Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

Answer to Problem 14.51P

The range of output voltage is,

  $4\text{mV}\le v_O\le 76\text{mV}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Input offset voltage is $V_{OS}=3\text{mV}$

Average input bias current is $I_B=0.4\mu \text{A}$

Offset bias current is $I_{OS}=0.06\mu \text{A}$

  $v_I=0$ and $R=0$

Calculation:

As,

  $\begin{array}{l}{I}_B=\frac{I_{B1}+I_{B2}}{2}\\ I_{B1}+I_{B2}=2I_B\\ I_{B1}+I_{B2}=2\times 0.4\times {10}^{-6}\\ I_{B1}+I_{B2}=0.8\times {10}^{-6}\mathrm{......}\left(1\right)\end{array}$

And

  $\begin{array}{l}{I}_{OS}=|I_{B1}-I_{B2}|\\ I_{B1}-I_{B2}=\pm I_{OS}\\ I_{B1}-I_{B2}=\pm 0.06\times {10}^{-6}\mathrm{.....}\left(2\right)\end{array}$

Adding (1) and (2)

  $\begin{array}{l}{I}_{B1}+I_{B2}+I_{B1}-I_{B2}=0.8\times {10}^{-6}\pm 0.06\times {10}^{-6}\\ 2I_{B1}=0.8\times {10}^{-6}\pm 0.06\times {10}^{-6}\\ I_{B1}=\frac{0.8\times {10}^{-6}\pm 0.06\times {10}^{-6}}{2}\\ =\frac{0.8\times {10}^{-6}+0.06\times {10}^{-6}}{2}\text{or}\frac{0.8\times {10}^{-6}-0.06\times {10}^{-6}}{2}\\ \Rightarrow I_{B1}=0.43\mu \text{A}\text{or}0.37\mu \text{A}\end{array}$

From (1)

  $I_{B1}=0.8\times {10}^{-6}-I_{B2}$

For $I_{B1}=0.43\mu \text{A}$

  $\begin{array}{l}{I}_{B1}=0.8\times {10}^{-6}-I_{B2}\\ I_{B1}=0.8\times {10}^{-6}-0.43\times {10}^{-6}\\ \Rightarrow I_{B1}=0.37\mu \text{A}\end{array}$

For $I_{B1}=0.37\mu \text{A}$

  $\begin{array}{l}{I}_{B1}=0.8\times {10}^{-6}-I_{B2}\\ I_{B1}=0.8\times {10}^{-6}-0.37\times {10}^{-6}\\ \Rightarrow I_{B1}=0.43\mu \text{A}\\ \therefore I_{B1}=0.37\mu \text{A}\text{or}{I}_{B1}=0.43\mu \text{A}\end{array}$

Now, the modified circuit with offset voltage and bias current is shown below.

  

KCL at non-inverting terminal,

  $\begin{array}{l}\frac{V_Y-\left(v_I+V_{OS}\right)}{R}+I_{B2}=0\\ \frac{V_Y-\left(v_I+V_{OS}\right)}{R}=-I_{B2}\\ V_Y-\left(v_I+V_{OS}\right)=-I_{B2}R\\ V_Y=v_I+V_{OS}-I_{B2}R\end{array}$

Now, for ideal op-amp $V_X=V_Y$

  $V_X=v_I+V_{OS}-I_{B2}R\mathrm{.....}\left(3\right)$

Now, KCL at inverting terminal,

  $\begin{array}{l}\frac{V_X}{10\text{k}}+\frac{V_X-v_O}{100\text{k}}+I_{B1}=0\\ \frac{10V_X+V_X-v_O}{100\text{k}}+I_{B1}=0\\ 10V_X+V_X-v_O=-I_{B1}\times 100\text{k}\\ 11V_X-v_O=-I_{B1}\times 100\text{k}\\ v_O=11V_X+\left(I_{B1}\times 100\text{k}\right)\end{array}$

Putting $V_X$ from (3)

  $v_O=11\left(v_I+V_{OS}-I_{B2}R\right)+\left(I_{B1}\times 100\text{k}\right)\mathrm{.....}\left(4\right)$

Now,

For $v_I=0$ and $R=0$

  $\begin{array}{l}{v}_O=11\left(0+V_{OS}-I_{B2}\left(0\right)\right)+\left(I_{B1}\times 100\text{k}\right)\\ v_O=11V_{OS}+\left(I_{B1}\times 100\text{k}\right)\end{array}$

For $I_{B1}=0.43\mu \text{A}\text{and}{V}_{OS}\text{=3}\text{mV}$

  $\begin{array}{l}{v}_O\left(\text{max}\right)=11\left(3\text{m}\right)+\left(0.43\mu \times 100\text{k}\right)\\ v_O\left(\text{max}\right)=33\times {10}^{-3}+43\times {10}^{-3}\\ \Rightarrow v_O\left(\text{max}\right)=76\text{mV}\end{array}$

For $I_{B1}=0.37\mu \text{A}\text{and}{V}_{OS}\text{=-3}\text{mV}$

  $\begin{array}{l}{v}_O\left(\text{min}\right)=11\left(-3\text{m}\right)+\left(0.37\mu \times 100\text{k}\right)\\ v_O\left(\text{min}\right)=\left(-33\times {10}^{-3}\right)+\left(37\times {10}^{-3}\right)\\ \Rightarrow v_O\left(\text{min}\right)=4\text{mV}\end{array}$

0So, the range of output voltage is,

  $\begin{array}{l}{v}_O\left(\text{min}\right)\le v_O\le v_O\left(\text{max}\right)\\ \Rightarrow 4\text{mV}\le v_O\le 76\text{mV}\end{array}$

b.

