Gain enhancement in Sallen-Key circuit

In the equal-component Sallen-Key low-pass filter the amplifier gain is fixed by $Q$ ($K = 3 - 1/Q$), so the dc gain is fixed too. Gain enhancement raises the overall gain above $K$ while keeping $\omega_0$ and $Q$ unchanged.

Basic idea

The RC network only "sees" the voltage at the (+) input and the voltage fed back through $C_1$. Suppose the op-amp output is made larger by a factor $1/\beta$, but only the fraction $\beta V_o$ is fed back to $C_1$, with the effective gain $\beta\mu$ set equal to $K$. The RC network then works exactly as before, while the output is $1/\beta$ times larger.

Circuit

              C1
          +---||-------------------+
          |                        |
Vin--R1---+A--R2--+B--(+)          |
                  |     OA(mu)--+--|--- Vo
                 C2             |  |
                  |             Rc |
                 GND            +--+ Vo' = beta*Vo
                                |
                                Rd
                                |
                               GND
 OA(mu): non-inverting amp, mu = 1 + RB/RA

The op-amp output $V_o$ drives a divider $R_c$, $R_d$; capacitor $C_1$ is returned to the tap $V_o' = \beta V_o$ where $\beta = \dfrac{R_d}{R_c + R_d} < 1$ (keep $R_c\parallel R_d$ small compared with the impedance of $C_1$, or include it in the design).

Derivation

Let the non-inverting amplifier gain be $\mu = 1 + R_B/R_A$, so $V_o = \mu V_B$ and $V_o' = \beta\mu V_B$. The RC network now sees an "effective gain" $\beta\mu$ from node B to the point where $C_1$ returns. Writing the node equations exactly as for the normal circuit with $V_o$ replaced by $V_o'$:

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

For the required $Q$, set $\beta\mu = K = 3 - 1/Q$ (equal-component case). Then $\omega_0$ and $Q$ are the same as before, and the output is

$$\frac{V_o}{V_{in}} = \frac{1}{\beta}\cdot\frac{V_o'}{V_{in}} \;\Rightarrow\; \text{dc gain} = \frac{K}{\beta} > K$$

Design steps

1. Design the normal equal-component Sallen-Key for $\omega_0$ and $Q$; this gives $K = 3 - 1/Q$.
2. For a required dc gain $H > K$, choose $\beta = K/H$.
3. Set the op-amp gain $\mu = 1 + R_B/R_A = H$ and choose the divider $R_d/(R_c+R_d) = \beta$.

Example: Butterworth ($Q = 0.707$, $K = 1.586$) with required gain $H = 5$: $\beta = 1.586/5 = 0.317$, $\mu = 5$ ($R_B = 4R_A$), and the divider is $R_c = 2.15R_d$ (e.g. $R_d = 1$ k$\Omega$, $R_c = 2.15$ k$\Omega$).

For gain reduction (the opposite case) the input resistor $R_1$ is split into a voltage divider instead.