Chapter 10, Problem 10.83P

A BJT amplifier with active load is shown in Figure P10.83. The circuit contains emitter resistors $R_E$ and a load resistor $R_L$ . (a) Derive theexpression for the output resistance looking into the collector of $Q_2$ .(b) Using the small-signal equivalent circuit, derive the equation for thesmallsigna1 voltage gain. Express the relationship in a form similar toEquation (10.94).

To determine

To derive:An expression for the output resistance ${\mathrm{R}}_{\mathrm{O}}$ looking into the collector of transistor ${\mathrm{Q}}_2$

Answer to Problem 10.83P

  ${\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}({\mathrm{R}}_{\mathrm{E}}\parallel {\mathrm{r}}_{\mathrm{\pi }2})\right]$

Explanation of Solution

Given:

The given circuit is,

  

Calculation:

The small-signal equivalent circuit of BJT amplifier is as shown,

  

Apply KCL at base terminal of ${\mathrm{Q}}_1$

  ${\mathrm{I}}_{\mathrm{x}1}=\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}{\mathrm{V}}_{\mathrm{\pi }1}+\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}\left(1\right)$

  ${\mathrm{V}}_{\mathrm{x}1}$ input voltage of ${\mathrm{Q}}_1$

  ${\mathrm{r}}_{\mathrm{o}1}$ output resistance of ${\mathrm{Q}}_1$

  ${\mathrm{R}}_1$ input resistance of ${\mathrm{Q}}_1$

  ${\mathrm{r}}_{\mathrm{\pi }1}$ input resistance of hybrid parameter

  ${\mathrm{g}}_{\mathrm{m}1}$ trans-conductance of ${\mathrm{Q}}_1$

Hybrid parameters p,

  ${\mathrm{V}}_{\mathrm{\pi }1}$ = ${\mathrm{V}}_{\mathrm{x}1}$

Substitute ${\mathrm{V}}_{\mathrm{\pi }1}$ = ${\mathrm{V}}_{\mathrm{x}1}$ in $\mathrm{e}\mathrm{q}1$

  ${\mathrm{I}}_{\mathrm{x}1}=\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}{\mathrm{V}}_{\mathrm{x}1}+\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}$

Now rearrange the equation for output resistance $\frac{1}{{\mathrm{R}}_{\mathrm{o}1}}$

  $\frac{{\mathrm{I}}_{\mathrm{x}1}}{{\mathrm{V}}_{\mathrm{x}1}}=\frac{1}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}(1)+\frac{1}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}\left(2\right)$

Expression for output resistance

  $\frac{1}{{\mathrm{R}}_{\mathrm{o}1}}=\frac{{\mathrm{I}}_{\mathrm{x}1}}{{\mathrm{V}}_{\mathrm{x}1}}\left(3\right)$

Substitute equation (3) in equation (2)

  $\frac{1}{{\mathrm{R}}_{\mathrm{o}1}}=\frac{1}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}(1)+\frac{1}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}\left(4\right)$

Now the small-signal equivalent circuit is modified for output resistance as shown,

  

Apply KVL for input voltage at above circuit,

  ${\mathrm{V}}_{\mathrm{x}}={\mathrm{I}}_{\mathrm{x}}{\mathrm{r}}_{\mathrm{o}2}-{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}{\mathrm{V}}_{\mathrm{\pi }2}+{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}\left(5\right)$

Expression for hybrid parameter.

  ${\mathrm{V}}_{\mathrm{\pi }2}=-{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}\left(6\right)$

Put ${\mathrm{V}}_{\mathrm{\pi }2}=-{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}$ in equation (5),

  ${\mathrm{V}}_{\mathrm{x}}={\mathrm{I}}_{\mathrm{x}}{\mathrm{r}}_{\mathrm{o}2}-{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}(-{\mathrm{I}}_{\mathrm{x}}{\mathrm{R}}_{\mathrm{E}}\text{'})+{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}$

Rearrange the equation for output resistance,

  $\frac{{\mathrm{V}}_{\mathrm{x}}}{{\mathrm{I}}_{\mathrm{x}}}={\mathrm{r}}_{\mathrm{o}2}-{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}(-{\mathrm{R}}_{\mathrm{E}}\text{'})+\mathrm{R}{\text{'}}_{\mathrm{E}}\left(7\right)$

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}+{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}{\mathrm{R}}_{\mathrm{E}}\text{'}+\mathrm{R}{\text{'}}_{\mathrm{E}}\\ {\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}{\mathrm{R}}_{\mathrm{E}}\text{'}+\frac{\mathrm{R}{\text{'}}_{\mathrm{E}}}{{\mathrm{r}}_{\mathrm{o}2}}\right]\\ {\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{R}}_{\mathrm{E}}\text{'}\left({\mathrm{g}}_{\mathrm{m}2}+\frac{1}{{\mathrm{r}}_{\mathrm{o}2}}\right)\right]\left(8\right)\end{array}$

