Sallen-Key LPF with ωo 1000 rad/s, Q 0.866, gain 2

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: $\omega_o = 1000$ rad/s, $Q = 0.866$, gain 2

Equal-element design would give $K = 3 - 1/Q = 3 - 1.1547 = 1.845$, which is not 2. So set $K = 2$ ($R_B = R_A$) and $R_1 = R_2 = 1$, and find unequal capacitors (normalized $\omega_o = 1$):

$$\begin{aligned} \omega_o^2 &= \frac{1}{C_1C_2} = 1 \Rightarrow C_2 = \frac{1}{C_1} \\ \frac{1}{Q} &= \frac{2}{C_1} + \frac{1-2}{C_2} = \frac{2}{C_1} - C_1 = 1.1547 \\ C_1^2 &+ 1.1547\,C_1 - 2 = 0 \\ C_1 &= \frac{-1.1547 + \sqrt{1.3333 + 8}}{2} = 0.9502\ \text{F}, \quad C_2 = 1.0524\ \text{F} \end{aligned}$$

Check: $2/0.9502 - 1/1.0524 = 2.1048 - 0.9502 = 1.1547 = 1/Q$.

Denormalization: $k_f = 1000$; choose $k_m = 10^4$ so the capacitors are near 0.1 μF:

$$\begin{aligned} R_1 = R_2 &= 1 \times 10^4 = 10\ \text{k}\Omega \\ C_1 &= \frac{0.9502}{1000\times10^4} = 95.0\ \text{nF} \\ C_2 &= \frac{1.0524}{1000\times10^4} = 105.2\ \text{nF} \end{aligned}$$

Gain $K = 1 + R_B/R_A = 2$: take $R_A = R_B = 10\ \text{k}\Omega$.

Answer: $R_1 = R_2 = R_A = R_B = 10\ \text{k}\Omega$, $C_1 = 95.0$ nF, $C_2 = 105.2$ nF. This gives $\omega_o = 1000$ rad/s, $Q = 0.866$ and DC gain 2.