Chapter 6, Problem 6.45P

Consider the circuit in Figure P6.45. The transistor parameters are $\beta =120$ and $V_A=\infty$ . Repeat parts (a)−(d) of Problem 6.44.

Figure P6.45

To determine

The value of the I_CQ and V_CEQ .

Answer to Problem 6.45P

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\text{=2}\text{.09}\text{mA}\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}=3.69\text{V}\end{array}$

Explanation of Solution

Given:

The circuit is given as:

  

Calculation:

Let the BJT be the single node, then by applying Kirchhoff’s current law, the quiescent emitter current ${\mathrm{I}}_{\mathrm{E}\mathrm{Q}}$ , the quiescent base current ${\mathrm{I}}_{\mathrm{B}\mathrm{Q}}$ and the quiescent collector current ${\mathrm{I}}_{\mathrm{C}\mathrm{Q}}$ are expressed as:

  ${\mathrm{I}}_{\mathrm{E}\mathrm{Q}}={\mathrm{I}}_{\mathrm{B}\mathrm{Q}}+{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}$ ……………………….(1)

In Common-Emitter mode:

The relation between the quiescent collector current is ${\mathrm{I}}_{\mathrm{C}\mathrm{Q}}$ and the quiescent base current is ${\mathrm{I}}_{\mathrm{B}\mathrm{Q}}$ .

  ${\mathrm{I}}_{\mathrm{C}\mathrm{Q}}=\mathrm{\beta }{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}\mathrm{...............}\left(2\right)$

For dc analysis,

All the capacitors act as an open circuit and all the ac sources act as short circuits.

DC analysis:

Redrawing the circuit by open circuit, the capacitors, and short circuit the ac sources.

  

Evaluating the Thevenin resistance ${\mathrm{R}}_{\mathrm{T}\mathrm{H}}$ by using voltage division rule:

  $\begin{array}{c}{\mathrm{R}}_{\mathrm{T}\mathrm{H}}={\mathrm{R}}_1\parallel {\mathrm{R}}_2\\ =\frac{{\mathrm{R}}_1\times {\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}\\ {\mathrm{R}}_{\mathrm{T}\mathrm{H}}=\left(\frac{10\times 10}{10+10}\right)\mathrm{k}\mathrm{\Omega }\\ {\mathrm{R}}_{\mathrm{T}\mathrm{H}}=5\mathrm{k}\mathrm{\Omega }\end{array}$

Evaluating the Thevenin voltage, ${\mathrm{V}}_{\mathrm{T}\mathrm{H}}$ :

  $\begin{array}{c}{\mathrm{V}}_{\mathrm{T}\mathrm{H}}=\frac{-10}{{\mathrm{R}}_1+{\mathrm{R}}_2}{\mathrm{R}}_2\\ =\left(\frac{-10}{10\mathrm{k}\mathrm{\Omega }+10\mathrm{k}\mathrm{\Omega }}\right)\times 10\mathrm{k}\mathrm{\Omega }\\ =-5\text{V}\end{array}$

Drawing the modified circuit:

  

Applying the Kirchhoff s voltage law in the base-emitter loop,

  $\begin{array}{c}-{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}{\mathrm{R}}_{\mathrm{T}\mathrm{H}}-\mathrm{V}{\mathrm{B}\mathrm{E}\left(\mathrm{o}\mathrm{n}\right)}_{}-{\mathrm{I}}_{\mathrm{E}\mathrm{Q}}{\mathrm{R}}_{\mathrm{E}}+10-{\mathrm{V}}_{\mathrm{T}\mathrm{H}}=0\\ -{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}{\mathrm{R}}_{\mathrm{T}\mathrm{H}}-{\mathrm{V}}_{\mathrm{B}\mathrm{E}\left(\mathrm{o}\mathrm{n}\right)}-\left(1+\mathrm{\beta }\right){\mathrm{I}}_{\mathrm{B}\mathrm{Q}}{\mathrm{R}}_{\mathrm{E}}+10-{\mathrm{V}}_{\mathrm{T}\mathrm{H}}=0\\ {\mathrm{I}}_{\mathrm{B}\mathrm{Q}}=\frac{10-{\mathrm{V}}_{\mathrm{T}\mathrm{H}}-{\mathrm{V}}_{\mathrm{B}\mathrm{E}\left(\mathrm{o}\mathrm{n}\right)}}{{\mathrm{R}}_{\mathrm{T}\mathrm{H}}+\left(1+\mathrm{\beta }\right){\mathrm{R}}_{\mathrm{E}}}\end{array}$

