Concept

Distributed element — where it appears

An element whose resistance, capacitance or inductance is spread along its length, so that different parts of it see different voltages. No single lumped value stands for it at every frequency, and each reading of it takes a different weighted average of what is there.

Named by 3 essays across 2 fields — each of them below, with the objects they name alongside it.

With 100% of its capacitance to its mounting, a 141.4 Ω helix is 1.37° out at 100 MHz, not 2.7 × 10⁻³°. computed by solving, not by drawing. The angle of the impedance of a 141.4 Ω resistor — √(L/C) for 8 nH of lead and 0.4 pF — against frequency: the three-element part with all 0.4 pF across its film (dashed), and a helix of 32 turns with the same totals (solid), 0% of the capacitance between adjacent turns and 100% from the turns to the part's mounting, with one end grounded. At √(L/C) the three-element part has no first-order angle and rises as the cube of frequency. The helix is not flat there: its capacitance to the mounting counts only a third, so its own flat resistance is √(3L/C) = 244.9 Ω, and at 141.4 Ω its angle is first order in frequency — 1.37° at 100 MHz against 2.7 × 10⁻³°.

The resistor that is a helix

A spiral-cut film resistor is a helix, and its capacitance is distributed along the cut rather than lumped across the ends. Where it goes decides whether the three-element model survives. Capacitance between adjacent turns bridges each turn's own piece of film, and a helix of any number of turns built that way is the three-element part exactly — to 8×10⁻¹³ at 64 turns. Capacitance from the turns to what the part is mounted over is different: its current returns through only part of the film, it weighs (4N² − 1)/(12N²) of its value, a third in the limit, and the resistance with the flattest angle moves from √(L/C) = 141.4 Ω to √(3L/C) = 245 Ω. At 141.4 Ω such a part is 1.37° out at 100 MHz where the lumped model says 0.0027°. The third is a first-order rule; above about a gigahertz the helix is a line.

frequency · Real resistor

A megohm part with one picofarad to its mounting, heavier at one end: 0.433 pF with that end driven and 0.233 pF turned round, 1.56° against 0.84° at 10 kHz. computed by solving, not by drawing: the angle of the impedance of a 1 MΩ resistor as 200 sections, with 1 pF in total from its film to its mounting tapered as c(x) ∝ 1 + a(1 − 2x), a = 0.6, read at the terminal with the far end grounded. Driving the end nearer the mounting (solid) it presents 0.433 pF and 1.56° at 10 kHz; turned round (dashed), 0.233 pF and 0.84°. The first-order weight is ∫c(x)(1 − x)²dx, which is C(1/3 + a/6) and C(1/3 − a/6). The part's total capacitance has not changed, nor has its resistance, and neither of the readings is that total.

The resistor turned round

A resistor's stray capacitance is usually drawn as one capacitor across its ends, and a stray to the part's mounting is not one number. Read at the terminal with the far end grounded it is weighted by the square of the distance from ground, so a uniform stray counts as a third of itself, and turned round a part with a tapered stray reads differently: 0.433 pF one way and 0.233 pF the other from one picofarad in all. In the current through the part the weight is x(1 − x), a sixth, and the current reaching the far end is 3 dB down at ωRC = 9.18 rather than at one. Two readings of one part separate the total from the taper, and a distributed stray's phase never reaches a capacitor's: it stops at 45 degrees.

frequency · Real resistor $The same 15 pF spread along 2 kΩ of line, on a 1 ns edge: with 1 of its branches copied the output is flat to 1.07%, against 11.2% with one trimmer alone. computed by solving, not by drawing: the response to a 1 ns edge of a ten-to-one divider whose lower arm is 100 pF direct and 15 pF along a 2 kΩ resistive line, with one trimmer (dashed) and with the first 1 branch of the line's exact partial-fraction expansion copied into the upper arm (solid). The expansion's time constants are 12.2 ns, 1.35 ns, 486 ps and so on, as 1/(2k − 1)². Inverted numerically by Talbot's method, and the line itself checked against a netlist of sections.$

The branches a square root needs

A resistive cable's late charge is a continuum and not one branch, and the partial fractions of its tanh are an exact description of it: 81.1 per cent of the capacitance behind a time constant of 12.2 ns, 9.01 per cent behind a ninth of that, 3.24 behind a twenty-fifth, all through the same kilohm. So a divider's upper arm can copy it one piece at a time. On a 1 ns edge one trimmer alone is 11.2 per cent out, one copied branch 1.07, two 0.252 and four 0.0353. The first branch does most of the work and the uncopied capacitance falls only as one over the count; what saves the rest is that a branch faster than the edge is averaged away. Faster edges ask for more branches, and the tolerance on the first ends the series before the arithmetic does.

instruments · Compensation

Named alongside it

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

Lumped element Model range Parasitics Attenuator probe Characteristic impedance Compensated divider Component tolerance Design tradeoff Lead inductance Measurement condition Phasor Residues

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