Chapter 14, Problem 14.61P

a.

To determine

Maximum value of tolerance of resistor ‘ x’for given ${\text{CMRR}}_{\text{dB}}$ .

Answer to Problem 14.61P

Maximum value of tolerance of resistor is $x_{\text{max}}=0.474\%$

Explanation of Solution

Given:

The given difference amplifier circuit is,

  

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

Minimum ${\text{CMRR}}_{\text{dB}}$ is $50\text{dB}$ .

Calculation:

Circuit with voltage and resistance notation is,

  

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

  $\begin{array}{l}{\text{CMRR}}_{\text{dB}}=20{\log }_{10}\left(\text{CMRR}\right)\\ \frac{{\text{CMRR}}_{\text{dB}}}{20}={\log }_{10}\left(\text{CMRR}\right)\\ \text{CMRR}={10}^{\left(\frac{{\text{CMRR}}_{\text{dB}}}{20}\right)}\\ \text{CMRR}={10}^{\left(\frac{50}{10}\right)}\\ \text{CMRR}=316.228\end{array}$

KCL at $V_Y$ node,

  $\begin{array}{l}\frac{V_Y}{R_4}+\frac{V_Y-v_{l2}}{R_3}=0\\ \frac{R_3{V}_Y+R_4\left(V_Y-v_{l2}\right)}{R_4{R}_3}=0\\ R_3{V}_Y+R_4\left(V_Y-v_{l2}\right)=0\\ \left(R_3+R_4\right)V_Y-R_4{v}_{l2}=0\\ V_Y=\frac{R_4}{R_3+R_4}{v}_{l2}\end{array}$

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

  $\therefore V_X=\frac{R_4}{R_3+R_4}{v}_{l2}\mathrm{.......}\left(1\right)$

KCL at $V_X$ node;

  $\begin{array}{l}\frac{V_X-v_{l1}}{R_1}+\frac{V_X-v_0}{R_2}=0\\ \frac{R_2\left(V_X-v_{l1}\right)+R_1\left(V_X-v_0\right)}{R_1{R}_2}=0\\ R_2\left(V_X-v_{l1}\right)+R_1\left(V_X-v_0\right)=0\\ \left(R_1+R_2\right)V_X-R_2{v}_{l1}-R_1{v}_0=0\end{array}$

Now put $V_X$ from (1)

  $\begin{array}{l}\left(R_1+R_2\right)\left(\frac{R_4}{R_3+R_4}{v}_{l2}\right)-R_2{v}_{l1}-R_1{v}_0=0\\ R_1{v}_0=\left(R_1+R_2\right)\left(\frac{R_4}{R_3+R_4}\right)v_{l2}-R_2{v}_{l1}\\ v_0=\left(\frac{R_1+R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)v_{l2}-\frac{R_2}{R_1}{v}_{l1}\\ v_0=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)v_{l2}-\frac{R_2}{R_1}{v}_{l1}\end{array}$

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

  $\begin{array}{l}{v}_{l1}=v_{cm}-\frac{v_d}{2}\\ v_{l2}=v_{cm}+\frac{v_d}{2}\\ v_0=\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)\left(v_{cm}+\frac{v_d}{2}\right)-\frac{R_2}{R_1}\left(v_{cm}-\frac{v_d}{2}\right)\mathrm{.....}\left(2\right)\end{array}$

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,

  $\begin{array}{l}\text{CMRR=}\left|\frac{A_d}{A_{cm}}\right|\\ \text{CMRR=}\frac{\frac{1}{2}\left[\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)+\frac{R_2}{R_1}\right]}{\left(1+\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3+R_4}\right)-\frac{R_2}{R_1}}\\ \text{CMRR=}\frac{\frac{1}{2}\left[\frac{R_4}{R_3}\left(1+\frac{R_2}{R_1}\right)\frac{1}{\left(1+\frac{R_4}{R_3}\right)}+\frac{R_2}{R_1}\right]}{\frac{R_4}{R_3}\left(1+\frac{R_2}{R_1}\right)\frac{1}{\left(1+\frac{R_4}{R_3}\right)}-\frac{R_2}{R_1}}\mathrm{.........}\left(3\right)\end{array}$

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.

