Why gain enhancement is needed in Sallen-Key biquad

Why gain enhancement is needed

In the Sallen-Key low-pass biquad the gain $K$ is not a free design parameter: it also sets $Q$.

$$\frac{\omega_o}{Q} = \frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}$$

• Equal-element design: $K = 3 - 1/Q$, so $K$ always lies between 1 and 3 (e.g. 1.586 for Butterworth).
• Unity-gain design (Design 2): $K = 1$.

Once $Q$ is chosen, the passband gain is fixed. If the specification needs a larger gain (say 10), the circuit cannot give it directly, so the gain must be enhanced without changing $\omega_o$ and $Q$. (If the gain is too large, it is reduced with an input voltage divider instead.)

Gain enhancement in the Sallen-Key low-pass biquad

The idea is to let the op-amp amplify by a larger factor $\mu$ while the feedback capacitor $C_1$ still sees only the voltage $K V_B$ that the design needs.

1. Set the op-amp's non-inverting gain to $\mu = 1 + R_B/R_A$, larger than the design value $K$.
2. Connect a resistive divider $R_x$, $R_y$ across the output. Return $C_1$ to the tap of this divider (not to $V_o$). The tap voltage is

$$V_t = \beta V_o, \qquad \beta = \frac{R_y}{R_x + R_y}$$

3. Choose $\beta = K/\mu$, so that $V_t = \beta\mu V_B = K V_B$.

               C1
        +------||-----+ tap (beta*Vo)
        |             |
Vin -R1-+- R2 -+------|+\       Rx
               |      |  >--+--/\/--+
              === C2 +|-/   |       |
               |     |      Vo     Ry
              GND  RA/RB           |
                   network        GND

Analysis: the node equations are the same as in the normal circuit, with $V_o$ replaced by $\beta V_o = K V_B$ at $C_1$. So the denominator (and therefore $\omega_o$ and $Q$) is unchanged:

$$\frac{V_o}{V_{in}} = \frac{\mu\,\omega_o^2}{s^2 + (\omega_o/Q)s + \omega_o^2}, \qquad \mu = \frac{K}{\beta}$$

The passband gain rises from $K$ to $K/\beta$.

Design steps:

1. Design the normal circuit for the required $\omega_o$ and $Q$, which gives $K$.
2. Required gain $H$: set $\mu = H$, i.e. $R_B/R_A = H - 1$.
3. Divider ratio $\beta = K/H$.
4. Keep $R_x \parallel R_y$ much smaller than the reactance $1/(\omega_o C_1)$ so the divider is not loaded by $C_1$.

Example: Butterworth equal-element design ($K = 1.586$) needing $H = 10$: $R_B/R_A = 9$ and $\beta = 0.1586$.

A simpler alternative is to cascade the biquad with a non-inverting amplifier of gain $H/K$, at the cost of one more op-amp.