Sallen-Key LPF at 10 kHz, largest C 0.01 μF

Transfer function of the Sallen-Key low-pass biquad

The Sallen-Key low-pass biquad uses one op-amp wired as a non-inverting amplifier of gain $K = 1 + R_B/R_A$, two resistors $R_1, R_2$ in series, a capacitor $C_1$ fed back from the output to node A, and a grounded capacitor $C_2$ at node B.

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

Derivation (ideal op-amp, so $V_o = K V_B$ and no current enters the + input):

Node B: the current through $R_2$ flows into $C_2$:

$$\frac{V_A - V_B}{R_2} = sC_2 V_B \;\Rightarrow\; V_A = V_B(1 + sR_2C_2)$$

Node A (KCL):

$$\frac{V_{in} - V_A}{R_1} = \frac{V_A - V_B}{R_2} + sC_1 (V_A - V_o)$$

Substituting $V_A$ and $V_B = V_o/K$ and simplifying:

$$T(s) = \frac{V_o}{V_{in}} = \frac{\dfrac{K}{R_1R_2C_1C_2}}{s^2 + s\left(\dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-K}{R_2C_2}\right) + \dfrac{1}{R_1R_2C_1C_2}}$$

Comparing with $T(s) = \dfrac{K\omega_o^2}{s^2 + (\omega_o/Q)s + \omega_o^2}$:

$$\begin{aligned} \omega_o &= \frac{1}{\sqrt{R_1R_2C_1C_2}} \\ \frac{\omega_o}{Q} &= \frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2} \end{aligned}$$

DC gain $= K$.

Design: $f_c = 10$ kHz, largest $C = 0.01\ \mu$F, overall gain 1

Assume a Butterworth response (Table 1): $T(s) = \dfrac{1}{s^2 + \sqrt2 s + 1}$, so $\omega_o = 1$ and $Q = 0.7071$.

Gain 1 needs $K = 1$ ($R_B = 0$, $R_A$ open: a voltage follower). This is the unity-gain design (Design 2). Take $R_1 = R_2 = 1\ \Omega$:

$$\begin{aligned} \omega_o^2 &= \frac{1}{C_1C_2} = 1 \\ \frac{\omega_o}{Q} &= \frac{2}{C_1} + 0 = \sqrt2 \end{aligned} \;\Rightarrow\; C_1 = 2Q = 1.414\ \text{F},\; C_2 = \frac{1}{2Q} = 0.7071\ \text{F}$$

The largest capacitor is $C_1$ (the feedback capacitor).

Scaling:

$$\begin{aligned} k_f &= 2\pi\times 10^4 = 62\,831.85 \\ k_m &= \frac{1.414}{62\,831.85 \times 0.01\times10^{-6}} = 2250.8 \\ C_1 &= 0.01\ \mu\text{F}, \quad C_2 = \frac{0.7071}{k_f k_m} = 0.005\ \mu\text{F} \\ R_1 = R_2 &= 2250.8\ \Omega \approx 2.25\ \text{k}\Omega \end{aligned}$$

Answer: $R_1 = R_2 = 2.25\ \text{k}\Omega$, $C_1 = 0.01\ \mu$F (output feedback), $C_2 = 5$ nF (to ground), op-amp as a voltage follower. DC gain = 1, $f_c = 10$ kHz, $Q = 0.707$.