Chapter 9, Problem 9.65P

The circuit in Figure P9.65 is a representation of the common-mode and differential-input signals to a difference amplifier. The output voltage can bewritten as $v_O=A_d{v}_d+A_{cm}{v}_{cm}$ where $A_d$ is the differential-mode gain and $A_{cm}$ is the common-mode gain.

(a) Setting $v_d=0$ , show that the common-mode gain is given by
$A_{cm}=\frac{\left(\frac{R_4}{R_3}-\frac{R_2}{R_1}\right)}{\left(1+R_4/R_3\right)}$

(b) Determine $A_{cm}$ if $R_1=10.4k\Omega ,R_2=62.4k\Omega ,R_3=9.6k\Omega ,$ and $R_4=86.4k\Omega$ . (c) Determine the maximum value of $\left|A_{cm}\right|$ if $R_1=20k\Omega \pm 1\%,R_2=80k\Omega \pm 1\%,R_3=20k\Omega \pm 1\%,$ and $R_4=80k\Omega \pm 1\%$ .

To determine

To show: The expression for the common mode gain is $A_{CM}=\frac{\frac{R_4}{R_3}-\frac{R_2}{R_1}}{1+\frac{R_4}{R_3}}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  

The expression for the voltage $v_1$ is given by

  $v_1=v_2$

The expression for the value of the voltage $v_{I1}$ is given by

  $v_{I1}=v_{CM}+\frac{v_d}{2}$

The expression for the value of the voltage $v_{I2}$ is given by

  $v_{I2}=v_{CM}-\frac{v_d}{2}$

The expression for the voltage $v_2$ by voltage division rule is given by

  $v_2=\left(\frac{R_4}{R_3+R_4}\right)v_{I2}$

Apply KCL at the inverting terminal.

  $\begin{array}{l}\frac{v_{I1}-v_1}{R_1}=\frac{v_{I1}-v_O}{R_2}\\ v_O=\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)v_{I2}\right)-\left(\frac{R_2}{R_1}\right)v_{I1}\end{array}$

Substitute $v_{CM}+\frac{v_d}{2}$ for $v_{I1}$ and $v_{CM}+\frac{v_d}{2}$ for $v_{I1}$ in the above equation.

  $\begin{array}{c}{v}_O=\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)\left(v_{CM}-\frac{v_d}{2}\right)\right)-\left(\frac{R_2}{R_1}\right)\left(v_{CM}+\frac{v_d}{2}\right)\\ =\left[\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\right)\right]v_{CM}-\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)-\left(\frac{R_2}{R_1}\right)\right]\frac{v_d}{2}\end{array}$

Substitute $0$ for $v_d$ in the above equation.

  $\begin{array}{c}{v}_O=\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)\left(v_{CM}-\frac{v_d}{2}\right)\right)-\left(\frac{R_2}{R_1}\right)\left(v_{CM}+\frac{v_d}{2}\right)\\ =\left[\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\right)\right]v_{CM}-\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)-\left(\frac{R_2}{R_1}\right)\right]\frac{0}{2}\\ =\left[\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\right)\right]v_{CM}\end{array}$

The expression for the output voltage is given by

  $v_O=A_d{v}_d+A_{CM}{v}_{CM}$

Substitute $0$ for $v_d$ in the above equation.

  $\begin{array}{l}{v}_O=A_d\left(0\right)+A_{CM}{v}_{CM}\\ A_{CM}=\frac{v_O}{v_{CM}}\end{array}$

Substitute $\left[\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\right)\right]$ for $v_O$ in the above equation.

  $\begin{array}{c}{A}_{CM}=\frac{\left[\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\right)\right]v_{CM}}{v_{CM}}\\ =\left[\left(1+\frac{R_2}{R_1}\right)\left(\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\right)\right]\\ =\frac{R_1{R}_3\left(\frac{R_4}{R_3}-\frac{R_2}{R_1}\right)}{R_1{R}_3\left(1+\frac{R_4}{R_3}\right)}\\ =\frac{\frac{R_4}{R_3}-\frac{R_2}{R_1}}{1+\frac{R_4}{R_3}}\end{array}$

Conclusion:

Therefore, theexpression for the common mode gain is $A_{CM}=\frac{\frac{R_4}{R_3}-\frac{R_2}{R_1}}{1+\frac{R_4}{R_3}}$ .

To determine

The value of common mode voltage gain.

Answer to Problem 9.65P

The value of the common mode voltage gain is $0.3$ .

Explanation of Solution

Calculation:

The expression for the value of common voltage gain is given by

  $A_{CM}=\frac{\frac{R_4}{R_3}-\frac{R_2}{R_1}}{1+\frac{R_4}{R_3}}$

Substitute $10.4\text{k}\Omega$ for $R_1$ , $62.4\text{k}\Omega$ for $R_2$ , $9.6\text{k}\Omega$ for $R_3$ for $86.4\text{k}\Omega$ for $R_4$ in the above equation.

  $\begin{array}{c}{A}_{CM}=\frac{\frac{86.4\text{k}\Omega }{9.6\text{k}\Omega }-\frac{62.4\text{k}\Omega }{10.4\text{k}\Omega }}{1+\frac{86.4\text{k}\Omega }{9.6\text{k}\Omega }}\\ =0.3\end{array}$

Conclusion:

Therefore, the value of the common mode voltage gain is $0.3$ .

To determine

The value of common mode voltage gain.

Answer to Problem 9.65P

The value of the common mode voltage gain is $0.0315$ .

Explanation of Solution

Calculation:

The expression for the value of common voltage gain is given by

  $A_{CM}=\frac{\frac{R_4}{R_3}\left(1-\frac{R_2}{R_1}\frac{R_4}{R_3}\right)}{\frac{R_4}{R_3}\left(\frac{R_3}{R_4}+1\right)}$

For $A_{CM}$ to be maximum, the denominator must be maximum and $\frac{R_2}{R_1}\frac{R_4}{R_3}$ must be minimum.

Substitute $\left(20+1\%\right)\text{k}\Omega$ for $R_1$ , $\left(80-1\%\right)\text{k}\Omega$ for $R_2$ , $\left(20-1\%\right)\text{k}\Omega$ for $R_3$ and $\left(80+1\%\right)\text{k}\Omega$ for $R_4$ in the above equation.

  $\begin{array}{c}{A}_{CM}=\frac{\frac{\left(80+1\%\right)\text{k}\Omega }{R_3}\left(1-\frac{\left(\left(80-1\%\right)\text{k}\Omega \right)}{\left(\left(20+1\%\right)\text{k}\Omega \right)}\frac{\left(80+1\%\right)\text{k}\Omega }{\left(20-1\%\right)\text{k}\Omega }\right)}{\frac{\left(80+1\%\right)\text{k}\Omega }{\left(20-1\%\right)\text{k}\Omega }\left(\frac{R_3}{\left(80+1\%\right)\text{k}\Omega }+1\right)}\\ =\frac{1}{0.245+1}\left[1-\left(3.921\right)\left(0.245\right)\right]\\ =0.0315\end{array}$

Conclusion:

Therefore, the value of the common mode voltage gain is $0.0315$ .