Chapter 14, Problem 14.50P

A.

To determine

Output voltage due to input bias current and worst case output voltage.

Answer to Problem 14.50P

Output voltage due to input bias current is

  $v_O=0$ and worst case output voltage including the effect of input offset current is

  $v_O=-0.01\text{V}$

Explanation of Solution

Given:

The given circuit is:

  

Input bias current $I_B=0.8\mu \text{A}$

Input offset current $I_{OS}=0.2\mu \text{A}$

  $\begin{array}{l}{R}_1=R_2=50\text{k}\Omega \\ R_3=25\text{k}\Omega \end{array}$

Let the general circuit with input bias currents $I_{B1}$ and $I_{B2}$ be,

  

Now, using superposition determine $v_O$ as function of $I_{B1}$ and $I_{B2}$ .

For  $I_{B2}=0$

  $V_Y=V_X=0$ and

The output voltage due to $I_{B1}$ is

  $v_O\left(I_{B1}\right)=I_{B1}{R}_2\mathrm{......}\left(1\right)$

For  $I_{B1}=0$

  $V_Y=-I_{B2}{R}_3=V_X$

Since, $v_O=\left(1+\frac{R_2}{R_1}\right)V_X$

The output voltage due to $I_{B2}$ is

  $v_O\left(I_{B2}\right)=-I_{B2}{R}_3\left(1+\frac{R_2}{R_1}\right)\mathrm{......}\left(2\right)$

So, the vet output voltage due to both $I_{B1}$ and $I_{B2}$ is

  $v_O=I_{B1}{R}_2-I_{B2}{R}_3\left(1+\frac{R_2}{R_1}\right)\mathrm{......}\left(3\right)$

If $I_{B1}=I_{B2}=I_B$ , then$v_O=I_B{R}_2-I_B{R}_3\left(1+\frac{R_2}{R_1}\right)$

So, the output voltage due to bias current $I_B$ is

  $v_O=I_B\left[R_2-R_3\left(1+\frac{R_2}{R_1}\right)\right]\mathrm{.......}\left(4\right)$

Now putting the values,

  $\begin{array}{l}{v}_O=0.8\times {10}^{-6}\left[50\times {10}^3-25\times {10}^3\left(1+\frac{50\times {10}^3}{50\times {10}^3}\right)\right]\\ =0.8\times {10}^{-6}\left[50\times {10}^3-25\times {10}^3\times 2\right]\\ =0.8\times {10}^{-6}\left[50\times {10}^3-50\times {10}^3\right]\\ \Rightarrow v_O=0\end{array}$

Now, for worst case output

As,

  $\begin{array}{l}{I}_B=\frac{I_{B1}+I_{B2}}{2}\\ I_{B1}+I_{B2}=2I_B\\ I_{B1}+I_{B2}=2\times 0.8\times {10}^{-6}\\ I_{B1}+I_{B2}=1.6\times {10}^{-6}\mathrm{.....}\left(5\right)\end{array}$

And

  $\begin{array}{l}{I}_{OS}=|I_{B1}-I_{B2}|\\ I_{B1}-I_{B2}=\pm I_{OS}\\ I_{B1}-I_{B2}=\pm 0.2\times {10}^{-6}\mathrm{....}\left(6\right)\end{array}$

Adding (5) and (6)

  $\begin{array}{l}{I}_{B1}+I_{B2}+I_{B1}-I_{B2}=1.6\times {10}^{-6}\pm 0.2\times {10}^{-6}\\ 2I_{B1}=1.6\times {10}^{-6}\pm 0.2\times {10}^{-6}\\ \therefore I_{B1}=\frac{1.8\times {10}^{-6}}{2}\text{or}\frac{1.4\times {10}^{-6}}{2}\\ \Rightarrow I_{B1}=0.9\mu \text{A}\text{or}\text{0}\text{.7}\mu \text{A}\end{array}$

From (5)

  $I_{B2}=1.6\times {10}^{-6}-I_{B1}$

For $I_{B1}=0.7\mu \text{A}$

  $\begin{array}{l}{I}_{B2}=1.6\times {10}^{-6}-0.7\times {10}^{-6}\\ I_{B2}=0.9\mu \text{A}\end{array}$

For $I_{B1}=0.9\mu \text{A}$

  $\begin{array}{l}{I}_{B2}=1.6\times {10}^{-6}-0.9\mu \text{A}\\ I_{B2}=0.7\mu \text{A}\end{array}$

Assume $I_{B1}=0.7\mu \text{A}\text{and}{I}_{B2}=\text{0}\text{.9}\mu \text{A}$

For the given circuit

  $\begin{array}{l}{R}_2=R_1=50\text{k}\Omega \\ R_3=25\text{k}\Omega \end{array}$

Putting the values in (3)

  $\begin{array}{l}{v}_O=I_{B1}{R}_2-I_{B2}{R}_3\left(1+\frac{R_2}{R_1}\right)\\ v_O=\left(0.7\times {10}^{-6}\right)\left(50\times {10}^3\right)-\left(0.9\times {10}^{-6}\right)\left(25\times {10}^3\right)\left(1+\frac{50\times {10}^3}{50\times {10}^3}\right)\\ v_O=0.035-0.045\end{array}$

So, including the effect of input offset current, the worst case output voltage is

  $v_O=-0.01\text{V}$ .

B.

To determine

Output voltage due to input bias current and worst case output voltage.

Answer to Problem 14.50P

Output voltage due to input bias current is

  $v_O=-1.56\text{V}$ and worst case output voltage including the effect of input offset current is

  $v_O=-1.765\text{V}$

Explanation of Solution

Given:

The given circuit is:

  

Input bias current $I_B=0.8\mu \text{A}$

Input offset current $I_{OS}=0.2\mu \text{A}$

  $\begin{array}{l}{R}_1=R_2=50\text{k}\Omega \\ R_3=1\text{M}\Omega \end{array}$

Considering equation 4.

  $v_O=I_B\left[R_2-R_3\left(1+\frac{R_2}{R_1}\right)\right]$

  $\begin{array}{l}\therefore v_O=0.8\times {10}^{-6}\left[50\times {10}^3-1\times {10}^6\left(1+\frac{50\times {10}^3}{50\times {10}^3}\right)\right]\\ =0.8\times {10}^{-6}\left[50\times {10}^3-1\times {10}^6\times 2\right]\\ =0.8\times {10}^{-6}\left[50\times {10}^3-2\times {10}^6\right]\\ =0.8\times {10}^{-6}\left[0.05\times {10}^6-2\times {10}^6\right]\\ =0.8\times {10}^{-6}\left[-1.95\times {10}^6\right]\\ \Rightarrow v_O=-1.56\text{V}\end{array}$

Now, for worst case output voltage

Assume $I_{B1}=0.7\mu \text{A}\text{and}{I}_{B2}=\text{0}\text{.9}\mu \text{A}$

For the given circuit

  $\begin{array}{l}{R}_2=R_1=50\text{k}\Omega \\ R_3=1\text{M}\Omega \end{array}$

Putting the values in (3)

  $\begin{array}{l}{v}_O=I_{B1}{R}_2-I_{B2}{R}_3\left(1+\frac{R_2}{R_1}\right)\\ v_O=\left(0.7\times {10}^{-6}\right)\left(50\times {10}^3\right)-\left(0.9\times {10}^{-6}\right)\left(1\times {10}^6\right)\left(1+\frac{50\times {10}^3}{50\times {10}^3}\right)\\ v_O=0.035-1.8\end{array}$

So, including the effect of input offset current, the worst case output voltage is

  $v_O=-1.765\text{V}$ .