Excess gain compensation in Sallen-Key circuit

In the equal-component Sallen-Key low-pass design the amplifier gain is fixed by the required $Q$: $K = 3 - 1/Q$. The filter's dc gain therefore equals $K$, which is often more than the gain wanted. The excess gain is removed without disturbing $\omega_0$ and $Q$ by replacing the input resistor $R_1$ with a voltage divider (gain reduction).

Sallen-Key low-pass circuit

                 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 $\dfrac{\omega_0}{Q} = \dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-K}{R_2C_2}$. For the equal-component design ($R_1 = R_2 = R$, $C_1 = C_2 = C$): $\omega_0 = 1/RC$ and $\omega_0/Q = (3-K)/RC$, so $K = 3 - 1/Q$.

Compensation by input voltage divider

Suppose the required gain is $H < K$. Replace $R_1$ by two resistors $R_{1a}$ (series) and $R_{1b}$ (to ground):

Vin--R1a--+--> node A        Thevenin equivalent:
          |                  Vth = alpha*Vin
         R1b                 Rth = R1a || R1b = R1
          |
         GND

The Thevenin equivalent seen from node A is a source $\alpha V_{in}$ in series with $R_{1a}\parallel R_{1b}$, where

$$\alpha = \frac{R_{1b}}{R_{1a} + R_{1b}}, \qquad R_{1a}\parallel R_{1b} = \frac{R_{1a}R_{1b}}{R_{1a}+R_{1b}}$$

If we make $R_{1a}\parallel R_{1b} = R_1$, the network seen by node A is unchanged, so $\omega_0$ and $Q$ are unchanged, and the transfer function becomes

$$\frac{V_o}{V_{in}} = \alpha\,T(s) \;\Rightarrow\; \text{dc gain} = \alpha K = H \;\Rightarrow\; \alpha = \frac{H}{K}$$

Solving the two conditions:

$$R_{1a} = \frac{R_1}{\alpha}, \qquad R_{1b} = \frac{R_1}{1-\alpha}$$

Check: $R_{1a}\parallel R_{1b} = \dfrac{R_1^2/[\alpha(1-\alpha)]}{R_1/[\alpha(1-\alpha)]} = R_1$.

Example

For a 2nd order Butterworth filter ($Q = 0.7071$), $K = 3 - 1.414 = 1.586$. For unity overall gain ($H = 1$): $\alpha = 1/1.586 = 0.6306$, so $R_{1a} = R_1/0.6306 = 1.586R_1$ and $R_{1b} = R_1/0.3694 = 2.707R_1$. With $R_1 = 10$ k$\Omega$: $R_{1a} = 15.86$ k$\Omega$, $R_{1b} = 27.07$ k$\Omega$.

Notes

• Only the input resistor changes; $C_1$, $C_2$, $R_2$ and the gain resistors $R_A$, $R_B$ stay the same.
• This method can only reduce gain ($\alpha < 1$). If more gain is needed, gain enhancement (feeding $C_1$ from a tap of an output divider) or a separate amplifier stage is used.