Consider ${\mathrm{g}}_{\mathrm{m}2}\gg \frac{1}{{\mathrm{r}}_{\mathrm{o}2}}$ for trans-conductance, rewrite equation (8)

  ${\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}{\mathrm{R}}_{\mathrm{E}}\text{'}\right]\left(9\right)$

  ${\mathrm{R}}_{\mathrm{E}}\text{'}={\mathrm{R}}_{\mathrm{E}}\parallel {\mathrm{r}}_{\mathrm{\pi }2}\left(10\right)$

Substitute the value of equation (10) in (9)

  ${\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}({\mathrm{R}}_{\mathrm{E}}\parallel {\mathrm{r}}_{\mathrm{\pi }2})\right]$ is the derived expression for output resistance.

Conclusion:

  ${\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}({\mathrm{R}}_{\mathrm{E}}\parallel {\mathrm{r}}_{\mathrm{\pi }2})\right]$

To determine

To find:An expression for small-signal voltage gain.

Answer to Problem 10.83P

  ${\mathrm{A}}_{\mathrm{v}}=-({\mathrm{g}}_{\mathrm{m}\mathrm{O}})({\mathrm{r}}_{\mathrm{o}}\parallel {\mathrm{R}}_{\mathrm{L}}\parallel {\mathrm{r}}_{\mathrm{o}2})$

Explanation of Solution

Given:

The given circuit is,

  

Calculation:

The small-signal equivalent circuit of BJT amplifier is as shown,

  

Apply KCL at base terminal of ${\mathrm{Q}}_1$

  ${\mathrm{I}}_{\mathrm{x}1}=\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}{\mathrm{V}}_{\mathrm{\pi }1}+\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}\left(1\right)$

  ${\mathrm{V}}_{\mathrm{x}1}$ input voltage of ${\mathrm{Q}}_1$

  ${\mathrm{r}}_{\mathrm{o}1}$ output resistance of ${\mathrm{Q}}_1$

  ${\mathrm{R}}_1$ input resistance of ${\mathrm{Q}}_1$

  ${\mathrm{r}}_{\mathrm{\pi }1}$ input resistance of hybrid parameter

  ${\mathrm{g}}_{\mathrm{m}1}$ trans-conductance of ${\mathrm{Q}}_1$

Hybrid parameters p ,

  ${\mathrm{V}}_{\mathrm{\pi }1}$ = ${\mathrm{V}}_{\mathrm{x}1}$

Substitute ${\mathrm{V}}_{\mathrm{\pi }1}$ = ${\mathrm{V}}_{\mathrm{x}1}$ in equation (1)

  ${\mathrm{I}}_{\mathrm{x}1}=\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}{\mathrm{V}}_{\mathrm{x}1}+\frac{{\mathrm{V}}_{\mathrm{x}1}}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}$

Now rearrange the equation for output resistance $\frac{1}{{\mathrm{R}}_{\mathrm{o}1}}$

  $\frac{{\mathrm{I}}_{\mathrm{x}1}}{{\mathrm{V}}_{\mathrm{x}1}}=\frac{1}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}(1)+\frac{1}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}\left(2\right)$

Expression for output resistance

  $\frac{1}{{\mathrm{R}}_{\mathrm{o}1}}=\frac{{\mathrm{I}}_{\mathrm{x}1}}{{\mathrm{V}}_{\mathrm{x}1}}\left(3\right)$

Substitute equation (3) in equation (2)

  $\frac{1}{{\mathrm{R}}_{\mathrm{o}1}}=\frac{1}{{\mathrm{r}}_{\mathrm{\pi }1}}+{\mathrm{g}}_{\mathrm{m}1}(1)+\frac{1}{{\mathrm{r}}_{\mathrm{o}1}\parallel {\mathrm{R}}_1}\left(4\right)$

Now the small-signal equivalent circuit is modified for output resistance as shown,

Apply KVL for input voltage at above circuit,

  ${\mathrm{V}}_{\mathrm{x}}={\mathrm{I}}_{\mathrm{x}}{\mathrm{r}}_{\mathrm{o}2}-{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}{\mathrm{V}}_{\mathrm{\pi }2}+{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}\left(5\right)$

Expression for hybrid parameter.

