5.04 Active Filter Realization and Sallen-Key RC Networks

5.03 Chebyshev & Elliptic Filter Approximations | 5.05 Analog Frequency Transformations

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5.04 Active Filter Realization & Sallen-Key RC Networks

Core Idea

Passive filter topologies containing resistors, capacitors, and inductors ($RLC$) become highly bulky, expensive, and lossy at low frequencies (e.g., audio or medical ranges) because low-frequency inductors require large physical iron cores. Active filter realization eliminates inductors entirely by using operational amplifiers (op-amps) alongside resistors and capacitors ($RC$) [5.1]. The Sallen-Key topology is the classic second-order active filter structure used to implement complex conjugate poles without inductors, providing excellent frequency selectivity, high input impedance, and zero loading effects [1.298, 4.46].

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1. Passive vs. Active Filter Realizations

Before deriving the active filter equations, it is critical for examinations to understand why modern signal processing relies heavily on active realizations rather than classical passive $RLC$ circuits [1.41].

1.1 Limitations of Passive $RLC$ Filters

1. Bulky and Heavy Inductors: At low frequencies ($f<100\text{ kHz}$), the required inductance values are on the order of millihenries or henries. Achieving these values with passive coils requires large winding counts and heavy ferromagnetic cores, which are physically impractical [1.41].
2. High Losses (Low $Q$): Physical inductors have significant winding resistance. This internal resistance dissipates power, degrades the sharpness of the filter (lowers the quality factor $Q$), and deviates the circuit from its ideal mathematical model.
3. Insertion Loss (No Gain): Passive filters cannot amplify signals. The maximum gain of a passive network is always less than or equal to $1$ ($0\text{ dB}$).
4. Severe Loading Effects: Passive filters are highly sensitive to source and load impedances. Connecting a load resistor directly to a passive filter alters its pole-zero locations, distorting the frequency response and necessitating complex impedance-matching networks.

1.2 Advantages of Active $RC$ Realizations

• Inductor Elimination: Active filters replace bulky inductors with a combination of active op-amps and passive $RC$ elements, significantly reducing weight, cost, and physical size [1.41].
• High Input & Low Output Impedance: Op-amps act as physical isolating buffers. They draw negligible current from the source (infinite input impedance) and can drive heavy loads easily (near-zero output impedance), completely eliminating loading effects and allowing simple cascaded designs.
• Integrated Gain: Active filters can easily provide voltage or power gain ($A_v\ge 1$) by configuring the feedback path of the op-amp.

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2. Sallen-Key Low-Pass Filter Topology

The Sallen-Key filter (introduced by R. P. Sallen and E. L. Key in 1955) is the most popular active filter topology for realizing second-order ($2\text{nd}$-order) filter blocks [1.298].

2.1 Circuit Schematic

A unity-gain second-order active low-pass Sallen-Key filter consists of two resistors ($R_1,R_2$), two capacitors ($C_1,C_2$), and an op-amp configured as a voltage follower {unity-gain buffer} [4.46]:

                     C1 (Feedback Path)
               +-----------||------------+
               |                         |
               |   R1          R2        |    |\
  Vi(s) o-----+---/\/\/---v1---/\/\/--+--v2---| \
                                      |       |  \____o Vo(s)
                                    ----- C2  |- /  |
                                    -----     | /   |
                                      |       |/    |
                                     GND            |
                                      |             |
                                      +-------------+ (Unity Gain)

2.2 Functional Path Breakdown

• $R_1$ and $R_2$: Form the primary resistive path carrying the input signal toward the non-inverting terminal.
• $C_2$: Acts as a low-pass element that shunts high-frequency signals directly to ground [4.46].
• $C_1$: Provides a positive feedback path from the low-impedance op-amp output back to the intermediate node $v_1$ [4.46]. This feedback boosts the signal near the cutoff frequency, creating the desired sharp “knee” (resonance) in the frequency response.

