Series RLC Circuit Calculator

The frequency where a coil and a capacitor cancel each other out

Every coil paired with a capacitor has one frequency at which their opposing reactances are equal and annul each other. That frequency is f₀ = 1 / (2π√(LC)), and it is the number this calculator hands back first. Give it a resistance, an inductance and a capacitance, say whether the three parts sit in one loop or side by side, and it returns the resonant frequency, the quality factor Q, the bandwidth, the impedance at resonance and the half-power edges. A 1.5 mH coil with a 4.7 µF capacitor and 8 Ω of speaker resistance, for example, resonates at 1.896 kHz with a Q of 2.23.

It also runs the other way. Tell it the frequency you want and one of the two parts, and it works out the other one, then lists the values you can actually buy from the E12 ladder next to the frequency each one really delivers.

Problem: the parts list is right and the notch still lands somewhere else

A filter design gives you an inductance and a capacitance to four decimal places. The parts drawer does not. You buy the nearest thing on the shelf, the circuit works, and the corner sits somewhere near where the design asked for it. Nobody tells you how near.

The second half of the problem is the choice of topology. The same three components in one loop and the same three components side by side give the same resonant frequency and then behave in opposite ways: one collapses to its minimum impedance at resonance, the other climbs to its maximum. Reading a series answer while building a parallel tank is a mistake that costs a whole evening, because the arithmetic looks right the entire time.

Solution: pick the unknown, then read the frequency off the stock shelf

The calculator is built around the thing you are missing rather than around the formula. Three modes cover the cases that actually come up on a bench.

You have the parts, you want the frequency. Enter R, L and C and read f₀, Q, the bandwidth and the two half-power edges. Add a signal frequency and it also gives the impedance and the phase angle at that exact point.

You have the coil, you want the capacitor. Enter the target frequency and the inductance. A 10 mH coil aimed at 1 kHz calls for 2.533 µF, which nobody stocks, so the table puts 2.7 µF beside it and admits that the result lands at 968.6 Hz, off by 3.1%.

You have the capacitor, you want the coil. Enter the target and the capacitance. 180 pF aimed at a 455 kHz intermediate frequency asks for 679.7 µH, which is why 680 µH is the value printed on so many of those cans.

Filling in the boxes for each of the three jobs

Circuit type - series when the three parts form a single loop that the signal passes through, parallel when they hang across the same two nodes. A speaker crossover leg is series; an oscillator tank or a trap across a line is parallel.
What to find - the resonant frequency, the capacitor for a target, or the inductor for a target. The form hides whatever the chosen job does not need.
Resistance with its unit - in a series circuit this is everything resistive in the loop added together, including the driver and the coil's own wire. In a parallel tank it is the damping resistor or the load across it, and values in the tens of kilohms are normal, so the unit selector accepts Ω, kΩ and MΩ.
Inductance with its unit - nH, µH, mH or H. Radio work lives at the top of that list, audio crossovers in the middle, mains chokes at the bottom.
Capacitance with its unit - pF, nF or µF, matching how the part is actually marked.
Target frequency with its unit, in the two design modes - Hz, kHz or MHz.
Signal frequency, optional, in the first mode - a single frequency at which you want the impedance and the phase, rather than the resonant point.
Read the results. The headline tile carries the answer for the job you picked, the small tiles carry Q, bandwidth and the impedance at resonance, and the table underneath shows either the circuit across the band or the buyable values.

Series or parallel: the choice that flips every answer

Both arrangements resonate at the same frequency, because f₀ depends only on L and C. Everything downstream of that point is reversed, including what the resistance does to the sharpness. In a series circuit a bigger R spoils the Q; in a parallel one a bigger R improves it.

At resonance	Series R + L + C	Parallel R with L and C
Impedance	Falls to its minimum, Z = R	Rises to its maximum, Z = R
Current from the source	Peaks	Drops to its lowest
Quality factor	Q = X₀ / R, so less resistance is sharper	Q = R / X₀, so more resistance is sharper
What gets multiplied	Voltage across L and across C, by Q	Current circulating inside the tank, by Q
Typical use	Crossover leg, band-pass path, series trap to ground	Oscillator tank, notch across a line, antenna trap
Set R to zero	Unbounded Q, a model limit rather than a circuit	A dead short, 0 Ω at every frequency

Six circuits run through the calculator

A two-way speaker crossover leg. Series, 8 Ω driver, 1.5 mH, 4.7 µF. The pass band is centered on 1.896 kHz with X₀ = 17.86 Ω and Q = 2.23, so the bandwidth is 848.8 Hz and the half-power edges fall at 1.518 kHz and 2.367 kHz. Gentle on purpose: a crossover is meant to hand over, not to ring.

A radio-frequency tank. Series, 50 Ω, 100 µH, 100 pF. Resonance lands at 1.592 MHz, right in the middle of the medium-wave band, with X₀ = 1 kΩ, Q = 20 and a bandwidth of 79.58 kHz. In the PL version this circuit could not be entered at all, because the only units on offer were millihenries and microfarads.

