Chapter 11, Problem 11.70P

(a)

To determine

The value of the output resistance.

Answer to Problem 11.70P

The value of the output resistance $R_O=73.2\text{k}\Omega$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The expression for the input current is given by,

  $I_{\pi }=\frac{V_x-V_{\pi 4}}{r_{O2}}+g_{m2}{V}_{\pi 2}+g_m{V}_{\pi 4}+\frac{V_x}{r_{O4}}$ ....................(1)

The expression to determine the value of the transconductance is given by,

  $\begin{array}{l}{g}_{m2}{V}_{\pi 2}+\frac{V_x-V_{\pi 4}}{r_{O2}}=\frac{V_{\pi 4}}{r_{\pi 4}+r_{\pi 2}}\\ \frac{V_x-V_{\pi 4}}{r_{O2}}=\frac{V_{\pi 4}}{r_{\pi 4}+r_{\pi 2}}-g_{m2}{V}_{\pi 2}\end{array}$

The expression for the relation for the voltage for hybrid pie parameter.

  $V_{\pi 4}=-V_{\pi 2}$

Substitute $\frac{V_{\pi 4}}{r_{\pi 4}+r_{\pi 2}}-g_{m2}{V}_{\pi 2}$ for $\frac{V_x-V_{\pi 4}}{r_{O2}}$ in equation (1).

  $\begin{array}{l}{I}_{\pi }=\frac{V_{\pi 4}}{r_{\pi 4}\left|\right|r_{\pi 2}}-g_{m2}{V}_{\pi 2}+g_{m2}{V}_{\pi 2}+g_m{V}_{\pi 4}+\frac{V_x}{r_{O4}}\\ I_{\pi }=\frac{V_{\pi 4}}{r_{\pi 4}\left|\right|r_{\pi 2}}+g_m{V}_{\pi 4}+\frac{V_x}{r_{O4}}\\ \frac{V_x}{r_{O2}}=V_{\pi 4}\left[\frac{1}{r_{\pi 4}\left|\right|r_{\pi 2}}+\frac{1}{r_{O2}}+g_{m2}\right]\end{array}$ ....................(2)

The value of the collector current $I_{C4}$ is calculated as,

  $\begin{array}{c}{I}_{C4}=\left(\frac{\beta }{1+\beta }\right)\frac{I_Q}{2}\\ =\left(\frac{120}{120+0}\right)\frac{1\text{mA}}{2}\\ =0.496\text{mA}\end{array}$

The value of the collector current $I_{C2}$ is calculated as,

  $\begin{array}{c}{I}_{C2}=\left(\frac{\beta }{1+\beta }\right)\frac{I_Q}{2}\left(\frac{1}{1+\beta }\right)\\ =\left(\frac{120}{120+0}\right)\frac{1\text{mA}}{2}\left(\frac{1}{1+120}\right)\\ =0.0041\text{mA}\end{array}$

The expression to determine the expression for the diffusion resistance is given by,

  $r_{\pi 2}=\frac{\beta V_T}{I_{C2}}$

Substitute $0.026\text{V}$ for $V_T$ , $120$ for $\beta$ and $0.0041\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{\left(120\right)\left(0.026\text{V}\right)}{0.0041\text{mA}}\\ =761\text{k}\Omega \end{array}$

The value of the transconductance is calculated as,

  $g_{m2}=\frac{I_{C2}}{0.026\text{V}}$

Substitute $0.0041\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{0.0041\text{mA}}{0.026\text{V}}\\ =0.158\text{mA}\end{array}$

The expression to determine the value of the resistance $r_{O2}$ is given by,

  $r_{O2}=\frac{V_A}{I_{C2}}$

Substitute $100\text{V}$ for $V_A$ and $0.0041\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{r}_{O2}=\frac{100\text{V}}{0.0041\text{mA}}\\ =24.4\text{M}\Omega \end{array}$

The expression to determine the expression for the diffusion resistance is given by,

  $r_{\pi 4}=\frac{\beta V_T}{I_{C4}}$

Substitute $0.026\text{V}$ for $V_T$ , $120$ for $\beta$ and $0.496\text{mA}$ for $I_{C4}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 4}=\frac{\left(120\right)\left(0.026\text{V}\right)}{0.496\text{mA}}\\ =6.29\text{k}\Omega \end{array}$

The value of the transconductance is calculated as,

  $g_{m4}=\frac{I_{C4}}{0.026\text{V}}$

Substitute $0.496\text{mA}$ for $I_{C4}$ in the above equation.

