Circuit Response
every model has an edge, and the figure carries it
Concept

Loop area — where it appears

The area enclosed between a current's outward path and its return, which sets the inductance of the pair and how much either radiates. It appears in what a current radiates and in what a neighbouring loop picks up, which is why a return path directly under its own track matters.

Named by 7 essays across 2 fields — each of them below, with the objects they name alongside it.

Matching 50 Ω to 200 Ω with 51.7 mm of 100.0 Ω line. computed 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%.

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.

lines · Matching The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres.

Where the current comes back

The return current under a track spreads out at low frequency and runs directly beneath it at high. Estimated from two paths chosen in advance, the change is a corner at 106 kilohertz for any track two hundred micrometres above a half-milliohm plane, with neither the length of the track nor its width in it. Solved, it is a band three decades wide rather than a corner, and the width enters it after all — but the length is still absent and the stack-up still sets the top, which is the part of the argument a designer needs.

lines · Return path Three nanohenries of copper move the peak to 5.63 MHz and raise it to 1.29 Ω. computed by solving, not by drawing. The same two capacitors, with and without the inductance of the way to them: one nanohenry of mounting loop per part and two nanohenries of plane between the bank and the load. The dashed curve is the bank as the rung below drew it, peaking at 1.187 Ω at 6.52 MHz; the solid one is what the load sees, peaking at 1.293 Ω at 5.63 MHz. The peak moves down because the branch that is inductive at that frequency got more inductive, and it rises for the same reason. Above about twenty megahertz the two part company entirely: the bank is still falling toward its parts' own resistances and the load is rising on two nanohenries that no capacitor is across.

The capacitor that is not where the load is

The rung below this one connects two capacitors to a load through nothing, and says so. Put three nanohenries of ordinary copper in — one of mounting loop per part and two of plane between the bank and the load — and the anti-resonance moves down to 5.63 megahertz and up to 1.29 ohms, a probe touching the ceramic reads a twelfth of what the load sees at 16.7 megahertz and three and a half times too much at 8.35, and the twentieth capacitor is worse than the second.

power · Decoupling The bank reaches 1.500 mΩ and the load sees 141.5 mΩ at that same frequency. computed by solving, not by drawing. One bulk part and 20 ceramics, with a nanohenry of mounting loop each and two nanohenries of plane between the bank and the load, solved once per frequency and read at both nodes. The dashed curve is the bank's own node — what a probe on the parts measures. The solid one is the load. The bank's least impedance is 1.500 mΩ at 11.3 MHz, and at that frequency the load sees 141.5 mΩ, which is 94.3 times more, against 141.5 mΩ of plane reactance at that frequency. Whatever the parts do, the load's reading cannot fall below the reactance of the copper in front of them, and the parts reach their best by moving up the frequency axis into it.

The floor and the ceiling move apart

A decoupling bank is judged by two numbers — the lowest impedance it reaches and the highest frequency at which it still meets its target — and with no copper between the parts and the load both improve together as capacitors are added, 5.000 milliohms down to 1.500 and 19.8 megahertz up to 162. Three nanohenries of ordinary board separate them. The bank's own floor still falls 3.33 times while the load's falls 1.49, and the ceiling read at the parts climbs to 82.4 megahertz while the load's peaks at 8.06 and falls to 5.05. At twenty parts the two nodes disagree by a factor of 94 about the same solve.

power · Decoupling The return under a track gathers from 26.3 kHz to 1.42 MHz, not at one frequency. computed by solving, not by drawing, on a cross-section of a 50 mm plane cut into 120 strips, each with its resistance and its partial inductance to every other strip and to the track. The solid curve is the share of the return current inside one track-height of the point beneath the track; the second is the share inside ten heights. At direct current the return spreads evenly — 0.8 per cent within one height — and far above the band it is the image-current distribution, 48.7 per cent, which the closed form gives as 48.7. Between them it gathers across three decades: a tenth of the way by 26.3 kHz, half by 283 kHz, nine tenths by 1.42 MHz, shaded. The resistance of the path equals its reactance at 1.59 kHz, where the return has not yet moved. The single corner estimated from a path three track-widths wide and a parallel-plate inductance is 106 kHz, 28 per cent of the way through the band.

The corner that is three decades wide

Where the current comes back put the change in a return current's path at 106 kilohertz, from a low-frequency path assumed three track-widths wide and an inductance taken from a parallel-plate formula. Solved across a plane cut into a hundred and twenty strips, the loop's resistance equals its reactance at 1.59 kilohertz, where the current has not moved at all, and the return then gathers beneath the track over three decades — half of the way by 283 kilohertz, nine tenths by 1.42 megahertz. The single corner is a point about a quarter of the way through a band.

lines · Return path A track needs about three heights of copper beside it, and it is the resistance that says so. computed by solving, not by drawing at 100 MHz, each point a strip solve of its own on a 50 mm plane of the same area, moved sideways. The horizontal axis is where the track's centre sits relative to the plane's edge, in units of the track's height above it; negative is a track hanging past the edge with no copper beneath it. With the centre directly over the edge the loop's inductance is 1.161 times its centred value and its resistance 2.90 times, because the return has to crowd into the last few hundred micrometres of copper. Three heights in, the inductance is 1.006 times and the resistance 1.09; ten heights in, both are within 0.7 per cent. Three heights past the edge the inductance is 1.82 times. At direct current every point on this axis is exactly one, because the copper has been moved and not removed.

Where the plane runs out

The corner that is three decades wide solved a return current over a plane that extends well past the track on both sides. Where it does not, the two costs arrive at opposite ends of the band: at direct current a plane that ends under the track costs 27 per cent of inductance and not one part in a million of resistance, and above the band it costs 16 per cent of inductance and 199 per cent of resistance. Three track-heights of copper beside the track removes almost all of both, and the number three has no millimetres in it — sixteen times the whole cross-section gives the same ratios to a part in a billion.

lines · Return path Two tracks 1 mm apart share all of the plane's resistance at direct current and 14.5% of it above the band. computed by solving, not by drawing, across a 50 mm plane cut into 239 strips, with two tracks 200 µm above it and 1 mm apart — 5.0 heights. One track carries the current and the voltage along the other's loop is measured. The shared resistance, as a fraction of the driven loop's own, is 1 at direct current, where both returns spread across the whole plane and share its 10 mΩ/m; it falls through a half at 183 kHz and settles at 0.1454 above 10.0 MHz, where each return has gathered under its own track. The shared inductance is 0.398 of the loop's own at direct current and 0.0341 above the band.

Two returns in one plane

Two tracks over one plane share the whole of its resistance at direct current, however far apart they are routed: 10 milliohms a metre on a fifty-millimetre plane, from tracks a millimetre apart or ten. The sharing ends across the same band a single return gathers over, and it ends sooner the farther apart the tracks are — through a half at 525 kilohertz for tracks three heights apart and at 9.88 kilohertz for fifty. Above the band what is left is the overlap of two image distributions, 4h²/(4h² + d²): 14.5 per cent of the resistance at a millimetre, 0.68 at five. And in the middle of the band the two loops' mutual inductance changes sign.

lines · Return path

Named alongside it

The objects these essays reach for when they reach for this one.

ParasiticsReturn currentSkin effectDecouplingModel rangeVerificationCurrent densityEquivalent series inductanceImpedance matchingMeasurement conditionSelf-resonanceCommon-impedance

All concepts