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3. Step-by-Step Nodal Derivation of the Transfer Function

To secure full marks on analytical exam questions, you must prove the system function of the Sallen-Key circuit from foundational circuit principles using node-voltage analysis [4.46]:

Step 1: Establish Node Voltages & Op-Amp Constraints

Let the non-inverting input node of the op-amp be $v_2(t)$, and the intermediate node between $R_1$ and $R_2$ be $v_1(t)$. Because the op-amp is configured as an ideal unity-gain voltage follower [4.46]: $v_2(s)=v_o(s)$

Since the input terminals of an ideal op-amp draw zero current, the current through capacitor $C_2$ is exactly equal to the current through $R_2$. We write the current through $C_2$ in the $s$-domain as [4.46]: $I_{C2}(s)=s{C}_2{v}_2(s)=s{C}_2{v}_o(s)$

Step 2: Formulate Node Equation at $v_2$

Applying Kirchhoff’s Current Law (KCL) at node $v_2$ [4.46]: $\frac{v_1(s)-v_2(s)}{R_2}=I_{C2}(s)$

Substitute $v_2(s)=v_o(s)$ and $I_{C2}(s)=s{C}_2{v}_o(s)$ into this expression [4.46]: $\frac{v_1(s)-v_o(s)}{R_2}=s{C}_2{v}_o(s)$

Multiply by $R_2$ and isolate $v_1(s)$ [4.46]: $v_1(s)-v_o(s)=s{R}_2{C}_2{v}_o(s)$ $v_1(s)=(1+s{R}_2{C}_2)v_o(s)\text{— [Equation 1]}$

Step 3: Formulate Node Equation at $v_1$

Applying KCL at the intermediate node $v_1$ (sum of outgoing currents must equal zero) [4.46]: $I_{R1}(s)+I_{R2}(s)+I_{C1}(s)=0$ $\frac{v_1(s)-v_i(s)}{R_1}+\frac{v_1(s)-v_o(s)}{R_2}+s{C}_1(v_1(s)-v_o(s))=0$

Multiply the entire equation by $R_1{R}_2$ to clear the denominators: $R_2(v_1(s)-v_i(s))+R_1(v_1(s)-v_o(s))+s{R}_1{R}_2{C}_1(v_1(s)-v_o(s))=0$

Regroup the terms to isolate the input voltage $v_i(s)$ on one side: $R_2{v}_i(s)=(R_1+R_2+s{R}_1{R}_2{C}_1)v_1(s)-(R_1+s{R}_1{R}_2{C}_1)v_o(s)\text{— [Equation 2]}$

Step 4: Substitute and Eliminate $v_1(s)$

Now, substitute the expression for $v_1(s)$ from Equation 1 into Equation 2 [4.46]: $R_2{v}_i(s)=(R_1+R_2+s{R}_1{R}_2{C}_1)\cdot (1+s{R}_2{C}_2)v_o(s)-(R_1+s{R}_1{R}_2{C}_1)v_o(s)$

Divide both sides by $R_2$ to make the algebra easier to trace [4.47]: $v_i(s)=\left[1+\frac{R_1}{R_2}+s{R}_1{C}_1\right](1+s{R}_2{C}_2)v_o(s)-\left[\frac{R_1}{R_2}+s{R}_1{C}_1\right]v_o(s)$

Expand the product of the two binomial terms: $\left(1+\frac{R_1}{R_2}+s{R}_1{C}_1\right)(1+s{R}_2{C}_2)=1+\frac{R_1}{R_2}+s{R}_1{C}_1+s{R}_2{C}_2+s{R}_1{C}_2+s^2{R}_1{R}_2{C}_1{C}_2$

Substitute this expansion back into the input equation [4.47]: $v_i(s)=\left[\left(1+\frac{R_1}{R_2}+s{R}_1{C}_1+s{R}_2{C}_2+s{R}_1{C}_2+s^2{R}_1{R}_2{C}_1{C}_2\right)-\left(\frac{R_1}{R_2}+s{R}_1{C}_1\right)\right]v_o(s)$

Notice how the term $\frac{R_1}{R_2}$ and the term $s{R}_1{C}_1$ subtract out completely! This simplifies the equation to [4.47]: $v_i(s)=\left[s^2{R}_1{R}_2{C}_1{C}_2+s(R_1+R_2)C_2+1\right]v_o(s)$

Step 5: Formulate the Final System Transfer Function

The continuous-time system function $H(s)$ is the ratio of output to input [4.45, 4.47]: $H(s)=\frac{v_o(s)}{v_i(s)}=\frac{1}{s^2(R_1{R}_2{C}_1{C}_2)+s(R_1+R_2)C_2+1}$

