Chapter 14, Problem 14.38P

In the circuit in Figure P14.38, the offset voltage of each op−amp is ±3 mV.
(a) Determine the possible range in output voltages ${\upsilon }_{O1}$ and ${\upsilon }_{O2}$ for ${\upsilon }_I=0$ .

(b) Repeat part (a) for ${\upsilon }_I=10\text{mV}$ . (c) Repeat part (a) for ${\upsilon }_I=100\text{mV}$ .

(d) Design offset voltage compensation circuit(s) to adjust both ${\upsilon }_{O1}$ and ${\upsilon }_{O2}$ to zero when ${\upsilon }_I=0$ .

Figure P14.38

To determine

The possible range in output voltages $v_{o1}$ and $v_{o2}$ for the given $v_I$ .

Answer to Problem 14.38P

The possible range in output voltages $v_{o1}$ and $v_{o2}$ for $v_I=0$ are

  $-0.033\le v_{o1}\le 0.033\text{V}$

  $-0.165\le v_{o2}\le 0.165\text{V}$

Explanation of Solution

Given:

Given circuit:

  

Given offset voltage, $V_{os}=\pm 3\text{mV}$

  $v_I=0$

Calculation:

Now the input voltage is, $v_i+V_{os}$

For the first op-amp which is a non-inverting amplifier, the gain is,

  $\frac{v_{o1}}{v_1+V_{os}}=1+\frac{100\text{k}}{10\text{k}}$

  $\frac{v_{o1}}{v_1+V_{os}}=1+10$

  $\begin{array}{l}{v}_{o1}=11\left(v_I+V_{os}\right)\hfill \\ v_{o1}=11\left(v_I\pm 3\text{m}\right)\hfill \end{array}$

The possible range in the output voltage $v_{o1}$ is,

  $11\left(v_I-3\text{m}\right)\le v_{o1}\le 11\left(v_I+3\text{m}\right)\mathrm{............}(1)$

For the second op-amp which is an inverting amplifier, the gain is,

  $\begin{array}{l}\frac{v_{o2}}{v_{o1}}=-\frac{50\text{k}}{10\text{k}}\hfill \\ \frac{v_{o2}}{v_{o1}}=-5\hfill \\ v_{o2}=-5v_{o1}\hfill \end{array}$

Substitute $v_{o1}$ in the above equation

  $\begin{array}{l}{v}_{O2}=-5\times 11\left(v_I\pm 3\text{m}\right)\hfill \\ v_{O2}=-55\left(v_I\pm 3\text{m}\right)\hfill \end{array}$

The possible range in the output voltage $v_{o2}$ is,

  $-55\left(v_1+3\text{m}\right)\le v_{o2}\le -55\left(v_1-3\text{m}\right)\mathrm{.............}(2)$

Given $v_I=0$

Above circuit can be represented as below

  

From equation $(1),$ the possible range in the output voltage $v_{o1}$ is,

  $11(0-3m)\le v_{ot}\le 11(0+3m)$

  $-33m\le v_{o1}\le 33m$

  $-33\times {10}^{-3}\le v_{o1}\le 33\times {10}^{-3}$

  $\therefore -0.033\le v_{o1}\le 0.033\text{V}$

From equation (2) , the possible range in the output voltage $v_{o2}$ is,

  $-55(0+3m)\le v_{o2}\le -55(0-3m)$

  $-165\text{m}\le v_{o2}\le 165\text{m}$

  $-165\times {10}^{-3}\le v_{o2}\le 165\times {10}^{-3}$

  $\therefore -0.165\le v_{o2}\le 0.165\text{V}$

To determine

The possible range in output voltages $v_{o1}$ and $v_{o2}$ for the given $v_I$

Answer to Problem 14.38P

The possible range in output voltages $v_{o1}$ and $v_{o2}$ for $v_I=10\text{mV}$

  $0.077\le v_{o1}\le 0.133\text{V}$

  $-0.715\le v_{o2}\le -0.385\text{V}$

Explanation of Solution

Given:

Given circuit:

  

Given offset voltage, $V_{os}=\pm 3\text{mV}$

  $v_I=10\text{mV}$

Calculation:

Now the input voltage is, $v_i+V_{os}$

For the first op-amp which is a non-inverting amplifier, the gain is,

  $\frac{v_{o1}}{v_1+V_{os}}=1+\frac{100\text{k}}{10\text{k}}$

  $\frac{v_{o1}}{v_1+V_{os}}=1+10$

  $\begin{array}{l}{v}_{o1}=11\left(v_I+V_{os}\right)\hfill \\ v_{o1}=11\left(v_I\pm 3\text{m}\right)\hfill \end{array}$

The possible range in the output voltage $v_{o1}$ is,

  $11\left(v_I-3\text{m}\right)\le v_{o1}\le 11\left(v_I+3\text{m}\right)\mathrm{............}(1)$

For the second op-amp which is an inverting amplifier, the gain is,

  $\begin{array}{l}\frac{v_{o2}}{v_{o1}}=-\frac{50\text{k}}{10\text{k}}\hfill \\ \frac{v_{o2}}{v_{o1}}=-5\hfill \\ v_{o2}=-5v_{o1}\hfill \end{array}$

Substitute $v_{o1}$ in the above equation

  $\begin{array}{l}{v}_{O2}=-5\times 11\left(v_I\pm 3\text{m}\right)\hfill \\ v_{O2}=-55\left(v_I\pm 3\text{m}\right)\hfill \end{array}$

