An active low-pass filter has R_f = 100\ k and a feedback capacitor of 1.0\ nF. Find the cut-off frequency. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: State two advantages of an active filter over a passive RC filter.

Advantages (up to 4 marks, 2 each): 1. An active filter can provide **gain** in the pass band (a passive RC filter can only attenuate, never amplify). 2. The op-amp **buffers** the output, presenting a low output resistance and a high input resistance, so the filter does not load the source or get loaded by the next stage (a passive filter's response shifts when loaded). Other valid points: a sharper roll-off is achievable by cascading stages without interaction, and no bulky inductors are needed for low-frequency filters. Markers reward any two of: gain in the pass band, buffering / no loading, sharper roll-off by cascading, or avoiding inductors.

EnglandElectronicsSyllabus dot point

How does adding an op-amp to a filter give gain, a sharper roll-off and no loading?

Active filters: op-amp low-pass and high-pass filters, the cut-off frequency, pass-band gain, band-pass filters, and the advantages over passive filters.

An Eduqas A-Level Electronics answer on active filters: op-amp low-pass and high-pass filters with the cut-off frequency and pass-band gain, band-pass filters made by cascading them, and the advantages of active filters over passive ones (gain, buffering and a sharper roll-off).

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Eduqas wants you to analyse active op-amp low-pass and high-pass filters, find the cut-off frequency and the pass-band gain, make a band-pass filter by cascading them, and state the advantages of active filters over passive ones. Active filters shape signals while also amplifying and buffering them.

The answer

Active low-pass and high-pass filters

Cut-off frequency and pass-band gain

Band-pass filters

Advantages over passive filters

Designing an active high-pass filter

Design an active high-pass filter with a pass-band gain of $-5$ and a cut-off frequency of $100\ \text{Hz}$ , using a $10\ \text{k}\Omega$ input resistor in series with the capacitor.

step 1: Set the pass-band gain

$A_v = -\frac{R_f}{R_\text{in}} = -5$ , so with $R_\text{in} = 10\ \text{k}\Omega$ , $R_f = 50\ \text{k}\Omega$ .

step 2: Set the cut-off frequency

$f_c = \frac{1}{2\pi R_\text{in} C}$ , so $C = \frac{1}{2\pi f_c R_\text{in}} = \frac{1}{2\pi \times 100 \times 10000} = \frac{1}{6.28 \times 10^{6}} = 1.6 \times 10^{-7}\ \text{F} = 160\ \text{nF}$ .

step 3: Choose preferred values

$R_f = 47\ \text{k}\Omega$ (giving a gain of about $-4.7$ ) and $C = 150\ \text{nF}$ (giving $f_c \approx 106\ \text{Hz}$ ) are close standard values.

Examples in context

Active filters are everywhere in signal processing: a band-pass filter selects the voice band in a communication system, a low-pass filter is the anti-aliasing filter before an analogue-to-digital converter and the reconstruction filter after a digital-to-analogue converter, and tone controls in an audio amplifier are active filters. Their gain and buffering are exactly why they replace passive filters in almost all modern analogue design.

Try this

Q1. An active low-pass filter has $R_f = 100\ \text{k}\Omega$ and a feedback capacitor of $1.0\ \text{nF}$ . Find the cut-off frequency. [2 marks]

Cue. $f_c = \frac{1}{2\pi \times 100000 \times 1.0 \times 10^{-9}} = 1.6\ \text{kHz}$ .

Q2. State how a band-pass filter is constructed from active stages. [2 marks]

Cue. Cascade a high-pass stage (lower cut-off) and a low-pass stage (upper cut-off).

Q3. State one advantage of an active filter over a passive RC filter. [1 mark]

Cue. It can provide gain in the pass band (or it buffers the signal so it is not loaded).

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20216 marksAn active low-pass filter is built from an inverting amplifier with a

4.7\ \text{k}\Omega

input resistor and a

47\ \text{k}\Omega

feedback resistor with a

3.3\ \text{nF}

capacitor in parallel with the feedback resistor. Calculate the pass-band gain and the cut-off frequency.

Show worked answer →

Pass-band gain (up to 3 marks): at low frequency the capacitor's reactance is very high, so the feedback is set by $R_f$ , giving the inverting gain $A_v = -\dfrac{R_f}{R_\text{in}} = -\dfrac{47}{4.7} = -10$ .

Cut-off frequency (up to 3 marks): the cut-off is set by the feedback resistor and capacitor: $f_c = \dfrac{1}{2\pi R_f C} = \dfrac{1}{2\pi \times 47000 \times 3.3 \times 10^{-9}} = \dfrac{1}{9.75 \times 10^{-4}} = 1.0 \times 10^{3}\ \text{Hz} = 1.0\ \text{kHz}$ .

Markers reward the pass-band gain $-10$ , the cut-off formula $f_c = \frac{1}{2\pi R_f C}$ , and the value $1.0\ \text{kHz}$ .

Eduqas 20194 marksState two advantages of an active filter over a passive RC filter.

Show worked answer →

Advantages (up to 4 marks, 2 each):

An active filter can provide gain in the pass band (a passive RC filter can only attenuate, never amplify).
The op-amp buffers the output, presenting a low output resistance and a high input resistance, so the filter does not load the source or get loaded by the next stage (a passive filter's response shifts when loaded).
Other valid points: a sharper roll-off is achievable by cascading stages without interaction, and no bulky inductors are needed for low-frequency filters.

Markers reward any two of: gain in the pass band, buffering / no loading, sharper roll-off by cascading, or avoiding inductors.

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)