Substituting $\text{5kΩ}\text{for}{\text{R}}_{\text{TH}}\text{,-5V}\text{for}{\text{V}}_{\text{TH}}\text{,120}\text{for}\text{β}\text{,2kΩfor}{\text{R}}_{\text{E}}\text{,0}\text{.7V}\text{for}{\text{V}}_{\text{BE}\left(\text{on}\right)}$

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}=\left[\frac{10-5-0.7}{\left(5\times {10}^3\right)+\left(1+120\right)\times \left(2\times {10}^3\right)}\right]\\ =0.0174\text{mA}\end{array}$

From equation (2):

  ${\mathrm{I}}_{\mathrm{C}\mathrm{Q}}=\mathrm{\beta }{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}$

Substituting $\text{120}\text{for}\text{β}\text{,and0}\text{.174mA}{\text{forI}}_{\text{BQ}}\text{.}$

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\text{=120×0}\text{.0174mA}\\ \text{=2}\text{.09mA}\end{array}$

Therefore, the value of ${\mathrm{I}}_{\mathrm{C}\mathrm{Q}}$ is $\text{2}\text{.09mA}$ .

From equation (1) and (2):

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}=\mathrm{\beta }{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}\\ {\mathrm{I}}_{\mathrm{E}\mathrm{Q}}={\mathrm{I}}_{\mathrm{B}\mathrm{Q}}+{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\end{array}$

Substituting $\frac{{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}}{\mathrm{\beta }}\text{for}{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}$ ,

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{E}\mathrm{Q}}=\frac{{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}}{\mathrm{\beta }}+{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\\ {\mathrm{I}}_{\mathrm{E}\mathrm{Q}}=\left(\frac{1+\mathrm{\beta }}{\mathrm{\beta }}\right){\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\end{array}$

Substitute $\text{120}\text{for}\text{β}\text{,and}\text{2}\text{.09mA}\text{for}{\text{I}}_{\text{CQ}}\text{.}$

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{E}\mathrm{Q}}\text{=}\left(\frac{\text{1+120}}{\text{120}}\right)\text{×2}\text{.09mA}\\ \text{=2}\text{.11mA}\end{array}$

Drawing the modified circuit:

  

Applying Kirchhoff s voltage law to the loop,

  $\begin{array}{l}-{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}{\mathrm{R}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}-{\mathrm{I}}_{\mathrm{E}\mathrm{Q}}{\mathrm{R}}_{\mathrm{E}}+10=0\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}=10-{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}{\mathrm{R}}_{\mathrm{C}}-{\mathrm{I}}_{\mathrm{E}\mathrm{Q}}{\mathrm{R}}_{\mathrm{E}}\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}=10-\left({\mathrm{R}}_{\mathrm{C}}+\frac{1+\mathrm{\beta }}{\mathrm{\beta }}{\mathrm{R}}_{\mathrm{E}}\right)\times {\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\left\{\sin \mathrm{c}\mathrm{e}{\mathrm{I}}_{\mathrm{E}\mathrm{Q}}=\left(\frac{1+\mathrm{\beta }}{\mathrm{\beta }}\right){\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\right\}\end{array}$

Substituting $\text{120}\text{for}\text{β}\text{,1kΩ}\text{for}{\text{R}}_{\text{C}}\text{,2kΩ}\text{for}{\text{R}}_{\text{E}}\text{and}\text{2}\text{.09mA}\text{for}{\text{I}}_{\text{CQ}}\text{.}$

  $\begin{array}{c}{\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}=\left\{10-\left(1+\frac{1+120}{120}\times 2\right)\text{k}\mathrm{\Omega }\times 2.09\text{mA}\right\}\\ =3.69\text{V}\end{array}$

Therefore, the value of ${\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}$ is 3.69V.

To determine

To plot: The dc and the ac load lines.

Explanation of Solution

Given:

The circuit is given as:

  

Calculation:

Applying Kirchhoff s voltage law around the collector-emitter loop in figure 1.

  $\begin{array}{l}-{\mathrm{I}}_{\mathrm{C}}{\mathrm{R}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{C}\mathrm{E}}-{\mathrm{I}}_{\mathrm{E}}{\mathrm{R}}_{\mathrm{E}}+10=0\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}}=10-{\mathrm{I}}_{\mathrm{C}}{\mathrm{R}}_{\mathrm{C}}-{\mathrm{I}}_{\mathrm{E}}{\mathrm{R}}_{\mathrm{E}}\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}}=10-\left\{{\mathrm{R}}_{\mathrm{C}}+\left(\frac{1+\mathrm{\beta }}{\mathrm{\beta }}\right){\mathrm{R}}_{\mathrm{E}}\right\}{\mathrm{I}}_{\mathrm{C}}\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}}=10-\left\{1\times {10}^3+\left(\frac{1+120}{120}\right)2\times {10}^3\right\}\left({\mathrm{I}}_{\mathrm{C}}\right)\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}}=10-\left(3.02\mathrm{k}\mathrm{\Omega }\right){\mathrm{I}}_{\mathrm{C}}\mathrm{........................}\left(3\right)\end{array}$

