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Resonant circuits and filters

An inductor and a capacitor disagree about frequency, and at one frequency they cancel each other exactly. That is how a radio pulls one station out of a crowded sky, and a resistor and capacitor on their own will throw away whichever end of the spectrum you do not want.

Year 13AQA 3.13.2

Builds on Capacitors and energy stored and Forced vibrations and resonance.

IN THIS TOPIC

Find the resonant frequency of an LC circuit, and say what cancels at it.
Use Q and bandwidth together, and predict what extra resistance does to a resonance curve.
Tell band-pass from band-stop and high-pass from low-pass, and find an RC filter's cut-off.

WHAT YOU PROBABLY THINK

A tuned circuit responds at one frequency and rejects everything else.

Where the reactances cancel

An inductor opposes a changing current, and it opposes it harder the faster the current changes, so its reactance $2\pi fL$ rises with frequency. A capacitor does the opposite. It passes a rapidly alternating current easily and blocks a slow one, so its reactance $1/2\pi fC$ falls as the frequency rises. Put the two in series and somewhere between the extremes their reactances are equal in size. They also act in opposite senses, so at that frequency they cancel and leave only the circuit's resistance to limit the current.

There the current from a given driving voltage is at its largest, and the circuit is at resonance. Setting the two reactances equal and solving for the frequency gives the point where it happens:

f_{0} = \frac{1}{2\pi\sqrt{LC}}

ON YOUR DATA SHEET

FIG. 1Current against frequency for 1.0 mH with 100 nF. Both curves peak at 15.9 kHz, but the low-resistance circuit peaks sharply while the 25 Ω circuit is lower and far broader.

Both L and C sit under a square root, so a factor of four in either shifts the resonant frequency by a factor of two. Tuning exploits that gentleness. A variable capacitor swinging over a range of nine to one covers a frequency range of three to one, which is enough for a whole broadcast band on one knob.

WORKED EXAMPLE

Checking the marked peak

The circuit drawn above has L = 1.0 mH and C = 100 nF. Verify the resonant frequency marked on the graph.

$LC$ = 1.0 × 10⁻³ × 100 × 10⁻⁹ = 1.0 × 10⁻¹⁰, and the square root of that is 1.0 × 10⁻⁵ s.

$f_{0}$ = 1/(2π × 1.0 × 10⁻⁵) = 1.59 × 10⁴ Hz = 15.9 kHz.

Take the root before dividing, and convert the prefixes before either. Millihenries and nanofarads left unconverted are where most of the marks in this calculation are lost.

Sharpness, and what resistance costs

The two curves in that figure are the same L and C with different resistance, and between them they say everything about the Q factor. Measure the peak's width between the two frequencies at which the current has fallen to $1/\sqrt{2}$ of its peak value, the points where the power delivered is half its peak. That width is the bandwidth $f_{B}$ , and Q compares it against the resonant frequency itself:

Q = \frac{f_{0}}{f_{B}}

ON YOUR DATA SHEET

A high Q means a narrow peak, so the circuit is fussy about frequency. Resistance is what ruins it. Energy sloshes between the inductor's magnetic field and the capacitor's electric field every cycle, and resistance takes a bite out of the store each time round, so Q measures how small that bite is. For a series circuit it works out as $2\pi f_{0}L/R$ . The cyan curve has 10 Ω and a Q of 10; raise the resistance to 25 Ω and Q drops to 4.

In bandwidth those two are 1.6 kHz and 4.0 kHz about the same 15.9 kHz centre. The peak did not move sideways, since $f_{0}$ depends on L and C alone. It fell, and it spread.

Two filters from one tuned circuit

Where the output is taken decides what the circuit does. Take it across the resistor and the response peaks at $f_{0}$ , since that is where the current, and so the pd across the resistor, is largest. That is a band-pass filter, passing a band of width $f_{B}$ centred on resonance. Take the output across the series inductor and capacitor instead and the result is the mirror image. Their combined reactance vanishes at resonance, so the output there falls to zero while everything well away from resonance gets through. That is a band-stop filter, or a notch.

FIG. 2The two responses of one tuned circuit. The band-pass peaks at the resonant frequency; the band-stop is its mirror, near full output far from resonance and zero at the resonant frequency itself.

Both have work to do. A receiver's band-pass keeps one station and drops its neighbours. A band-stop is aimed at a single offender, the 50 Hz mains hum crawling along an audio line, or an interfering transmitter camped near the frequency you actually want.

The opening claim was wrong, and it needs to be. No tuned circuit passes a single frequency, and one that came close would be useless, because a real signal is a band of frequencies rather than a line. The band has to be wide enough to carry it. Wind Q too high and a broadcast arrives with its top notes shaved off.

