Chapter 9, Problem D9.18P

Design an amplifier system with three inverting op-amps circuits in cascadesuch that the overall closed-loop voltage gain is $A_v=v_O/v_I=-300$ . Themaximum resistance is limited to $200k\Omega$ and the minimum resistance islimited to $20k\Omega$ . In addition, the maximum current in any resistor is to belimited to $60\mu A$ when $v_O=6V$ .

To determine

The design parameters for an inverting op-amp circuit.

Answer to Problem D9.18P

The value of the resistance $R_2$ is $100\text{k}\Omega$ , $R_3$ is $20\text{k}\Omega$ , $R_4$ is $120\text{k}\Omega$ , $R_5$ is $20\text{k}\Omega$ and the value of the resistance $R_6$ is $100\text{k}\Omega$ .

Explanation of Solution

Calculation:

The required diagram for the inverting op amp with feedback is shown below.

The required diagram is shown in Figure 1.

  

The expression for the value of the voltage $v_{11}$ is given by,

  $v_{11}=v_{21}$

Substitute $0$ for $v_{21}$ in the above equation.

  $v_{11}=0$

The expression for the voltage $v_{22}$ is given by,

  $v_{22}=v_{12}$

Substitute $0$ for $v_{22}$ in the above equation.

  $v_{12}=0$

The expression for the value of the voltage $v_{13}$ is given by.

  $v_{13}=v_{23}$

Substitute $0$ for $v_{23}$ in the above equation.

  $v_{13}=0$

The expression for the output voltage of the first op-amp is given by,

  $v_{O1}=-v_I\left(\frac{R_2}{R_1}\right)$

The expression for the second voltage of the first op amp is given by,

  $v_{O2}=-v_{O1}\left(\frac{R_2}{R_4}\right)$

Substitute $-v_I\left(\frac{R_2}{R_1}\right)$ for $v_{O1}$ in the above equation.

  $v_{O2}=v_I\left(\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3}\right)$

The expression for the voltage gain is given by,

  $A_v=\frac{v_O}{v_I}$

Substitute $300$ for $A_v$ and $6\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{l}300=\frac{6\text{V}}{v_I}\\ v_I=0.02\text{V}\end{array}$

Apply KCL at the negative terminal of the right most op-amp.

  $\begin{array}{l}\frac{v_{O2}-0}{R_5}=\frac{0-v_O}{R_6}\\ v_O=\frac{-v_{O2}{R}_6}{R_5}\\ \frac{v_O}{v_{O2}}=\frac{-R_6}{R_5}\end{array}$

Substitute $v_I\left(\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3}\right)$ for $v_{O2}$ in the above equation.

  $\begin{array}{l}\frac{v_O}{v_I\left(\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3}\right)}=\frac{-R_6}{R_5}\\ \frac{v_O}{v_I}=\frac{-R_6}{R_5}\left(\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3}\right)\end{array}$

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

  $A_v=\frac{-R_6}{R_5}\left(\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3}\right)$

Substitute $300$ for $A_v$ in the above equation.

  $300=\frac{-R_6}{R_5}\left(\frac{R_2}{R_1}\right)\left(\frac{R_4}{R_3}\right)$

The expression for the value of the current $R_{6,\min }$ is given by,

  $R_{6,\min }=\frac{v_O}{i_{6,\max }}$

Substitute $60\text{μA}$ for $i_{6,\max }$ and $6\text{V}$ for $v_O$ in the above equation.

  $\begin{array}{c}{R}_{6,\min }=\frac{6\text{V}}{60\text{μA}}\\ =0.1\text{M}\Omega \end{array}$

The expression for the value off the voltage $i_{5,\max }$ is given by,

  $i_{5,\max }=i_{6,\max }$

Substitute $60\text{μA}$ for $i_{6,\max }$ in the above equation.

