First-order Lowpass Butterworth Filter

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The lowpass filter is a filter that allows the signal with the frequency is lower than the cutoff frequency and attenuates the signals with the frequency is more than cutoff frequency.

In the first-order filter, the number of reactive components is only one. The below figure shows the circuit diagram of the first-order lowpass Butterworth filter.

The low pass Butterworth filter is an active Low pass filter as it consists of the op-amp. This op-amp operates on non-inverting mode. Hence, the gain of the filter will decide by the resistor R₁ and R_F. And the cutoff frequency decides by R and C.

Applying the voltage divider rule at point Va reveals the voltage across the capacitor, calculated as follows:

$V_a = \frac{-jX_C}{R-jX_C} V_{in}$

$V_a = \frac{-j(\frac{1}{2\pi f C})}{R-j (\frac{1}{2\pi f C})} V_{in}$

$V_a = \frac{-j}{2\pi fRC - j} V_{in}$

$V_a = \frac{V_{in}}{1-\frac{2 \pi fRC}{j}}$

$V_a = \frac{V_{in}}{1+j2 \pi fRC}$

Because of the non-inverting configuration of an op-amp,

$V_0 = \left( 1+ \frac{R_f}{R_1} \right) V_a$

$V_0 = \left( 1+ \frac{R_f}{R_1} \right) \frac{V_{in}}{1+j2 \pi fRC}$

$\frac{V_0}{V_a} = \frac{A_f}{1+j\frac{f}{f_c}}$

Where,

$A_f = 1 + \frac{R_F}{R_1}$

A_f = Gain of filter in Passband

$f_c = \frac{1}{2 \pi RC}$

f_c = Cutoff Frequency
f = Operating Frequency

$\frac{V_0}{V_a} = \left|\frac{V_0}{V_a} \right|\angle \phi$

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$\left|\frac{V_0}{V_a} \right| = \frac{A_f}{\sqrt{1+j \left( \frac{f}{f_c} \right) ^2}}$

$\phi = - \tan^{-1} \left( \frac{f}{f_c} \right)$

At very low frequency, f<<f_c
$\left|\frac{V_0}{V_a} \right| \approx A_f (Constant)$
At cutoff frequency, f= f_c
$\left|\frac{V_0}{V_a} \right| = \frac{A_f}{\sqrt{2}} = 0.707A_f$
At high frequency, f> f_c
$\left|\frac{V_0}{V_a} \right| < A_f$

The below figure shows the frequency response of first-order lowpass Butterworth filter.

Second-order Butterworth Filter

The second-order Butterworth filter consists of two reactive components. The circuit diagram of a second-order low pass Butterworth filter is as shown in the below figure.

In this type of filter, resistor R and R_F are the negative feedback of op-amp. And the cutoff frequency of the filter decides by R₂, R₃, C₂, and C₃.

The second-order lowpass Butterworth filter consists of two back-to-back connected RC networks. And R_L is the load resistance.

Ezoic

First-order and second-order Butterworth filters are very important. Because we can get higher-order Butterworth filter by just cascading of the first-order and second-order Butterworth filters.

Let’s analyse the circuit of second-order Butterworth filter,

Apply Kirchhoff’s Current Law at point V₁.

$I_1 = I_2 + I_3$

(1) $\begin{equation*} \frac{V_{in}-V_1}{R_2} = \frac{V_1-V_0}{\frac{1}{sC_2}}+\frac{V_1-V_a}{R_3} \end{equation*}$

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Using potential divider rule at point V_a

$V_a = V_1\left[ \frac{\frac{1}{sC_3}}{R_3+\frac{1}{sC_3}} \right]$

$V_a = V_1\left[ \frac{\frac{1}{sC_3}}{\frac{R_3{sC_3} +{1}}{sC_3}} \right]$

$V_a = \frac{V_1}{1+sR_3C_3}$

$V_1 = V_a (1+sR_3C_3)$

Put the value of V₁ in equation-(1)

$\frac{V_{in}-V_a (1+sR_3C_3)}{R_2} = \frac{V_a (1+sR_3C_3)-V_0}{\frac{1}{sC_2}} + \frac{V_a (1+sR_3C_3)-V_a}{\frac{1}{R_3}}$

$\frac{V_{in}}{R_2} -\frac{V_a (1+sR_3C_3)}{R_2} = \frac{V_a (1+sR_3C_3)}{\frac{1}{sC_2}} -\frac{V_0}{\frac{1}{sC_2}} + \frac{V_a (1+sR_3C_3)}{R_3}-\frac{V_a}{R_3}$

$\frac{V_{in}}{R_2} + \frac{V_0}{\frac{1}{sC_2}} = \frac{V_a (1+sR_3C_3)}{\frac{1}{sC_2}} + \frac{V_a (1+sR_3C_3)}{R_2} + \frac{V_a (1+sR_3C_3)}{R_3} -\frac{V_a}{R_3}$

$\frac{V_{in}}{R_2} +V_0 sC_2 = V_a\left[sC_2(1+sR_3C_3) + \frac{(1+sR_3C_3)}{R_2} + \frac{(1+sR_3C_3)}{R_3} - \frac{1}{R_3} \right]$

$\frac{V_{in} +V_0 sC_2 R_2}{R_2} = V_a\left[ \frac{ R_3 R_2 sC_2(1+sR_3C_3) + R_3(1+sR_3C_3) + R_2(1+sR_3C_3) - R_2}{R_2R_3} \right]$

$R_3(V_{in} +V_0 sC_2 R_2) = V_a\left[ R_3 R_2 sC_2(1+sR_3C_3) + R_3(1+sR_3C_3) + R_2(1+sR_3C_3) - R_2\right]$

