Circuit Response
every model has an edge, and the figure carries it
Lines, where a wire has a length

A quarter wave, and the path the current takes back

A line a quarter of a wavelength long, whose impedance is the geometric mean of the two it joins, matches them exactly — reflecting 5×10⁻¹⁷ of what arrives, which is the arithmetic's floor. At one frequency. Seventeen per cent either side of it the reflection is back to a tenth, and that band is the whole of what the technique is worth.
16 min read 7 figures Every model has an edgeThe constants decideThe schematic is a label

Two facts about a quarter-wavelength of line are usually stated together and are worth separating, because one of them is an identity and the other is a technique with a bandwidth.

The identity is that a line a quarter of a wavelength long inverts whatever is on its far end about the square of its own characteristic impedance: Z_in = Z₀²/Z_L. A short circuit becomes an open circuit, an open becomes a short, and a load becomes something else entirely.

The technique is what falls out of that: to join two lines of different impedance without reflection, put a quarter-wavelength of line between them whose impedance is the geometric mean of the two.

Matching 50 Ω to 200 Ω with 51.7 mm of 100.0 Ω linecomputed by solving, not by drawing at 261 frequencies. The reflection at the design frequency is 4.6e-17 — nothing, to the arithmetic — against 0.600 for the bare junction, which throws 36% of the power back. It stays under 0.1 from 0.914 to 1.086 of that frequency, a band of 17.1%.00.2000.4000.6000.8000.50011.50frequency ÷ the frequency it was cut forreflection |Γ|the bare junction: 0.600|Γ| = 0.1, or 20.0 dB return lossexact here only17.1% of a bandsolved, then checked — the band scanned, not approximatedexact at one frequency, 17% of band
Fig. 1 Fifty ohms matched to two hundred by 51.7 mm of hundred-ohm line, swept from a fiftieth to twice the frequency it was cut for. At the design frequency the reflection is 5×10⁻¹⁷ — nothing, to the arithmetic — against 0.600 for the bare junction. The shaded region is where it stays under a tenth, which is 17.1% of the design frequency. The slider is the load impedance.

Exact, and only there

The reflection at the design frequency is 4.6×10⁻¹⁷. That is not a small number in the sense of being negligible; it is the residue of a complex division and it means the match is exact.

Twenty per cent below the design frequency it is 0.14. Twenty per cent above it, 0.13. Half the design frequency and it is 0.36 — most of the way back to the 0.60 the bare junction reflects with nothing between the two lines at all.

So the useful statement about a quarter-wave transformer is not the identity but a band, and the figure measures it rather than quoting the standard approximate expression for it. Scanning the solved reflection and bisecting for the two frequencies where it crosses a tenth gives 914 MHz and 1,086 MHz for a design frequency of 1 GHz: a fractional bandwidth of 17.1%.

That band depends strongly on how large a transformation is being attempted, which is the slider’s argument.

It is worth saying what “reflection” costs, because a coefficient between zero and one is easy to read as a percentage of something and it is a percentage of the wrong thing. Γ is a voltage ratio, so the power reflected is Γ². A bare fifty-to-two-hundred junction reflects 0.600 of the voltage, which is 36% of the power; a bare fifty-to-a-thousand junction reflects 0.905, which is 82%. The transformer takes both to nothing at the design frequency and to Γ = 0.1 — one per cent of the power — at the edges of the band this page measures.

A tenth is a common enough threshold that it deserves a name: it corresponds to a return loss of 20 dB, which is what a specification usually asks for and what the shaded regions in the figures mark.

The band narrows as the transformation grows

Matching fifty ohms to seventy-five — a 1.5:1 step, and the mismatch between two common cable impedances — gives a bare reflection of 0.200 and a band of 65.6%. Matching fifty to a hundred is 0.333 and 36.7%. Fifty to two hundred is 0.600 and 17.1%. Fifty to a thousand is 0.905 and 6.0%.

The pattern is clear enough to be worth naming: the harder the transformation, the narrower the band over which one section achieves it. A single quarter-wave section is a good match over a wide band when there was not much wrong in the first place, and a narrow-band trick when there was.

What is done about that in practice is to use several sections in series, each making part of the transformation, with impedances chosen so that their individual reflections cancel over a band rather than at a point. That is a genuinely different design problem — it is filter design, with the reflection as the quantity being shaped — and it is named here and not built, because it needs machinery this site does not have.

