R_c+R_L\right)C_c}\\ C_C& =\frac{1}{2\pi f_c\left(R_c+R_L\right)}\end{array}$

Substitute $7.65\text{kΩ}$ for $R_c,20\text{kΩ}$ for $R_L,$ and $50\text{Hz}$ for $f_c$ in the equation

  $C_C=\frac{1}{2\pi (50)\left[\left(7.65\times {10}^3\right)+\left(20\times {10}^3\right)\right]}$

  $\begin{array}{l}=\frac{1}{2\pi (50)\left(27.65\times {10}^3\right)}\hfill \\ =1.15\times {10}^{-7}\hfill \\ =0.115\mu \text{F}\hfill \end{array}$

Thus, the value of capacitance, $C_C$ when $f_c=50\text{Hz}$ is $0.115\mu \text{F}$

Determine the value of capacitance, $C_E$ when $f_E=10\text{Hz}$

  $f_E=\frac{1+\beta }{2\pi \left(r_{\pi }+R_i\right)C_c}$

  $C_E=\frac{1+\beta }{2\pi f_E\left(r_n+R_i\right)}$

Substitute $15.73\text{kΩ}$ for $r_{\pi },10\text{kΩ}$ for $R_i,120$ for $\beta ,$ and $10\text{Hz}$ for $f_E$ in the equation.

  $C_E=\frac{1+120}{2\pi (10)\left[\left(15.73\times {10}^3\right)+\left(10\times {10}^3\right)\right]}$

  $=\frac{121}{2\pi (10)\left(25.73\times {10}^3\right)}$

  $=0.0748\times {10}^{-3}$

  $=74.8\mu \text{F}$

Thus, the value of capacitance, $C_E$ when $f_E=10\text{Hz}$ is $74.8\mu \text{F}$