$R_3 V_{in} +V_0 sC_2 R_2 R_3 = V_a\left[ (1+sR_3C_3) \left( R_3 R_2 sC_2 + R_3 + R_2 \right) - R_2\right]$

$V_a = \frac{ R_3 V_{in} +V_0 sC_2 R_2 R_3 }{ (1+sR_3C_3) \left( R_3 R_2 sC_2 + R_3 + R_2 \right) - R_2 }$

Because of the non-inverting configuration of an op-amp,

$V_0 = A_f V_a$

Where,

$A_f = 1+\frac{R_f}{R_1} = Gain \, of \, filter \, in \, passband$

$V_0 = A_f\left[ \frac{ R_3 V_{in} +V_0 sC_2 R_2 R_3 }{ (1+sR_3C_3) \left( R_3 R_2 sC_2 + R_3 + R_2 \right) - R_2 } \right]$

$V_0-\frac{A_f V_0 sC_2 R_2 R_3}{(1+sR_3C_3) \left( R_3 R_2 sC_2 + R_3 + R_2 \right) - R_2 } = \frac{A_f R_3 V_{in}}{ (1+sR_3C_3) \left( R_3 R_2 sC_2 + R_3 + R_2 \right) - R_2 }}$

$V_0 \left[(1+sR_3C_3) (R_3 R_2 sC_2 + R_3 + R_2) - R_2 - A_f sC_2 R_2 R_3 \right] = A_f R_3 V_{in}$

$\frac{V_0}{V_{in}} = \frac{A_f R_3}{ \left[(1+sR_3C_3) (R_3 R_2 sC_2 + R_3 + R_2) - R_2 - A_f sC_2 R_2 R_3 \right]}$

Rearrange this equation,

$\frac{V_0}{V_{in}} = \frac{A_f R_3}{ \left[(1+sR_3C_3) (R_2+R_3+sR_2R_3C_2) - R_2 - sA_fR_2 R_3C_2 \right]}$

$\frac{V_0}{V_{in}} = \frac{A_f R_3}{ \left[ (R_2+R_3+sR_2R_3C_2+sR_2R_3C_3+ sR_3^2 C_3+s^2R_2R_3^2C_2C_3)- R_2 - sA_fR_2 R_3C_2 \right]}$

$\frac{V_0}{V_{in}} = \frac{A_f R_3}{ s^2R_2R_3^2C_2C_3+s(R_2R_3C_2 + R_2R_3C_3 + R_3^2C_3 - A_fR_2 R_3C_2) + R_3}$

$\frac{V_0}{V_{in}} = \frac{A_f R_3}{R_2R_3^2C_2C_3 \left(s^2 + s\frac{R_2R_3C_2 + R_2R_3C_3 + R_3^2C_3 - A_fR_2 R_3C_2}{ R_2R_3^2C_2C_3} + \frac{R_3}{R_2R_3^2C_2C_3}\right) }$

$\frac{V_0}{V_{in}} = \frac{A_f }{R_2R_3C_2C_3 \left(s^2 + s\frac{R_2C_2 + R_2C_3 + R_3C_3 - A_fR_2C_2}{ R_2R_3C_2C_3} + \frac{1}{R_2R_3C_2C_3}\right) }$

$\frac{V_0}{V_{in}} = \frac{\frac{A_f}{ R_2R_3C_2C_3}}{\left(s^2 + s\frac{R_2C_2 + R_2C_3 + R_3C_3 - A_fR_2C_2}{ R_2R_3C_2C_3} + \frac{1}{R_2R_3C_2C_3}\right) }$

Compare this equation with the standard form transfer function for second-order Butterworth filter. And that is,

$\frac{V_0}{V_{in}} = \frac{A}{s^2+ 2 \zeta \omega_c s+ \omega_c^2}$

By comparing above equations, we can find the equation of cutoff frequency and overall gain for the second-order lowpass Butterworth filter.

The gain of filter is,

$A_{max} = \frac{A_f}{R_2R_3C_2C_3}$

And the Cutoff frequency of filter is ,

$\omega_c^2 = \frac{1}{R_2R_3C_2C_3}$

$\omega_c = \frac{1}{\sqrt{R_2R_3C_2C_3}}$

$f_c = \frac{1}{2\pi \sqrt{R_2R_3C_2C_3}}$

Now, if we consider the value of R₂ is same as R₃ and the value of C₂ is same as C₃.

$R_2 = R_3 = R \quad and \quad C_2 = C_3 = C$

$f_c = \frac{1}{2\pi R C}$

Now if we put above values in transfer function,

$\frac{V_0}{V_{in}} = \frac{\frac{A_f}{R^2C^2}}{s^2+s\frac{RC+RC+RC-A_f RC}{R^2C^2}+ \frac{1}{ R^2C^2 }}$

$\omega_c = \frac{1}{RC}$

$\frac{V_0}{V_{in}} = \frac{A_f \omega^2}{s^2+s(3-A_f)\omega+ \omega^2}}$

From above equation, the quality factor Q is equal to,

$Q = \frac{1}{3-A_f}$

We can say that, the quality factor is only depends on the gain of filter. And the value of gain should not more than 3. If the value of gain is more than 3, the system will be unstable.

The value of quality factor is 0.707 for the Butterworth filter. And if we put this value in equation of quality factor, we can find the value of gain.

$0.707 = \frac{1}{3-A_f}$

$A_f = 1.586$

$1 + \farc{R_f}{R_1} = 1.586$

$\farc{R_f}{R_1} = 0.586$

While designing the second-order Butterworth filter above relation must be satisfy. The frequency response of this filter is as shown in below figure.

Frequency Response of Second-order Low Pass Butterworth Filter

First-order Lowpass Butterworth Filter

Second-order Butterworth Filter

Published by howdy