Matching 50 Ω to 1000 Ω with 51.7 mm of 223.6 Ω linecomputed by solving, not by drawing at 261 frequencies. The reflection at the design frequency is 1.3e-16 — nothing, to the arithmetic — against 0.905 for the bare junction, which throws 82% of the power back. It stays under 0.1 from 0.970 to 1.030 of that frequency, a band of 6.0%.00.2500.5000.75010.50011.50frequency ÷ the frequency it was cut forreflection |Γ|the bare junction: 0.905|Γ| = 0.1, or 20.0 dB return lossexact here only6.0% of a bandsolved, then checked — the band scanned, not approximatedexact at one frequency, 6% of band
Fig. 2 Fifty ohms matched to a thousand. The bare junction reflects 0.905 — eighty-two per cent of the power comes straight back — and the transformer takes that to nothing at one frequency and to a tenth over 6.0% of it. The same technique, the same exactness, and a band three times narrower than the four-to-one case, because the reflection has three times as far to travel.

The other exactness, and what breaks it

The identity also says that a quarter-wave line with a short circuit at its far end presents an open circuit at its near end. That is exact in a lossless line, and it cannot be exact in a real one for a reason that needs no arithmetic at all: an open circuit carries no current, and a line that dissipates power must be carrying some.

Putting attenuation into the line model and evaluating the same expression with a complex propagation constant gives the size of it. A quarter-wave shorted stub in a lossless line has an input impedance limited only by how exactly the length is cut. With 0.2 dB per metre of attenuation the same stub presents about 38 kΩ, and with 1 dB per metre about 8 kΩ.

Those are large impedances and for most purposes they are open circuits. They are also finite, and the value is set by the loss rather than by anything about the geometry — which means the quantity that decides how good a resonator a length of line makes is the one property the ideal model leaves out entirely.

This is the same shape as the capacitor that is an inductor: a component behaves as its symbol says up to a point set by something the symbol does not show, and the number is computable from the thing that was omitted.

Impedance of a series RLC of Q = 4, measured by driving itOne ampere is forced into the terminals at each frequency and the resulting voltage is the impedance. The minimum is 7.91 Ω at 5.03 kHz.1101001k10k1001k10k100k1Mfrequency (hertz)impedance magnitude (ohms)reactances cancel at 5.03 kHz7.91 Ωsolved, then checked — one ampere in, 201 frequenciesnot a component value: what the pair does
Fig. 3 An impedance swept over frequency. A quarter-wave stub’s input impedance does the same thing at every odd multiple of its design frequency and the opposite at every even one, so a length of line used as a resonator has an infinite series of resonances rather than one — which is the property that makes it useful as a filter and dangerous as an accident.

Where the current goes back

The second half of this essay is about a question the first half did not have to ask, because a coaxial cable answers it by construction: where does the return current flow?

On a circuit board it is not obvious. A track runs over a plane, and the plane is a sheet of copper that the current could return through anywhere. What decides is a competition between two costs.

Resistance is minimised by spreading out. A wide path has more copper in parallel and less resistance, so at low frequency the return current fans out across the plane.

Inductance is minimised by staying close. The loop the current encloses is the area between the outgoing track and its return, and inductance is proportional to that area, so a return directly beneath the track has the least. At high frequency the reactance dominates the resistance and the current has effectively no choice.

The return under 10.0 cm of track, 200 µm above the planecomputed by solving, not by drawing from the geometry. The path's resistance is 83.3 mΩ and its inductance 125.7 nH, so resistance decides below 106 kHz and inductance above it. Above the crossover the current runs directly beneath the track and encloses 20.0 mm² rather than 50.0 mm². A milliamp round that loop at 100 MHz radiates -1.1 dBµV/m at three metres.10m1.0e+2m1101001k1k10k100k1M10M100Mfrequency (hertz)impedance of the return path (ohms)resistance: the wide pathreactance: the tight pathR = ωL at 106 kHzsolved, then checked — geometry, not a tableleast resistance below 106 kHz
Fig. 4 The two competing costs against frequency, for ten centimetres of track two hundred micrometres above the plane. The resistance of the spread path is 83.3 mΩ, the inductance of the tight path is 126 nH, and the crossover — where the reactance of the second equals the resistance of the first — is at 105.5 kHz. The second marked edge is where the small-loop expression for what the loop radiates stops applying, which is 1.68 GHz for this geometry.

A hundred kilohertz, which is nothing

The crossover comes out at 105.5 kHz for that geometry, and the number is worth pausing on because it is so much lower than most readers expect.

