Chapter 7, Problem 7.41P

In the common−source amplifier in Figure 7.25(a) in the text, a source bypass capacitor is to be added between the source terminal and ground potential. The circuit parameters are $R_S=\text{3}\text{.2kΩ}$ , $R_D=\text{10kΩ}$ , $R_L=\text{20kΩ}$ , and $C_L=10\text{pF}$ . The transistor parameters are $V_{TP}=-2\text{V}$ $K_P\text{=0}{\text{.25mA/V}}^2$ , and $\lambda =0$ . (a) Derive the small−signal voltage gain expression, as a function of s, that describes the circuit behavior in the high−frequency range. (b) What is the expression for the time constant associated with the upper 3dB frequency? (c) Determine the time constant, upper 3dB frequency, and small−signal midband voltage gain.

To determine

To derive: The small signal voltage gain expression.

Answer to Problem 7.41P

The expression for small signal voltage gain:

  $A_v=\frac{-g_m\left(R_D𑨈R_L\right)}{\left[1+g_m\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\right]\left(\frac{1}{1+s\left(R_D𑨈R_L\right)C_L}\right)}$

Explanation of Solution

Given:

The circuit parameter is given as:

  $\begin{array}{l}{R}_S=3.2\text{k}\Omega \\ R_D=10\text{k}\Omega \\ R_L=20\text{k}\Omega \\ C_L=10\text{pF}\end{array}$

The transistor parameter are given as:

  $\begin{array}{l}{V}_{TP}=-2\text{V}\\ K_P=0.25{\text{mA/V}}^2\\ \lambda =0\end{array}$

Drawing the small signal model of the circuit with the source bypass capacitor:

  

Applying the Ohm’s law to the drain terminal:

  $V_0=g_m{V}_{sg}\left(R_D𑨈R_L𑨈\frac{1}{s{C}_L}\right)$

Evaluating the input voltage $V_i$ :

  $\begin{array}{l}{V}_i=-V_{sg}-g_m{V}_{sg}\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\\ =-V_{sg}\left[1+g_m\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\right]\end{array}$

Evaluating the ratio of output voltage to the input voltage:

  $\begin{array}{l}\frac{V_0}{V_i}=\frac{-g_m\left(R_D𑨈R_L𑨈\frac{1}{s{C}_L}\right)}{\left[1+g_m\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\right]}\\ =\frac{-g_m\left(\left(R_D𑨈R_L\right)\times \frac{1}{s{C}_L}\right)}{\left[1+g_m\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\right]\left(\left(R_D𑨈R_L\right)+\frac{1}{s{C}_L}\right)}\\ =\frac{-g_m\left(R_D𑨈R_L\right)}{\left[1+g_m\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\right]\left(\frac{1}{1+s\left(R_D𑨈R_L\right)C_L}\right)}\mathrm{...........}\left(1\right)\end{array}$

Therefore, the expression for small signal voltage gain:

  $A_v=\frac{-g_m\left(R_D𑨈R_L\right)}{\left[1+g_m\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\right]\left(\frac{1}{1+s\left(R_D𑨈R_L\right)C_L}\right)}$

To determine

The expression for the time constant associated with the upper 3dB frequency.

Answer to Problem 7.41P

The expression for the time constant associated with the upper 3 Db frequency is:

  $\tau =\left(R_D𑨈R_L\right)C_L$ .

Explanation of Solution

Given:

The circuit parameter is given as:

  $\begin{array}{l}{R}_S=3.2\text{k}\Omega \\ R_D=10\text{k}\Omega \\ R_L=20\text{k}\Omega \\ C_L=10\text{pF}\end{array}$

The transistor parameter are given as:

  $\begin{array}{l}{V}_{TP}=-2\text{V}\\ K_P=0.25{\text{mA/V}}^2\\ \lambda =0\end{array}$

The expression for time constant $\tau$ :

  $\begin{array}{l}\tau =R_{eq}{C}_{eq}\\ \tau =\left(R_D𑨈R_L\right)C_L\mathrm{.........}\left(2\right)\end{array}$

Here, the expression for the time constant associated with the upper 3 Db frequency is:

  $\tau =\left(R_D𑨈R_L\right)C_L$

To determine

The time constant, upper 3 dB frequency and the small signal midband voltage gain.

Answer to Problem 7.41P

The small signal mid-band voltage gain is -4.7.

