Chapter 8, Problem 8.47P

To determine

The value of the output resistance $R_O$ .

Answer to Problem 8.47P

Thevalue of output resistance is $7.46\Omega$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  

The expression for the value of base current of transistor $Q_3$ is given by,

  $I_{B_3}=I_{E_2}$

The expression for the value of the collector current for transitory $Q_3$ is given by,

  $I_{C_3}\left(1+\beta \right)I_{C_2}$

The expression for the value of the base current for transistor $Q_3$ is given by,

  $I_{C_3}=\left(\frac{1+{\beta }_n}{{\beta }_n}\right)I_{C_2}$

The expression for the value of the base current for the transistor $Q_2$ is given by,

  $I_{B_2}=\frac{I_{C_2}}{{\beta }_n}$

The expression for the value of $I_{B_2}$ is given by,

  $I_{B_2}=\left(\frac{{\beta }_P}{1+{\beta }_P}\right)I_{E_1}$

The expression for the value of the collector current through the transistor $Q_2$ is given by,

  $I_{C_3}=\left(1+{\beta }_n\right)I_{C_2}$

The expression for the value of the collector current for the transistor $I_{C_2}$ is given by,

  $I_{C_2}={\beta }_n\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_1}$ ........ (1)

The expression for the value of the collector current for transistor $Q_3$ is given by,

  $I_{C_3}=\left(1+{\beta }_n\right)\left({\beta }_n\right)\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_1}$ ........ (2)

The conversion from $\text{mA}$ into $\text{A}$ is given by,

  $1\text{mA}={\text{10}}^{-3}\text{A}$

The conversion from $\text{4}\text{mA}$ into $\text{A}$ is given by,

  $\text{4}\text{mA}=4\times {\text{10}}^{-3}\text{A}$

The expression for the value of the quiescent current is given by,

  $I_Q=I_{C3}+I_{C2}+I_{E1}$

Substitute $\left(1+{\beta }_n\right)\left({\beta }_n\right)\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_1}$ for $I_{C_3}$ and ${\beta }_n\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_1}$ for $I_{C_2}$ in the above equation.

  $I_Q=\left(1+{\beta }_n\right)\left({\beta }_n\right)\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_1}+{\beta }_n\left(\frac{{\beta }_n}{1+{\beta }_p}\right)I_{E_1}+I_{E1}$

Substitute $50$ for ${\beta }_n$ , $4\times {\text{10}}^{-3}\text{A}$ for $I_Q$ and $50$ for ${\beta }_p$ in the above equation.

  $\begin{array}{l}4\times {\text{10}}^{-3}\text{A}=\left(1+50\right)\left(50\right)\left(\frac{10}{10+1}\right)I_{E_1}+50\left(\frac{50}{1+10}\right)I_{E_1}+I_{E1}\\ I_{E_1}=1.692\times {10}^{-6}\text{A}\end{array}$

The expression for the value of the collector current $I_{C_1}$ is given by,

  $I_{C_1}=\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{E_1}$

Substitute $10$ for ${\beta }_p$ and $1.692\times {10}^{-6}\text{A}$ for $I_{E_1}$ in the above equation.

  $\begin{array}{c}{I}_{C_1}=\left(\frac{10}{1+10}\right)\left(1.692\times {10}^{-6}\text{A}\right)\\ =1.538\times {10}^{-6}\text{A}\end{array}$

Substitute $50$ for, ${\beta }_n$

  $10$ for ${\beta }_p$ and $1.692\times {10}^{-6}\text{A}$ for $I_{E_1}$ in equation (1).

  $\begin{array}{c}{I}_{C_2}=50\left(\frac{10}{1+10}\right)1.692\times {10}^{-6}\text{A}\\ =\text{76}\text{.9}\times {10}^{-6}\text{A}\end{array}$

Substitute $50$ for, ${\beta }_n$

  $10$ for ${\beta }_p$ and $1.692\times {10}^{-6}\text{A}$ for $I_{E_1}$ in equation (2).

  $\begin{array}{c}{I}_{C_3}=\left(1+50\right)\left(50\right)\left(\frac{10}{1+10}\right)1.692\times {10}^{-6}\text{A}\\ =\text{3}\text{.92}\times {10}^{-3}\text{A}\end{array}$

The expression for the value of $V_{{\pi }_1}$ is given by,

  $V_{{\pi }_1}=\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x$

The expression for the value of $V_{{\pi }_2}$ is given by,

  $V_{{\pi }_2}=g_{m_1}{V}_{{\pi }_1}{r}_{{\pi }_2}$ ........ (3)

The expression for the value of $V_{{\pi }_3}$ is given by,

  $V_{{\pi }_3}=\left[g_{m_1}{V}_{{\pi }_1}+g_{m_2}{V}_{{\pi }_1}{g}_{m_1}\right]r_{{\pi }_3}$

Substitute $\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x$ for $V_{{\pi }_1}$ and $\left[g_{m_1}{V}_{{\pi }_1}+g_{m_2}{V}_{{\pi }_1}{g}_{m_1}\right]r_{{\pi }_3}$ for $V_{{\pi }_3}$ in the above equation.