Above the crossover the return current runs directly beneath the outgoing track. That is not a high-frequency phenomenon in any useful sense — 105 kHz is audio, near enough — so for essentially everything a digital board does, the return path is determined by inductance and the current is directly under the track whether the designer intended it or not.

Two practical consequences follow, and both are about what happens when the current is prevented from taking that path.

A slot in the plane is a detour. Cut a slot across the plane under a track and the return current cannot go straight; it goes around, and the loop area becomes the area of the detour rather than the area of the track’s own height. A ten-centimetre detour around a slot turns a twenty-square-millimetre loop into something a hundred times larger.

A change of layer is a discontinuity. A track that passes through a via to another layer takes its return current with it, and unless there is a path for the return to change layers too — a nearby via between the two planes — it must find one somewhere else, which is another detour of the same kind.

Neither of those is visible on a schematic. Both are decided by the geometry of a copper sheet that the schematic represents by a symbol, and the site’s habit that a schematic is a label rather than a drawing has an unusually literal application here: the layout is the circuit.

What the loop radiates, and where the model stops

The loop area matters because it is what radiates. A small current loop’s far field rises as the square of the frequency and in proportion to the enclosed area, which is why the two consequences above are stated in terms of area rather than in terms of length.

For the geometry drawn — twenty square millimetres, a milliamp, a hundred megahertz, three metres away — the field is −1.1 dBµV/m. Lift the track to 1.6 mm above the plane and the loop is 160 mm², eight times larger, and the field rises by the same factor to 16.9 dBµV/m. The Class B limit at three metres is 30 dBµV/m in that part of the spectrum, so the difference between a well-routed track and a badly routed one is a substantial fraction of the whole allowance, from geometry alone.

The expression used is the standard small-loop result, and it has a validity condition that is computed and drawn rather than assumed: the loop’s perimeter must be a small fraction of a wavelength. At a tenth of a wavelength, the two-hundred-micrometre case gives out at 1.68 GHz and the 1.6 mm case at 593 MHz. Above those the calculation is not slightly wrong, it is answering a different question, and the figure refuses to apply it there rather than extrapolating.

The return under 10.0 cm of track, 1600 µm above the planecomputed by solving, not by drawing from the geometry. The path's resistance is 83.3 mΩ and its inductance 1005.3 nH, so resistance decides below 13.2 kHz and inductance above it. Above the crossover the current runs directly beneath the track and encloses 160.0 mm² rather than 190.0 mm². A milliamp round that loop at 100 MHz radiates 16.9 dBµV/m at three metres.10m1.0e+2m1101001k1001k10k100k1M10Mfrequency (hertz)impedance of the return path (ohms)resistance: the wide pathreactance: the tight pathR = ωL at 13.2 kHzsolved, then checked — geometry, not a tableleast resistance below 13.2 kHz
Fig. 5 The same track 1.6 mm above the plane rather than 0.2 mm. The inductance is eight times larger, so the crossover falls to 13.2 kHz — the return is inductance-determined from almost any frequency of interest — and the loop area is eight times larger, so the radiated field is eighteen decibels higher. One dimension of the board’s stack-up, and eighteen decibels.

Why the two costs are the ones that compete

There is a step in the return-path argument that deserves its own paragraph, because it is the part most often stated as a slogan.

The claim is not that current “prefers” one path. Current divides between every available path in inverse proportion to their impedances, always, and both paths carry some at every frequency. What changes with frequency is the ratio, and it changes because one path’s impedance is a resistance and the other’s is a reactance: one is flat and the other rises linearly.

Below the crossover the spread path’s 83.3 mΩ is smaller than the tight path’s ωL, so most of the current spreads. Above it the reactance exceeds the resistance and the split reverses. At the crossover itself the two are equal by definition, and the current divides roughly evenly between paths that are physically far apart — which is the least convenient case for anyone trying to predict what the board will radiate.

The number that decides it, R/2πL, has the form of a corner frequency because that is what it is: the return path is a one-pole divider between two impedances, and its crossover is the pole. Nothing about the phenomenon needs vocabulary the rest of this collection does not already have.

Two things follow that a slogan does not give. The crossover moves with the geometry rather than with the signal — a thinner track raises the resistance and moves it up, a greater height raises the inductance and moves it down — and the transition is gradual, spanning about a decade either side, so a board carrying signals near a hundred kilohertz has a return current that is genuinely undecided.

The two halves of the page

The quarter-wave transformer and the return path look like separate subjects and they are the same one, which is why they share an essay.

