Chapter 7, Problem 7.40P

The parameters of the transistor in the circuit in Figure P7.40 are $\beta =100$ , $V_{BE}\text{(on)=0}\text{.7V}$ , and $V_A=\infty$ . Neglect the capacitance effects of the transistor. (a) Draw the three equivalent circuits that represent the amplifier in the low−frequency range, midband range, and the high frequency range. (b) Sketch the Bode magnitude plot. (c) Determine the values of ${\left|A_m\right|}_{dB}$ , $f_L$ , and $f_H$ .

Figure P7.40

To determine

To draw: Three equivalent circuits that represent the amplifier in the low frequency range, mid-band range and the high frequency range.

Answer to Problem 7.40P

Three equivalent circuits that represent the amplifier in the low frequency range, mid-band range and the high frequency range are shown in Figure 1, 2 and 3 respectively.

Explanation of Solution

Given:

The diagram is given as:

  

Calculation:

Calculate the value of current ${\mathrm{I}}_{\mathrm{C}\mathrm{Q}}$ from the dc analysis of the circuit. In dc analysis, the dc sources are connected to the ground and the capacitors are open-circuited as they offer very high impedance to the dc signal. Figure 1 shows the modified circuit for the dc analysis.

  

Applying Kirchhoff s voltage law in the above loop:

  $12-{\mathrm{R}}_{\mathrm{B}}{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}-{\mathrm{V}}_{\mathrm{B}\mathrm{E}}=0$

Here,

  ${\mathrm{I}}_{\mathrm{B}\mathrm{Q}}$ is the quiescent base current.

  ${\mathrm{V}}_{\mathrm{B}\mathrm{E}}$ is the base to emitter voltage.

Substituting $\text{1MΩ}\text{for}{\mathrm{R}}_{\mathrm{B}}\text{and}\text{0}\text{.7V}\text{for}{\mathrm{V}}_{\mathrm{B}\mathrm{E}}$ in the above equation,

  $\begin{array}{l}12-\left(1\times {10}^6\right){\mathrm{I}}_{\mathrm{B}\mathrm{Q}}-0.7=0\\ {\mathrm{I}}_{\mathrm{B}\mathrm{Q}}=\frac{12-0.7}{{10}^6}\\ =11.3\text{μA}\end{array}$

The quiescent collector current ${\mathrm{I}}_{\mathrm{C}\mathrm{Q}}$ :

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

Here,

  $\mathrm{\beta }$ is the common-emitter current gain.

Substituting $\text{100for}\mathrm{\beta }\text{and11}\text{.3μA}\text{for}{\mathrm{I}}_{\mathrm{B}\mathrm{Q}}$ in the equation:

  $\begin{array}{l}{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}=100\times 11.3\mathrm{\mu }\mathrm{A}\\ =1.13\text{mA}\end{array}$

Evaluating the resistance ${\mathrm{r}}_{\mathrm{\pi }}$ as follows:

  ${\mathrm{r}}_{\mathrm{\pi }}=\frac{\mathrm{\beta }{\mathrm{V}}_{\mathrm{T}}}{{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}}$

Here,

  ${\mathrm{V}}_{\mathrm{T}}$ is the thermal voltage and has a value of 26V.

  $\begin{array}{l}{\mathrm{r}}_{\mathrm{\pi }}=\frac{100\times 26\times {10}^{-3}}{1.13\times {10}^{-3}}\\ =2300.88\\ =2.3\times {10}^3\\ =2.3\text{kΩ}\end{array}$

Evaluating the transconductance ${\mathrm{g}}_{\mathrm{m}}$ as follows:

  ${\mathrm{g}}_{\mathrm{m}}=\frac{{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}}{{\mathrm{V}}_{\mathrm{T}}}$

Substituting $\text{1}\text{.13mA}\text{for}{\mathrm{I}}_{\mathrm{C}\mathrm{Q}}\text{and}\text{26mV}\text{for}{\mathrm{V}}_{\mathrm{T}}$ .

  $\begin{array}{l}{\text{g}}_{\text{m}}\text{=}\frac{\text{1}\text{.13mA}}{\text{26mV}}\\ \text{=0}\text{.0434A/V}\end{array}$

Figure 2 shows the low −frequency small signal transistor with the output resistance ${\mathrm{r}}_0$ assumed as infinity. The circuit consists of a coupling capacitor ${\mathrm{C}}_{\mathrm{C}}$ and resistor ${\mathrm{R}}_{\mathrm{B}}$ and the load capacitors are open-circuited.

  

Figure 1

In the mid-frequency range, the coupling and bypass capacitors are short-circuited and the load capacitors are open-circuited.

It shows the mid-frequency small signal transistor with the output resistance ${\mathrm{r}}_0$ assumed as infinity.Figure 3shows the mid-frequency small signal transistor with the output resistance ${\mathrm{r}}_0$ assumed as infinity.

