4th order Butterworth LPF with two MFB biquads

Butterworth poles (n = 4) lie at ±22.5° and ±67.5° from the negative real axis on the unit circle:

$$T_N(s) = \frac{1}{(s^2 + 1.8478s + 1)(s^2 + 0.7654s + 1)}$$

so the two biquads have $Q_1 = 1/1.8478 = 0.5412$ and $Q_2 = 1/0.7654 = 1.3066$, with $\omega_0$ = 1000 rad/s.

MFB lowpass biquad

              +----------[R2]-----------+
              |                         |
              |       +----[C2]----+    |
              |       |            |    |
 Vi o--[R1]---+--[R3]-+----(-)     |    |
              |              A ----+----+--o Vo
             [C1]     GND---(+)
              |
             GND

For this multiple-feedback (MFB) lowpass biquad:

$$T(s) = \frac{V_o}{V_i} = \frac{-\dfrac{1}{R_1R_3C_1C_2}}{s^2 + \dfrac{s}{C_1}\left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right) + \dfrac{1}{R_2R_3C_1C_2}}, \qquad T(0) = -\frac{R_2}{R_1}$$

Normalised design ($\omega_0 = 1$, unity DC gain): take $R_1 = R_2 = R_3 = 1\ \Omega$. Then $\omega_0^2 = 1/(C_1C_2) = 1$ and $\omega_0/Q = 3/C_1$, so

$$C_1 = 3Q, \qquad C_2 = \frac{1}{3Q}$$

Scaling

$\omega_0$ = 1000 rad/s. In each section set $C_1$ (the larger capacitor) to 0.1 μF: $k_z = C_{1n}/(\omega_0 \times 10^{-7})$; then $R_1 = R_2 = R_3 = k_z$ and $C_2 = C_{2n}/(k_z\omega_0)$.

Check for section 2: $\omega_0 = 1/\sqrt{R^2C_1C_2} = 1/\sqrt{(39.2\text{k})^2(10^{-7})(6.509 \times 10^{-9})} \approx 1000$ rad/s.

Final circuit

 Vi -->[MFB 1, Q=0.5412]-->[MFB 2, Q=1.3066]--> Vo
         R = 16.24k             R = 39.2k
         C1 = 0.1uF             C1 = 0.1uF
         C2 = 37.94nF           C2 = 6.509nF

Each section has DC gain −1, so the cascade has DC gain +1 (unity) and half-power frequency 1000 rad/s.

Answer: section 1: $R_1 = R_2 = R_3$ = 16.24 kΩ, $C_1$ = 0.1 μF, $C_2$ = 37.94 nF; section 2: $R_1 = R_2 = R_3$ = 39.2 kΩ, $C_1$ = 0.1 μF, $C_2$ = 6.509 nF.