Both are about the fact that above the frequency where a circuit is electrically large, the geometry is the circuit. A quarter-wavelength of line is a component whose value is a length. A return path’s inductance is an area. Neither has a symbol on a schematic, neither appears in a netlist, and both are decided by decisions that look like drawing rather than like design.

The frequency at which that transition happens is the one Kirchhoff’s own frequency computes, and this field is what is on the far side of it. What this page adds is that the transition is not a single frequency for all purposes: the return path crosses over at a hundred kilohertz, the lumped model gives out at four megahertz, and the small-loop radiation model gives out at one and a half gigahertz. Three boundaries, four decades apart, belonging to the same ten centimetres of copper.

10.0 cm of track, solved as a lumped circuit and as a lineThe two agree to 0.030% at 3.97 MHz, where the track is one degree long, and to 30.1% at 143 MHz, where it is a tenth of a wavelength. Above that the lumped model is not approximately right; it is describing a different object.1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz
Fig. 6 The lumped and distributed models of the same track, from the limits field. The crossover marked on it is the middle of the three boundaries above. Below it a netlist describes the copper; above it the copper has to be described by its shape, and this field is the vocabulary for doing that.

One number this page does not compute

The radiated field here comes from a closed-form expression rather than from a solve, and that makes it the only quantity in this field that is quoted rather than measured. It is worth being explicit about, because everything around it is solved and the difference is easy to miss.

The expression is the standard small-loop result: a loop of area A carrying a current I produces a field at distance r proportional to f²AI/r. Its derivation assumes the current is uniform around the loop, which is what “electrically small” means, and the figure carries the frequency at which that assumption fails. What the figure does not do is derive the coefficient, and it could not without machinery — a field solver — that is a different subject from the one this site is about.

The honest description is therefore that the loop area is computed from the geometry, the validity boundary is computed from the geometry, and the constant between them is taken. That is one borrowed number in a collection that computes everything else, and stating which one it is seems better than letting it blend in.

A 100 nF capacitor, and what it is above 14.5 MHzThe dashed line is 1/(ωC), which is what the symbol means. The solid line is the same part with 30 mΩ of series resistance and 1.2 nH of series inductance, solved. They part company at 4.69 MHz and by a decade above resonance the part's impedance is 101× what its capacitance predicts.10m1.0e+2m1101001k10k10k100k1M10M100M1Gfrequency (hertz)impedance magnitude (ohms)1/(ωC), the symbol's promise10% off above 4.69 MHzinductive above 14.5 MHz30 mΩ — the floor the resistance setssolved, then checked — the part as three elementsa capacitor below 14.5 MHz, an inductor above
Fig. 7 A capacitor’s impedance against frequency, from the frequency field, with the resonance its own lead inductance gives it. The parallel with this page is exact: a component behaves as its symbol says below a frequency set by a geometric property the symbol does not show, and above it behaves as something else entirely. A capacitor becomes an inductor; a length of line becomes a transformer; a plane becomes a loop.

Where this goes next

00.50011.5002468time, in propagation delaysvoltsthe divider: 1.0000 V0.833 V: the source against the linethe far endthe near endfirst arrival, 4.83 nssolved, then checked — a series against a dividernothing at the far end before 4.83 ns The staircase in time A source driving a metre of cable does not know what is on the far end of it for 4.83 nanoseconds. What it drives into during that time is decided by the cable's characteristic impedance and nothing else — and when the far end finally answers, the answer comes back as a staircase whose limit is the resistive divider the circuit was going to be all along. 1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz Kirchhoff's own frequency The current law says the current entering a node equals the current leaving it at the same instant, which assumes the signal crosses the circuit in no time. It crosses at about two-thirds the speed of light, so the law has a frequency of its own — set by nothing but the physical size of the board. 10m1.0e+2m1101001k10k10k100k1M10M100M1Gfrequency (hertz)impedance magnitude (ohms)1/(ωC), the symbol's promise10% off above 4.69 MHzinductive above 14.5 MHz30 mΩ — the floor the resistance setssolved, then checked — the part as three elementsa capacitor below 14.5 MHz, an inductor above The capacitor that is an inductor A hundred-nanofarad capacitor follows 1/(2πfC) for four decades and then turns round and climbs. Above 14.5 MHz it is an inductor, and a decade past that its impedance is ninety-nine times what its capacitance predicts — all of it caused by about a nanohenry of lead and via that nobody chose and nobody drew.