Sallen-Key LP derivation; MFB design of 5/(s²+1.2s+1)

Sallen-Key LP biquad and its 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 $= K$.

MFB low-pass design for $T(s) = 5/(s^2 + 1.2s + 1)$

The multiple-feedback (MFB, Rauch) low-pass circuit uses an inverting op-amp: $R_1$ from $V_{in}$ to node A, $C_1$ from A to ground, $R_3$ from A to $V_o$, $R_2$ from A to the inverting input, and $C_2$ from the inverting input to $V_o$.

              +-----R3-----------+
              |                  |
              |      +---||--+   |
              |      |   C2  |   |
Vin --R1--+--(A)--R2-+--|-\   |   |
          |             |  >--+---+-- Vo
         === C1    GND--|+/
          |
         GND

Its transfer function is

$$T(s) = \frac{-\dfrac{1}{R_1R_2C_1C_2}}{s^2 + \dfrac{s}{C_1}\left(\dfrac1{R_1} + \dfrac1{R_2} + \dfrac1{R_3}\right) + \dfrac{1}{R_2R_3C_1C_2}}, \qquad |H_{DC}| = \frac{R_3}{R_1}$$

Given: $\omega_o = 1$, $\omega_o/Q = 1.2$ ($Q = 0.833$), $H = 5$.

For real resistors the capacitor ratio must satisfy $C_1/C_2 \ge 4Q^2(1+H) = 4(0.694)(6) = 16.67$. Take $C_2 = 1$ F and $C_1 = 16.67$ F (the minimum, which gives equal roots):

$$\begin{aligned} R_2R_3 &= \frac{1}{C_1C_2} = 0.06 \;\Rightarrow\; R_3 = \frac{0.06}{R_2}, \qquad R_1 = \frac{R_3}{5} = \frac{0.012}{R_2} \\ \frac{1}{C_1}\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right) &= 1.2 \;\Rightarrow\; \frac{1}{16.67}\left(100R_2 + \frac{1}{R_2}\right) = 1.2 \;\Rightarrow\; 100R_2 + \frac{1}{R_2} = 20 \\ (10R_2 - 1)^2 &= 0 \;\Rightarrow\; R_2 = 0.1\ \Omega,\; R_3 = 0.6\ \Omega,\; R_1 = 0.12\ \Omega \end{aligned}$$

(The last step multiplies the previous equation by $R_2$.)

Check: $\dfrac{1}{R_1R_2C_1C_2} = \dfrac{1}{0.12\times0.1\times16.67} = 5$, $\dfrac{1}{C_1}(8.33 + 10 + 1.67) = 1.2$, $\dfrac{1}{R_2R_3C_1C_2} = 1$. Correct.

Practical values: the given $T(s)$ is normalized, so assume $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and $k_m = 10^4$:

Answer: $R_1 = 0.12$, $R_2 = 0.1$, $R_3 = 0.6\ \Omega$, $C_1 = 16.67$, $C_2 = 1$ F (normalized); with $k_m = k_f = 10^4$: $R_1 = 1.2\ \text{k}\Omega$, $R_2 = 1\ \text{k}\Omega$, $R_3 = 6\ \text{k}\Omega$, $C_1 = 166.7$ nF, $C_2 = 10$ nF. The gain is $-5$; the minus sign is a 180° phase shift. Follow the stage with a unity-gain inverter if $+5$ is needed.