Resonant circuits and filters

Where the reactances cancel

An inductor opposes a changing current, and it opposes it harder the faster the current changes, so its reactance 2πfL$2\pi fL$2πfL rises with frequency. A capacitor does the opposite. It passes a rapidly alternating current easily and blocks a slow one, so its reactance 1/2πfC$1/2\pi fC$1/2πfC 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:

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.

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/√2$1/\sqrt{2}$1/√(2) of its peak value. Power goes as the square of the current, so at those two frequencies the power delivered has dropped to half its peak, and they are known as the half-power points. The width between them is the bandwidth f_B$f_{B}$f_B, written Δf$\Delta f$Δf where a symbol for a width is wanted, and Q compares it against the resonant frequency itself:

In that second notation the same statement reads Q = f₀/Δf$Q = f_{0}/\Delta f$Q = f_0/Δf. 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 2πf₀L/R$2\pi f_{0}L/R$2πf_0L/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 f₀$f_{0}$f_0 depends on L and C alone. It fell, and it spread.

Side by side instead of end to end

Put the inductor and the capacitor in parallel rather than in series and the circuit turns inside out. Both branches now share the same pd, so each carries a current of its own, and those two currents are in antiphase: the capacitor's leads the applied voltage by a quarter of a cycle and the inductor's lags it by a quarter. At the frequency where the two reactances are equal in size the two branch currents are equal in size as well, so they cancel where the branches meet and the current drawn from the supply falls to a minimum. The pair presents its largest impedance at resonance, which is why the parallel circuit is called a rejector and the series one an acceptor.

Nothing inside the loop has stopped. Charge sloshes round and round between the capacitor and the inductor, and the current circulating in that loop is Q times the current the supply has to provide, because the supply makes up only what the loop's resistance takes each cycle. That factor of Q is not a slogan but a ratio of two impedances: at resonance the pair as a whole presents Q times the reactance of either branch on its own, so for the same applied pd the branch current is Q times the supply current.

Feed such a circuit through a series resistance, or from a transistor's drain, and the pd across it peaks sharply at resonance, since that is where its impedance is greatest. The resonant frequency is the same 1/2π√LC$1/2\pi\sqrt{LC}$1/2π√(LC) as before, so none of the tuning arithmetic changes. What changes is which quantity peaks: the current in the series circuit, the voltage in the parallel one.

Both circuits are a mass on a spring in different clothing, and the analogy is worth keeping. Inductance plays the part of the mass, since it opposes a change of current exactly as inertia opposes a change of velocity. The capacitor is the spring, storing energy when it is charged as the spring stores it when it is stretched, and resistance is the damping. Energy shuttles between the magnetic and electric stores, each peaking twice per cycle, just as it shuttles between kinetic and potential in the oscillator, and the driving voltage plays the part of the hand that pushes the mass. A high Q is a lightly damped spring: a tall narrow response, and a long ring after the drive is removed.

Two filters from one tuned circuit

Where the output is taken determines what the circuit does. Take it across the resistor and the response peaks at f₀$f_{0}$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$f_{B}$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.

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.

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.

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/√2$1/\sqrt{2}$1/√(2), about 0.71, of its full value. Setting 1/2πf C$1/2\pi f C$1/2πf C equal to R gives it:

For the drawn pair f_c$f_{c}$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.