  $V_{{\pi }_3}=\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x\left[g_{m_1}\left(V_{{\pi }_1}\right)+g_{m_2}{g}_{m_1}{r}_{{\pi }_2}\right]r_{{\pi }_3}{V}_x$

Substitute $\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x$ for $V_{{\pi }_1}$ in equation (3).

  $V_{{\pi }_2}=g_{m_1}\left(\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x\right)r_{{\pi }_2}$

The expression for the value of $r_{{\pi }_1}$ is given by,

  $r_{{\pi }_1}=\frac{\left(\beta \right)V_T}{I_{CQ}}$

Substitute $10$ for $\beta$ , $0.026\text{V}$ for $V_T$ and $1.534\text{A}$ for $I_{CQ}$ in the above equation.

  $\begin{array}{c}{r}_{{\pi }_1}=\frac{\left(10\right)\left(0.026\text{V}\right)}{1.534\text{A}}\\ =169\times {10}^{-3}\Omega \end{array}$

The expression for the value of current $I_x$ is given by,

  $I_x=g_{m_3}{V}_{{\pi }_3}+g_{m_2}{V}_{{\pi }_2}+g_{m_1}{V}_{{\pi }_1}+\frac{V_x}{r_{{\pi }_1}+z}$

Substitute $\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x\left[g_{m_1}\left(V_{{\pi }_1}\right)+g_{m_2}{g}_{m_1}{r}_{{\pi }_2}\right]r_{{\pi }_3}{V}_x$ for $V_{{\pi }_3}$ , $g_{m_1}\left(\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x\right)r_{{\pi }_2}$ for $V_{{\pi }_2}$ and $\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x$ for $V_{{\pi }_1}$ in the above equation.

  $\begin{array}{c}{I}_x=\left[\begin{array}{l}{g}_{m_3}\left(\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x\left[g_{m_1}+g_{m_2}{g}_{m_1}{r}_{{\pi }_2}\right]r_{{\pi }_3}{V}_x\right)+g_{m_2}\left(g_{m_1}\left(\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x\right)r_{{\pi }_2}\right)\\ +g_{m_1}\left(\frac{r_{{\pi }_1}}{r_{{\pi }_1}+z}{V}_x\right)+\frac{V_x}{r_{{\pi }_1}+z}\end{array}\right]\\ =\frac{\left({\beta }_1+{\beta }_1{\beta }_2\right){\beta }_3}{r_{{\pi }_1}+z}{V}_x+\frac{\beta {\beta }_2}{r_{{\pi }_1}+z}+\frac{{\beta }_1}{r_{{\pi }_1}+z}{V}_x+\frac{V_x}{r_{{\pi }_1}+z}\end{array}$

The expression to determine the value of $R_O$ is given by,

  $R_O=\frac{V_x}{I_x}$

Substitute $1\text{V}$ for $V_x$ and $\frac{\left({\beta }_1+{\beta }_1{\beta }_2\right){\beta }_3}{r_{{\pi }_1}+z}{V}_x+\frac{\beta {\beta }_2}{r_{{\pi }_1}+z}+\frac{{\beta }_1}{r_{{\pi }_1}+z}{V}_x+\frac{V_x}{r_{{\pi }_1}+z}$ for $I_x$ in the above equation.

  $R_O=\frac{1}{\frac{\left({\beta }_1+{\beta }_1{\beta }_2\right){\beta }_3}{r_{{\pi }_1}+z}{V}_x+\frac{\beta {\beta }_2}{r_{{\pi }_1}+z}+\frac{{\beta }_1}{r_{{\pi }_1}+z}{V}_x+\frac{V_x}{r_{{\pi }_1}+z}}$

Substitute $1\text{V}$ for $V_x$ , $169\times {10}^{-3}\Omega$ for $r_{{\pi }_1}$ , $10$ for ${\beta }_1$ , $50$ for ${\beta }_2$ and $25\times {10}^3\Omega$ for $z$ in the above equation.

  $\begin{array}{c}{R}_O=\left[\frac{1}{\begin{array}{l}\frac{\left(10+10\left(50\right)\right){\beta }_3}{25\times {10}^3\Omega +169\times {10}^{-3}\Omega }\left(1\text{V}\right)+\frac{10\left(50\right)}{169\times {10}^{-3}\Omega +25\times {10}^3\Omega }\\ +\frac{10}{169\times {10}^{-3}\Omega +25\times {10}^3\Omega }\left(1\text{V}\right)+\frac{\left(1\text{V}\right)}{169\times {10}^{-3}\Omega +25\times {10}^3\Omega }\end{array}}\right]\\ =7.46\Omega \end{array}$

Conclusion:

Therefore, the value of output resistance is $7.46\Omega$ .