A two-stage passive RC cascade is degraded by interstage loading. Buffering each stage restores the ideal product but adds one op-amp per stage. The Sallen-Key topology produces the same second-order response, with adjustable damping, from a single op-amp.
Every second-order low-pass on this page reduces to a standard form with natural frequency $\omega_0$ and quality factor $Q$:
$\omega_0$ sets the corner frequency. $Q$ sets the response shape near the corner. The two parameters are independent: $\omega_0$ is fixed first, then $Q$ is chosen to trade passband flatness against settling time and transition sharpness.
The denominator factors to give the pole pair
The pair lies on a circle of radius $\omega_0$ for every $Q$; the angle off the negative real axis is set by $Q$. Three regimes follow:
The right panel shows the response to a unit step. Three quantities summarise the underdamped case, with $\zeta = 1/(2Q)$:
The slider drives both panels. Reference values used later: $Q = 0.5$ (passive cascade limit), $Q = 1/\sqrt{3} \approx 0.58$ (Bessel), $Q = 1/\sqrt{2} \approx 0.71$ (Butterworth), and $Q > 1$ (Chebyshev, peaking).
Cascading two first-order RC sections is the simplest way to obtain a steeper roll-off. If the stages were isolated, the transfer function would be $1/(1+sRC)^{2}$, with a $-40\,\text{dB/decade}$ asymptote and a repeated pole at $s = -1/RC$. In practice the second stage draws current from the first capacitor and the isolation assumption fails.
Solving the two-node network with $R_1 = R_2 = R$ and $C_1 = C_2 = C$:
against the ideal product
The middle coefficient increases from $2$ to $3$. The repeated pole at $s = -1/RC$ splits into two real poles at $s = (-3 \pm \sqrt{5})/(2RC)$, approximately $-0.382/RC$ and $-2.618/RC$. The slow pole pulls the passband edge inward; the fast pole moves outward but does not compensate.
Any second-order all-pole RC network has a quality factor bounded by
The bound is reached by a coincident real pole pair, i.e. the ideal product above. Loading reduces $Q$ further, so the passband sags well before the cutoff: at $\omega = 1/RC$ the magnitude has already fallen below $-6\,\text{dB}$. A sharper corner requires complex conjugate poles, which a purely RC network cannot produce.
An ideal voltage follower has infinite input impedance and zero output impedance. Inserting one between the RC sections forces the input current of the second stage to zero, so the first stage's output is independent of the load. The product rule then holds exactly:
This is the ideal product from the previous section. The asymptotic roll-off is $-40\,\text{dB/decade}$ and the $-3\,\text{dB}$ frequency is $\omega_{-3} = \omega_{c}\sqrt{2^{1/2} - 1} \approx 0.64\,\omega_{c}$. The passband no longer sags.
A second-order section now contains two op-amps: one buffer between the stages and a second at the output. The active device count scales linearly with order. The composite transfer function is still a product of coincident real poles, so the corner remains soft ($Q = \tfrac{1}{2}$) regardless of order. The buffered cascade restores the ideal product but does not sharpen the corner.
The Sallen-Key low-pass retains the two resistors and two capacitors of the passive network and adds one op-amp wired as a unity-gain buffer at the output. The structural change is the routing of $C_1$, which now connects from internal node $A$ to $V_{\text{out}}$ rather than to ground. At low frequency $C_1$ is effectively open and the network reduces to the passive cascade. At high frequency $C_1$ is a short, feeding the op-amp output back into node $A$.
The feedback is positive: the op-amp drives node $A$ in the same direction as $V_{\text{in}}$. Its strength, set by $C_1$, cancels the loading interaction of the passive cascade and places a complex pole pair. The same op-amp also provides the output buffering, so a single device performs both isolation and pole placement where the buffered cascade required two.
Applying KCL at node $A$ and at $V_{+}$, with $V_{\text{out}} = V_{+}$ for the unity-gain buffer:
Matching the standard form $H(s) = 1\big/\bigl(s^{2}/\omega_0^{2} + s/(Q\omega_0) + 1\bigr)$ gives
Setting $R_1 = R_2 = R$ reduces this to $\omega_0 = 1\big/\bigl(R\sqrt{C_1 C_2}\bigr)$ and $Q = \tfrac{1}{2}\sqrt{C_1/C_2}$. $\omega_0$ is fixed by the geometric mean of the capacitors and $Q$ by their ratio. Feedback through $C_1$ removes the $Q \le \tfrac{1}{2}$ constraint of the passive cascade, so $Q$ can take any value in the standard range.
The plot overlays three responses with $\omega_0 = 1\,\text{rad/s}$: the loaded passive cascade, the buffered cascade $1/(1+sRC)^{2}$, and a Sallen-Key tuned to $Q = 1/\sqrt{2}$ (Butterworth). The Sallen-Key trace stays flat past the corner where the buffered cascade has already dropped, then rolls off at the same $-40\,\text{dB/decade}$. A single op-amp produces a sharper corner than two op-amps in the buffered topology.
With $R_1 = R_2 = R$ and $\omega_0 = 1\,\text{rad/s}$ fixed, the capacitors satisfy $C_1 C_2 = 1/(R^{2}\omega_0^{2})$. Sweeping the ratio $C_1/C_2$ moves the pole pair along a vertical line in the s-plane, varying $Q$ without changing $\omega_0$.
The plot shows the Sallen-Key response (amber) against three reference responses:
The slider sets the Sallen-Key $Q$. At $Q = 0.5$ the curve coincides with the buffered cascade of section 03, since both have coincident real poles. Increasing $Q$ flattens the passband and then introduces a peak. A single topology realises all three classical responses, with the capacitor ratio as the only distinguishing parameter.