Active Low Pass Filter: Design and Applications

Contents

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Key learnings:

Active Low Pass Filter Definition: An active low pass filter allows low-frequency signals to pass while blocking higher frequencies, essential for various electronic applications.
Component Significance: The use of an operational amplifier (Op-Amp) is critical for adjusting frequency response and enhancing signal quality.
Filter Design: Active low pass filters can be designed in various orders, with the first order typically involving a capacitor and an Op-Amp.
Voltage Gain Understanding: The voltage gain of an active low pass filter decreases as the input signal frequency approaches the cutoff frequency.
Applications: These filters are widely used in audio systems, digital converters, and biomedical devices to ensure optimal signal fidelity and functionality.

What is an Active Filter?

An active filter combines resistors and capacitors with one or more powered active circuit components, commonly a transistor or an operational amplifier (op amp). The active device can provide buffering or voltage gain while the circuit shapes the frequency response.

The component values, circuit topology and closed-loop gain set the filter poles, zeros and quality factor. The op amp must reproduce that intended response over the required signal range.

An op amp can give the filter high input impedance, low output impedance and controlled gain through its feedback loop. Real performance also depends on gain-bandwidth product, slew rate, input and output range, noise, offset, load drive, component tolerances and stability.

General - Diagram - for- Active -Filter — General Diagram for Active Filter

What is an Active Low Pass Filter?

An Active Low Pass Filter passes signals below its transition region with a defined gain and increasingly attenuates higher frequencies. Its op amp can isolate the source from the load and provide gain, but it does not produce an ideal brick-wall stopband.

Amplifier limits are part of active-filter design. Continuous-time RC filters and switched-capacitor filters are two implementation families, but they use different design methods. Filter order is not limited to eight; practical order depends on the required attenuation, phase response, noise, tolerance and complexity.

A buffered first-order circuit has the same ideal pole as its passive RC section. Depending on the amplifier configuration, its passband voltage gain can be unity, greater than unity or inverting.

An ordinary low-pass response includes DC. A first-order pole is 3.0103 dB below its passband gain at its pole frequency and then approaches a 20 dB-per-decade slope. For a general higher-order response, “cutoff” depends on the chosen approximation and specification. The sections below discuss 1^st and 2^nd order circuits.

A high pass filter attenuates low frequencies and passes higher frequencies within its design range. Bandpass filters pass a bounded frequency band while attenuating frequencies below and above it.

First Order Active Low Pass Filter

A first-order active low pass filter has one pole. The common voltage-mode circuit places a resistor with a capacitor before or within an op-amp stage. An inductor can form a passive RL low-pass network, but it is not part of the RC op-amp circuits shown here. A passive filter can have insertion loss and its response can change with loading.

Adding an op amp can buffer the RC network and set a predictable closed-loop gain. Both inverting and non-inverting active low-pass topologies are possible. The image below shows a simple RC circuit followed by an op-amp buffer.

First -Order -Active- Low -Pass -Filter -with -frequency -response — First Order Active Low Pass Filter with the frequency response

The RC network presents its low-pass output to a unity-gain buffer. The buffer’s high input impedance reduces loading of the RC section, while its low output impedance reduces the effect of the next stage. Stability still depends on the selected op amp and load.

With the op amp connected as a voltage-follower, the ideal DC gain is one. This circuit supplies impedance transformation rather than voltage amplification. A non-inverting or inverting stage can provide voltage gain, subject to bandwidth, noise and output-swing limits.

Active - Low - Pass - Filter - Circuit — Active Low Pass Filter Circuit

First Order Active Low Pass Filter with Amplification

If the application requires passband voltage gain above one, the buffer can be replaced by the non-inverting amplifier shown below.

At low frequency, the capacitor’s impedance is high and the signal reaches the amplifier. As frequency rises, the capacitor shunts more of the input toward the reference node, so the output falls relative to the passband level. The amplifier sets the passband gain; the capacitor does not increase it.

For an ideal non-inverting amplifier, the passband gain is one plus the ratio of the feedback resistor R₂ to the resistor R₃ connected from the inverting input to its reference. The labels must match the circuit drawing.

