Chapter 15, Problem 58E

(a) Design a two-stage op amp filter circuit with a bandwidth of 1000 rad/s, a low-frequency cutoff of 100 rad/s, and a voltage gain of 20 dB. (b) Verify your design with an appropriate LTspice simulation.

To determine

Design a two-stage op amp filter circuit with a bandwidth of $1000\frac{\text{rad}}{\text{s}}$, a low cutoff frequency of $100\frac{\text{rad}}{\text{s}}$ and a voltage gain of $20\text{dB}$.

Explanation of Solution

Given data:

The value of the bandwidth $\left(B\right)$ is $1000\frac{\text{rad}}{\text{s}}$.

The value of the lower cutoff frequency $\left({\omega }_L\right)$ is $100\frac{\text{rad}}{\text{s}}$.

The value of the voltage gain $\left(A_{\max }\right)$ is $20\text{dB}$.

Formula used:

Write the expression to calculate the impedance of the passive elements resistor and capacitor.

$Z_R=R$        (1)

$Z_C=\frac{1}{sC}$        (2)

Here,

$\omega$ is the angular frequency,

$R$ is the value of the resistor, and

$C$ is the value of the capacitor.

Calculation:

The two-stage op amp filter circuit can be obtained in a band-pass filter by cascading the low pass filter and high pass filter.

In order to design the band-pass filter of gain $20\text{dB}$, the low-pass filter with a gain of $10\text{dB}$ and higher cutoff frequency of the band-pass is connected in series with the high-pass filter with a gain of $10\text{dB}$ and lower cutoff frequency of the band-pass.

In two-stage op amp filter, the low pass op amp filter is the first stage and the output of first stage filter is given as an input of the second stage high-pass op amp filter.

Low-pass filter design:

For low-pass filter, the voltage gain $\left(A_{\max }\right)$ is $10\text{dB}$.

The active low-pass op amp filter is drawn as Figure 1.

The s-domain circuit of the Figure 1 is drawn as Figure 2 using the equations (1) and (2).

Write the general expression to calculate the transfer function of the circuit in Figure 2.

$H\left(s\right)=\frac{V_1}{V_{\text{in}}}$        (3)

Here,

$V_1$ is the output response of the system, and

$V_{\text{in}}$ is the input response of the system.

Use nodal analysis on node $V_{+}$ in Figure 2.

$\begin{array}{l}\frac{V_{+}-V_{\text{in}}}{R_2}+\frac{V_{+}}{\left(\frac{1}{s{C}_1}\right)}=0\\ \frac{V_{+}}{R_2}-\frac{V_{\text{in}}}{R_2}+s{C}_1{V}_{+}=0\\ V_{+}\left(\frac{1}{R_2}+s{C}_1\right)=\frac{V_{\text{in}}}{R_2}\\ V_{+}\left(\frac{1+s{R}_2{C}_1}{R_2}\right)=\frac{V_{\text{in}}}{R_2}\end{array}$

Rearrange the above equation to find $V_{+}$.

$\begin{array}{c}{V}_{+}=\frac{V_{\text{in}}}{R_2}\left(\frac{R_2}{1+s{R}_2{C}_1}\right)\\ =V_{\text{in}}\left(\frac{1}{1+s{R}_2{C}_1}\right)\end{array}$

Use nodal analysis on node $V_{-}$ in Figure 2.

$\begin{array}{l}\frac{V_{-}-0}{R_1}+\frac{V_{-}-V_1}{R_{f_1}}=0\\ \frac{V_{-}}{R_1}+\frac{V_{-}}{R_{f_1}}-\frac{V_1}{R_{f_1}}=0\\ V_{-}\left(\frac{1}{R_1}+\frac{1}{R_{f_1}}\right)=\frac{V_1}{R_{f_1}}\\ V_{-}\left(\frac{R_{f_1}+R_1}{R_1{R}_{f_1}}\right)=\frac{V_1}{R_{f_1}}\end{array}$

Rearrange the above equation to find $V_{-}$.

$\begin{array}{c}{V}_{-}=\frac{V_1}{R_{f_1}}\left(\frac{R_1{R}_{f_1}}{R_{f_1}+R_1}\right)\\ =V_1\left(\frac{R_1}{R_{f_1}+R_1}\right)\end{array}$

It is known that for an ideal operational amplifier,

$V_{+}=V_{-}$

Substitute $V_{\text{in}}\left(\frac{1}{1+s{R}_2{C}_1}\right)$ for $V_{+}$ and $V_1\left(\frac{R_1}{R_{f_1}+R_1}\right)$ for $V_{-}$ in above equation.