To determine

Range of output voltage for the given specifications.

Answer to Problem 14.51P

The range of output voltage is,

  $-39\text{mV}\le v_O\le 39\text{mV}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Input offset voltage is $V_{OS}=3\text{mV}$

Average input bias current is $I_B=0.4\mu \text{A}$

Offset bias current is $I_{OS}=0.06\mu \text{A}$

  $v_I=0$ and $R=9.09\text{k}\Omega$

Calculation:

From equation (4)

  $v_O=11\left(v_I+V_{OS}-I_{B2}R\right)+\left(I_{B1}\times 100\text{k}\right)$

For $v_I=0$ and $R=9.09\text{k}\Omega$

  $\begin{array}{l}{v}_O=11\left(0+V_{OS}-I_{B2}\left(9.09\text{k}\right)\right)+\left(I_{B1}\times 100\text{k}\right)\\ v_O=11V_{OS}-99.99\text{k}\times {\text{I}}_{B2}+\left(I_{B1}\times 100\text{k}\right)\end{array}$

Putting the values,

For $I_{B1}=0.43\mu \text{A,}{I}_{B2}=0.37\mu \text{A}\text{and}{V}_{OS}\text{=3}\text{mV}$

  $\begin{array}{l}{v}_O\left(\text{max}\right)=11\left(3\text{m}\right)-\left(99.99\text{k}\times \text{0}\text{.37}\mu \right)+\left(0.43\mu \times 100\text{k}\right)\\ v_O\left(\text{max}\right)=33\times {10}^{-3}-36.99\times {10}^{-3}+43\times {10}^{-3}\\ \Rightarrow v_O\left(\text{max}\right)=39\text{mV}\end{array}$

For $I_{B1}=0.37\mu \text{A,}{I}_{B2}=0.43\mu \text{A}\text{and}{V}_{OS}\text{=}-\text{3}\text{mV}$

  $\begin{array}{l}{v}_O\left(\text{max}\right)=11\left(-3\text{m}\right)-\left(99.99\text{k}\times \text{0}\text{.43}\mu \right)+\left(0.37\mu \times 100\text{k}\right)\\ v_O\left(\text{max}\right)=\left(-33\times {10}^{-3}\right)-\left(42.99\times {10}^{-3}\right)+\left(37\times {10}^{-3}\right)\\ \Rightarrow v_O\left(\text{max}\right)=-39\text{mV}\end{array}$

So, the range of output voltage is,

  $\begin{array}{l}{v}_O\left(\text{min}\right)\le v_O\le v_O\left(\text{max}\right)\\ \Rightarrow -39\text{mV}\le v_O\le 39\text{mV}\end{array}$

c.

To determine

Range of output voltage.

Answer to Problem 14.51P

The range of output voltage is,

  $2.161\text{V}\le v_O\le 2.239\text{V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Input offset voltage is $V_{OS}=3\text{mV}$

Average input bias current is $I_B=0.4\mu \text{A}$

Offset bias current is $I_{OS}=0.06\mu \text{A}$

  $v_I=0.2\text{V}$ and $R=9.09\text{k}\Omega$

Calculation:

From equation (4)

  $v_O=11\left(v_I+V_{OS}-I_{B2}R\right)+\left(I_{B1}\times 100\text{k}\right)$

For $v_I=0.2\text{V}$ and $R=9.09\text{k}\Omega$

  $\begin{array}{l}{v}_O=11\left(0.2+V_{OS}-I_{B2}\left(9.09\text{k}\right)\right)+\left(I_{B1}\times 100\text{k}\right)\\ v_O=2.2+11V_{OS}-99.99\text{k}\times {\text{I}}_{B2}+\left(I_{B1}\times 100\text{k}\right)\end{array}$

Putting the values,

For $I_{B1}=0.43\mu \text{A,}{I}_{B2}=0.37\mu \text{A}\text{and}{V}_{OS}\text{=3}\text{mV}$

  $\begin{array}{l}{v}_O\left(\text{max}\right)=2.2+11\left(3\text{m}\right)-\left(99.99\text{k}\times \text{0}\text{.37}\mu \right)+\left(0.43\mu \times 100\text{k}\right)\\ v_O\left(\text{max}\right)=2.2+33\times {10}^{-3}-36.99\times {10}^{-3}+43\times {10}^{-3}\\ \Rightarrow v_O\left(\text{max}\right)=2.239\text{V}\end{array}$

For $I_{B1}=0.37\mu \text{A,}{I}_{B2}=0.43\mu \text{A}\text{and}{V}_{OS}\text{=}-\text{3}\text{mV}$

  $\begin{array}{l}{v}_O\left(\text{max}\right)=2.2+11\left(-3\text{m}\right)-\left(99.99\text{k}\times \text{0}\text{.43}\mu \right)+\left(0.37\mu \times 100\text{k}\right)\\ v_O\left(\text{max}\right)=2.2+\left(-33\times {10}^{-3}\right)-\left(42.99\times {10}^{-3}\right)+\left(37\times {10}^{-3}\right)\\ v_O\left(\text{max}\right)=2.2-39\times {10}^{-3}\text{V}\\ \Rightarrow v_O\left(\text{max}\right)=2.161\text{V}\end{array}$

So, the range of output voltage is,

  $\begin{array}{l}{v}_O\left(\text{min}\right)\le v_O\le v_O\left(\text{max}\right)\\ \Rightarrow 2.161\text{V}\le v_O\le 2.239\text{V}\end{array}$