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Analog12 min read
Active Filters
Combine op-amps, resistors and capacitors to shape frequency response without using inductors.
Mental model
An active filter is a frequency-aware gatekeeper with an assistant: the RC network decides what should pass, while the op-amp buffers or amplifies the result.
R and C form the input low-pass filter. The op-amp buffers and amplifies that filtered voltage; Rf and Rg set passband gain to 1 + Rf/Rg. All grounds are common; op-amp supply pins are omitted.Scroll sideways to view the complete circuit.UNDERSTAND THE BEHAVIOR
01
A passive RC filter loses signal level and can be disturbed by the load connected after it. Adding an op-amp can isolate stages, restore gain and create steeper responses.
02
Filter order counts the number of energy-storage poles. Each pole eventually contributes about −20 dB/decade, so a second-order low-pass can approach −40 dB/decade.
03
Component ratios also set damping or Q. Two filters with the same cutoff can have different peaking and transient behavior, so choosing Butterworth, Bessel or another response is a design decision.
Why it matters
Active filters remove sensor noise, limit signal bandwidth and prevent aliasing before an ADC.
Core ideas
Active filters can provide gain as well as filtering.
The op-amp prevents the load from strongly changing the RC response.
Real op-amp bandwidth must be comfortably above the filter cutoff.
Key equationsFirst-order cutoff: fc = 1 / (2πRC)Non-inverting passband gain: Av = 1 + Rf / Rg
3Select the standard value 1.6 kΩ; the actual cutoff is about 995 Hz.
4If Av = 2 is needed, choose equal feedback resistors, for example Rf = Rg = 10 kΩ.
Common mistake
Do not select an op-amp whose gain-bandwidth product is only equal to the cutoff. Leave substantial margin.
RememberActive filters can provide gain as well as filtering.
DEEPER EXPLANATION
What actually changes at the cutoff frequency?
Consider a first-order low-pass made from a series resistor and a capacitor to ground, with output taken across the capacitor. At slow changes, the capacitor has ample time to charge, so the output follows the input. At high frequency, the capacitor presents a lower reactance and more of the signal appears across the resistor instead.
The characteristic time is τ = RC. For a sinusoid, the corner frequency is fc = 1/(2πRC). At this frequency the output amplitude of the ideal unloaded RC low-pass is 1/√2, or about 0.707, of its low-frequency amplitude. This is approximately −3.01 dB, and it does not mean the signal disappears above fc.
An active filter adds an amplifier and can provide gain, buffering or a higher-order response. Its actual corner or natural frequency depends on the particular circuit and component placement. The simple RC equation is a useful starting point only when that first-order section and its loading assumptions match your circuit.
Walk through another example
Use R = 10 kΩ and C = 10 nF = 0.00000001 F. The time constant is RC = 0.0001 s.
fc = 1/(2π × 0.0001) ≈ 1,591.5 Hz. At that frequency a 1 V sine amplitude becomes about 0.707 V across the capacitor in the ideal unloaded model.
Doubling C to 20 nF doubles RC, so the corner falls to roughly 795.8 Hz; you can predict the direction before calculating.
Check your intuition
Sketch a low-pass response: nearly flat at low frequency, down 3 dB at its corner, and approximately 20 dB lower per decade well above that corner for one pole.
Where this model can mislead you
Do not enter 10 for a 10 nF capacitor when the calculator asks for farads. Do not assume the same cutoff for a loaded RC network or an arbitrary multi-pole active filter.
TEST YOUR UNDERSTANDING
Three quick checks
Choose one answer per question, then check your work. Each answer includes an explanation.
Practice and check your understanding
Try each question before opening the worked answer.
1. For a first-order RC section with R = 10 kΩ and C = 10 nF, find fc.
Show answer and reasoning
Convert C to 10 × 10⁻9 F. fc = 1/(2πRC) ≈ 1591.5 Hz. This formula applies to this first-order RC section, not every higher-order active-filter topology.
2. What does −3 dB mean for the voltage amplitude relative to the passband?
Show answer and reasoning
The ratio is 10^(−3/20) ≈ 0.708, approximately 1/√2. This corresponds to roughly half the power when comparing the same resistance.