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

Lead inductance — where it appears

The inductance of the wire, track or via that connects a component, which nobody draws and which decides what the component does at high frequency. It is what turns a capacitor into an inductor above its self-resonance, and a few nanohenries is enough to do it in the tens of megahertz.

Named by 8 essays across 6 fields — each of them below, with the objects they name alongside it.

A 10 kΩ resistor, and the 19.3 MHz it is one below. computed by solving, not by drawing. The dashed line is R, which is what the symbol means. The solid line is the same part with 8.0 nH of lead inductance in series and 0.40 pF across the body, solved as a three-element network and checked against the closed form for the same three elements to 3.3e-16. It is ten per cent below its own value by 19.3 MHz, and which of the two parasitics does that depends on the resistance: the shunt capacitance wins above 91.02 Ω and the lead inductance below it. The slider is the resistance, and the departure frequency it moves is not monotonic — it rises a decade per decade of resistance, peaks near 91.02 Ω at 2.00 GHz, and falls a decade per decade after that.

The resistor that is only a resistor

A capacitor becomes an inductor above a frequency its leads decide, and an inductor becomes a capacitor. The third member of that family is the one nobody draws, and it is the only one whose edge is not monotonic in its own value: a ten-megohm resistor stops being one at 19 kilohertz, a ten-ohm resistor at 92 megahertz, and between them sits a resistance whose impedance is flat to fourth order — 91.02 ohms here, which is the square root of L over C divided by the root of one plus root two.

frequency · Real resistor A follower with 1000 pF on it looks like -1182 Ω of negative resistance. computed by solving, not by drawing. The impedance looking into the base of an emitter follower carrying 5.0 mA, with 1000 pF on its emitter. The real part is negative from 1.25 MHz upward and reaches -1182 Ω at 3.40 MHz: the load's reactance multiplied by a complex current gain, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it — the total loop reactance passes through zero at a frequency the inductance chooses, and the loop resistance there goes negative above 74.9 nH with 10 Ω of source, which is a few centimetres of wire. A hundred ohms of source raises that to 913 nH: the repair is a resistor in the base, and it works by making the source worse.

The input that pushes back

An emitter follower with a capacitor on its emitter has a negative resistance looking into its base — 1182 ohms of it at 3.4 megahertz for a nanofarad, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it: with ten ohms of source the loop goes unstable above 74.9 nanohenries, which is seven centimetres of wire. The repair is the opposite of the instinct — a hundred ohms of source raises the threshold to 913 nanohenries, so the fix for a follower that oscillates is to make the thing driving it worse.

semiconductors · Emitter follower 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 with no inductance in the way, 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

A decoupling bank is usually drawn as two capacitors joined to a load through nothing, which is a stated simplification. 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 Flat in angle at √(L/C), 141.42 Ω, and flat in size at 91.018 Ω. computed by solving, not by drawing, at 127 resistances on the closed form the network was checked against. The upper curve is the frequency at which the part's size is 1% away from R, the lower one the frequency at which its angle reaches 1°. Both are V-shaped and their points are in different places: the angle's first-order term vanishes at √(L/C) = 141.42 Ω, bisected on the measured slope to ten figures, and the size's second-order term at 91.018 Ω. The widest 1° band is 1.13 GHz, at 150.69 Ω; the widest 1% band is 1.66 GHz, at 95.806 Ω. At 91.018 Ω, flattest in size, the angle reaches 1° by 53.9 MHz. The flat line is 139 MHz, where a 6 mm body is one degree long: it binds the angle's edge from 118.72 Ω to 168.88 Ω and the size's from 43.947 Ω to 359.53 Ω.

The resistor that is right in size and wrong in angle

A resistor's impedance departs from its value in size as the square of frequency and in angle as the first power, so the angle always leaves first: at ten milliohms and at a megohm alike, where the angle has reached a degree the size is still only 152 parts per million out. One time constant, L/R − RC, sets that degree — 4.00 nanoseconds and 695 kilohertz at ten kilohms, 800 nanoseconds and 3.47 kilohertz for a ten-milliohm shunt. The resistance flattest in angle is exactly √(L/C), 141.42 ohms, the value the size question rejected; at the 91.02 ohms flattest in size the angle reaches a degree by 53.9 megahertz, and no resistance is flat in both.

frequency · Real resistor 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 In the stopband the series-first ladder filter is an inductance — 5.24 R₀ at three times the corner — and its dual a capacitance, 0.191 R₀: the same filter, opposite loads. computed by solving, not by drawing. The input impedance of the order-5 0.5 dB Chebyshev ladder filter realised with a series inductor first and with a shunt capacitor first, both terminated in the design resistance R₀, against frequency from a tenth of the corner to ten times it. In the passband both sit near R₀ and are nearly resistive. Above the corner they part: at 1.2 times the corner 1.33 R₀ at 89.4° against 0.749 R₀ at −89.4°; 2 times the corner 3.19 R₀ at 90.0° against 0.314 R₀ at −90.0°; 3 times the corner 5.24 R₀ at 90.0° against 0.191 R₀ at −90.0°; 5 times the corner 8.90 R₀ at 90.0° against 0.112 R₀ at −90.0°. The product of the two is R₀² at every frequency to 2e-15, which is what duality means for an input impedance. The slider is the order.