YOUR TURN

How fussy must a radio be

Medium-wave stations sit 9 kHz apart. A receiver tuned to 909 kHz has to keep its own station and reject the next one along. Estimate the Q its tuned circuit needs.

Show the working

The bandwidth it can afford is about the channel spacing, 9 kHz.

Q = $f_{0}/f_{B}$ = 909/9 = about 100, an ordinary figure for a coil and capacitor.

Check a designed bandwidth against the channel spacing rather than making it as narrow as the components allow. Sharper than 9 kHz here would start cutting into the station's own sidebands.

Filters without an inductor

Most filtering is done with a resistor and a capacitor and nothing else. Wire the pair in series across the signal and take the output from one of them. Across the capacitor, whose reactance is large at low frequency and small at high, low frequencies survive and high ones are shorted away, giving a low-pass filter. Across the resistor the two roles swap, and the same components become a high-pass filter.

Neither has a sharp edge. The response slides away over a decade or more, so the agreed marker of where a filter starts to act is the cut-off frequency, the frequency at which the reactance equals the resistance and the output has fallen to $1/\sqrt{2}$ , about 0.71, of its full value. Setting $1/2\pi f C$ equal to R gives it:

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

NOT ON THE DATA SHEET: LEARN IT

FIG. 3One 1.6 kΩ resistor and one 0.10 μF capacitor, read two ways. The low-pass and high-pass curves cross at 0.71 of full output, at a cut-off just under 1.0 kHz.

For the drawn pair $f_{c}$ = 1/(2π × 1600 × 0.10 × 10⁻⁶) = 995 Hz, about a kilohertz. Speech below that passes the low-pass version almost untouched, while a 10 kHz hiss, ten times the cut-off, comes out at a tenth of its size.

TRY IT UNSEEN

Killing the hum

Speech from 300 Hz upward shares a line with 50 Hz mains hum. Choose the filter, and with R = 10 kΩ find the capacitor that puts the cut-off at 150 Hz. Estimate how much of the hum survives.

Show the working

A high-pass filter, since the wanted signal lies above the interference.

C = 1/(2π $f_{c}$ R) = 1/(2π × 150 × 10 × 10³) = 1.1 × 10⁻⁷ F, about 0.11 μF.

At 50 Hz the frequency is a third of the cut-off, and the high-pass response there is 0.32, so about a third of the hum gets through. Speech at 300 Hz and above passes at 0.89 or better, so the wanted signal is barely touched.

THE EXAM BIT

Convert the prefixes, multiply, then take the square root. Millihenries and nanofarads mishandled in that order account for most of the lost marks on $f_{0}$ .
Q and bandwidth travel together. $f_{B} = f_{0}/Q$ , and the passband runs half a bandwidth either side of the resonant frequency, so quote a range when one is asked for.
Adding resistance never moves the resonant frequency. Say that first, then say the peak is lower and broader because Q has fallen.
Name the output terminals in a filter answer. Output across the capacitor is low-pass, across the resistor is high-pass, and the cut-off is where the output has dropped to 0.71 of full size.

CHECK YOURSELF

A tuned circuit uses L = 2.0 mH with C = 47 nF, and its Q factor is 25. Find the resonant frequency and the bandwidth, give the range of frequencies passed, and state what happens to each if resistance is added to the circuit.

Show a hint

Root LC first. Q compares the resonant frequency with the width of the peak.

Show the answer

$LC$ = 2.0 × 10⁻³ × 47 × 10⁻⁹ = 9.4 × 10⁻¹¹, whose square root is 9.70 × 10⁻⁶ s.

$f_{0}$ = 1/(2π × 9.70 × 10⁻⁶) = 16.4 kHz.

$f_{B} = f_{0}/Q$ = 16.4/25 = 0.66 kHz, so the circuit passes roughly 16.1 kHz to 16.7 kHz.

Extra resistance leaves $f_{0}$ exactly where it is, since that depends on L and C only. It lowers Q, so the peak falls and the bandwidth widens.

At resonance the two reactances cancel, the current peaks, and f0 depends on L and C alone.

Q is the resonant frequency over the bandwidth, so resistance lowers Q, broadens the peak and lets in more of what you did not want.

WORKBOOK

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Find the resonant frequency of an LC circuit, and say what cancels at it.
Use Q and bandwidth together, and predict what extra resistance does to a resonance curve.
Tell band-pass from band-stop and high-pass from low-pass, and find an RC filter's cut-off.

Open the full revision checklist to track your progress across the whole unit.