To write this in standard quadratic second-order form, divide the numerator and denominator by the lead coefficient $R_1{R}_2{C}_1{C}_2$ [4.47]: $H(s)=\frac{\frac{1}{R_1{R}_2{C}_1{C}_2}}{s^2+s\frac{R_1+R_2}{R_1{R}_2{C}_1}+\frac{1}{R_1{R}_2{C}_1{C}_2}}$

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4. Parameter Mapping & Filter Design

For standard active filter design, we map our derived transfer function directly to the standard $2\text{nd}$-order LPF model [1.51, 1.296]: $H(s)=\frac{{\omega }_0^2}{s^2+2\zeta {\omega }_{0}s+{\omega }_0^2}=\frac{{\omega }_0^2}{s^2+\frac{{\omega }_0}{Q}s+{\omega }_0^2}$

Where:

• ${\omega }_0$ is the undamped natural angular frequency {cutoff frequency} [1.296].
• $\zeta$ is the damping ratio [1.296].
• $Q$ is the Quality Factor governing peaking at the cutoff band [1.296].

By comparing the coefficients of the Sallen-Key transfer function, we establish the following design relations:

4.1 Natural Cutoff Frequency (${\omega }_0$)

${\omega }_0^2=\frac{1}{R_1{R}_2{C}_1{C}_2}⟹{\omega }_0=\frac{1}{\sqrt{R_1{R}_2{C}_1{C}_2}}$

4.2 Quality Factor ($Q$)

$\frac{{\omega }_0}{Q}=\frac{R_1+R_2}{R_1{R}_2{C}_1}⟹Q=\frac{{\omega }_0\cdot R_1{R}_2{C}_1}{R_1+R_2}=\frac{\sqrt{R_1{R}_2{C}_1{C}_2}}{(R_1+R_2)C_2}=\frac{\sqrt{\frac{C_1}{C_2}}\cdot \sqrt{R_1{R}_2}}{R_1+R_2}$

4.3 High-Yield Equal-Component Simplification

If we set $R_1=R_2=R$ and $C_1=C_2=C$: ${\omega }_0=\frac{1}{\sqrt{R^2{C}^2}}=\frac{1}{RC}$ $Q=\frac{\sqrt{\frac{C}{C}}\cdot \sqrt{R^2}}{R+R}=\frac{R}{2R}=0.5$

The Equal-Component Limit

Designing with $R_1=R_2$ and $C_1=C_2$ limits the Quality Factor to $Q=0.5$. This represents an over-damped response with a very soft roll-off, which cannot approximate a sharp Butterworth response ($Q=1/\sqrt{2}\approx 0.707$) or Chebyshev peaking. Therefore, practical active filters must be designed with unequal components [1.51].

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5. ECE 2108 Laboratory Connection & MATLAB Codes

In ECE 2108 (Signal & Systems Laboratory) Experiment 5, students must implement continuous state-space models and observe their step, magnitude, and phase responses [1.245, 1.260].

5.1 Nodal-to-State-Space Conversion

To analyze the Sallen-Key filter in MATLAB, we represent our derived $2\text{nd}$-order system using a companion state-space model [1.260]: $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)+\mathbf{BU}(t)$ $\mathbf{Y}(t)=\mathbf{CX}(t)+\mathbf{DU}(t)$

For the Sallen-Key parameters $b_0=\frac{1}{R_1{R}_2{C}_1{C}_2}$ and $a_1=\frac{R_1+R_2}{R_1{R}_2{C}_1}$: $\mathbf{A}=\left[\begin{array}{cc}0& 1\\ -b_0& -a_1\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{cc}{b}_0& 0\end{array}\right],\mathbf{D}=\left[\begin{array}{c}0\end{array}\right]$

5.2 Lab Manual MATLAB Script [1.262, 1.263]

Below is the standard, fully documented MATLAB script to model, plot, and verify the Sallen-Key low-pass filter:

% ECE 2108 Experiment 5: Sallen-Key Active Low-Pass Filter Analysis
clc; clear all; close all;
 