The intermediate-frequency can inside a superheterodyne receiver. Series, 20 Ω, 680 µH, 180 pF. That pair gives 454.9 kHz, a Q of 97.18 and a bandwidth of 4.681 kHz, which is about as wide as an amplitude-modulated voice channel needs.

A damped oscillator tank. Parallel, 10 kΩ across 10 mH and 1 µF. Same 1.592 kHz as the series pairing of the same parts, but the impedance now peaks at 10 kΩ, the Q is 100 and the bandwidth collapses to 15.92 Hz. Lower that resistor and the tank gets blunter, which is the opposite of what the series row does.

A mains-frequency trap. Series, 2 Ω, 0.5 H, 20 µF. It sits at 50.33 Hz with a Q of 79.06 and a bandwidth of only 0.6366 Hz. A trap that narrow is unforgiving: a capacitor 10% away from its marking moves the notch further than the whole bandwidth.

A subsonic filter ahead of a woofer. Series, 47 Ω, 100 mH, 10 µF. Resonance at 159.2 Hz, Q = 2.13, bandwidth 74.8 Hz. The broad skirt is the point here, because a sharp filter in the bass range rings audibly.

The cheat card: what Q buys and what it costs

Q is the single number that decides how selective the circuit is, and the bandwidth follows from it directly as BW = f₀ / Q. The half-power edges are not simply half a bandwidth on either side of f₀; they sit at f₀(√(1 + 1/4Q²) ± 1/2Q), which is why the lower edge is always a little closer to resonance than the upper one. The gap only matters below a Q of about 5, where the table's last column shows it plainly.

Q	Bandwidth as a share of f₀	What it feels like	Edges, lower and upper
0.5	200%	Barely resonant, a broad hump	0.414 and 2.414 × f₀
1	100%	A gentle shelf, no ringing	0.618 and 1.618 × f₀
2.23	44.8%	Crossover territory	0.801 and 1.249 × f₀
10	10%	A clear peak you can hear or see	0.951 and 1.051 × f₀
100	1%	Tuned circuit, needs trimming	0.995 and 1.005 × f₀

Read the third row against the crossover above: a Q of 2.23 puts the edges at 0.801 and 1.249 times the center, which is exactly the 1.518 kHz to 2.367 kHz span the calculator returned for it.

Questions that arrive with a parts list

Why does swapping series for parallel leave the frequency unchanged?

Because resonance is the point where X₁ = X₊, and both reactances depend only on L, C and the frequency. The resistor sets how sharply the circuit reacts, never where it reacts. That is also why the same 10 mH and 1 µF give 1.592 kHz in both arrangements above while the impedance at that point goes from 100 Ω to 10 kΩ.

Can the voltage across the coil really exceed the supply?

In a series circuit at resonance, yes, by a factor of Q. Feed the crossover leg above with 10 V and the coil and the capacitor each see about 22.3 V, which is harmless there. Push the same trick at a Q of 79.06, as in the mains trap, and a 10 V drive puts close to 791 V across parts rated for far less. This is the most common way a resonant circuit destroys itself.

What resistance should I type if I do not have a resistor in the circuit?

There is always one. In a series loop it is the wire resistance of the coil plus whatever the circuit drives; measuring the coil with a meter at direct current gets you most of the way. In a parallel tank it is the load across it. Typing 0 is allowed, but the calculator will tell you what that means rather than quietly returning an answer: an unbounded Q in series, and a dead short in parallel.

Why is the capacitance it asks for never a value anyone sells?

Because the formula does not know about parts drawers. Stock values follow the E12 ladder, twelve steps per decade, which is why the design modes put the three values below and above the exact answer next to it and print the frequency each one actually produces. The 2.533 µF the arithmetic wants becomes a 2.7 µF you can buy, at the price of landing 3.1% low.

Does the real circuit match these numbers?

The frequency, closely. The Q, rarely. This model treats the coil as pure inductance and the capacitor as pure capacitance, while a real coil carries winding resistance that rises with frequency and a real capacitor has an equivalent series resistance of its own. Both losses drag the measured Q below the calculated one, and by how much depends on the parts rather than on the arithmetic. The frequency survives because it depends on L and C, and those two are the parameters parts makers control best.

What is the phase angle telling me?

Which of the two parts is currently winning. Below resonance in a series circuit the capacitor dominates, the angle is negative and the circuit behaves capacitively; above resonance the coil takes over and the angle turns positive. At resonance it passes through zero, which is the electrical definition of the tuning point. Enter a signal frequency and the calculator names the character in words alongside the number.

Why does the coil stop working above a certain frequency?

Every winding has stray capacitance between its turns, so a coil is itself a resonant circuit. Above its own self-resonant frequency it stops behaving as an inductor and starts behaving as a capacitor, and no external capacitor will make it tune. Datasheets quote that limit; a calculation that ignores it, this one included, will happily return a frequency the part cannot reach.

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