  $\begin{array}{c}{g}_{m4}=\frac{0.496\text{mA}}{0.026\text{V}}\\ =19.08\text{mA}/\text{V}\end{array}$

The expression to determine the value of the resistance $r_{O4}$ is given by,

  $r_{O4}=\frac{V_A}{I_{C4}}$

Substitute $100\text{V}$ for $V_A$ and $0.496\text{mA}$ for $I_{C4}$ in the above equation.

  $\begin{array}{c}{r}_{O4}=\frac{100\text{V}}{0.496\text{mA}}\\ =202\text{k}\Omega \end{array}$

Substitute $24.4\text{M}\Omega$ for $r_{O2}$ , $202\text{k}\Omega$ for $r_{O4}$ , $6.29\text{k}\Omega$ for $r_{\pi 4}$ , $761\text{k}\Omega$ for $r_{\pi 2}$ and $0.158\text{mA}$ for $g_{m2}$ in equation (2)

  $\begin{array}{l}\frac{V_x}{202\text{k}\Omega }=V_{\pi 4}\left[\frac{1}{6.29\left|\right|r_{\pi 2}}+\frac{1}{24.4\text{M}\Omega }+0.158\text{mA}\right]\\ \frac{V_x}{202\text{k}\Omega }=\left[\frac{1}{6.29\left|\right|761\text{k}\Omega }+\frac{1}{24.4\text{M}\Omega }+0.158\text{mA}\right]\\ V_{\pi 4}=\left[\frac{V_x}{\left(0.318\right)r_{O2}}\right]\end{array}$

The expression for the input current $I_x$ is given by,

  $I_x=\frac{V_x}{r_{O1}}+\frac{V_x}{r_{O4}}+V_{\pi 4}\left(g_{m4}-g_{m2}-\frac{1}{r_{O2}}\right)$

Substitute $\frac{V_x}{0.318r_{O2}}$ for $V_{\pi 4}$ in the above equation.

  $I_x=\frac{V_x}{r_{O1}}+\frac{V_x}{r_{O4}}+\frac{V_x}{0.318r_{O2}}\left(g_{m4}-g_{m2}-\frac{1}{r_{O2}}\right)$

Substitute $24.4\text{M}\Omega$ for $r_{O2}$ , $202\text{k}\Omega$ for $r_{O4}$ , $19.08\text{mA}/\text{V}$ for $g_{m4}$ and $0.158\text{mA}$ for $g_{m2}$ in the above equation.

  $\begin{array}{c}{I}_x=\frac{V_x}{r_{O1}}+\frac{V_x}{r_{O4}}+\frac{V_x}{0.318r_{O2}}\left(19.08\text{mA}/\text{V}-0.158\pi -\frac{1}{24.4\text{M}\Omega }\right)\\ \frac{V_x}{I_x}=135\times 10\text{k}\Omega \end{array}$

The expression to determine the output resistance is given by,

  $R_{o2}=\frac{V_x}{I_x}$

Substitute $135\times 10\text{k}\Omega$ for $\frac{V_x}{I_x}$ in the above equation.

  $R_{o2}=\frac{V_x}{I_x}$

The value of the quiescent current is given by,

  $\begin{array}{c}{I}_C=\frac{I_Q}{2}\\ =\frac{1\text{mA}}{2}\\ =0.5\text{mA}\end{array}$

The small signal output resistance is calculated as,

  $\begin{array}{c}{r}_{o6}=\frac{\beta }{I_C}\\ =\frac{80}{0.5\text{mA}}\\ =160\text{k}\Omega \end{array}$

The value of the output resistance is calculated as,

  $\begin{array}{c}{R}_O=r_{O2}\left|\right|r_{O6}\\ =\frac{\left(135\text{k}\Omega \right)\left(160\text{k}\Omega \right)}{\left(135\text{k}\Omega +160\text{k}\Omega \right)}\\ =73.2\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the output resistance $R_O=73.2\text{k}\Omega$ .