$V_{\text{in}}\left(\frac{1}{1+s{R}_2{C}_1}\right)=V_1\left(\frac{R_1}{R_{f_1}+R_1}\right)$

Rearrange the above equation to find $\frac{V_1}{V_{\text{in}}}$.

$\begin{array}{c}\frac{V_1}{V_{\text{in}}}=\left(\frac{1}{1+s{R}_2{C}_1}\right)\left(\frac{R_{f_1}+R_1}{R_1}\right)\\ =\left(\frac{1}{1+s{R}_2{C}_1}\right)\left(\frac{R_{f_1}}{R_1}+\frac{R_1}{R_1}\right)\\ =\left(\frac{1}{1+s{R}_2{C}_1}\right)\left(1+\frac{R_{f_1}}{R_1}\right)\end{array}$

Substitute $\left(\frac{1}{1+s{R}_2{C}_1}\right)\left(1+\frac{R_{f_1}}{R_1}\right)$ for $\frac{V_1}{V_{\text{in}}}$ in equation (3) to find $H\left(s\right)$.

$H\left(s\right)=\left(\frac{1}{1+s{R}_2{C}_1}\right)\left(1+\frac{R_{f_1}}{R_1}\right)$        (4)

Therefore, the equation (4) is the transfer function of the active low-pass filter that is the product of the transfer function of general low-pass filter and the gain of the non-inverting amplifier.

From equation (4), the gain of non-inverting amplifier is,

$\frac{V_1}{V_{\text{in}}}=\left(1+\frac{R_{f_1}}{R_1}\right)$        (5)

Write the general expression to calculate the voltage gain in dB.

$A_{\max }=20{\log }_{10}\left(\frac{V_1}{V_{\text{in}}}\right)$

Substitute $10\text{dB}$ for $A_{\max }$ in above equation.

$10\text{dB}=20{\log }_{10}\left(\frac{V_1}{V_{\text{in}}}\right)$

Rearrange the above equation to find $\frac{V_1}{V_{\text{in}}}$.

$\begin{array}{c}\frac{V_1}{V_{\text{in}}}={10}^{\left(\frac{10\text{dB}}{20}\right)}\\ =3.16\end{array}$

Assume the resistor, $R_1=100\Omega$.

Substitute $100\Omega$ for $R_1$ and $3.16$ for $\frac{V_1}{V_{\text{in}}}$ in equation

n (5).

$\begin{array}{l}3.16=\left(1+\frac{R_{f_1}}{100\Omega }\right)\\ 3.16-1=\frac{R_{f_1}}{100\Omega }\\ 2.16=\frac{R_{f_1}}{100\Omega }\end{array}$

Rearrange the above equation to find $R_{f_1}$.

$\begin{array}{c}{R}_{f_1}=2.16\left(100\Omega \right)\\ =216\Omega \end{array}$

Write the expression to calculate the corner frequency of the filter circuit shown in Figure 1.

${\omega }_c=\frac{1}{R_2{C}_1}$        (6)

Assume the capacitor, $C_1=1\mu \text{F}$.

Write the expression to calculate the bandwidth of the filter.

$B={\omega }_H-{\omega }_L$

It is known that the corner frequency ${\omega }_c$ of the low-pass filter is equal to the higher cutoff frequency of the bandpass filter. Therefore, the above equation becomes,

$B={\omega }_c-{\omega }_L$

Substitute $1000\frac{\text{rad}}{\text{s}}$ for $B$ and $100\frac{\text{rad}}{\text{s}}$ for ${\omega }_L$ in above equation

$1000\frac{\text{rad}}{\text{s}}={\omega }_c-100\frac{\text{rad}}{\text{s}}$

Rearrange the above equation to find ${\omega }_c$.

$\begin{array}{c}{\omega }_c=1000\frac{\text{rad}}{\text{s}}+100\frac{\text{rad}}{\text{s}}\\ =1100\frac{\text{rad}}{\text{s}}\end{array}$

Substitute $1\mu \text{F}$ for $C_1$ and $1100\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in equation (6).

$\begin{array}{l}1100\frac{\text{rad}}{\text{s}}=\frac{1}{R_2\left(1\mu \text{F}\right)}\\ 1100\frac{\text{1}}{\text{s}}=\frac{1}{R_2\left(1\times {10}^{-6}\text{F}\right)}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Rearrange the above equation to find $R_2$.

