Active Low Pass Filter Interview Questions

Q1. What is the difference between an active and a passive low pass filter?

An active low pass filter uses an op-amp along with resistors and capacitors to provide gain and buffering, whereas a passive filter uses only R, L, or C components and cannot provide gain. A first-order active LPF using the UA741 op-amp with RF = 10kΩ and R1 = 10kΩ gives a passband gain of 2 (6dB) while also filtering. The active filter's op-amp output impedance is near zero, meaning it can drive loads without the cutoff frequency shifting — something a passive RC filter cannot do.

Follow-up: What is the main disadvantage of an active filter compared to a passive filter?

Q2. How do you calculate the cutoff frequency of a first-order active low pass filter?

The cutoff frequency fc = 1/(2πRC), where R and C are the filter components in the feedback or input network — the same formula as a passive RC filter. For R = 15.9kΩ and C = 10nF, fc = 1/(2π × 15900 × 10×10⁻⁹) = 1kHz. The op-amp sets the gain independently and does not affect the cutoff frequency when properly designed, which is a key advantage over passive designs.

Follow-up: How would you design an active LPF with fc = 500Hz and a passband gain of 10?

Q3. What is the roll-off rate of a first-order active low pass filter?

A first-order active LPF has a roll-off rate of −20dB/decade (or −6dB/octave) above the cutoff frequency, identical to a passive first-order RC filter. At 10kHz, a filter with fc = 1kHz would attenuate the signal by 20dB. Adding another RC stage (second-order) doubles the roll-off to −40dB/decade, which is why multiple-order filters are used when sharper attenuation is needed.

Follow-up: How many dB/decade does a second-order Butterworth LPF roll off above the cutoff frequency?

Q4. What is a Sallen-Key topology and why is it used for second-order active filters?

The Sallen-Key topology uses a single op-amp (typically in unity-gain configuration) with two RC networks to realize a second-order filter response. It is widely used because it requires minimal components, the op-amp is used as a unity-gain buffer (voltage follower) so offset and drift are minimized, and it is easy to cascade for higher orders. A Sallen-Key LPF with R1=R2=10kΩ and C1=C2=10nF gives fc ≈ 1.59kHz with Butterworth Q=0.707.

Follow-up: How does the quality factor Q affect the transient response of a Sallen-Key filter?

Q5. How do you achieve a second-order Butterworth response in an active LPF?

A Butterworth response requires setting Q = 1/√2 ≈ 0.707, which in a Sallen-Key configuration means the op-amp gain must be set to 1.586 (achieved with RF/R1 = 0.586). For equal-value Sallen-Key with R = 10kΩ and C = 10nF, adding RF = 5.86kΩ and R1 = 10kΩ as gain resistors achieves Butterworth Q. The Butterworth filter gives the flattest possible passband response with no ripple.

Follow-up: What is the Chebyshev filter and how does it trade passband flatness for roll-off steepness?

Q6. What op-amp parameters are critical when designing an active LPF for high frequencies?

The op-amp's gain-bandwidth product (GBW) must be much larger than the filter's operating frequency multiplied by the closed-loop gain. For a 100kHz LPF with a gain of 10, the op-amp needs a GBW of at least 1MHz; the TL071 with GBW = 3MHz is adequate, but the UA741 with GBW = 1MHz would be too slow. Slew rate must also be considered — for a 10V peak output at 100kHz, slew rate must exceed 2π × 100k × 10 ≈ 6.28V/µs.

Follow-up: What distortion arises if the op-amp's slew rate is insufficient for the filter's operating frequency and signal amplitude?

Q7. How does the gain of an active LPF affect the cutoff frequency in a non-inverting configuration?

In a properly designed active LPF, the passband gain and cutoff frequency are independently set — gain is determined by the feedback resistors and cutoff by the RC network. However, if the op-amp's finite GBW is a limiting factor, increasing the closed-loop gain reduces the available bandwidth, effectively shifting the usable operating range. A TL072 based filter designed for gain=100 and fc=10kHz would require GBW of at least 1MHz, which the TL072 (GBW=3MHz) can only marginally support.

Follow-up: How would you modify the design to maintain the cutoff frequency when increasing the passband gain?

Q8. What is the phase response of a first-order active LPF at its cutoff frequency?

