Chapter 7, Problem 7.23P

Consider the circuit shown in Figure P7.23. (a) Write the transfer function $T\left(s\right)=V_o\left(s\right)/V_i\left(s\right)$ . Assume $\lambda =0$ for the transistor. (b) Determine the expression for the time constant associated with the input portion of the circuit. (c) Determine the expression for the time constant associated with theoutput portion of the circuit.

Figure P7.23

To determine

The transfer function $T\left(s\right)=V_o\left(s\right)/V_i\left(s\right)$ .

Answer to Problem 7.23P

The expression for transfer function of the circuit is,

  $T\left(s\right)=\left(\frac{g_m}{1+g_m{R}_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\left[\frac{s{R}_D{R}_L{C}_C}{s\left(R_D+R_L\right)C_C+1}\right]$ .

Explanation of Solution

Given:

The circuit is given as:

  

Calculation:

Refer to the given circuit:

Drawing the small signal equivalent circuit using the simplified hybrid $\pi$ model:

  

Now drawing the frequency domain equivalent circuit of above figure:

  

Applying the nodal analysis to derive the expression for $\frac{V_{gs}\left(s\right)}{V_i\left(s\right)}$ associated with the input portion of the circuit:

  $\begin{array}{c}\frac{-V_{gs}\left(s\right)-V_i\left(s\right)}{R_S}+\frac{-V_{gs}\left(s\right)}{\frac{1}{s{C}_i}}-g_m{V}_{gs}\left(s\right)=0\\ -V_{gs}\left(s\right)\left[\frac{1}{R_S}+s{C}_i+g_m\right]-\frac{V_i\left(s\right)}{R_s}=0\\ \frac{V_{gs}\left(s\right)}{V_i\left(s\right)}=\frac{1}{R_S\left[\frac{1}{R_S}+s{C}_i+g_m\right]}\\ =\frac{-1}{1+R_s\left(g_m+s{C}_i\right)}\\ =\frac{-\frac{1}{g_m}}{\frac{1}{g_m}+R_s\frac{1}{g_m}\left(g_m+s{C}_i\right)}\\ \frac{V_{gs}\left(s\right)}{V_i\left(s\right)}=\frac{-\frac{1}{g_m}}{\frac{1}{g_m}+R_s+s\frac{1}{g_m}{R}_s{C}_i}\mathrm{............}\left(1\right)\end{array}$

Solving the above equation:

  $\begin{array}{c}\frac{V_{gs}\left(s\right)}{V_i\left(s\right)}=\left(\frac{-\left(\frac{1}{g_m}\right)}{\frac{1}{g_m}+R_s}\right)\frac{1}{\left(1+s\frac{\left(\frac{1}{g_m}\right)R_s}{\frac{1}{g_m}+R_s}{C}_i\right)}\\ =\left(\frac{-\left(\frac{1}{g_m}\right)}{\frac{1}{g_m}+R_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\end{array}$

Applying the Kirchhoff s current law at node $V_1\left(s\right)$ :

  $\begin{array}{l}{g}_m{V}_{gs}\left(s\right)+\frac{V_1\left(s\right)-V_0\left(s\right)}{\frac{1}{s{C}_C}}+\frac{V_1\left(s\right)}{R_D}=0\\ g_m{V}_{gs}\left(s\right)+\left[\frac{1}{R_D}+s{C}_C\right]V_1\left(s\right)-s{C}_C{V}_0\left(s\right)=\mathrm{0..........}\left(2\right)\end{array}$

Applying the Kirchhoff s current law at output node:

  $\begin{array}{l}\frac{V_0\left(s\right)-V_1\left(s\right)}{\frac{1}{s{C}_C}}+\frac{V_0\left(s\right)}{R_L}=0\\ V_0\left(s\right)\left[s{C}_C+\frac{1}{R_L}\right]=s{C}_C{V}_1\left(s\right)\\ V_1\left(s\right)=V_0\left(s\right)\frac{1}{s{C}_C}\left[s{C}_C+\frac{1}{R_L}\right]\\ V_1\left(s\right)=\left[1+\frac{1}{s{C}_C{R}_L}\right]V_0\left(s\right)\\ \end{array}$

Evaluating the expression for $\frac{V_0\left(s\right)}{V_{gs}\left(s\right)}$ associated with the output portion of the circuit:

Substituting $\left[1+\frac{1}{s{C}_C{R}_L}\right]V_0\left(s\right)$ for $V_1\left(s\right)$ in equation (2):