$\begin{array}{c}{R}_2=\frac{1}{\left(1\times {10}^{-6}\text{F}\right)\left(1100\frac{\text{1}}{\text{s}}\right)}\\ =\frac{1}{\left(1100\times {10}^{-6}\right)\text{F}\cdot \left(\frac{\text{1}}{\text{s}}\right)}\\ =\frac{1}{\left(1100\times {10}^{-6}\right)\left(\frac{\text{s}}{\Omega }\right)\cdot \left(\frac{\text{1}}{\text{s}}\right)}\left\{\because 1\text{F}=\frac{1\text{s}}{1\Omega }\right\}\\ =909\Omega \end{array}$

For first stage low-pass filter,

$R_1=100\Omega$

$R_2=909\Omega$

$R_{f_1}=216\Omega$

$C_1=1\mu \text{F}$

High-pass filter design:

For high-pass filter, the voltage gain $\left(A_{\max }\right)$ is $10\text{dB}$.

The active high-pass op amp filter is drawn as Figure 3.

The s-domain circuit of the Figure 3 is drawn as Figure 4 using the equations (1) and (2).

Write the general expression to calculate the transfer function of the circuit in Figure 4.

$H\left(s\right)=\frac{V_{\text{out}}}{V_1}$        (7)

Here,

$V_{\text{out}}$ is the output response of the system, and

$V_1$ is the input response of the system.

Use nodal analysis on node $V_{+}$ in Figure 4.

$\begin{array}{l}\frac{V_{+}-V_1}{\left(\frac{1}{s{C}_2}\right)}+\frac{V_{+}}{R_4}=0\\ s{C}_2{V}_{+}-s{C}_2{V}_1+\frac{V_{+}}{R_4}=0\\ V_{+}\left(s{C}_2+\frac{1}{R_4}\right)=s{C}_2{V}_1\\ V_{+}\left(\frac{1+s{R}_4{C}_2}{R_4}\right)=s{C}_2{V}_1\end{array}$

Rearrange the above equation to find $V_{+}$.

$\begin{array}{c}{V}_{+}=s{C}_2\left(\frac{R_4}{1+s{R}_4{C}_2}\right)V_1\\ =\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)V_1\end{array}$

Use nodal analysis on node $V_{-}$ in Figure 4.

$\begin{array}{l}\frac{V_{-}-0}{R_3}+\frac{V_{-}-V_{\text{out}}}{R_{f_2}}=0\\ \frac{V_{-}}{R_3}+\frac{V_{-}}{R_{f_2}}-\frac{V_{\text{out}}}{R_{f_2}}=0\\ V_{-}\left(\frac{1}{R_3}+\frac{1}{R_{f_2}}\right)=\frac{V_{\text{out}}}{R_{f_2}}\\ V_{-}\left(\frac{R_{f_2}+R_3}{R_3{R}_{f_2}}\right)=\frac{V_{\text{out}}}{R_{f_2}}\end{array}$

Rearrange the above equation to find $V_{-}$.

$\begin{array}{c}{V}_{-}=\frac{V_{\text{out}}}{R_{f_2}}\left(\frac{R_3{R}_{f_2}}{R_{f_2}+R_3}\right)\\ =V_{\text{out}}\left(\frac{R_3}{R_{f_2}+R_3}\right)\end{array}$

It is known that for an ideal operational amplifier,

$V_{+}=V_{-}$

Substitute $\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)V_1$ for $V_{+}$ and $V_{\text{out}}\left(\frac{R_3}{R_{f_2}+R_3}\right)$ for $V_{-}$ in above equation.

$\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)V_1=V_{\text{out}}\left(\frac{R_3}{R_{f_2}+R_3}\right)$

Rearrange the above equation to find $\frac{V_{\text{out}}}{V_1}$.

$\begin{array}{c}\frac{V_{\text{out}}}{V_1}=\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)\left(\frac{R_{f_2}+R_3}{R_3}\right)\\ =\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)\left(\frac{R_{f_2}}{R_3}+\frac{R_3}{R_3}\right)\\ =\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)\left(1+\frac{R_{f_2}}{R_3}\right)\end{array}$

Substitute $\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)\left(1+\frac{R_{f_2}}{R_3}\right)$ for $\frac{V_{\text{out}}}{V_1}$ in equation (7) to find $H\left(s\right)$.

$H\left(s\right)=\left(\frac{s{R}_4{C}_2}{1+s{R}_4{C}_2}\right)\left(1+\frac{R_{f_2}}{R_3}\right)$        (8)

Therefore, the equation (8) is the transfer function of the active high-pass filter that is the product of the transfer function of general high-pass filter and the gain of the non-inverting amplifier.

