BMEN90033 · Week 8
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BMEN90033 · WEEK 8 · SALLEN-KEY FILTERS

A second-order response from a single op-amp.

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.

passive cascade buffered cascade Sallen-Key complex poles Q tuning
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01the Q parameter

The quality factor and second-order damping

Every second-order low-pass on this page reduces to a standard form with natural frequency $\omega_0$ and quality factor $Q$:

$$H(s) \;=\; \frac{1}{\dfrac{s^{2}}{\omega_0^{\,2}} \;+\; \dfrac{s}{Q\,\omega_0} \;+\; 1}.$$

$\omega_0$ 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.

Pole locations.

The denominator factors to give the pole pair

$$s \;=\; -\,\frac{\omega_0}{2Q} \;\pm\; j\,\omega_0\sqrt{\,1 - \frac{1}{4Q^{2}}\,}.$$

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:

Time-domain response.

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).

The same $Q$ governs the time-domain overshoot and the frequency-domain magnitude peak. At $Q = 2$ the magnitude peaks near $\omega_0$ by about $6\,\text{dB}$ and the step overshoot is $44\%$. Both descriptions reflect the same pole pair.
Q 1.00
underdamped · complex conjugate pair · ringing
s-plane · pole pair on |s| = ω0
step response · y(t) for a unit step
02passive cascade

Interstage loading in a passive RC cascade

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 $C_1 = C_2 = C$:

$$H_{\text{loaded}}(s) \;=\; \frac{1}{1 + 3sRC + s^{2}R^{2}C^{2}},$$

against the ideal product

$$H_{\text{ideal}}(s) \;=\; \frac{1}{(1+sRC)^{2}} \;=\; \frac{1}{1 + 2sRC + s^{2}R^{2}C^{2}}.$$

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.

Bound on the corner sharpness.

Any second-order all-pole RC network has a quality factor bounded by

$$Q \;\le\; \tfrac{1}{2}.$$

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.

The plot shows the loaded response (red) against the ideal product (blue). The ideal curve is itself softer than Butterworth, since two coincident real poles already give $Q = \tfrac{1}{2}$. The loaded curve lies below it across the passband.
bode response · loaded passive 2RC vs ideal product
03buffered active cascade

The buffered RC cascade

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:

$$H_{\text{buffered}}(s) \;=\; \frac{1}{(1+sRC)^{2}}.$$

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.

Cost of the buffer.

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 plot compares the buffered cascade (blue) against the passive baseline (faded red). The buffered curve coincides with the ideal product. The hardware difference is one op-amp per interstage boundary, with its supply and bias network.
bode response · buffered cascade vs passive
04the Sallen-Key topology

The Sallen-Key low-pass topology

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.

Transfer function.

Applying KCL at node $A$ and at $V_{+}$, with $V_{\text{out}} = V_{+}$ for the unity-gain buffer:

$$H(s) \;=\; \frac{V_{\text{out}}(s)}{V_{\text{in}}(s)} \;=\; \frac{1}{1 + sC_2(R_1+R_2) + s^{2}\,R_1 R_2 C_1 C_2}.$$

Matching the standard form $H(s) = 1\big/\bigl(s^{2}/\omega_0^{2} + s/(Q\omega_0) + 1\bigr)$ gives

$$\omega_0 = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}, \qquad Q = \frac{\sqrt{R_1 R_2 C_1 C_2}}{C_2(R_1+R_2)}.$$

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.

Comparison.

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.

The passive network is rearranged, not extended. The single positive-feedback path through $C_1$ is the only structural difference from the buffered cascade of section 03, and is responsible for both the impedance isolation and the placement of the complex pole pair.
bode response · passive vs buffered vs Sallen-Key
05tuning the response

Tuning Q via the capacitor ratio

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.

The $-3\,\text{dB}$ frequency stays near $\omega_0$ across the Bessel-to-Butterworth range and departs as $Q$ exceeds unity. For Butterworth the $-3\,\text{dB}$ point is exactly $\omega_0$; for Chebyshev the corner is defined by the passband ripple specification rather than the half-power point.
Q (Sallen-Key) 0.71 passive cascade Bessel / Butterworth / Cheby
s-plane · Sallen-Key pole pair
bode response · Sallen-Key vs references