Chapter 6, Problem 6.51P

In the circuit shown in Figure P6.51, determine the range in small−signal voltage gain $A_{\upsilon }={\upsilon }_o/{\upsilon }_s$ and current gain $A_i=i_o/i_s$ if $\beta$ is in the range $75\le \beta \le 150$ .

Figure P6.51

To determine

The range in small-signal voltage gain $A_V=\frac{v_o}{v_s}$ and current gain $A_i=\frac{i_o}{i_s}$ .

Answer to Problem 6.51P

The final range of the voltage gain $A_V$ is

  $\therefore 0.789\le A_V\le 0.811$

The final range of the current gain $A_i$ is

  $\therefore 17\le A_i\le 19.6$

Explanation of Solution

Given:

Range of $\beta$ is $75\le \beta \le 150$

The given circuit is:

  

From above circuit, considering BJT s single node, then by KCL, Quiescent emitter current $I_{EQ}$ , the quiescent base current $I_{BQ}$ and the quiescent collector current $I_{CQ}$ are related as

  $I_{EQ}=I_{BQ}+I_{CQ}\mathrm{..........}\left(1\right)$

In CE mode, $I_{CQ}$ and $I_{BQ}$ are related as

  $I_{CQ}=\beta I_{BQ}\mathrm{..........}\left(2\right)$

Now, DC analysis of the given circuit:

Reduce source V_s to zero and open the capacitor as shown below:

  

The Theveninresistance is:

  $\begin{array}{l}{R}_{TH}=R_1\left|\right|R_2\\ =\left(\frac{60\times 40}{60+40}\right)\text{k}\Omega \\ \text{=}{R}_{TH}=24\text{k}\Omega \mathrm{............}\left(3\right)\end{array}$

Modify the circuit as:

  

Therefore, Thevenin voltage from above circuit is

  $\begin{array}{l}{V}_{TH}=\frac{V_{CC}}{R_1+R_2}\times R_2\\ =\left(\frac{10}{40+60}\right)\text{V}\\ \therefore V_{TH}=6\mathrm{..........}\left(4\right)\end{array}$

Modify the circuit as:

  

Applying KCL in the base-emitter loop to determine $I_{BQ}$

  $\begin{array}{l}-I_{BQ}{R}_{TH}-V_{BE}\left(\text{on}\right)-I_{EQ}{R}_E+V_{TH}=0\\ \Rightarrow -I_{BQ}{R}_{TH}-V_{BE}\left(\text{on}\right)-\left(1+\beta \right)I_{BQ}{R}_E+V_{TH}=0\\ \therefore I_{BQ}=\frac{V_{TH}-V_{BE}\left(\text{on}\right)}{R_{TH}+\left(1+\beta \right)R_E}\mathrm{..........}(5)\end{array}$

From (3), (4) and (5) and $\beta =75$

  $\begin{array}{l}{I}_{BQ}=\left[\frac{6-0.7}{\left\{24+\left(1+75\right)\times 5\right\}\times {10}^3}\right]\text{A}\\ \Rightarrow I_{BQ}=\left[\frac{5.3}{404\times {10}^3}\right]\text{A}\\ \therefore I_{BQ}=13.12\mu \text{A}\mathrm{.........}(6)\end{array}$

From equation (2) and $\beta =75$

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\\ \Rightarrow I_{CQ}=75\times 13.12\mu \text{A}\\ \therefore I_{CQ}=0.984\text{mA}\mathrm{..........}\left(7\right)\end{array}$

Now, small signal analysis of the given circuit:

Reduce dc voltage sources to zero, dc current source to open and capacitors to short.

  

Diffusion resistance $r_R$ and $\beta =75$

  $\begin{array}{l}{r}_R=\frac{\beta V_T}{I_{CQ}}\\ \Rightarrow r_R=\frac{75\times 0.026\text{V}}{0.984\times {10}^{-3}\text{A}}\\ \Rightarrow r_R=1.98\times {10}^3\Omega \\ \therefore r_R=1.98\text{k}\Omega \mathrm{..........}\left(8\right)\end{array}$

Input resistance and $\beta =75$

  $\begin{array}{l}{R}_{ib}=r_R+\left(1+\beta \right)\left(R_E\left|\right|R_L\right)\\ \Rightarrow R_{ib}=\left\{1.98\_\left(1+75\right)\left(5\left|\right|1\right)\right\}\text{k}\Omega \\ \therefore R_{ib}=65.31\text{k}\Omega \mathrm{.........}\left(9\right)\end{array}$