From equation (8), the gain of non-inverting amplifier is,

$\frac{V_{\text{out}}}{V_1}=1+\frac{R_{f_2}}{R_3}$        (9)

Write the general expression to calculate the voltage gain in dB.

$A_{\max }=20{\log }_{10}\left(\frac{V_{\text{out}}}{V_1}\right)$

Substitute $10\text{dB}$ for $A_{\max }$ in above equation.

$10\text{dB}=20{\log }_{10}\left(\frac{V_{\text{out}}}{V_1}\right)$

Rearrange the above equation to find $\frac{V_{\text{out}}}{V_1}$.

$\begin{array}{c}\frac{V_{\text{out}}}{V_1}={10}^{\left(\frac{10\text{dB}}{20}\right)}\\ =3.16\end{array}$

Assume the resistor, $R_3=100\Omega$.

Substitute $100\Omega$ for $R_3$ and $3.16$ for $\frac{V_{\text{out}}}{V_1}$ in equation (9).

$\begin{array}{l}3.16=1+\frac{R_{f_2}}{\left(100\Omega \right)}\\ 3.16-1=\frac{R_{f_2}}{\left(100\Omega \right)}\\ 2.16=\frac{R_{f_2}}{\left(100\Omega \right)}\end{array}$

Rearrange the above equation to find $R_{f_2}$.

$\begin{array}{c}{R}_{f_2}=2.16\left(100\Omega \right)\\ =216\Omega \end{array}$

Write the expression to calculate the corner frequency of the filter circuit shown in Figure 4.

${\omega }_c=\frac{1}{R_4{C}_2}$        (10)

Assume the capacitor, $C_2=1\mu \text{F}$.

It is known that the corner frequency ${\omega }_c$ of the high-pass filter is equal to the lower cutoff frequency $\left({\omega }_L\right)$ of the bandpass filter. Therefore, ${\omega }_L={\omega }_c$.

Given that, ${\omega }_L=100\frac{\text{rad}}{\text{s}}$

Substitute $1\mu \text{F}$ for $C_2$ and $100\frac{\text{rad}}{\text{s}}$ for ${\omega }_c$ in equation (10).

$\begin{array}{l}100\frac{\text{rad}}{\text{s}}=\frac{1}{R_4\left(1\mu \text{F}\right)}\\ 100\frac{\text{1}}{\text{s}}=\frac{1}{R_4\left(1\times {10}^{-6}\text{F}\right)}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Rearrange the above equation to find $R_4$.

$\begin{array}{c}{R}_4=\frac{1}{\left(1\times {10}^{-6}\text{F}\right)\left(100\frac{\text{1}}{\text{s}}\right)}\\ =\frac{1}{\left(100\times {10}^{-6}\right)\text{F}\cdot \left(\frac{\text{1}}{\text{s}}\right)}\\ =\frac{1}{\left(100\times {10}^{-6}\right)\left(\frac{\text{s}}{\Omega }\right)\cdot \left(\frac{\text{1}}{\text{s}}\right)}\left\{\because 1\text{F}=\frac{1\text{s}}{1\Omega }\right\}\\ =10\times {10}^3\Omega \end{array}$

Simplify the above equation to find $R_4$.

$R_4=10\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}$

For second stage high-pass filter,

$R_3=100\Omega$

$R_4=10\text{k}\Omega$

$R_{f_2}=216\Omega$

$C_2=1\mu \text{F}$

The two-stage op amp band-pass filter circuit is drawn as Figure 5.

Conclusion:

Thus, the two-stage op amp filter circuit with a bandwidth of $1000\frac{\text{rad}}{\text{s}}$, a low cutoff frequency of $100\frac{\text{rad}}{\text{s}}$ and a voltage gain of $20\text{dB}$ is designed.

To determine

Verify the two-stage op amp filter circuit design with an appropriate LTspice simulation.

Explanation of Solution

Calculation:

Create the new schematic in LTspice and draw the circuit in Figure 5 as shown in Figure 6 by entering the corresponding values of resistors and capacitors.

Using SPICE Directive in Edit menu, mention the command .ac dec 1000 0.1 100k and .lib opamp.sub as shown in Figure 7 and Figure 8.

Use Label Net option and enter the Vout. After adding all the commands mentioned above, the circuit is shown as in Figure 9.

Now run the simulation and place the probe at resistor $R_{f_2}$ to calculate the output voltage, and the plot is shown in Figure 10.

Refer to Figure 10 the thick red line indicates magnitude response, the dotted red line indicates phase response, and the black dotted line represents the corner frequency of the magnitude response.

Conclusion:

Thus, the design of the two-stage op amp filter is verified by using the LTspice simulation.