The load a filter is in its stopband

A fifth-order Chebyshev ladder filter can begin with a series inductor or, as its dual, with a shunt capacitor, and between its design resistances the two transmit identically. In the stopband they are opposite loads: at three times the corner the first is 5.24 R₀ of inductance and the second 0.191 R₀ of capacitance, their product R₀² at every frequency. That stops mattering only while the source is the design resistance. From a tenth of R₀ the shunt-first ladder filter rejects 55.9 dB at three times the corner instead of 64.0, draws 14 times the passband current at five times the corner against the series-first ladder filter's 0.24, and with a lead inductance of a tenth of R₀ over the corner's angular frequency grows a hole in its stopband — −33.8 dB where a matched source gives −54.0. A soft source does the mirror image to the series-first ladder filter. The first element has to be the one the source cannot short or open.

filters · Filter termination A 10 µs lead-inductance resonance holds the start at t⁴ until √12·τr = 34.6 µs, at a damping of 0.1 or any other. Expanded about infinity, and by residues after. The start of the step through two buffered RC sections (τ = 1 ms and 0.5 ms) on logarithmic axes: bare, a line of slope two (dashed), and with a buffered series resistance and lead inductance into a stray capacitance of natural time constant τr = 10 µs and damping 0.1 (solid). The relative degree rises by two, so the start is the bare step times t²/(12τr²) — the faint line, checked on the expansion to three per cent a hundredth of τr in — until the two meet at √12·τr = 34.6 µs. The damping does not enter that coefficient: it is the inductance and the capacitance, through τr² = LC, and nothing else. After it the curve rejoins the bare one, displaced by a delay that the damping does set.

The delay an inductance does not add

Lead inductance against a stray capacitance is a resonance, and far above a network's bandwidth it raises the relative degree by two: a two-section step starts as t⁴ instead of t² until √12·τr — 34.6 µs for a 10 µs resonance — at any damping, because the coefficient is LC and nothing else. After that the resonance moves a half-way threshold by R·C, its first moment, and the inductance is not in it. It moves every threshold by R·C only at ζ = 1/√2, where its second moment is a pure delay's, and below that its variance is negative: at ζ = 0.1 the 1% point moves by 1.48 R·C while the 10–90 rise time gets shorter. The ring that reaches the output is the network's own gain at the resonance, 2.0 × 10⁻⁴ at ω0τ = 100, and the quality factor decides only how long it lasts. Lightly damped, the start is a staircase that touches the bare step once a period and never overtakes it.

transients · Value theorems A 150 nH ground lead on a ten-to-one probe, at a 50 Ω source: the screen shows 49.4% of overshoot ringing at 121 MHz on an edge that has none. computed by solving, not by drawing, twice. A 1 V step from a 50 Ω source, watched through a ten-to-one probe with 11.5 pF at its tip, its ground returned through 150 nH of lead — about fifteen centimetres — and the scope reading the tip against the lead's end (solid, from residues; dots, a trapezoidal march, peak 1.4942). The same probe with its ground at the circuit's ground (dashed). The node itself steps cleanly. With the lead the tip capacitance and the lead inductance resonate at 1/(2π√(LC)) = 121 MHz, damped only by the source: ζ = (Rs/2)√(C/L) = 0.219, and the display overshoots by 49.4%.

The ring the ground lead adds

Every earlier measurement of a probe returned it to the circuit's ground directly. A real ten-to-one probe returns through fifteen centimetres of clip lead, about 150 nH, and its 11.5 pF tip and that lead are a series resonance at 121 MHz across the node being watched. The only damping it has is the source's own resistance, ζ = (Rs/2)·√(C/L), so the screen shows 49.4% of overshoot on a clean step from 50 Ω, 87.1% from 10 Ω, and none above 228 Ω — the stiffer the node, the worse the picture. Solved by residues and by a march of the netlist, the overshoot follows the second-order law at every source. A spring tip's 5 nH rings at 664 MHz and not at all from 50 Ω; a 150 Ω resistor in series with the tip brings a clip lead under 2% at the price of a 3.65 ns rise.

instruments · Probe loading

Named alongside it

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

ParasiticsModel rangeCharacteristic impedanceLumped elementMeasurement conditionDamping ratioInput capacitanceReactanceSelf-resonanceSeries resonanceStep responseAttenuator probe

All concepts