Explanation of Solution

Given:

The circuit parameter is given as:

  $\begin{array}{l}{R}_S=3.2\text{k}\Omega \\ R_D=10\text{k}\Omega \\ R_L=20\text{k}\Omega \\ C_L=10\text{pF}\end{array}$

The transistor parameter are given as:

  $\begin{array}{l}{V}_{TP}=-2\text{V}\\ K_P=0.25{\text{mA/V}}^2\\ \lambda =0\end{array}$

The equation for the time constant is given as:

  $\tau =\left(R_D𑨈R_L\right)C_L$

Substituting the known values in the above equation:

  $\begin{array}{l}\tau =\left(10\times {10}^3𑨈20\times {10}^3\right)\times 10\times {10}^{-12}\\ =\left(\frac{10\times 20\times {10}^{-12}}{10+20}\right)\times 10\times {10}^{-12}\\ =\left(\frac{10\times 20\times {10}^{-12}}{30}\right)\times 10\times {10}^{-12}\\ =6.67\times {10}^{-8}s\end{array}$

Hence, the value of time constant, $\tau =6.67\times {10}^{-8}\text{s}$ .

Now, evaluating the upper $\text{3dB}$ frequency $f_H$ :

  $f_H=\frac{1}{2\pi \tau }$

Substitute $6.67\times {10}^{-8}$ for $\tau$ .

  $\begin{array}{c}{f}_H=\frac{1}{2\pi \times 6.67\times {10}^{-8}}\\ =2.386\text{MHz}\end{array}$

Hence, the value of upper $\text{3dB}$ frequency is $\text{2}\text{.386MHz}$ .

Applying the Kirchhoff s voltage law to the outer loop:

  $\begin{array}{l}-5+I_D{R}_S+V_{sg}=0\\ -5+K_P{R}_s{\left(V_{sg}+V_{TP}\right)}^2+V_{sg}=0\\ K_P{R}_s{\left(V_{sg}+V_{TP}\right)}^2+V_{sg}=5\end{array}$

Substituting the known values:

  $\begin{array}{l}0.25\times {10}^{-3}\times 3.2\times {10}^3{\left(V_{sg}-2\right)}^2+V_{sg}=5\\ 0.8\left(V_{sg}{}^2-4V_{sg}+4\right)+V_{sg}=5\\ 0.8V_{sg}{}^2-2.2V_{sg}+3.2=5\\ 0.8V_{sg}{}^2-2.2V_{sg}-1.8=0\\ V_{sg}=3.14V\end{array}$

Evaluating the value of current $I_{DQ}$ :

  $I_{DQ}=K_P{\left(V_{sg}+V_{TP}\right)}^2$

Substituting the known values:

  $\begin{array}{l}{I}_{DQ}=0.25\times {10}^{-3}{\left(3.14-2\right)}^2\\ =0.497\text{mA}\end{array}$

Evaluating the transconductance $g_m$ :

  $g_m=2\sqrt{K_P{I}_{DQ}}$

Substituting the known values in the above equation:

  $\begin{array}{c}{g}_m=2\sqrt{0.25\times {10}^{-3}\times 0.497\times {10}^{-3}}\\ =0.705\text{mA/V}\end{array}$

Since, the voltage gain is given as:

  $A_v=\frac{-g_m\left(R_D𑨈R_L\right)}{\left[1+g_m\left(R_s𑨈\left(\frac{1}{s{C}_s}\right)\right)\right]\left(\frac{1}{1+s\left(R_D𑨈R_L\right)C_L}\right)}$

Evaluating the mid-band voltage gain $\left|A_v\right|$ :

  $\begin{array}{c}\left|A_v\right|=\frac{-g_m\left(R_D𑨈R_L\right)}{\left[1+g_m\left(R_s𑨈\left(\frac{1}{s\infty }\right)\right)\right]\left(\frac{1}{1+s\left(R_D𑨈R_L\right)\infty }\right)}\\ =-g_m\left(R_D𑨈R_L\right)\end{array}$

Substituting the known values in the above equation:

  $\begin{array}{c}\left|A_v\right|=0.705\times {10}^{-3}\left(10\times {10}^3𑨈20\times {10}^3\right)\\ =0.705\times {10}^{-3}\left(\frac{10\times 20\times {10}^3}{10+20}\right)\\ =-4.7\end{array}$

Hence, the small signal mid-band voltage gain is -4.7.