Sallen-Key LPF; obtaining HPF; 12 kHz design

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 $= K$.

Obtaining a high-pass filter from the low-pass one

Use the RC-CR transformation: replace every resistor $R_i$ by a capacitor of value $1/R_i$ and every capacitor $C_j$ by a resistor of value $1/C_j$ (normalized values, $\omega_o = 1$). The op-amp gain resistors $R_A, R_B$ set only a ratio, so they stay unchanged. This replaces $s$ by $1/s$, so for the normalized function

$$T_{LP}(s) = \frac{K}{s^2 + \frac{1}{Q}s + 1} \;\Rightarrow\; T_{HP}(s) = T_{LP}\!\left(\frac{1}{s}\right) = \frac{K s^2}{s^2 + \frac{1}{Q}s + 1}$$

Frequency scaling by $\omega_o$ then gives $\dfrac{K s^2}{s^2 + (\omega_o/Q)s + \omega_o^2}$. In the high-pass circuit, capacitors sit in the series positions and resistors in the feedback and grounded positions.

Design: Butterworth LPF at 12 kHz

$T_2(s) = \dfrac{1}{s^2 + \sqrt2 s + 1}$: $\omega_o = 1$, $Q = 0.7071$.

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

$$\frac{1}{Q} = 1 + 1 + (1 - K) = 3 - K \;\Rightarrow\; K = 3 - \frac{1}{Q}$$ $$K = 3 - 1.414 = 1.586, \quad R_B/R_A = 0.586$$

Scaling (choose $C = 0.01\ \mu$F):

$$\begin{aligned} k_f &= 2\pi \times 12\,000 = 75\,398.2\ \text{rad/s} \\ k_m &= \frac{1}{75\,398.2 \times 10^{-8}} = 1326.3 \\ R_1 = R_2 &= 1.326\ \text{k}\Omega \end{aligned}$$

$R_A = 10\ \text{k}\Omega$, $R_B = 5.86\ \text{k}\Omega$.

Answer: $R_1 = R_2 = 1.33\ \text{k}\Omega$, $C_1 = C_2 = 0.01\ \mu$F, $R_A = 10\ \text{k}\Omega$, $R_B = 5.86\ \text{k}\Omega$; $f_c = 12$ kHz, DC gain 1.586.

If unity gain is needed (as in $T_2(s)$), use an input divider with $\alpha = 1/1.586 = 0.631$: $R_{1a} = 1.326/0.631 = 2.10\ \text{k}\Omega$ and $R_{1b} = 1.326/0.369 = 3.59\ \text{k}\Omega$.