  $\begin{array}{c}{g}_m{V}_{gs}\left(s\right)+\left[\frac{1}{R_D}+s{C}_C\right]\left[1+\frac{1}{s{C}_C{R}_L}\right]V_0\left(s\right)-s{C}_C{V}_0\left(s\right)=0\\ g_m{V}_{gs}\left(s\right)+\left[\frac{1}{R_D}+\frac{1}{s{C}_C{R}_D{R}_L}+s{C}_C+\frac{1}{R_L}\right]V_0\left(s\right)-s{C}_C{V}_0\left(s\right)=0\\ g_m{V}_{gs}\left(s\right)+\left[\frac{1}{R_D}+\frac{1}{s{C}_C{R}_D{R}_L}+\frac{1}{R_L}\right]V_0\left(s\right)=0\\ \frac{1}{R_D{R}_L}\left[R_L+\frac{1}{s{C}_C}+R_D\right]V_0\left(s\right)=-g_m{V}_{gs}\left(s\right)\\ V_0\left(s\right)=\frac{-g_m{V}_{gs}\left(s\right)R_D{R}_L}{\left[R_L+\frac{1}{s{C}_C}+R_D\right]}\\ \frac{V_0\left(s\right)}{V_{gs}\left(s\right)}=-\frac{g_m{R}_L{R}_D}{R_D+R_L+\frac{1}{s{C}_C}}\end{array}$

The expression for transfer function of the circuit is given as:

  $\frac{V_0\left(s\right)}{V_i\left(s\right)}=\frac{V_0\left(s\right)}{V_{gs}\left(s\right)}\times \frac{V_{gs}\left(s\right)}{V_i\left(s\right)}$

  $\begin{array}{c}=-\left[\frac{g_m{R}_L{R}_D}{R_D+R_L+\frac{1}{s{C}_C}}\right]\left(\frac{-\frac{1}{g_m}}{\frac{1}{g_m}+R_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\\ =\left(\frac{1}{\frac{1}{g_m}+R_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\left[\frac{s{R}_D{R}_L{C}_C}{s\left(R_D+R_L\right)C_C+1}\right]\\ \frac{V_0\left(s\right)}{V_i\left(s\right)}=\left(\frac{g_m}{1+g_m{R}_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\left[\frac{s{R}_D{R}_L{C}_C}{s\left(R_D+R_L\right)C_C+1}\right]\frac{\left(R_D+R_L\right)}{\left(R_D+R_L\right)}\end{array}$

Solving the above equation:

  $\begin{array}{c}\frac{V_0\left(s\right)}{V_i\left(s\right)}=\left(\frac{g_m}{1+g_m{R}_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\left[\frac{s{R}_D{R}_L{C}_C}{s\left(R_D+R_L\right)C_C+1}\right]\frac{\left(R_D+R_L\right)}{\left(R_D+R_L\right)}\\ T\left(s\right)=\left(\frac{g_m}{1+g_m{R}_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\left[\frac{s{R}_D{R}_L{C}_C}{s\left(R_D+R_L\right)C_C+1}\right]\left(R_D\parallel R_L\right)\\ =\left(\frac{g_m\left(R_D\parallel R_L\right)}{1+g_m{R}_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\left[\frac{s\left(R_D+R_L\right)C_C}{1+s\left(R_D+R_L\right)C_C}\right]\\ =\left(\frac{g_m\left(R_D\parallel R_L\right)}{1+g_m{R}_s}\right)\frac{1}{\left[1+s{\tau }_i\right]}\left[\frac{s{\tau }_c}{1+s{\tau }_c}\right]\end{array}$

So,

  $\begin{array}{l}{\tau }_c=\left(R_D+R_L\right)C_{C}and\\ {\tau }_i=\left(\frac{1}{g_m}\parallel R_s\right)C_i\mathrm{....................}\left(3\right)\end{array}$

Therefore, the expression for transfer function of the circuit is,

  $T\left(s\right)=\left(\frac{g_m}{1+g_m{R}_s}\right)\frac{1}{\left[1+s\left(\frac{1}{g_m}\parallel R_s\right)C_i\right]}\left[\frac{s{R}_D{R}_L{C}_C}{s\left(R_D+R_L\right)C_C+1}\right]$ .

To determine

The expression for the time constant associating with the input portion of the circuit.

Answer to Problem 7.23P

The expression of the time constant associated with the input portion of the circuit:

  ${\tau }_i$ is $\left(\frac{1}{g_m}\parallel R_s\right)C_i$

Explanation of Solution

Given:

The circuit is given as:

  

Calculation:

Evaluating the expression for the time constant associated with the input portion of the circuit.

From the equation (3) , the time constant associated with the input portion of the circuit is,

  ${\tau }_i=\left(\frac{1}{g_m}\parallel R_s\right)C_i$

Hence, the expression of the time constant associated with the input portion of the circuit . ${\tau }_i$ is $\left(\frac{1}{g_m}\parallel R_s\right)C_i$ .

To determine

The expression for the time constant associating with the output portion of the circuit.

Answer to Problem 7.23P

The expression of the time constant associated with the output portion of the circuit:

  ${\tau }_c=\left(R_D+R_L\right)C_C$

Explanation of Solution

Given:

The circuit is given as:

  

Calculation:

Evaluating the expression for the time constant associated with the input portion of the circuit.

From equation (3), the time constant associated with the input portion of the circuit is expressed as:

  ${\tau }_c=\left(R_D+R_L\right)C_C$

Therefore, the expression for the time constant associated with the output portion of the circuit: ${\tau }_c=\left(R_D+R_L\right)C_C$ .