The possible range in the output voltage $v_{o2}$ is,

  $-55\left(v_1+3\text{m}\right)\le v_{o2}\le -55\left(v_1-3\text{m}\right)\mathrm{.............}(2)$

Given $v_I=0$

Above circuit can be represented as below

  

From equation $(1),$ the possible range in the output voltage $v_{o1}$ is,

  $11(10\text{m}-3\text{m})\le v_{o1}\le 11(10\text{m}+3\text{m})$

  $11(7\text{m})\le v_{o1}\le 11(13\text{m})$

  $77\text{m}\le v_{o1}\le 143\text{m}$

  $77\times {10}^{-3}\le v_{o1}\le 143\times {10}^{-3}$

  $\therefore 0.077\le v_{o1}\le 0.133\text{V}$

From equation (2), the possible range in the output voltage $v_{o2}$ is,

  $-55(10\text{m}+3\text{m})\le v_{o2}\le -55(10\text{m}-3\text{m})$

  $-55(13\text{m})\le v_{o2}\le -55(7\text{m})$

  $-715\text{m}\le v_{o2}\le -385\text{m}$

  $-715\times {10}^{-3}\le v_{o2}\le -385\times {10}^{-3}$

  $\therefore -0.715\le v_{o2}\le -0.385\text{V}$

To determine

The possible range in output voltages $v_{o1}$ and $v_{o2}$ for the given value of $v_I$ .

Answer to Problem 14.38P

The possible range in output voltages $v_{o1}$ and $v_{o2}$ for $v_I=100\text{mV}$

  $1.067\le v_{o1}\le 1.133\text{V}$

  $-5.665\le v_{o2}\le -5.335\text{V}$

Explanation of Solution

Given:

Given circuit:

  

Given offset voltage, $V_{os}=\pm 3\text{mV}$

The given value is $v_I=100\text{mV}$

Calculation:

Now the input voltage is, $v_i+V_{os}$

For the first op-amp which is a non-inverting amplifier, the gain is,

  $\frac{v_{o1}}{v_1+V_{os}}=1+\frac{100\text{k}}{10\text{k}}$

  $\frac{v_{o1}}{v_1+V_{os}}=1+10$

  $\begin{array}{l}{v}_{o1}=11\left(v_I+V_{os}\right)\hfill \\ v_{o1}=11\left(v_I\pm 3\text{m}\right)\hfill \end{array}$

The possible range in the output voltage $v_{o1}$ is,

  $11\left(v_I-3\text{m}\right)\le v_{o1}\le 11\left(v_I+3\text{m}\right)\mathrm{............}(1)$

For the second op-amp which is an inverting amplifier, the gain is,

  $\begin{array}{l}\frac{v_{o2}}{v_{o1}}=-\frac{50\text{k}}{10\text{k}}\hfill \\ \frac{v_{o2}}{v_{o1}}=-5\hfill \\ v_{o2}=-5v_{o1}\hfill \end{array}$

Substitute $v_{o1}$ in the above equation

  $\begin{array}{l}{v}_{O2}=-5\times 11\left(v_I\pm 3\text{m}\right)\hfill \\ v_{O2}=-55\left(v_I\pm 3\text{m}\right)\hfill \end{array}$

The possible range in the output voltage $v_{o2}$ is,

  $-55\left(v_1+3\text{m}\right)\le v_{o2}\le -55\left(v_1-3\text{m}\right)\mathrm{.............}(2)$

Given $v_I=100\text{mV}$

Above circuit can be represented as below

  

From equation $(1),$ the possible range in the output voltage $v_{o1}$ is,

  $11(100\text{m}-3\text{m})\le v_{o1}\le 11(100\text{m}+3\text{m})$

  $11(97\text{m})\le v_{o1}\le 11(103\text{m})$

  $1067\text{m}\le v_{o1}\le 1133\text{m}$

  $1067\times {10}^{-3}\le v_{o1}\le 1133\times {10}^{-3}$

  $\therefore 1.067\le v_{o1}\le 1.133\text{V}$

From equation (2) , the possible range in the output voltage $v_{o2}$ is,

  $-55(100\text{m}+3\text{m})\le v_{o2}\le -55(100\text{m}-3\text{m})$

  $-55(103\text{m})\le v_{o2}\le -55(97\text{m})$

  $-5665\text{m}\le v_{o2}\le -5335\text{m}$

  $-5665\times {10}^{-3}\le v_{o2}\le -5335\times {10}^{-3}$

  $\therefore -5.665\le v_{o2}\le -5.335\text{V}$

To determine

To design: The offset voltage compensation circuit(s) to adjust both $v_{o1}$ and $v_{o2}$ to zero for the given $v_I$ .

Answer to Problem 14.38P

The offset voltage compensation circuit is shown in Figure 1.

Explanation of Solution

Given:

Given circuit:

  

Given offset voltage, $V_{os}=\pm 3\text{mV}$

  $v_I=0$

Calculation:

The output voltages for the given input voltage $v_I=0$ are

  $-0.033\le v_{o1}\le 0.033\text{V}$

  $-0.165\le v_{o2}\le 0.165\text{V}$

The adjusting compensation circuit for the range of output voltages as well as input voltage is shown below.

  

Figure 1

  $\text{Where}{V}^{+}=10\text{V}\text{and}{V}^{-}=-10\text{V}$