$DC= 1 +\frac{R_{2}}{R_{3}}$

First Order Low Active Pass Filter Inverted Configuration

The illustrated inverting low-pass filter uses a 741 op amp in an eight-pin package. Its passband transfer has a negative sign, which corresponds to 180-degree inversion for a sinusoid. Supply rails must satisfy the data sheet, input common-mode range and required output headroom; they are not chosen from gain alone.

In this example, pin 7 receives V⁺ = +12 V and pin 4 receives V^– = -12 V, both within the 741’s permitted total supply range. A gain magnitude of 10 does not prove that these rails provide adequate headroom: the input amplitude, DC offsets, load, bandwidth and the 741’s non-rail-to-rail output limits must also be checked. The AC output is inverted relative to the input.

Inverting -amplifier- configuration — Inverting amplifier configuration

First Order Low Active Pass Filter Non-Inverted Configuration

This non-inverting example also uses a 741. The input and output are in phase in the passband, while the RC network before the op amp creates the first-order low-pass pole.

The op amp amplifies the filtered signal with an ideal gain set by R2 and R1. Its high input impedance reduces loading of the RC network but does not change the capacitor’s reactance. Actual stability depends on the op amp, feedback network, source and load.

Non-inverting -configuration — Non-inverting configuration

First Order Low Pass Filter Voltage Gain

For the first-order non-inverting circuit, the magnitude response is:

$Voltage\ Gain\ (A_{v}) =\frac{V_{out}}{V_{in}} = \frac{A_{F}}{\sqrt{1+\left(\frac{f}{f_{c}}\right)^{2}}}$

where,

V_in is the input voltage
V_out is the output voltage
A_f is the passband gain of the filter (1+R₂/R₁)
f is the frequency of the input signal in Hertz
f_c is the cutoff frequency in Hertz

Above the pole, the ideal first-order magnitude approaches a slope of -20 dB per decade, not a fixed 20 dB loss for any increase in frequency. Let f be the operating frequency and f_c the pole frequency.

At low frequency

$\begin{align*} \frac{V_{out}}{V_{in}} = A_{f} \end{align*}$

When operating frequency is equal to cut off frequency

$\begin{align*} \frac{A_{f}}{\sqrt{2}} =0.707 A_{f} \end{align*}$

And at high frequency

$\begin{align*} \frac{V_{out}}{V_{in}} < A_{f}\end{align*}$

At frequencies well below the pole, the ideal magnitude approaches Af. At f = fc, it is Af/√2, or about 0.707Af. Well above the pole, it approaches zero. Voltage-gain magnitude may be expressed in decibels or dB as follows.

$A_{v} (dB) = 20 log_{10}\left [ V_{out}/V_{in} \right ]$

Frequency -Response -Curve — Frequency Response Curve

First Order Active Low Pass Filters Transfer Function

A transfer function describes the output-to-input ratio of a linear time-invariant control system in the Laplace domain under zero initial conditions. Setting s = jω gives its sinusoidal steady-state frequency response.

Magnitude and phase versus frequency show how the filter changes sinusoidal components. Time-domain checks are also needed when settling, overshoot, clipping or slew rate matters.

The voltage transfer function is the ratio of the output and input Laplace transforms under the stated linear model:

Block -Diagram -of - Transfer -Function — Block Diagram of a Transfer Function

$H(s)=\frac{V_{0}(s)}{V_{i}(s)}$

Where V0(s) and Vi(s) are the output and input voltages and s is the complex Laplace transform variable.

The resistor-capacitor network below is a passive single-pole low-pass section. An active circuit follows it with a buffer or gain stage.

Transfer - function - circuit — Transfer function circuit

The transfer function of the above circuit can be given as

$H(s)= \frac{1/sC}{R+1/sC} =\frac{1}{RcS+1} = \frac{1/RC}{s+1/RC}$

Substitute s = jω to obtain the complex frequency response. The protected legacy equation below incorrectly places magnitude bars around a complex value; its left side should read H(jω).

$|H(\omega)|= \frac{1/RC}{j\omega+1/RC}$

The magnitude of this unity-gain RC transfer function is 1/√(1 + (RCω)²). The protected legacy equation below omits parentheses around RCω and uses inconsistent letter case.