The input resistance $R_i$ and $\beta =75$

  $\begin{array}{l}\Rightarrow R_i=\left(40\left|\right|60\left|\right|65.31\right)\text{k}\Omega \\ \therefore R_i=17.55\text{k}\Omega \mathrm{..........}\left(10\right)\\ A_V=\frac{\left(1+\beta \right)\left(R_E\left|\right|R_L\right)}{R_{ib}}\times \left(\frac{R_i}{R_i+R_S}\right)\\ \Rightarrow A_V=\frac{\left(1+75\right)\left(5\left|\right|1\right)}{65.31}\times \left(\frac{17.55}{17.55+4}\right)\\ \therefore A_V=0.789\text{with}\beta =75\end{array}$

Now, small signal current gain $A_i$ with $\beta =75$

  $\begin{array}{l}{A}_i=\frac{\left(1+\beta \right)\left(R_1\left|\right|R_2\right)}{\left\{\left(R_1\left|\right|R_2\right)+R_{ib}\right\}}\times \left(\frac{R_E}{R_E+R_L}\right)\\ \Rightarrow A_i=\frac{\left(1+75\right)\left(40\left|\right|60\right)}{\left\{\left(40\left|\right|60\right)+65.31\right\}}\times \left(\frac{5}{5+1}\right)\\ \therefore A_i=17\end{array}$

From (3), (4) and (5) and $\beta =150$

  $\begin{array}{l}{I}_{BQ}=\left[\frac{6-0.7}{\left\{24+\left(1+150\right)\times 5\right\}\times {10}^3}\right]\text{A}\\ \Rightarrow I_{BQ}=\left[\frac{5.3}{779\times {10}^3}\right]\text{A}\mathrm{..........}\left(11\right)\\ \end{array}$

From equation (2) and $\beta =150$

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\\ \Rightarrow I_{CQ}=150\times \frac{5.3}{779\times {10}^3}\text{A}\\ \therefore I_{CQ}=1\text{mA}\mathrm{..........}\left(12\right)\end{array}$

Diffusion resistance $r_R$ and $\beta =150$

  $\begin{array}{l}{r}_R=\frac{\beta V_T}{I_{CQ}}\\ \Rightarrow r_R=\frac{150\times 0.026\text{V}}{1\times {10}^{-3}\text{A}}\\ \Rightarrow r_R=3.9\times {10}^3\Omega \\ \therefore r_R=3.9\text{k}\Omega \mathrm{..........}\left(13\right)\end{array}$

Input resistance $R_{ib}$ and $\beta =150$

  $\begin{array}{l}{R}_{ib}=r_R+\left(1+\beta \right)\left(R_E\left|\right|R_L\right)\\ \Rightarrow R_{ib}=\left\{3.9+\left(1+150\right)\left(5\left|\right|1\right)\right\}\text{k}\Omega \\ \therefore R_{ib}=129.73\text{k}\Omega \mathrm{.........}\left(14\right)\end{array}$

The input resistance $R_i$ and $\beta =150$

  $\begin{array}{l}\Rightarrow R_i=\left(40\left|\right|60\left|\right|129.73\right)\text{k}\Omega \\ \therefore R_i=20.25\text{k}\Omega \mathrm{..........}\left(15\right)\\ A_V=\frac{\left(1+\beta \right)\left(R_E\left|\right|R_L\right)}{R_{ib}}\times \left(\frac{R_i}{R_i+R_S}\right)\\ \Rightarrow A_V=\frac{\left(1+150\right)\left(5\left|\right|1\right)}{129.73}\times \left(\frac{17.55}{20.25+4}\right)\\ \therefore A_V=0.811\text{with}\beta =150\end{array}$

Now, small signal current gain $A_i$ with $\beta =150$

  $\begin{array}{l}{A}_i=\frac{\left(1+\beta \right)\left(R_1\left|\right|R_2\right)}{\left\{\left(R_1\left|\right|R_2\right)+R_{ib}\right\}}\times \left(\frac{R_E}{R_E+R_L}\right)\\ \Rightarrow A_i=\frac{\left(1+150\right)\left(40\left|\right|60\right)}{\left\{\left(40\left|\right|60\right)+129.73\right\}}\times \left(\frac{5}{5+1}\right)\\ \therefore A_i=19.6\end{array}$

The final range of the voltage gain $A_V$ is

  $\therefore 0.789\le A_V\le 0.811$

The final range of the current gain $A_i$ is

  $\therefore 17\le A_i\le 19.6$