Sallen-Key 2nd order Butterworth (design 2 method)

In Design 2 (unity-gain design) of the Sallen-Key low-pass biquad, the op-amp is a voltage follower ($K = 1$) and the two resistors are equal; the capacitors are then chosen to set $Q$.

              C1 (feedback)
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         |
                         RA
                         |
                        GND

(For $K = 1$: $R_B = 0$ and $R_A$ is removed.)

Transfer function with K = 1:

$$T(s) = \frac{\dfrac{1}{R_1R_2C_1C_2}}{s^2 + s\left(\dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1}\right) + \dfrac{1}{R_1R_2C_1C_2}}$$

Normalized design ($\omega_o = 1$): 2nd-order Butterworth $T(s) = \dfrac{1}{s^2 + \sqrt2 s + 1}$, $Q = 0.7071$. With $R_1 = R_2 = 1\ \Omega$:

$$\begin{aligned} \omega_o^2 &= \frac{1}{C_1C_2} = 1 \\ \frac{\omega_o}{Q} &= \frac{2}{C_1} = \sqrt2 \\ \Rightarrow C_1 &= 2Q = 1.414\ \text{F}, \quad C_2 = \frac{1}{2Q} = 0.7071\ \text{F} \end{aligned}$$

Capacitor ratio $C_1/C_2 = 4Q^2 = 2$.

Denormalization: no frequency is given, so assume $f_c = 1$ kHz and make the largest capacitor $C_1 = 0.01\ \mu$F:

$$\begin{aligned} k_f &= 2\pi\times1000 = 6283.19 \\ k_m &= \frac{1.414}{6283.19 \times 10^{-8}} = 22\,508 \\ R_1 = R_2 &= 22.5\ \text{k}\Omega \\ C_1 &= 0.01\ \mu\text{F}, \quad C_2 = \frac{0.7071}{6283.19 \times 22\,508} = 5\ \text{nF} \end{aligned}$$

Answer: $R_1 = R_2 = 22.5\ \text{k}\Omega$, $C_1 = 10$ nF, $C_2 = 5$ nF, unity-gain follower. This gives a Butterworth response with $f_c = 1$ kHz and DC gain 1.

Features of Design 2: the gain is 1, so no gain resistors are needed and the op-amp works as a follower (wide bandwidth). The follower gain is very close to 1, so gain error has little effect. In general $S^{Q}_{K} = 2Q^2$ (equal to 1 for a Butterworth filter), and the capacitor spread $C_1/C_2 = 4Q^2$ grows quickly at high $Q$.