$|H(\omega)|= |\frac{1}{RcS+1}|=|\frac{1}{jRC\omega+1}|= \frac{1}{\sqrt{1+RC(\omega)^{2}}}$

At the pole angular frequency, the resistor magnitude equals the capacitor reactance magnitude:

$\begin{align*} \omega = \omega_{c} \quad when \quad R = \frac{1}{\omega_{c}C} \end{align*}$

Therefore,

${\omega_{c}}= \frac{1}{RC}$

Using ω/ωc = f/fc gives the magnitude shown below. The protected legacy display is missing the closing magnitude bar after H(ω).

$Gain = |H(\omega)= \frac{1}{\sqrt{1+(\omega/\omega_{c})^{2}}} = \frac{1}{\sqrt{1+(f/f_{c})^{2}}}$

The correct phase shift is -tan⁻¹(ω/ωc). The two protected legacy displays below omit the leading minus sign.

$\varphi = tan^{-1}\left(\frac{\omega}{\omega_{c}}\right)$

The cut off frequency and phase-shift of the filter can be calculated as follows

$f_{c}= \frac{1}{2\pi RC}$

$\phi = tan^{-1}(2\pi fRC)$

Cascading buffered first-order and properly designed second-order stages can produce an nth-order response. The stage pole frequencies and quality factors must be chosen for the intended Butterworth, Bessel, Chebyshev or other approximation.

First Order Active Low Pass Filter Design And Example

Design the illustrated non-inverting first-order active low-pass filter for a passband gain of 10, a pole frequency of 175 Hz and an RC resistance of 20 kΩ. The complete input impedance remains frequency-dependent.

The voltage gain of the non-inverting amplifier is given as

$A_{f}=1+\frac{R_{2}}{R_{1}} =10$

Choose R1 = 1 kΩ and calculate R2 from the gain equation.

$R_{2}=(10-1)x R_{1}= 9x\times1k\Omega = 9k\Omega$

Thus R1 = 1 kΩ and R2 = 9 kΩ give an ideal passband gain of 10 V/V. In decibels, 20 log10(10) = 20 dB.

Using R = 20 kΩ and fc = 175 Hz, calculate C from the first-order pole equation:

$f_{c}= \frac{1}{2\pi RC}$

$C= \frac{1}{2\pi Rf_{c}}=\frac{1}{2\pi\times 175\times 20k\Omega}$ = 45.47nF

The calculated value is 45.47 nF. A standard 47 nF capacitor would move the nominal pole to about 169.3 Hz, before component tolerance and op-amp effects. The circuit and idealised response are shown below.

Active Low Pass Filter Circuit

A typical circuit for an active low pass filter is given below:

Active Low Pass Filter Frequency Response Curve

The frequency response curve for an active low pass filter is given below:

Frequency- Response- of -the- problem — Frequency Response of the problem

Non-Inverting Amplifier Filter

A simple non-inverting amplifier filter is given below:

Non-inverting -circuit- for-the- problem — Non-inverting circuit for the problem

Inverting Amplifier Filter

An equivalent inverting amplifier filter is given below:

Equivalent -inverting- circuit- for -the -problem — Equivalent inverting circuit for the problem

Second Order Active Low Pass Filter

A second-order low-pass filter has two poles. The Sallen-Key circuit shown here is also called a voltage-controlled voltage-source (VCVS) topology, but other second-order active topologies include multiple-feedback designs.

An ideal two-pole low-pass response approaches -40 dB per decade far above its poles. Quality factor Q controls its behaviour near cutoff. Cascading two buffered first-order stages produces one two-pole implementation with two real poles; a Butterworth response needs the prescribed pole locations.

Second -Order- Active -Low -Pass -Filter — Second-Order Active Low Pass Filter

For unloaded cascaded stages, the overall transfer function is the product of the stage transfer functions. Higher-order filters are commonly built from second-order sections, with an extra first-order section for an odd order. Each section needs the pole frequency and Q assigned by the selected approximation.

