Chapter 14, Problem 14.60P

The op−amp in the difference amplifier configuration in Figure P14.60 isideal. (a) If the tolerance of each resistor is ± 1.5%, determine the minimum value of ${\text{CMRR}}_{\text{dB}}$ . (b) Repeat part (a) if the tolerance of each resistor is±3%.

Figure P14.60

To determine

Minimum value of ${\text{CMRR}}_{\text{dB}}$ .

Answer to Problem 14.60P

  ${\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=39.96\text{dB}$

Explanation of Solution

Given:

The given difference amplifier circuit is shown below.

  

Tolerance of each resistor is $\pm \text{1}\text{.5\%}$ .

Calculation:

Circuit with voltage and resistance notation is,

  

KCL at $V_Y$ node,

  $\begin{array}{l}\frac{V_Y}{R_4}+\frac{V_Y-v_{I2}}{R_3}=0\\ \frac{R_3{V}_Y+R_4\left(V_Y-v_{I2}\right)}{R_4{R}_3}=0\\ R_3{V}_Y+R_4\left(V_Y-v_{I2}\right)=0\\ \left(R_3+R_4\right)V_Y-R_4{v}_{I2}=0\\ V_Y=\frac{R_4}{R_3+R_4}{v}_{I2}\end{array}$

As op-amp is ideal, so $V_X=V_Y$

  $\therefore V_X=\frac{R_4}{R_3+R_4}{v}_{I2}\mathrm{.......}\left(1\right)$

KCL at $V_X$ node,

  $\begin{array}{l}\frac{V_X-v_{I1}}{R_1}+\frac{V_X-v_0}{R_2}=0\\ \frac{R_2\left(V_X-v_{I1}\right)+R_1\left(V_X-v_0\right)}{R_1{R}_2}=0\\ R_2\left(V_X-v_{I1}\right)+R_1\left(V_X-v_0\right)=0\\ \left(R_1+R_2\right)V_X-R_2{v}_{I1}-R_1{v}_0=0\end{array}$

Now put $V_X$ from (1)

  $\begin{array}{l}\left(R_1+R_2\right)\left(\frac{R_4}{R_3+R_4}{v}_{I2}\right)-R_2{v}_{I1}-R_1{v}_0=0\\ R_1{v}_0=\left(R_1+R_2\right)\left(\frac{R_4}{R_3+R_4}\right)v_{I2}-R_2{v}_{I1}\\ v_0=\left(\frac{R_1+R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)v_{I2}-\frac{R_2}{R_1}{v}_{I1}\\ v_0=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)v_{I2}-\frac{R_2}{R_1}{v}_{I1}\end{array}$

Now,

If $v_d$ is differential mode input voltage and $v_{cm}$ is common-mode input voltage. Then,

  $\begin{array}{l}{v}_{I1}=v_{cm}-\frac{v_d}{2}\\ v_{I2}=v_{cm}+\frac{v_d}{2}\\ v_0=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)\left(v_{cm}+\frac{v_d}{2}\right)-\frac{R_2}{R_1}\left(v_{cm}-\frac{v_d}{2}\right)\mathrm{.....}\left(2\right)\end{array}$

For $v_d=0$ and $A_{cm}=\frac{v_0}{v_{cm}}$ equation (2) results,

  $\begin{array}{l}{v}_0=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)\left(v_{cm}+\frac{0}{2}\right)-\frac{R_2}{R_1}\left(v_{cm}-\frac{0}{2}\right)\\ v_0=\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\right]v_{cm}\\ \frac{v_0}{v_{cm}}=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\\ \therefore A_{cm}=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}\end{array}$

For $v_{cm}=0$ and $A_d=\frac{v_0}{v_d}$ equation (2) results,

  $\begin{array}{l}{v}_0=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)\left(0+\frac{v_d}{2}\right)-\frac{R_2}{R_1}\left(0-\frac{v_d}{2}\right)\\ v_0=\frac{1}{2}\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)+\frac{R_2}{R_1}\right]v_d\\ \frac{v_0}{v_d}=\frac{1}{2}\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)+\frac{R_2}{R_1}\right]\\ \therefore A_d=\frac{1}{2}\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)+\frac{R_2}{R_1}\right]\end{array}$

Now,

  $\begin{array}{l}\text{CMRR=}\left|\frac{A_d}{A_{cm}}\right|\\ \text{CMRR=}\frac{\frac{1}{2}\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)+\frac{R_2}{R_1}\right]}{\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}}\\ \text{CMRR=}\frac{\frac{1}{2}\left[\frac{R_4}{R_3}\left(1+\frac{R_2}{R_1}\right)\frac{1}{\left(1+\frac{R_4}{R_3}\right)}+\frac{R_2}{R_1}\right]}{\frac{R_4}{R_3}\left(1+\frac{R_2}{R_1}\right)\frac{1}{\left(1+\frac{R_4}{R_3}\right)}-\frac{R_2}{R_1}}\end{array}$

For minimum CMRR maximize the denominator i.e. $\frac{R_4}{R_3}$ will be maximum and $\frac{R_2}{R_1}$ will be minimum.

