Sallen-Key LPF for 1/(s²+0.765s+1)

Sallen-Key low-pass biquad

The Sallen-Key (positive-feedback, VCVS) biquad uses one op-amp as a non-inverting amplifier of gain $K$ with an RC network.

                 C1
          +------||------------+
          |                    |
Vin--R1---+A--R2--+B---(+)     |
                  |     OA >---+--- Vo
                 C2  +-(-)     |
                  |  |         RB
                 GND +---------+
                     |
                     RA
                     |
                    GND

Amplifier gain $K = 1 + R_B/R_A$, so $V_B = V_o/K$.

Derivation. Node B: $\dfrac{V_A - V_B}{R_2} = sC_2V_B \Rightarrow V_A = V_B(1 + sR_2C_2)$.

Node A:

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

Eliminating $V_A$ and $V_B$:

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

so $\omega_0 = \dfrac{1}{\sqrt{R_1R_2C_1C_2}}$ and $\dfrac{\omega_0}{Q} = \dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-K}{R_2C_2}$.

Design for $T(s) = \dfrac{1}{s^2 + 0.765s + 1}$

Step 1: Parameters.

$$\omega_0^2 = 1 \Rightarrow \omega_0 = 1\ \text{rad/s}, \qquad \frac{\omega_0}{Q} = 0.765 \Rightarrow Q = \frac{1}{0.765} = 1.307$$

Step 2: Meaning of the conditions. The capacitors are equal, $C_1 = C_2 = C = 0.01\ \mu$F. The "feedback resistors" are the gain-setting resistors $R_A$ and $R_B$ of the op-amp, so $R_A = R_B$ and

$$K = 1 + \frac{R_B}{R_A} = 2$$

Step 3: Find $R_1$ and $R_2$. With $C_1 = C_2 = C$ and $K = 2$, the $R_2$ terms cancel in the $\omega_0/Q$ expression:

$$\frac{\omega_0}{Q} = \frac{1}{C}\left(\frac{1}{R_1} + \frac{2-K}{R_2}\right) = \frac{1}{R_1C}, \qquad \omega_0 = \frac{1}{C\sqrt{R_1R_2}}$$ $$R_1 = \frac{Q}{\omega_0 C}, \qquad R_2 = \frac{1}{Q\,\omega_0 C}$$

For $\omega_0 = 1$ rad/s and $C = 0.01\ \mu$F $= 10^{-8}$ F: $R_1 = 1.307/10^{-8} = 130.7\ \text{M}\Omega$ and $R_2 = 0.765/10^{-8} = 76.5\ \text{M}\Omega$.

Because the given $T(s)$ is normalized ($\omega_0 = 1$ rad/s), these resistors are impractically large. Frequency scaling to a practical $\omega_0 = 10^4$ rad/s (same capacitors, same $K$) gives

$$R_1 = \frac{1.307}{10^4\times10^{-8}} = 13.07\ \text{k}\Omega, \qquad R_2 = \frac{0.765}{10^4\times10^{-8}} = 7.65\ \text{k}\Omega, \qquad R_A = R_B = 10\ \text{k}\Omega$$

Step 4: Gain. The dc gain of this circuit is $K = 2$, but the given $T(s)$ has unity dc gain. The excess gain is removed by splitting $R_1$ into a divider with $\alpha = 1/K = 0.5$: $R_{1a} = R_1/\alpha = 2R_1$ and $R_{1b} = R_1/(1-\alpha) = 2R_1$ (about 26.1 k$\Omega$ each for $\omega_0 = 10^4$ rad/s). This keeps $\omega_0$ and $Q$ unchanged.

Answer: $C_1 = C_2 = 0.01\ \mu$F, $R_A = R_B$ ($K = 2$), $R_1 = 130.7$ M$\Omega$, $R_2 = 76.5$ M$\Omega$ for $\omega_0 = 1$ rad/s (13.07 k$\Omega$ and 7.65 k$\Omega$ if scaled to $10^4$ rad/s). For unity dc gain, replace $R_1$ by $R_{1a} = R_{1b} = 2R_1$.

Alternative reading (equal $R_1 = R_2$). If "equal resistors" means $R_1 = R_2 = R$ with $C_1 = C_2 = C$, then $\omega_0 = 1/RC$ and $\omega_0/Q = (3-K)/RC$, so $K = 3 - 1/Q = 3 - 0.765 = 2.235$. This needs $R_B = 1.235R_A$ (e.g. $R_A = 10$ k$\Omega$, $R_B = 12.35$ k$\Omega$) and $R = 1/(\omega_0 C) = 100\ \text{M}\Omega$ for $\omega_0 = 1$ rad/s (10 k$\Omega$ for $\omega_0 = 10^4$ rad/s). For unity dc gain, $\alpha = 1/2.235 = 0.4474$, $R_{1a} = 2.235R_1$ and $R_{1b} = 1.810R_1$.