Third - Order - Active - Low - Pass - Filter - Configuration — Third Order Active Low Pass Filter Configuration

Fourth -Order -Active -Low- Pass- Filter- Configuration — Fourth Order Active Low Pass Filter Configuration

Second Order Active Low Pass Filter Voltage Gain

The passband gain of cascaded stages is their product. For two stages with gains of 10 V/V and 3 V/V:

$\begin{align*} A_{v}= A_{v1}\times A_{v2} \end{align*}$

$\begin{align*} A_{v}= 10\times 3 = 30\end{align*}$

If stage gains are expressed in dB, add them:

$\begin{align*} A_{v} (dB) = A_{v1} + A_{v2} \end{align*}$

Second Order Active Low Pass Filter Cutoff Frequency

For the component labelling used in the illustrated two-pole section, its natural frequency is:

$\begin{align*} f_{c}=\frac{1}{2\pi\sqrt{R_{3}R_{4}C_{1}C_{2}}} \end{align*}$

When R3 = R4 = R and C1 = C2 = C, this becomes:

$f_{c}=\frac{1}{2\pi RC}$

Two identical buffered first-order stages are each -3.0103 dB at their common pole frequency, so their combined response is -6.0206 dB there. That common pole is not the cascade’s overall -3 dB cutoff; the latter is about 0.6436 times the individual pole frequency.

Second Order Active Low Pass Filter Design And Example

Consider two buffered first-order stages with Rs1 = Rs2 = 15 kΩ and C1 = C2 = 100 nF. Let the first non-inverting gain use R1 = 1 kΩ and R2 = 9 kΩ, while the second uses R3 = 6 kΩ and R4 = 3 kΩ.

The pole frequency of each RC stage is:

$\begin{align*} f_{c}=\frac{1}{2\pi RC} \end{align*}$

(1) $\begin{align*} f_{c}=\frac{1}{2\pi \times 15\times 10^{3}\times 100\times 10^{-9}} \end{align*}$

$\begin{align*} f_{c}=106.10Hz\end{align*}$

The first-stage passband gain is:

$\begin{align*} A_{1} = 1 +(R_{2}/R_{1}) \end{align*}$

$\begin{align*} A_{1} = 1 + \frac{9\times 10^{3}}{1\times 10^{3}} \end{align*}$

$\begin{align*} A_{1} =10 \end{align*}$

The second-stage passband gain is:

$\begin{align*} A_{2} = 1 +(R_{4}/R_{3}) \end{align*}$

$\begin{align*} A_{2} = 1 + \frac{3\times 10^{3}}{6\times 10^{3}} \end{align*}$

$\begin{align*} A_{2} =1.5 \end{align*}$

The total passband gain is:

$\begin{align*} A_{v} = 10\times 1.5 \end{align*}$

$\begin{align*} A_{v} =15 \end{align*}$

The total passband gain in dB is:

(2) $\begin{align*} A_{v} = 20 log_{10}(15) \end{align*}$

(3) $\begin{align*} A_{v} = 23.52dB \end{align*}$

At the 106.10 Hz pole shared by both stages, their combined attenuation is 6.02 dB, so the gain is:

(4) $\begin{align*} \ Gain \ at 106.10Hz = 23.52dB -6.02dB = 17.50dB \end{align*}$

Active Low Pass Filter Applications

Active low-pass filters are used for audio crossovers and equalisation, sensor-signal conditioning, noise limiting and anti-aliasing before an analog-to-digital converter. An anti-alias filter must provide enough attenuation above the converter’s Nyquist frequency for the signal and sampling plan. A low-pass filter does not prevent acoustic echoes; echoes require acoustic, signal-processing or room-treatment measures.

Active Filtering in Automotive Audio Applications

Op amps are common building blocks in line-level automotive audio filters. They can set crossover response and gain before the power amplifier, but they cannot eliminate all interference. Power-supply noise, electromagnetic compatibility, grounding, layout, component tolerance and amplifier noise remain design concerns.

An active crossover divides the line-level signal into bands for suitable speaker drivers. A low-pass output can feed a subwoofer amplifier, while a high-pass output protects a smaller driver from excessive low-frequency excursion. The filter does not drive a high-power loudspeaker directly; amplifier power, driver limits, enclosure and crossover slopes must be designed together.

Active Low Pass Filters For Biomedical Applications

An ECG Monitoring System can use low-pass filtering to limit out-of-band noise before conversion or display. Its bandwidth affects waveform fidelity and must match the intended monitoring or diagnostic use and applicable medical-device requirements. A generic two-stage CMOS op amp does not establish suitability for a pacemaker or other implant; those devices require specialised safety, reliability and regulatory design.