  

Figure 2

In the high-frequency range, the coupling and bypass capacitors are short-circuited and the load capacitors are included.

  

Figure 3

Hence, the three equivalent circuits that represent the amplifier in the low-frequency range, mid-band range, and high-frequency range are plotted.

To determine

To sketch: The bode magnitude plot.

Answer to Problem 7.40P

The sketch of bode magnitude plot is shown in Figure 4.

Explanation of Solution

Given:

The diagram is given as:

  

Calculation:

Consider the values, calculated in part (a).

The effect of the coupling capacitor ${\mathrm{C}}_{\mathrm{C}}$ is such that the circuit behaves as a high pass filter and the effect of the load capacitor ${\mathrm{C}}_{\mathrm{L}}$ is such that the circuit behaves as a low pass filter.

The figure shows the bode plot for the circuit having a combination of a coupling capacitor and load capacitor:

  

Figure 4

Hence, the bode magnitude plot is sketched.

To determine

The values of the $|{\mathrm{A}}_{\mathrm{m}}{|}_{\mathrm{d}\mathrm{B}},{\mathrm{f}}_{\mathrm{L}}\text{and }{\mathrm{f}}_{\mathrm{H}}$ .

Answer to Problem 7.40P

The values are:

  $\begin{array}{l}|{\mathrm{A}}_{\mathrm{m}}{|}_{\mathrm{d}\mathrm{B}}=43.66\text{dB}\\ {\mathrm{f}}_{\mathrm{L}}=4.83\text{Hz}\\ {\mathrm{f}}_{\mathrm{H}}=3.15\text{MHz}\end{array}$ .

Explanation of Solution

Given:

The diagram is given as:

  

Calculation:

Evaluating the midband gain by short-circuiting the coupling and bypass capacitors and open-circuiting the load capacitors.

  $\begin{array}{l}\left|{\mathrm{A}}_{\mathrm{m}}\right|=\left|\frac{{\mathrm{V}}_0}{{\mathrm{V}}_{\mathrm{i}}}\right|\\ ={\mathrm{g}}_{\mathrm{m}}{\mathrm{r}}_{\mathrm{\pi }}\left({\mathrm{R}}_{\mathrm{C}}\left|\right|{\mathrm{R}}_{\mathrm{L}}\right)\left(\frac{{\mathrm{R}}_{\mathrm{B}}}{{\mathrm{R}}_{\mathrm{B}}+{\mathrm{r}}_{\mathrm{\pi }}}\right)\left(\frac{1}{{\mathrm{R}}_{\mathrm{S}}+\left({\mathrm{R}}_{\mathrm{B}}\left|\right|{\mathrm{r}}_{\mathrm{\pi }}\right)}\right)\end{array}$

Substituting the values,

  $\begin{array}{c}\left|{\mathrm{A}}_{\mathrm{m}}\right|=0.0434\times 2.3\times {10}^3\left(5.1\times {10}^3\left|\right|500\times {10}^3\right)\left(\frac{1\times {10}^6}{1000\times {10}^3+2.3\times {10}^3}\right)\left(\frac{1}{\left[1\times {10}^3+\left(1\times {10}^6\left|\right|2.3\times {10}^3\right)\right]}\right)\\ =0.09982\times {10}^3\left(\frac{5.1\times 500\times {10}^6}{5.1\times {10}^3+500\times {10}^3}\right)\left(\frac{1\times {10}^6}{1002.3\times {10}^3}\right)\left(\frac{1}{3.294\times {10}^3}\right)\\ =0.09982\times {10}^6\left(\frac{5.1\times 500}{505.1}\right)\left(\frac{1}{1002.3}\right)\left(\frac{1}{3.294}\right)\\ \left|{\mathrm{A}}_{\mathrm{m}}\right|=152.636\end{array}$

Evaluating the gain in dB:

  ${\left|{\mathrm{A}}_{\mathrm{m}}\right|}_{\mathrm{d}\mathrm{B}}=20{\log }_{10}\left|{\mathrm{A}}_{\mathrm{m}}\right|$

Substituting $152.636\text{for}\left|{\mathrm{A}}_{\mathrm{m}}\right|$ in the equation:

  $\begin{array}{l}{\left|{\mathrm{A}}_{\mathrm{m}}\right|}_{\mathrm{d}\mathrm{B}}=20{\log }_{10}152.636\\ =20\times 2.183\\ =43.66\text{dB}\end{array}$

Hence, the value of gain ${\left|{\mathrm{A}}_{\mathrm{m}}\right|}_{\mathrm{d}\mathrm{B}}$ is $\text{43}\text{.66dB}$ .