Substitute $\text{0V}\text{for}{\text{V}}_{\text{CE}}$ in equation (3):

  $\begin{array}{l}0=10-\left(3.02\mathrm{k}\mathrm{\Omega }\right){\mathrm{I}}_{\mathrm{C}}\\ {\mathrm{I}}_{\mathrm{C}}=\frac{10}{3.02}\times {10}^{-3}\mathrm{A}\\ =3.31\text{mA}\end{array}$

Substitute $\text{0A}\text{for}{\text{I}}_{\text{C}}$ in equation (3).

  $\begin{array}{l}{\mathrm{V}}_{\mathrm{C}\mathrm{E}}=10-\left(3.02\mathrm{k}\mathrm{\Omega }\right)\left(0\right)\\ =10\mathrm{V}\end{array}$

The required “Q” point co-ordinates are ${\text{I}}_{\text{CQ}}\text{=2}\text{.09mA}{\text{,V}}_{\text{CEQ}}\text{=3}\text{.69V}$ .

Draw the DC load line:

  

Drawing the AC equivalent circuit:

  

Evaluating the ac load ${\mathrm{r}}_{\mathrm{c}}$ :

  $\begin{array}{c}{\mathrm{r}}_{\mathrm{c}}={\mathrm{R}}_{\mathrm{E}}\parallel {\mathrm{R}}_{\mathrm{E}}\\ =\left(\text{2kΩ}\Vert \text{2kΩ}\right)\\ \text{=1kΩ}\end{array}$

The expression for the ac load line:

  ${\mathrm{I}}_{\mathrm{C}}=\left({\mathrm{I}}_{\mathrm{C}\mathrm{Q}}+\frac{{\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}}{{\mathrm{r}}_{\mathrm{c}}}\right)-\frac{{\mathrm{V}}_{\mathrm{C}\mathrm{E}}}{{\mathrm{r}}_{\mathrm{C}}}\mathrm{................}\left(4\right)$

Substitute $\text{0V}\text{for}{\text{V}}_{\text{CE}}$ in equation (4).

  $\begin{array}{c}{\mathrm{I}}_{\mathrm{C}}=\left({\mathrm{I}}_{\mathrm{C}\mathrm{Q}}+\frac{{\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}}{{\mathrm{r}}_{\mathrm{c}}}\right)-\left(0\right)\\ =\left(2.09+\frac{3.69}{1}\right)\text{mA}\\ \text{=5}\text{.78mA}\end{array}$

Substitute $\text{0V}\text{for}{\text{I}}_{\text{C}}$ in equation (4).

  $\begin{array}{c}0=\left({\mathrm{I}}_{\mathrm{C}\mathrm{Q}}+\frac{{\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}}{{\mathrm{r}}_{\mathrm{c}}}\right)-\frac{{\mathrm{V}}_{\mathrm{C}\mathrm{E}}}{{\mathrm{r}}_{\mathrm{C}}}\\ {\mathrm{V}}_{\mathrm{C}\mathrm{E}}={\mathrm{V}}_{\mathrm{C}\mathrm{E}\mathrm{Q}}+{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}{\mathrm{r}}_{\mathrm{C}}\\ =3.69+\left(\text{2}\text{.09mA}\right)\left(\text{1kΩ}\right)\\ \text{=5}\text{.78V}\end{array}$

Drawing the ac load line:

  

To determine

The small signal voltage gain.

Answer to Problem 6.45P

The small signal voltage gain.

Explanation of Solution

Given:

The circuit is given as:

  

Drawing the small signal analysis of the circuit by short circuiting the dual dc sources and the capacitors:

  

Evaluating the diffusion resistance,

  $\begin{array}{l}{\mathrm{r}}_{\mathrm{\pi }}=\frac{\mathrm{\beta }{\mathrm{V}}_{\mathrm{T}}}{{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}}\\ {\mathrm{r}}_{\mathrm{\pi }}=\frac{120\times 0.026\mathrm{V}}{2.09\times {10}^{-3}\mathrm{A}}\\ =1.49\times {10}^{-3}\mathrm{\Omega }\\ =1.49\text{kΩ}\end{array}$

Evaluating the small signal transistor output resistance ${\mathrm{r}}_0$ ,

  $\begin{array}{l}{\mathrm{r}}_0=\frac{{\mathrm{V}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}}\\ {\mathrm{r}}_0=\frac{\mathrm{\infty }}{{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}}\\ =\mathrm{\infty }\end{array}$

Evaluating the input resistance ${\mathrm{R}}_{\mathrm{i}\mathrm{b}}$ of the transistor:

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{i}\mathrm{b}}={\mathrm{r}}_{\mathrm{\pi }}+\left(1+\mathrm{\beta }\right)\left({\mathrm{r}}_0\left|\right|{\mathrm{R}}_{\mathrm{E}}\left|\right|{\mathrm{R}}_{\mathrm{L}}\right)\\ ={\mathrm{r}}_{\mathrm{\pi }}+\left(1+\mathrm{\beta }\right)\left({\mathrm{R}}_{\mathrm{E}}\left|\right|{\mathrm{R}}_{\mathrm{L}}\right)\{\sin \mathrm{c}\mathrm{e}{\mathrm{r}}_0=\mathrm{\infty }\}\\ =\left\{1.49+(1+120)\times (2||2)\right\}\mathrm{k}\mathrm{\Omega }\\ {\mathrm{R}}_{\mathrm{i}\mathrm{b}}=122.5\mathrm{k}\mathrm{\Omega }\mathrm{................}\left(5\right)\end{array}$

Calculate the input resistance ${\mathrm{R}}_{\mathrm{i}}$ .

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{i}}={\mathrm{R}}_1\left|\right|{\mathrm{R}}_2\left|\right|{\mathrm{R}}_{\mathrm{i}\mathrm{b}}\\ =\left(10\left|\right|10\left|\right|122.5\right)\mathrm{k}\mathrm{\Omega }\\ =\left(5\left|\right|122.5\right)\mathrm{k}\mathrm{\Omega }\\ =4.8\mathrm{k}\mathrm{\Omega }\end{array}$

The small-signal voltage gain is, ${\mathrm{A}}_{\mathrm{\nu }}$ is

  $\begin{array}{c}{\mathrm{A}}_{\mathrm{v}}=\frac{\left(1+\mathrm{\beta }\right)\left({\mathrm{R}}_{\mathrm{E}}\Vert {\mathrm{R}}_{\mathrm{L}}\right)}{{\mathrm{r}}_{\mathrm{\pi }}+\left(1+\mathrm{\beta }\right)\left({\mathrm{R}}_{\mathrm{E}}\Vert {\mathrm{R}}_{\mathrm{L}}\right)}\cdot \left(\frac{{\mathrm{R}}_1\Vert {\mathrm{R}}_2\Vert {\mathrm{R}}_{\mathrm{i}\mathrm{b}}}{{\mathrm{R}}_1\Vert {\mathrm{R}}_2\Vert {\mathrm{R}}_{\mathrm{i}\mathrm{b}}+{\mathrm{R}}_{\mathrm{S}}}\right)\\ =\frac{121\times \left(2\Vert 2\right)}{1.49+\left(121\times \left(2\Vert 2\right)\right)}\cdot \left(\frac{4.8}{4.8+5}\right)\\ =0.484\end{array}$

To determine

The output resistance.

Answer to Problem 6.45P

The output resistance is $32.5\mathrm{\Omega }$ .

Explanation of Solution

Given:

The circuit is given as

  

Calculation:

From equation 5,

  ${\mathrm{R}}_{\mathrm{i}\mathrm{b}}=122.5\mathrm{k}\mathrm{\Omega }$

Thus, the input resistance ${\mathrm{R}}_{\mathrm{i}\mathrm{b}}$ is $122.5\mathrm{k}\mathrm{\Omega }$ :

Calculate the output resistance, ${\mathrm{R}}_0$ :

  $\begin{array}{c}{\mathrm{R}}_0=\left(\frac{{\mathrm{r}}_{\mathrm{\pi }}+\left({\mathrm{R}}_1\left|\right|{\mathrm{R}}_2\left|\right|{\mathrm{R}}_{\mathrm{s}}\right)}{1+\mathrm{\beta }}\right)\left|\right|{\mathrm{R}}_{\mathrm{E}}\left|\right|{\mathrm{r}}_0\\ =\left(\frac{{\mathrm{r}}_{\mathrm{\pi }}+\left({\mathrm{R}}_1\left|\right|{\mathrm{R}}_2\left|\right|{\mathrm{R}}_{\mathrm{s}}\right)}{1+\mathrm{\beta }}\right)\left|\right|{\mathrm{R}}_{\mathrm{E}}\{\mathrm{s}\text{ince}{\mathrm{r}}_0=\mathrm{\infty }\}\\ =\left\{\left(\frac{1.49+\left(10\left|\right|10\left|\right|5\right)}{1+120}\right)\left|\right|2\right\}\times {10}^3\mathrm{\Omega }\\ =32.5\mathrm{\Omega }\end{array}$

Therefore, the output resistance ${\mathrm{R}}_0$ is $32.5\mathrm{\Omega }$ .