  ${\mathrm{V}}_{\mathrm{\pi }2}=-{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}\left(6\right)$

Put ${\mathrm{V}}_{\mathrm{\pi }2}=-{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}$ in equation (5),

  ${\mathrm{V}}_{\mathrm{x}}={\mathrm{I}}_{\mathrm{x}}{\mathrm{r}}_{\mathrm{o}2}-{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}(-{\mathrm{I}}_{\mathrm{x}}{\mathrm{R}}_{\mathrm{E}}\text{'})+{\mathrm{I}}_{\mathrm{x}}\mathrm{R}{\text{'}}_{\mathrm{E}}$

Rearrange the equation for output resistance,

  $\frac{{\mathrm{V}}_{\mathrm{x}}}{{\mathrm{I}}_{\mathrm{x}}}={\mathrm{r}}_{\mathrm{o}2}-{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}(-{\mathrm{R}}_{\mathrm{E}}\text{'})+\mathrm{R}{\text{'}}_{\mathrm{E}}\left(7\right)$

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}+{\mathrm{g}}_{\mathrm{m}2}{\mathrm{r}}_{\mathrm{o}2}{\mathrm{R}}_{\mathrm{E}}\text{'}+\mathrm{R}{\text{'}}_{\mathrm{E}}\\ {\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}{\mathrm{R}}_{\mathrm{E}}\text{'}+\frac{\mathrm{R}{\text{'}}_{\mathrm{E}}}{{\mathrm{r}}_{\mathrm{o}2}}\right]\\ {\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{R}}_{\mathrm{E}}\text{'}\left({\mathrm{g}}_{\mathrm{m}2}+\frac{1}{{\mathrm{r}}_{\mathrm{o}2}}\right)\right]\left(8\right)\end{array}$

Consider ${\mathrm{g}}_{\mathrm{m}2}\gg \frac{1}{{\mathrm{r}}_{\mathrm{o}2}}$ for trans-conductance, rewrite equation (8)

  ${\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}{\mathrm{R}}_{\mathrm{E}}\text{'}\right]\left(9\right)$

  ${\mathrm{R}}_{\mathrm{E}}\text{'}={\mathrm{R}}_{\mathrm{E}}\parallel {\mathrm{r}}_{\mathrm{\pi }2}\left(10\right)$

Substitute the value of equation 10 in equation 9,

  ${\mathrm{R}}_{\mathrm{O}}={\mathrm{r}}_{\mathrm{o}2}\left[1+{\mathrm{g}}_{\mathrm{m}2}({\mathrm{R}}_{\mathrm{E}}\parallel {\mathrm{r}}_{\mathrm{\pi }2})\right]$ derived expression for output resistance.

Now derive an expression for output voltage.

  ${\mathrm{V}}_{\mathrm{O}}=-({\mathrm{g}}_{\mathrm{m}\mathrm{O}}{\mathrm{V}}_{\mathrm{\pi }1})({\mathrm{r}}_{\mathrm{o}}\parallel {\mathrm{R}}_{\mathrm{L}}\parallel {\mathrm{r}}_{\mathrm{o}2})\left(11\right)$

Input voltage will be,

  ${\mathrm{V}}_{\mathrm{\pi }1}={\mathrm{V}}_{\mathrm{i}}\left(12\right)$

Put ${\mathrm{V}}_{\mathrm{\pi }1}=\mathrm{V}$ in equation 11,

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{O}}=-({\mathrm{g}}_{\mathrm{m}\mathrm{O}}{\mathrm{V}}_{\mathrm{i}})({\mathrm{r}}_{\mathrm{o}}\parallel {\mathrm{R}}_{\mathrm{L}}\parallel {\mathrm{r}}_{\mathrm{o}2})\\ \frac{{\mathrm{V}}_{\mathrm{O}}}{{\mathrm{V}}_{\mathrm{i}}}=-({\mathrm{g}}_{\mathrm{m}\mathrm{O}})({\mathrm{r}}_{\mathrm{o}}\parallel {\mathrm{R}}_{\mathrm{L}}\parallel {\mathrm{r}}_{\mathrm{o}2})\end{array}$

It is known that ,

  ${\mathrm{A}}_{\mathrm{v}}=\frac{{\mathrm{V}}_{\mathrm{O}}}{{\mathrm{V}}_{\mathrm{i}}}$

The small-signal voltage gain derived expression is,

  ${\mathrm{A}}_{\mathrm{v}}=-({\mathrm{g}}_{\mathrm{m}\mathrm{O}})({\mathrm{r}}_{\mathrm{o}}\parallel {\mathrm{R}}_{\mathrm{L}}\parallel {\mathrm{r}}_{\mathrm{o}2})$

Conclusion:

  ${\mathrm{A}}_{\mathrm{v}}=-({\mathrm{g}}_{\mathrm{m}\mathrm{O}})({\mathrm{r}}_{\mathrm{o}}\parallel {\mathrm{R}}_{\mathrm{L}}\parallel {\mathrm{r}}_{\mathrm{o}2})$