Chapter 9, Problem 9.13P

(a) In an inverting op-amp circuit, the nominal resistance values are $R_2=300k\Omega$ and $R_1=15k\Omega$ . The tolerance of each resistor is $\pm 5\%$ ,which means that each resistance can deviate from its nominal value by $\pm 5\%$ . What is the maximum deviation in the voltage gain from its nominalvalue? (b) Repeat part (a) if the resistor tolerance is reduced to $\pm 1\%$ .

To determine

The maximum deviation in the voltage gain from its nominal value for $\pm 5\%$ tolerance.

Answer to Problem 9.13P

The maximum deviation of the voltage gain from the nominal value is $-9.5\%\le A_{v,dev}\le +10.5\%$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

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

  $A_v=\frac{v_O}{v_I}$

The expression for the voltage $v_1$ is given by,

  $v_1=v_2$

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

  $v_1=0$

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

  $i_1=\frac{v_I}{R_1}$

The expression for the value of the current $i_2$ is given by,

  $i_2=\frac{v_1-v_O}{R_2}$

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

  $i_2=\frac{-v_O}{R_2}$

Substitute $i_1$ for $i_2$ in the above equation.

  $i_1=\frac{-v_O}{R_2}$

Substitute $\frac{v_I}{R_1}$ for $i_1$ in the above equation.

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

Substitute $300\text{k}\Omega$ for $R_2$ and $15\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{l}\frac{v_I}{R_1}=\frac{-v_O}{R_2}\\ \frac{v_O}{v_I}=\frac{-300\text{k}\Omega }{15\text{k}\Omega }\end{array}$

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

  $\begin{array}{c}{A}_v=\frac{-300\text{k}\Omega }{15\text{k}\Omega }\\ =-20\end{array}$

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

  $A_{v,\max }=-\frac{R_2+R_2\left(0.05\right)}{R_1-R_1\left(0.05\right)}$

Substitute $300\text{k}\Omega$ for $R_2$ and $15\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{A}_{v,\max }=-\frac{\left(300\text{k}\Omega \right)+\left(300\text{k}\Omega \right)\left(0.05\right)}{15\text{k}\Omega -\left(15\text{k}\Omega \right)\left(0.05\right)}\\ =-22.10\end{array}$

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

  $A_{v,\min }=-\frac{R_2-R_2\left(0.05\right)}{R_1+R_1\left(0.05\right)}$

Substitute $300\text{k}\Omega$ for $R_2$ and $15\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{A}_{v,\min }=-\frac{\left(300\text{k}\Omega \right)-\left(300\text{k}\Omega \right)\left(0.05\right)}{15\text{k}\Omega +\left(15\text{k}\Omega \right)\left(0.05\right)}\\ =-18.09\end{array}$

The expression for the deviation of the voltage gain from the maximum voltage is given by,

  $A_{v,dev1}=\left(\frac{A_{v,\max }-A_v}{A_v}\right)100$

Substitute $-22.10$ for $A_{v,\max }$ and $-20$ for $A_v$ in the above equation.

  $\begin{array}{c}{A}_{v,dev1}=\left(\frac{-22.10-\left(-20\right)}{-20}\right)100\\ =10.5\%\end{array}$

The expression for the minimum deviation of the voltage gain is given by,

  $A_{v,dev2}=\left(\frac{A_{v,\min }-A_v}{A_v}\right)100\%$

Substitute $-18.09$ for $A_{v,\min }$ and $-22.10$ for $A_{v,\max }$ in the above equation.

  $\begin{array}{c}{A}_{v,dev2}=\left(\frac{-18.09-\left(-20\right)}{-20}\right)100\%\\ =-9.5\%\end{array}$

The expression for the deviation in the voltage gain is given by,

  $-9.5\%\le A_{v,dev}\le +10.5\%$

Conclusion:

Therefore, the maximum deviation of the voltage gain from the nominal value is $-9.5\%\le A_{v,dev}\le +10.5\%$ .

To determine

The maximum deviation in the voltage gain from its nominal value for $\pm 1\%$ tolerance.

Answer to Problem 9.13P

The maximum deviation of the voltage gain from the nominal value is $-2\%\le A_{v,dev}\le +2\%$ .

Explanation of Solution

Calculation:

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

  $A_{v,\max }=-\frac{R_2+R_2\left(0.01\right)}{R_1-R_1\left(0.01\right)}$

Substitute $300\text{k}\Omega$ for $R_2$ and $15\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{A}_{v,\max }=-\frac{\left(300\text{k}\Omega \right)+\left(300\text{k}\Omega \right)\left(0.01\right)}{15\text{k}\Omega -\left(15\text{k}\Omega \right)\left(0.01\right)}\\ =-20.40\end{array}$

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

  $A_{v,\min }=-\frac{R_2-R_2\left(0.01\right)}{R_1+R_1\left(0.01\right)}$

Substitute $300\text{k}\Omega$ for $R_2$ and $15\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{A}_{v,\min }=-\frac{\left(300\text{k}\Omega \right)-\left(300\text{k}\Omega \right)\left(0.01\right)}{15\text{k}\Omega +\left(15\text{k}\Omega \right)\left(0.01\right)}\\ =-19.60\end{array}$

The expression for the deviation of the voltage gain from the maximum voltage is given by,

  $A_{v,dev1}=\left(\frac{A_{v,\max }-A_v}{A_v}\right)100$

Substitute $-20.40$ for $A_{v,\max }$ and $-20$ for $A_v$ in the above equation.

  $\begin{array}{c}{A}_{v,dev1}=\left(\frac{-20.40-\left(-20\right)}{-20}\right)100\\ =+2\%\end{array}$

The expression for the minimum deviation of the voltage gain is given by,

  $A_{v,dev2}=\left(\frac{A_{v,\min }-A_v}{A_v}\right)100\%$

Substitute $-19.60$ for $A_{v,\min }$ and $-22.10$ for $A_{v,\max }$ in the above equation.

  $\begin{array}{c}{A}_{v,dev2}=\left(\frac{-19.60-\left(-20\right)}{-20}\right)100\%\\ =-2\%\end{array}$

The expression for the deviation in the voltage gain is given by,

  $-2\%\le A_{v,dev}\le +2\%$

Conclusion:

Therefore, the maximum deviation of the voltage gain from the nominal value is $-2\%\le A_{v,dev}\le +2\%$ .