Sallen-Key 2nd order Butterworth LPF at 10 kHz

Sallen-Key low-pass circuit and transfer function

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: 2nd-order Butterworth, $f_c = 10$ kHz

From Table 1, the normalized 2nd-order Butterworth function is

$$T(s) = \frac{1}{s^2 + \sqrt{2}\,s + 1} \;\Rightarrow\; \omega_o = 1,\; Q = \frac{1}{\sqrt{2}} = 0.7071$$

For a Butterworth filter the half-power frequency equals $\omega_o$.

Equal-element design (Design 1): take normalized $R_1 = R_2 = 1\ \Omega$, $C_1 = C_2 = 1$ F. Then $\omega_o = 1$ rad/s and

$$\frac{1}{Q} = 1 + 1 + (1 - K) = 3 - K \;\Rightarrow\; K = 3 - \frac{1}{Q}$$ $$K = 3 - \sqrt{2} = 1.586 \;\Rightarrow\; \frac{R_B}{R_A} = K - 1 = 0.586$$

Denormalization:

$$\begin{aligned} k_f &= 2\pi \times 10\,000 = 62\,831.85\ \text{rad/s} \\ C &= \frac{C_n}{k_f k_m} \Rightarrow k_m = \frac{1}{62\,831.85 \times 0.01\times10^{-6}} = 1591.55 \\ R_1 = R_2 &= k_m \times 1 = 1591.55\ \Omega \approx 1.59\ \text{k}\Omega \end{aligned}$$

Choose $R_A = 10\ \text{k}\Omega$, so $R_B = 0.586 \times 10 = 5.86\ \text{k}\Omega$.

Answer: $R_1 = R_2 = 1.59\ \text{k}\Omega$, $C_1 = C_2 = 0.01\ \mu\text{F}$, $R_A = 10\ \text{k}\Omega$, $R_B = 5.86\ \text{k}\Omega$; this gives $f_c = 10$ kHz with $Q = 0.707$.

If unity overall gain is needed, use an input divider of gain $\alpha = 1/1.586 = 0.631$: $R_{1a} = 1.59/0.631 = 2.52\ \text{k}\Omega$ and $R_{1b} = 1.59/0.369 = 4.31\ \text{k}\Omega$.

Other reading of "equal feedback resistors": if it means $R_A = R_B$ (so $K = 2$) with $C_1 = C_2 = 0.01\ \mu$F, the coefficient of $s$ reduces to $1/(R_1C)$, so $Q = \sqrt{R_1/R_2}$. For $Q = 0.7071$ this needs $R_2 = 2R_1$, and $\omega_o = 1/(\sqrt{2}\,R_1C)$ gives $R_1 = 1.125\ \text{k}\Omega$, $R_2 = 2.25\ \text{k}\Omega$, $R_A = R_B = 10\ \text{k}\Omega$. The passband gain is then 2.