Chapter 11, Problem D11.71P

To determine

The design parameters for the BJT differential amplifier with the active load and the mode voltage gain.

Answer to Problem D11.71P

The value of the differential voltage gain is $704$ , and the value of the resistances to design the circuit is $R_{id}=3.044\text{M}\Omega$ and $R_O=73.2\text{k}\Omega$ .

Explanation of Solution

Given:

The given circuit 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,

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

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 $704$ , and the value of the resistances to design the circuit is $R_{id}=3.044\text{M}\Omega$ and $R_O=73.2\text{k}\Omega$ .