  $\begin{array}{l}{\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+x\%}{1-x\%}\right)\left(\frac{R_4}{R_3}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{1+\frac{x}{100}}{1-\frac{x}{100}}\right)\left(\frac{50\text{k}}{10\text{k}}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=\left(\frac{100+x}{100-x}\right)\left(\frac{50}{10}\right)\\ {\left(\frac{R_4}{R_3}\right)}_{\text{max}}=5\left(\frac{100+x}{100-x}\right)\end{array}$

and

  $\begin{array}{l}{\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+x\%}{1-x\%}\right)\left(\frac{R_2}{R_1}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{1+\frac{x}{100}}{1-\frac{x}{100}}\right)\left(\frac{50\text{k}}{10\text{k}}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=\left(\frac{100+x}{100-x}\right)\left(\frac{50}{10}\right)\\ {\left(\frac{R_2}{R_1}\right)}_{\text{min}}=5\left(\frac{100+x}{100-x}\right)\end{array}$

Therefore,

  $\text{CMRR=}\frac{\frac{1}{2}\left[5\left(\frac{100+x}{100-x}\right)\left(1+5\left(\frac{100-x}{100+x}\right)\right)\frac{1}{\left(1+5\left(\frac{100+x}{100-x}\right)\right)}+5\left(\frac{100-x}{100+x}\right)\right]}{5\left(\frac{100+x}{100-x}\right)\left(1+5\left(\frac{100-x}{100+x}\right)\right)\frac{1}{\left(1+5\left(\frac{100+x}{100-x}\right)\right)}-5\left(\frac{100-x}{100+x}\right)}$

  $\text{2CMRR=}\frac{5\left[\left(\frac{100+x}{100-x}\right)\left(1+5\left(\frac{100-x}{100+x}\right)\right)\frac{1}{\left(1+5\left(\frac{100+x}{100-x}\right)\right)}+\left(\frac{100-x}{100+x}\right)\right]}{5\left[\left(\frac{100+x}{100-x}\right)\left(1+5\left(\frac{100-x}{100+x}\right)\right)\frac{1}{\left(1+5\left(\frac{100+x}{100-x}\right)\right)}-\left(\frac{100-x}{100+x}\right)\right]}$

  $\text{2CMRR=}\frac{\left[\left(\frac{100+x}{100-x}\right)\left(\frac{600-4x}{100+x}\right)\frac{1}{\left(\frac{600+4x}{100-x}\right)}+\left(\frac{100-x}{100+x}\right)\right]}{\left[\left(\frac{100+x}{100-x}\right)\left(\frac{600-4x}{100+x}\right)\frac{1}{\left(\frac{600+4x}{100-x}\right)}-\left(\frac{100-x}{100+x}\right)\right]}$

  $\text{2CMRR=}\frac{\left[\left(\frac{600-4x}{600+4x}\right)+\left(\frac{100-x}{100+x}\right)\right]}{\left[\left(\frac{600-4x}{600+4x}\right)-\left(\frac{100-x}{100+x}\right)\right]}$

  $\text{2CMRR=}\frac{120000-8x^2}{400x}$

As x is very small. So neglect $x^2$

  $\begin{array}{l}\therefore \text{2CMRR=}\frac{120000}{400x}\\ x=\frac{120000}{400\times 2\text{CMRR}}\mathrm{.....}\left(4\right)\\ x=\frac{120000}{400\times 2\times 316.228}\\ \therefore x_{\text{max}}=0.474\%\end{array}$

b.

To determine

Maximum value of tolerance of resistor ‘ x’ for given ${\text{CMRR}}_{\text{dB}}$ .

Answer to Problem 14.61P

Maximum value of tolerance of resistor is $x_{\text{max}}=0.0267\%$

Explanation of Solution

Given:

The given difference amplifier circuit is;

  

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

Minimum ${\text{CMRR}}_{\text{dB}}$ is $75\text{dB}$ .

Calculation:

Minimum ${\text{CMRR}}_{\text{dB}}$ is $50\text{dB}$ .

  $\begin{array}{l}{\text{CMRR}}_{\text{dB}}=20{\log }_{10}\left(\text{CMRR}\right)\\ \frac{{\text{CMRR}}_{\text{dB}}}{20}={\log }_{10}\left(\text{CMRR}\right)\\ \text{CMRR}={10}^{\left(\frac{{\text{CMRR}}_{\text{dB}}}{20}\right)}\\ \text{CMRR}={10}^{\left(\frac{75}{10}\right)}\\ \text{CMRR}=5623.41\end{array}$

Using equation (4)

  $x=\frac{120000}{400\times 2\text{CMRR}}$

Put the value of $\text{CMRR}$ in above equation

  $\begin{array}{l}x=\frac{120000}{400\times 2\times 5623.41}\\ x=\frac{120000}{4498728}\\ \therefore x_{\text{max}}=0.0267\%\end{array}$