Chapter 9, Problem 9.1EP

Design an ideal inverting op-amp circuit such that the voltage gain is $A_v=-25$ . The maximum current in any resistor is to be limited to $10\mu A$ withthe input voltage in the range $-25\le v_t\le +25mV$ . (a) What are the values of $R_1$ and $R_2$ ? (b) What is the range of output voltage $v_0$ ? (Ans. (a) $R_1=2.5k\Omega ,R_1=62.5k\Omega$ ;(b) $-0.625\le v_0\le 0.625V$ )

To determine

The value of the resistance $R_1$ and $R_2$ .

Answer to Problem 9.1EP

Thevalue of the resistance $R_1$ is $62.5\text{k}\Omega$ and the value of the resistance $R_2$ is $62.5\text{k}\Omega$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The above op-amp is considered to be ideal.

The value of the voltage $v_2$ is given by,

  $v_2=0$

The expression for the input voltage is given by,

  $v_1=v_2$

Substitute $0$ for $v_2$ in the above equation.

  $v_1=0$

Apply KCL at negative terminal.

  $i_1=\frac{v_I-v_1}{R_1}$

Substitute $0$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{v_I-0}{R_1}\\ =\frac{v_I}{R_1}\end{array}$ ........ (1)

The expression for the value of the voltage gain is given by,

  $A_v=\frac{v_O}{v_I}$ ........ (2)

The expression for the value of the current $i_1$ is given by,

  $i_1=\frac{v_1-v_O}{R_2}$

Substitute $0$ for $v_1$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{0-v_O}{R_2}\\ =\frac{-v_O}{R_2}\end{array}$

Substitute $\frac{-v_O}{R_2}$ for $i_1$ in equation (1).

  $\begin{array}{l}\frac{-v_O}{R_2}=\frac{v_I}{R_1}\\ \frac{v_O}{v_I}=-\frac{R_2}{R_1}\end{array}$

Substitute $\frac{v_O}{v_I}$ for $A_v$ in the above equation.

  $A_v=-\frac{R_2}{R_1}$

Substitute $-25\text{mV}$ for $A_v$ in the above equation.

  $\begin{array}{l}-25\text{mV}=-\frac{R_2}{R_1}\\ 25R_1=R_2\end{array}$

Substitute $\left|25\text{mV}\right|$ for $A_v$ in equation (2).

  $\begin{array}{l}\frac{v_O}{v_I}\le \left|25\text{mV}\right|\\ v_O\le \left|25\text{mV}\left(v_I\right)\right|\end{array}$

Substitute $25\text{mV}$ for $\left|v_I\right|$ in the above equation.

  $\begin{array}{l}\left|v_O\right|\le \left|25\text{mV}\left(25\text{mV}\right)\right|\\ \left|v_O\right|\le \pm 625\text{mV}\end{array}$

The expression for the maximum value of the current is given by,

  ${\left|i_1\right|}_{\max }=10\text{μA}$

The expression for the maximum current is given by,

  ${\left|i_1\right|}_{\max }={\left|i_2\right|}_{\max }$

Substitute $10\text{μA}$ for ${\left|i_1\right|}_{\max }$ in the above equation.

  ${\left|i_2\right|}_{\max }=10\text{μA}$

The expression for the minimum value of the resistance $R_1$ is given by,

  $R_{1,\min }=\frac{\left|v_I\right|}{{\left|i_1\right|}_{\max }}$

Substitute $25\text{mV}$ for $\left|v_I\right|$ and $10\text{μA}$ for ${\left|i_1\right|}_{\max }$ in the above equation.

  $\begin{array}{c}{R}_{1,\min }=\frac{\left|25\text{mV}\right|}{10\text{μA}}\\ =62.5\text{k}\Omega \end{array}$

The expression for the minimum value of the resistance $R_2$ is given by,

  $R_{2,\min }=R_{1,\min }$

Substitute $62.5\text{k}\Omega$ for $R_{1,\min }$ in the above equation.

  $R_{2,\min }=62.5\text{k}\Omega$

Conclusion:

Therefore, the value of the resistance $R_1$ is $62.5\text{k}\Omega$ and the value of the resistance $R_2$ is $62.5\text{k}\Omega$ .

To determine

The value for the range of the output voltage $v_O$ .

Answer to Problem 9.1EP

The range of the output voltage is $-625\text{mV}\le \left|v_O\right|\le 625\text{mV}$ .

Explanation of Solution

Calculation:

The expression for the variation of the output voltage is given by,

  $\left|v_O\right|\le \pm 625\text{mV}$

From above the variation of the voltage is given by,

  $-625\text{mV}\le \left|v_O\right|\le 625\text{mV}$

Conclusion:

Therefore, the range of the output voltage is $-625\text{mV}\le \left|v_O\right|\le 625\text{mV}$ .