  $\begin{array}{l}{\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+1.5\%}{1-1.5\%}\right)\left(\frac{R_4}{R_3}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+\frac{1.5}{100}}{1-\frac{1.5}{100}}\right)\left(\frac{50\text{k}}{10\text{k}}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+0.015}{1-0.015}\right)\left(\frac{50}{10}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=5.152\end{array}$

and

  $\begin{array}{l}{\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+1.5\%}{1-1.5\%}\right)\left(\frac{R_2}{R_1}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+\frac{1.5}{100}}{1-\frac{1.5}{100}}\right)\left(\frac{50\text{k}}{10\text{k}}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+0.015}{1-0.015}\right)\left(\frac{50}{10}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=4.852\end{array}$

Therefore,

  $\begin{array}{l}{\text{CMRR}}_{\text{min}}\text{=}\frac{\frac{1}{2}\left[{\left(\frac{R_4}{R_3}\right)}_{\text{max}}\left(1+{\left(\frac{R_2}{R_1}\right)}_{\text{min}}\right)\frac{1}{\left(1+{\left(\frac{R_4}{R_3}\right)}_{\text{max}}\right)}+{\left(\frac{R_2}{R_1}\right)}_{\text{min}}\right]}{{\left(\frac{R_4}{R_3}\right)}_{\text{max}}\left(1+{\left(\frac{R_2}{R_1}\right)}_{\text{min}}\right)\frac{1}{\left(1+{\left(\frac{R_4}{R_3}\right)}_{\text{max}}\right)}-{\left(\frac{R_2}{R_1}\right)}_{\text{min}}}\mathrm{.....}(3)\\ {\text{CMRR}}_{\text{min}}\text{=}\frac{\frac{1}{2}\left[5.152(1+4.852)\frac{1}{1+5.152}+4.852\right]}{5.152(1+4.852)\frac{1}{1+5.152}-4.852}\\ {\text{CMRR}}_{\text{min}}\text{=}\frac{\frac{1}{2}\left[4.901+4.852\right]}{4.901-4.852}\\ {\text{CMRR}}_{\text{min}}=99.52\end{array}$

So, minimum value of ${\text{CMRR}}_{\text{dB}}$ is

  $\begin{array}{l}{\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=20{\log }_{10}\left(99.52\right)\\ {\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=20\times 1.998\\ \therefore {\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=39.96\text{dB}\end{array}$

To determine

Minimum value of ${\text{CMRR}}_{\text{dB}}$ .

Answer to Problem 14.60P

  ${\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=34\text{dB}$

Explanation of Solution

Given:

The given difference amplifier circuit is shown below.

  

Tolerance of each resistor is $\pm 3\text{\%}$ .

Calculation:

For tolerance of resistor $\pm 3\%$

As,

  $\text{CMRR=}\frac{\frac{1}{2}\left[\frac{R_4}{R_3}\left(1+\frac{R_2}{R_1}\right)\frac{1}{\left(1+\frac{R_4}{R_3}\right)}+\frac{R_2}{R_1}\right]}{\frac{R_4}{R_3}\left(1+\frac{R_2}{R_1}\right)\frac{1}{\left(1+\frac{R_4}{R_3}\right)}-\frac{R_2}{R_1}}$

For minimum CMRR maximize the denominator that is, $\frac{R_4}{R_3}$ will be maximum and $\frac{R_2}{R_1}$ will be minimum.

  $\begin{array}{l}{\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+3\%}{1-3\%}\right)\left(\frac{R_4}{R_3}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+\frac{3}{100}}{1-\frac{3}{100}}\right)\left(\frac{50\text{k}}{10\text{k}}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+0.03}{1-0.03}\right)\left(\frac{50}{10}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=5.309\end{array}$

and

  $\begin{array}{l}{\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+3\%}{1-3\%}\right)\left(\frac{R_2}{R_1}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+\frac{3}{100}}{1-\frac{3}{100}}\right)\left(\frac{50\text{k}}{10\text{k}}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+0.03}{1-0.03}\right)\left(\frac{50}{10}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=4.709\end{array}$

Putting the value in equation (3),

  $\begin{array}{l}{\text{CMRR}}_{\text{min}}\text{=}\frac{\frac{1}{2}\left[5.309(1+4.709)\frac{1}{1+5.309}+4.709\right]}{5.309(1+4.709)\frac{1}{1+5.309}-4.709}\\ {\text{CMRR}}_{\text{min}}\text{=}\frac{\frac{1}{2}\left[4.804+4.709\right]}{4.804-4.709}\\ {\text{CMRR}}_{\text{min}}=50.07\end{array}$

So, minimum value of ${\text{CMRR}}_{\text{dB}}$ is

  $\begin{array}{l}{\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=20{\log }_{10}\left(50.07\right)\\ {\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=20\times 1.7\\ \therefore {\text{CMRR}}_{\text{dB}}\left(\text{min}\right)=34\text{dB}\end{array}$