Chapter 14, Problem D14.47P

A.

To determine

Worst case output voltages $v_{01}$ and $v_{02}$ .

Answer to Problem D14.47P

Worst case output voltages are $v_{O1}=0.1\text{V}$ and $v_{O2}=-0.45\text{V}$

Explanation of Solution

Given:

The given circuit is:

  

  $I_{B1}=I_{B2}=1\mu \text{A}$ and $v_I=0$

As $v_I=0$ , the modified circuit is;

  

For first op-amp $V_{Y1}=0$

Also $V_{X1}=V_{Y1}=0$ due to virtual ground.

Now, KCL at input node

  $\begin{array}{l}\frac{V_{X1}}{10\text{k}}+\frac{V_{X1}-v_{O1}}{100\text{k}}+I_{B1}=0\\ \because V_{X1}=0\\ \therefore \frac{0}{10\text{k}}+\frac{0-v_{O1}}{100\text{k}}+I_{B1}=0\\ v_{O1}=I_{B1}\times 100\text{k}\\ v_{O1}\text{=1}\mu \times 100\text{k}\\ \text{=1}\times {\text{10}}^{-6}\times 100\times {10}^3\\ \Rightarrow v_{O1}=0.1\text{V}\end{array}$

For second op-amp $V_{Y2}=0$

Also $V_{X2}=V_{Y2}=0$ due to virtual ground.

Now, KCL at input node

  $\begin{array}{l}\frac{V_{X2}-v_{O1}}{10\text{k}}+\frac{V_{X2}-v_{O2}}{50\text{k}}+I_{B1}=0\\ \because V_{X2}=0\\ \therefore \frac{0-v_{O1}}{10\text{k}}+\frac{0-v_{O2}}{50\text{k}}+I_{B1}=0\\ \frac{-5v_{O1}-v_{O2}+50\text{k}{I}_{B1}}{50\text{k}}=0\\ -5v_{O1}-v_{O2}+50\text{k}{I}_{B1}=0\\ v_{O2}=-5v_{O1}+50\text{k}{I}_{B1}\\ v_{O2}\text{=-5}\left(0.1\right)+50\times {10}^3\times 1\times {10}^{-6}\\ \Rightarrow v_{O2}=-0.45\text{V}\end{array}$

B.

To determine

To design:

Input bias current compensation circuit.

Explanation of Solution

Given:

  $v_{O1}=v_{O2}=0$ and $v_I=0$

A simple compensation circuit can minimize the effect of bias currents in op-amp.

Consider the sketch ofinput bias current compensation circuit as,

  

Now, determine $v_{O1}\text{and}{v}_{O2}$ as a function of $I_{B1}\text{and}{I}_{B2}$ .

For first op-amp:

KCL at non-inverting node,

  $\begin{array}{l}\frac{V_{Y1}}{R_A}+I_{B2}=0\\ V_{Y1}=-I_{B2}{R}_A\end{array}$

As, $V_{X1}=V_{Y1}$

Therefore, $V_{X1}=-I_{B2}{R}_A\mathrm{.....}\left(1\right)$

KCL at inverting node,

  $\begin{array}{l}\frac{V_{X1}}{10\text{k}}+\frac{V_{X1}-v_{O1}}{100\text{k}}+I_{B1}=0\\ V_{X1}\left(\frac{1}{10\text{k}}+\frac{1}{100\text{k}}\right)-\frac{v_{O1}}{100\text{k}}+I_{B1}=0\\ \frac{v_{O1}}{100\text{k}}=V_{X1}\left(\frac{10+1}{100\text{k}}\right)+I_{B1}\\ v_{O1}=11V_{X1}+I_{B1}\times 100\text{k}\end{array}$

Putting $V_{X1}$ from (1)

  $v_{O1}=11\left(-I_{B2}{R}_A\right)+I_{B1}\times 100\text{k}\mathrm{.....}\left(2\right)$

Now,

If $I_{B!}=I_{B2}=I_B$ and the three resistance are adjusted to give $v_{O1}=0$ , then from (2)

  $\begin{array}{l}0=11\left(-I_B{R}_A\right)+I_B\times 100\text{k}\\ -11R_A+100\text{k}=0\\ R_A=\frac{100\text{k}}{11}\\ \Rightarrow R_A=9.09\text{k}\Omega \end{array}$

For second op-amp:

KCL at non-inverting node,

  $\begin{array}{l}\frac{V_{Y2}}{R_B}+I_{B2}=0\\ V_{Y2}=-I_{B2}{R}_B\end{array}$

As, $V_{X2}=V_{Y2}$

Therefore, $V_{X2}=-I_{B2}{R}_B\mathrm{.....}\left(3\right)$

KCL at inverting node,

  $\begin{array}{l}\frac{V_{X2}-v_{O1}}{10\text{k}}+\frac{V_{X2}-v_{O2}}{50\text{k}}+I_{B1}=0\\ V_{X2}\left(\frac{1}{10\text{k}}+\frac{1}{50\text{k}}\right)-\frac{v_{O1}}{10\text{k}}-\frac{v_{O2}}{50\text{k}}+I_{B1}=0\\ V_{X2}\left(\frac{5+1}{50\text{k}}\right)-\frac{v_{O1}}{10\text{k}}-\frac{v_{O2}}{50\text{k}}+I_{B1}=0\\ \frac{6V_{X2}-5v_{O1}-v_{O2}+50\text{k}\times I_{B1}}{50\text{k}}=0\\ 6V_{X2}-5v_{O1}-v_{O2}+50\text{k}\times I_{B1}=0\\ v_{O2}=6V_{X2}-5v_{O1}+50\text{k}\times I_{B1}\end{array}$

Putting $V_{X2}$ from (3)

  $\begin{array}{l}{v}_{O2}=6\left(-I_{B2}{R}_B\right)-5v_{O1}+50\text{k}\times I_{B1}\\ v_{O2}+5v_{O1}=-6I_{B2}{R}_B+50\text{k}\times I_{B1}\mathrm{.....}\left(4\right)\end{array}$

Now,

If $I_{B!}=I_{B2}=I_B$ and the three resistance are adjusted to give $v_{O1}=v_{O2}=0$ , then from (4)

  $\begin{array}{l}0+5\left(0\right)=-6I_B{R}_B+50\text{k}\times I_B\\ 0=I_B\left(-6R_B+50\text{k}\right)\\ R_B=\frac{50\text{k}}{6}\\ \Rightarrow R_B=8.33\text{k}\Omega \end{array}$