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Signals & Wireless (RF)

Q factor and bandwidth: how wide a tuned circuit passes

The bandwidth of a tuned circuit is how wide a range of frequencies it lets through, and its Q factor is how sharp it is: Q = centre frequency ÷ bandwidth. For a series RLC circuit, bandwidth ≈ R ÷ (2π × L), so at the same resonant frequency a smaller coil with a bigger capacitor gives a wider band. A receiver's band must be wide enough for a station's sidebands and narrow enough to keep the next station out.

The formulas

f0 = 1 ÷ (2π√(LC)) · bandwidth ≈ R ÷ (2πL) · Q = f0 ÷ bandwidth = XL ÷ R

L × C sets the centre; L against R sets the width. At 20 kHz with 100 Ω in the loop, a 10 mH coil has a reactance of 1257 Ω: Q ≈ 12.6, and a band 20 ÷ 12.6 ≈ 1.6 kHz wide. A 3.3 mH coil has 415 Ω: Q ≈ 4.2, about 4.8 kHz wide.

How wide must it be?

An AM station is three tones: the carrier, and two sidebands as far either side of it as the audio's pitch. A 1 kHz voice on a 20 kHz carrier sits at 19, 20 and 21 kHz. So the band must be at least twice the highest audio frequency wide. Music up to 2 kHz spreads from 18 to 22 kHz, and a 2.2 mH coil, about 7 kHz wide, gives it room.

Too narrow, too wide

With the 10 mH coil, 1.6 kHz wide, the sidebands 1 kHz out arrive at about 31% while the voice should be 50% deep: it comes out muffled, losing its highest notes first. Go wide with a 1 mH coil, about 16 kHz across, and 23% of the next station at 40 kHz gets in.

Live experiment

Room for the sidebands

An AM station, a 20 kHz carrier with sidebands at 19 and 21 kHz, goes through a series coil and capacitor tuned to 20 kHz. Pick the coil and the capacitor follows: 10 mH is too narrow and muffles the voice, 3.3 mH passes it and keeps the 40 kHz station out, 1 mH lets the neighbour in.

Try it yourself: free lesson

“Room for the sidebands” is lesson 2 of 6 in Airwave Atoll, Signals & Wireless (RF) World 3. Free with an account. Lessons unlock in order, so you’ll do the 15 before it first.

Width first, then tune

Pick L for the width, then C for the centre:

C = 1 ÷ (4π² f² L) = 1 ÷ (4π² × (20 kHz)² × 2.2 mH) ≈ 28.8 nF

Where you'll meet it

AM broadcast stations sit 9 kHz apart in Europe and 10 kHz apart in the Americas, so a receiver's band is about that wide. An FM channel is about 200 kHz. Quartz crystals reach Q in the tens of thousands, which is how narrow radio channels and accurate clocks are made, and a guitar string's long ring is a high Q too.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Room for the sidebands” is lesson 2 of 6 in Airwave Atoll, Signals & Wireless (RF) World 3. Free with an account. Lessons unlock in order, so you’ll do the 15 before it first.

Or start from the very first lesson, no account needed

Work it out

Practice questions

Three quick questions from the lessons. Pick an answer to check it.

  1. Question 1 of 3You keep L × C the same and use a smaller coil. The tuned circuit's band…
    Show the answer

    Answer: Stays at the same frequency and gets wider. L × C sets the frequency, so the tune stays put. A smaller coil means less reactance against the loop's resistance: a lower Q, a wider band.

    • Why not “Gets narrower”? A smaller coil has less reactance against the same resistance: a lower Q, so a wider band.
    • Why not “Disappears”? The circuit still resonates at the same frequency. Only its width changes.
    • Why not “Moves higher”? The centre depends on L × C, which hasn't changed, so the tune stays put.
  2. Question 2 of 3Music up to 2 kHz on a 20 kHz carrier. Its sidebands spread from…
    Show the answer

    Answer: 18 to 22 kHz. Each audio frequency appears above and below the carrier. 2 kHz of audio reaches 2 kHz either side.

    • Why not “19 to 21 kHz”? 19 to 21 kHz would only hold audio up to 1 kHz.
    • Why not “16 to 24 kHz”? That would be audio up to 4 kHz. 2 kHz of audio reaches 2 kHz either side.
    • Why not “20 to 22 kHz”? AM makes sidebands on both sides of the carrier, below it as well as above.
  3. Question 3 of 3At resonance a series LC in the signal path is…
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    Answer: Nearly a short: that frequency passes. The coil's +j and the capacitor's −j cancel, leaving only a little resistance. Every other frequency meets a reactance.

    • Why not “A battery”? An LC passes energy back and forth between coil and capacitor but supplies none of its own.
    • Why not “An open circuit”? That's a parallel LC at resonance. In series, the coil's and the capacitor's reactances cancel.
    • Why not “A big resistor”? At resonance only the circuit's small resistance is left, not a big one.
Keep practising: free lesson

“Room for the sidebands” is lesson 2 of 6 in Airwave Atoll, Signals & Wireless (RF) World 3. Free with an account. Lessons unlock in order, so you’ll do the 15 before it first.

Questions people ask

What is the Q factor?

A measure of how sharp a resonance is: the centre frequency divided by the bandwidth. For a series RLC circuit it's also the coil's reactance at resonance divided by the resistance. High Q means a narrow, peaky response; low Q a broad one.

How do you calculate the bandwidth of an RLC circuit?

For a series RLC circuit, the bandwidth between the half-power points is R ÷ (2π × L), in hertz. 100 Ω with a 3.3 mH coil gives about 4.8 kHz. It doesn't depend on C: C only sets where the band is centred.

How much bandwidth does an AM signal need?

Twice its highest audio frequency, one sideband each side of the carrier. Speech up to 4.5 kHz needs about 9 kHz, and music up to 5 kHz needs 10 kHz.

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