(b)

To determine

The value of the differential mode voltage gain.

Answer to Problem 11.70P

The value of the differential voltage gain is $704$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The required small signal circuit is shown in Figure 2

  

Figure 2

The expression for the transconductance is given by,

  $g_m^c=\frac{{\Delta }_i}{\frac{v_d}{2}}$

The expression for the differential voltage gain is given by,

  $\begin{array}{c}{A}_d=g_m^c{R}_O\\ =\frac{{\Delta }_i}{\frac{v_d}{2}}{R}_O\end{array}$

The expression for the change in current is given by,

  ${\Delta }_i=g_{m1}{V}_{\pi 1}+g_{m3}{V}_{\pi 3}$ ....................(3)

Consider the value of $V_{\pi 3}$ is given by,

  $V_{\pi 1}+V_{\pi 3}=\frac{v_d}{2}$ ....................(4)

Substitute the values in the above equation.

  $\begin{array}{c}{V}_{\pi 3}=r_{\pi 3}\left(g_{m1}+\frac{1}{r_{\pi 1}}\right)V_{\pi 1}\\ =6.29\text{k}\Omega \left(\frac{120+1}{761\text{k}\Omega }\right)V_{\pi 1}\\ =V_{\pi 1}\end{array}$

Substitute $V_{\pi 1}$ for $V_{\pi 3}$ in equation (4).

  $\begin{array}{l}{V}_{\pi 1}+V_{\pi 1}=\frac{v_d}{2}\\ V_{\pi 1}=\frac{v_d}{4}\end{array}$

Substitute $V_{\pi 1}$ for $V_{\pi 3}$ in equation (3).

  ${\Delta }_i=g_{m1}{V}_{\pi 1}+g_{m3}{V}_{\pi 1}$

Substitute $19.08\text{mA}/\text{V}$ for $g_{m1}$ , $\frac{v_d}{4}$ for $V_{\pi 1}$ in the above equation.

  $\begin{array}{c}{\Delta }_i=\left(19.08\text{mA}/\text{V}\right)\frac{v_d}{4}+\left(19.08\text{mA}/\text{V}\right)\frac{v_d}{4}\\ =9.62\left(\frac{v_d}{2}\right)\end{array}$

Substitute $9.62\left(\frac{v_d}{2}\right)$ for ${\Delta }_i$ in the equation for transconductance.

  $\begin{array}{c}{g}_m^c=\frac{9.62\left(\frac{v_d}{2}\right)}{\frac{v_d}{2}}\\ =9.62\end{array}$

The value of gain is given by,

  $\begin{array}{c}{A}_d=\left(9.62\right)\left(73.2\text{k}\Omega \right)\\ =704\end{array}$

The expression for the input resistor is given by,

  $\begin{array}{c}{R}_i=r_{\pi 1}+\left(1+\beta \right)r_{\pi 3}\\ =761\text{k}\Omega +\left(1+120\right)\left(6.29\text{k}\Omega \right)\\ =1522\text{k}\Omega \end{array}$

The value of expression for the differential input resistor is calculated as,

  $\begin{array}{c}{R}_{id}=2R_i\\ =2\left(1522\text{k}\Omega \right)\\ =3.044\text{M}\Omega \end{array}$

Conclusion:

Therefore, the value of the differential voltage gain is $704$ .

(c)

To determine

The value of the differential mode input resistance $R_{id}$ .

Answer to Problem 11.70P

The value of $R_{id}=3.044\text{M}\Omega$ .

Explanation of Solution

Given:

The given diagram is shown below.

  

Calculation:

The expression for the input resistor is given by,

  $\begin{array}{c}{R}_i=r_{\pi 1}+\left(1+\beta \right)r_{\pi 3}\\ =761\text{k}\Omega +\left(1+120\right)\left(6.29\text{k}\Omega \right)\\ =1522\text{k}\Omega \end{array}$

The value of expression for the differential input resistor is calculated as,

  $\begin{array}{c}{R}_{id}=2R_i\\ =2\left(1522\text{k}\Omega \right)\\ =3.044\text{M}\Omega \end{array}$

Conclusion:

Therefore, the value of $R_{id}=3.044\text{M}\Omega$ .