5th order Butterworth LPF with first-order and MFB sections

Butterworth poles (n = 5) lie on the unit circle at 0°, ±36° and ±72° from the negative real axis, so

$$T_N(s) = \frac{1}{(s+1)(s^2 + 0.618s + 1)(s^2 + 1.618s + 1)}$$

Each quadratic $s^2 + (1/Q)s + 1$ gives $Q = 1/0.618 = 1.618$ and $Q = 1/1.618 = 0.618$. The filter is one first-order section followed by two MFB biquads; $\omega_0 = 2\pi(1000) = 6283.2$ rad/s.

First-order section

An inverting lossy integrator: input $R_1$, feedback $R_f \parallel C$:

$$T_1(s) = -\frac{1/(R_1C)}{s + 1/(R_fC)}$$

Normalised: $R_1 = R_f = 1\ \Omega$, $C = 1$ F (pole at 1, gain 1). Scaling with $C$ = 0.1 μF: $R = 1/(\omega_0 C) = 1/(6283.2 \times 10^{-7})$ = 1.592 kΩ for both $R_1$ and $R_f$.

          +---[Rf]---+
          +---| C |--+
 Vi-[R1]--+--(-)     |
              A -----+--o
     GND --(+)

MFB biquads

              +----------[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: $\omega_0$ = 6283.2 rad/s. For each section choose $k_z$ so that its largest capacitor $C_1$ becomes 0.1 μF: $k_z = C_{1n}/(\omega_0 \times 10^{-7})$. Then all three resistors are $k_z$ and $C_2 = C_{2n}/(k_z\omega_0)$. (Each section is driven by an op-amp output, so sections may be scaled separately.)

Final circuit

 Vi -->[1st order]-->[MFB, Q=1.618]-->[MFB, Q=0.618]--> Vo
       R=1.592k        R=7.726k          R=2.951k
       C=0.1uF         C1=0.1uF          C1=0.1uF
                       C2=4.244nF        C2=29.09nF

Each stage has DC gain −1, so the overall DC gain magnitude is 1 (sign −1; add a unity inverter if a positive output is needed). Half-power frequency 1 kHz.

Answer: first-order: R = 1.592 kΩ, C = 0.1 μF; MFB 1 (Q = 1.618): R = 7.726 kΩ, C1 = 0.1 μF, C2 = 4.244 nF; MFB 2 (Q = 0.618): R = 2.951 kΩ, C1 = 0.1 μF, C2 = 29.09 nF.