  $i_{5,\max }=60\text{μA}$

The expression for the value of the current $i_{5,\max }$ is given by,

  $i_{5,\max }=\frac{-v_O\frac{R_{5,\max }}{R_{6,\min }}}{R_{5,\min }}$

Substitute $60\text{μA}$ for $i_{5,\max }$ , $0.1\text{M}\Omega$ for $R_{6,\min }$ and $6\text{V}$ for $v_O$ in the above equation.

  $60\text{μA}=\frac{-\left(6\text{V}\right)\frac{R_{5,\max }}{0.1\text{M}\Omega }}{R_{5,\min }}$

The expression to determine the range of the current is given by,

  $\left|i_2\right|=10\left|\frac{-v_{O1}}{R_2}\right|$

Substitute $-0.02\text{V}$ for $v_{O1}$ and $0.2\text{M}\Omega$ for $R_2$ in the above equation.

  $\begin{array}{l}\left|i_2\right|=10\left|\frac{-0.02\text{V}}{0.2\text{M}\Omega }\right|\\ \left|i_2\right|=1\text{μA}<60\text{μA}\end{array}$

The expression for the value of the mid stage gain is given by,

  $\frac{v_O}{v_{O1}}=-6$

Substitute $\left(-10\right)\left(0.02\text{V}\right)$ for $v_{O2}$ in the above equation.

  $\begin{array}{l}\frac{v_{O2}}{\left(-6\right)\left(-10\right)\left(0.02\text{V}\right)}=-6\\ v_O=1.2\text{V}\end{array}$

The expression for the minimum value of the resistance $R_{3,\min }$ is given by,

  $R_{3,\min }=\frac{v_{O2}}{60\text{μA}}$

Substitute $\left(10\right)\left(0.02\text{V}\right)$ for $v_{O2}$ in the above equation.

  $\begin{array}{c}{R}_{3,\min }=\frac{\left(10\right)\left(0.02\text{V}\right)}{60\text{μA}}\\ =3.33\text{k}\Omega \end{array}$

The maximum value of the resistance $R_3$ to be used is $20\text{k}\Omega$ .

The expression for the value of the resistance $R_4$ is given by,

  $R_4=6R_3$

Substitute $20\text{k}\Omega$ for $R_3$ in the above equation.

  $\begin{array}{c}{R}_4=6\left(20\text{k}\Omega \right)\\ \approx 120\text{k}\Omega \end{array}$

The expression for the gain of the voltage gain of right most amplifier is given by,

  $\frac{v_O}{v_{O2}}=-5$

Substitute $\left(10\right)\left(-6\right)\left(0.02\text{V}\right)$ for $v_{O2}$ in the above equation.

  $\begin{array}{c}{v}_O=-5\left(10\right)\left(-6\right)\left(0.02\text{V}\right)\\ =6\text{V}\end{array}$

The value for the current ${\left|i_5\right|}_{\text{max}}$ is given by,

  ${\left|i_5\right|}_{\text{max}}=60\text{μA}$

The expression for the value of the resistance $R_{5,\min }$ is given by,

  $R_{5,\min }={\left|\frac{v_{O2}}{\left|i_5\right|}\right|}_{\max }$

Substitute $60\text{μA}$ for ${\left|i_5\right|}_{\text{max}}$ and $\left(10\right)\left(-6\right)\left(0.02\text{V}\right)$ for $v_{O2}$ in the above equation.

  $\begin{array}{c}{R}_{5,\min }=\left|\frac{\left(10\right)\left(-6\right)\left(0.02\text{V}\right)}{\left|60\text{μA}\right|}\right|\\ =20\text{k}\Omega \end{array}$

The expression for the maximum value of the resistance $R_6$ is given by,

  $R_6=5R_5$

Substitute $20\text{k}\Omega$ for $R_5$ in the above equation.

  $\begin{array}{c}{R}_6=5\left(20\text{k}\Omega \right)\\ =100\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of the resistance $R_2$ is $100\text{k}\Omega$ , $R_3$ is $20\text{k}\Omega$ , $R_4$ is $120\text{k}\Omega$ , $R_5$ is $20\text{k}\Omega$ and the value of the resistance $R_6$ is . $100\text{k}\Omega$ .