At the cutoff frequency, a first-order active LPF introduces a phase shift of −45° due to the RC network, regardless of the op-amp gain. At frequencies well above fc, the phase asymptotically approaches −90°. This phase shift must be accounted for in feedback control systems — a sensor conditioning LPF with fc at the system bandwidth adds 45° phase lag, reducing phase margin and potentially causing oscillation.

Follow-up: How does cascading two identical first-order LPFs affect the overall phase response?

Q9. How would you convert an active LPF to an active HPF?

You swap the positions of R and C in the filter network — capacitor in series (input) and resistor to ground — converting the integrating action to differentiating. An active LPF with R=10kΩ, C=10nF (fc=1.59kHz) becomes a HPF with the same cutoff when R and C positions are swapped. The op-amp gain stage remains unchanged, and the cutoff frequency formula 1/(2πRC) still applies.

Follow-up: If you cascade an LPF and an HPF with different cutoff frequencies, what type of filter do you get?

Q10. What is the effect of component tolerance on the cutoff frequency of an active LPF?

Component tolerances in R and C directly affect the cutoff frequency since fc = 1/(2πRC) — a 5% error in both R and C together can shift fc by up to 10%. In production, using 1% metal-film resistors and C0G (NP0) capacitors minimizes drift; standard ±20% ceramic capacitors are unsuitable for precision filter design. For tight fc tolerance, trimming the resistor value or using a digitally programmable resistor (MCP41010) is a practical solution.

Follow-up: How would you test and verify the cutoff frequency of a manufactured active LPF board?

Q11. What is the voltage gain of an inverting active LPF at DC and at very high frequencies?

At DC (f = 0), the capacitor in the feedback is open, so the inverting gain is −RF/R1, determined entirely by the resistors. At very high frequencies, the capacitor shorts the feedback resistor, reducing gain to nearly zero. For an inverting LPF with R1 = 10kΩ, RF = 100kΩ, and CF = 1.59nF across RF, gain at DC = −10 and fc = 1/(2π×100k×1.59n) = 1kHz.

Follow-up: What is the advantage of an inverting active LPF over a non-inverting topology in a signal chain?

Q12. How is an active LPF used in an anti-aliasing filter before an ADC?

An anti-aliasing filter must remove all frequency components above half the ADC's sampling rate (Nyquist frequency) before sampling to prevent aliasing distortion. For a 10kHz sampling ADC like the ADS1115, an active LPF with fc = 4kHz ensures that signals above 5kHz are attenuated sufficiently. A second- or fourth-order active Butterworth LPF is typically used because it provides adequate roll-off with flat passband gain, preserving signal integrity up to fc.

Follow-up: What happens to the digitized signal if the anti-aliasing filter cutoff is set too high relative to the sampling rate?

Q13. How do you design an active LPF for a 4–20mA current loop signal conditioning circuit?

In a 4–20mA loop, the current is first converted to voltage using a precision shunt resistor (e.g., 250Ω gives 1–5V), and then an active LPF is used to remove high-frequency noise before feeding an ADC or controller. An INA198 current sense amplifier followed by a Sallen-Key LPF with fc = 10Hz is a typical design for slow industrial process variables. The very low fc is appropriate because 4–20mA signals represent slow process variables like temperature or pressure, and aggressive filtering removes 50/60Hz interference.

Follow-up: How would you choose the cutoff frequency for a signal conditioning LPF in a vibration monitoring application?

Q14. What is the quality factor Q in a second-order active LPF and what does it affect?

Q determines the shape of the filter response near the cutoff frequency — Q = 0.707 gives the flat Butterworth response, Q > 1 gives a peaked response (Chebyshev), and Q < 0.707 gives an overdamped response. In a Sallen-Key filter with Q = 2, you will see a 6dB peak just before fc, which can cause overshoot in step responses. For sensor signal conditioning in a Bosch automotive ECU application, Butterworth Q = 0.707 is preferred for its predictable step response.

Follow-up: How does Q affect the step response of a second-order active LPF?

Q15. How do you cascade multiple active LPF stages to achieve a higher-order filter?

Cascading n identical first-order active LPF stages gives an nth-order response, but the overall −3dB cutoff frequency shifts lower because each stage contributes attenuation at the original fc. To get a true nth-order Butterworth LPF, each stage must use different Q and fc values derived from Butterworth pole tables. For a 4th-order Butterworth LPF at 1kHz, two Sallen-Key stages with Q1=0.541 and Q2=1.307 and appropriately adjusted R and C values per stage are needed.

Follow-up: Why do you use pole tables rather than just cascading identical stages for a Butterworth response?