% 1. Define Component Values
R1 = 10e3;  % 10 kOhms
R2 = 10e3;  % 10 kOhms
C1 = 22e-9; % 22 nF
C2 = 10e-9; % 10 nF
 
% 2. Calculate State-Space Coefficients
b0 = 1 / (R1 * R2 * C1 * C2);
a1 = (R1 + R2) / (R1 * R2 * C1);
 
% 3. Define State Matrices (Control Canonic Form)
A = [0, 1; -b0, -a1];
B = [0; 1];
C = [b0, 0];
D = 0;
 
% 4. Initialize State-Space System
sys = ss(A, B, C, D);
 
% 5. Generate Plots
figure(1)
step(sys) % Plots the transient step response of the active circuit
title('Sallen-Key Filter: Step Response');
grid on;
 
figure(2)
bode(sys) % Plots the frequency magnitude (dB) and phase (degrees) response
title('Sallen-Key Filter: Bode Plot');
grid on;
 
% 6. System Verification Checks
stb = isstable(sys);
if stb == 1
    disp('System Status: Absolutely Stable');
else
    disp('System Status: Unstable');
end
 
% 7. Controllability Check
S = ctrb(sys);
if det(S) == 0
    disp('Controllability: Not Controllable');
else
    disp('Controllability: Fully Controllable');
end
 
% 8. Observability Check
V = obsv(sys);
if det(V) == 0
    disp('Observability: Not Observable');
else
    disp('Observability: Fully Observable');
end

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6. Common Mistakes That Cost Marks

The Op-Amp Output Node KCL Violation

When writing nodal equations for the Sallen-Key circuit, never apply KCL at the op-amp output node $v_o$! An ideal op-amp output terminal acts as a low-impedance voltage source that can supply or sink arbitrary amounts of current to maintain its output voltage. The current leaving the op-amp output is mathematically unknown, meaning KCL at node $v_o$ is unsolvable and leads to a zero-mark evaluation [4.46].

The Passive Cascade Confusion

Do not confuse a cascaded passive RC circuit (which is formed by simply connecting two passive RC stages) with the active Sallen-Key circuit [1.298].

• Cascading passive RC networks can only yield real, separated poles, which limits the maximum quality factor to $Q\le 0.5$ and rounds off the frequency response curve.
• The active Sallen-Key network uses positive feedback ($C_1$) to realize complex conjugate poles, allowing $Q\ge 0.707$ to create highly sharp Butterworth or Chebyshev low-pass transitions.

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7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 ECE 2107 Question [10 Marks]

Question: Obtain the system function of the second-order Sallen-Key low-pass filter circuit shown in the diagram and discuss its stability [4.46].

• Answer Plan:
  1. Draw the annotated schematic labeling node voltages $v_1$, $v_2$, and the output follower relation $v_2=v_o$ [4.46].
  2. Set up the node equation at $v_2$ to write $v_1(s)=(1+s{R}_2{C}_2)v_o(s)$ [4.46].
  3. Apply KCL at node $v_1$, substitute $v_1(s)$ to isolate $v_i(s)$ and show the step-by-step algebraic cancelation of the $R_1/R_2$ and $s{R}_1{C}_1$ terms [4.47].
  4. Write the final system function $H(s)=\frac{1}{s^2{R}_1{R}_2{C}_1{C}_2+s(R_1+R_2)C_2+1}$ [4.47].
  5. Discuss stability: Since all component values ($R_1,R_2,C_1,C_2$) are positive, all poles of $H(s)$ lie strictly in the left-half s-plane (LHP). Therefore, the Sallen-Key low-pass filter is absolutely stable [4.47].

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8. Self-Check Before Moving On

• Can you list the four main limitations of passive $RLC$ circuits at low frequencies? [1.41]
• Can you sketch the Sallen-Key unity-gain active low-pass schematic from memory? [4.46]
• Do you know how to derive the Sallen-Key transfer function without looking at the node substitutions? [4.46, 4.47]
• Why is an equal-component active filter ($R_1=R_2,C_1=C_2$) limited to $Q=0.5$? [1.296]
• Can you write the state-space matrices of a second-order Sallen-Key transfer function in Control Canonic Form? [1.260]

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 4), ECE 2108 Laboratory Manual (Experiment 5), Azmat Sir-2309008.pdf (Class Slides).