Evaluating the equivalent resistance $\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)$ associated with the coupling capacitors by substituting zero for the signal source $\left({\mathrm{v}}_{\mathrm{i}}\right)$ :

  ${\mathrm{R}}_{\mathrm{e}\mathrm{q}}=\left[{\mathrm{R}}_{\mathrm{S}}+\left({\mathrm{R}}_{\mathrm{B}}\left|\right|{\mathrm{r}}_{\mathrm{\pi }}\right)\right]$

Substituting $\text{1kΩ}\text{for}{\mathrm{R}}_{\mathrm{S}}\text{,1MΩ}\text{for}{\mathrm{R}}_{\mathrm{B}}\text{,and}\text{2}\text{.3kΩ}\text{for}{\mathrm{r}}_{\mathrm{\pi }}$ in the equation:

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{e}\mathrm{q}}=\left[1\mathrm{k}+\left(1\text{M||}2.3\mathrm{k}\right)\right]\\ =1\mathrm{k}+\frac{1000\times 2.3\mathrm{k}}{1000+2.3}\\ =1+2.294\\ =3.294\text{kΩ}\end{array}$

Evaluating the time constant ${\mathrm{\tau }}_{\mathrm{S}}$ associated with ${\mathrm{C}}_{\mathrm{C}}$ as:

  ${\mathrm{\tau }}_{\mathrm{S}}={\mathrm{R}}_{\mathrm{e}\mathrm{q}}{\mathrm{C}}_{\mathrm{C}}$

Substituting $\text{3}\text{.294kΩ}\text{for}{\mathrm{R}}_{\mathrm{e}\mathrm{q}}\text{and}\text{10μF}\text{for}{\mathrm{C}}_{\mathrm{C}}$ in the equation:

  $\begin{array}{l}{\mathrm{\tau }}_{\mathrm{S}}=3.294\times {10}^3\left(10\times {10}^{-6}\right)\\ =3.294\times {10}^{-2}\\ =32.94\text{ms}\end{array}$

Evaluating the equivalent resistance $\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)$ associated with the load capacitor:

  $\begin{array}{l}\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)=\left({\mathrm{R}}_{\mathrm{C}}\parallel {\mathrm{R}}_{{}_{\mathrm{L}}}\right)\\ =\frac{{\mathrm{R}}_{\mathrm{C}}\times {\mathrm{R}}_{\mathrm{L}}}{{\mathrm{R}}_{\mathrm{C}}+{\mathrm{R}}_{\mathrm{L}}}\end{array}$

Substituting $\text{5}\text{.1kΩ}\text{for}{\mathrm{R}}_{\mathrm{C}}\text{and}\text{500kΩ}\text{for}{\mathrm{R}}_{\mathrm{L}}$ in the equation:

  $\begin{array}{l}{\mathrm{R}}_{\mathrm{e}\mathrm{q}}=\frac{5.1\times 500}{5.1+500}\\ =5.048\mathrm{k}\mathrm{\Omega }\end{array}$

Evaluating the time constant ${\mathrm{\tau }}_{\mathrm{p}}$ associated with ${\mathrm{C}}_{\mathrm{L}}$ as:

  ${\mathrm{\tau }}_{\mathrm{p}}={\mathrm{R}}_{\mathrm{e}\mathrm{q}}{\mathrm{C}}_{\mathrm{L}}$

Substituting $\text{5}\text{.048kΩ}\text{for}{\mathrm{R}}_{\mathrm{e}\mathrm{q}}\text{and}\text{10pF}\text{for}{\mathrm{C}}_{\mathrm{L}}$ .

  $\begin{array}{l}{\mathrm{\tau }}_{\mathrm{p}}=5.048\times {10}^3\times 10\times {10}^{-12}\\ =5.048\times {10}^{-8}\\ =0.0504\mathrm{\mu }\mathrm{s}\\ \end{array}$

Evaluate the lower corner frequency ${\mathrm{f}}_{\mathrm{L}}$ as:

  ${\mathrm{f}}_{\mathrm{L}}=\frac{1}{2\mathrm{\pi }{\mathrm{\tau }}_{\mathrm{s}}}$

Substituting $\text{32}\text{.94ms}\text{for}{\mathrm{\tau }}_{\mathrm{s}}$ in the equation.

  $\begin{array}{l}{\mathrm{f}}_{\mathrm{L}}=\frac{1}{2\mathrm{\pi }\left(32.94\times {10}^{-3}\right)}\\ =4.831\text{Hz}\end{array}$

Evaluated the upper corner frequency:

  ${\mathrm{f}}_{\mathrm{H}}=\frac{1}{2\mathrm{\pi }{\mathrm{\tau }}_{\mathrm{p}}}$

Substitute $\text{0}\text{.0504μs}\text{for}{\mathrm{\tau }}_{\mathrm{p}}$ in the equation.

  $\begin{array}{l}{\mathrm{f}}_{\mathrm{L}}=\frac{1}{2\mathrm{\pi }\left(0.504\times {10}^{-6}\right)}\\ =3.158\text{MHz}\end{array}$