Concept

Parasitics — where it appears

The inductance, capacitance and resistance a component has by being an object, which decide where it stops behaving as the symbol drawn for it. They are what the realisability of a design is bounded by, and which of them binds depends on the impedance level the design is built at.

Named by 33 essays across 10 fields — each of them below, with the objects they name alongside it.

A 100 nF capacitor, and what it is above 14.5 MHz. The 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 99.0× what its capacitance predicts.

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.

frequency · Real capacitor

Where a switch is a switch: a band, and the 6.43 MHz at which it closes. computed by solving, not by drawing. A switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it is within 1.0% of being ideal only for loads between 49.5 Ω and 1.01 MΩ — 4.31 decades, and both edges are the same part. The upper edge is a frequency as well as a resistance, because the off-capacitance shunts the open switch: it falls a decade per decade above 318 Hz and meets the lower edge at 6.43 MHz, where the band closes and no load at all will do. Checked by scanning every load at 1.3 times that frequency and finding the best possible error to be 1.17%.

A band rather than an edge

Every other boundary in this collection is one-sided: a model is true below a frequency, or below an amplitude. A switch is a switch only for loads between 49.5 ohms and 1.01 megohms — bounded at both ends by the same part — and the upper end is a frequency as well as a resistance, so the band narrows as the frequency rises and shuts completely at 6.43 megahertz, above which no load resistance at all will do.

limits · Ideal switch

The millivolts in the wire

Ten millimetres of one-ounce copper is five milliohms and ten nanohenries, and if a hundred-milliamp load and a ten-millivolt sensor both return through it, half a millivolt of somebody else's current is added to the reading — five per cent of it, before anything has been amplified. Above 79.6 kilohertz the error rises a decade per decade with no ceiling, and shortening the shared run moves the whole curve down and the corner not at all.

instruments · Common-impedance

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

The edges that are lengths

Almost every boundary in this collection is a frequency or an amplitude, and both of those are things a circuit designer chooses. A handful are lengths — the 0.60 millimetres a gap's field reaches into a window, the 200 microns between a track and its plane, the 10 centimetres at which Kirchhoff's laws are a degree out — and they behave differently in one way that matters: nobody chooses them at the schematic, they are set by whoever builds the thing, and they appear in no netlist at all.

limits · Model edges

A 100 nF capacitor with 100 mΩ in series, written the other way round. computed by solving, not by drawing. At 100 kHz the series pair and the parallel pair are the same impedance to 8.7e-19 of itself — the arithmetic's floor, not a tolerance — with Rp = 2.533 kΩ against Rs = 0.100 Ω and Cp = 99.996 nF against Cs = 100 nF. Away from it they part company at a rate set by Q = 159.2: the substitution costs one per cent below 48.2 kHz and above 207 kHz, a band of 4.3 to one.

The same part written two ways

A capacitor's loss is quoted either as a resistance in series with it or as one across it, and the pair of expressions that converts between them is exact at one frequency and at no other. How wide the band is around that frequency is set entirely by the quality factor: an octave and a half at Q of eight, a thousand to one at Q of three thousand.

frequency · Series parallel

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

What a mistuned arm leaves at 1300 Hz. computed by solving, not by drawing. Neither curve is the null's depth — the null is still bottomless, it has simply moved — but the depth at the frequency the notch was designed for, which is the number a filter is bought for. One per cent components leave -32.9 dB if both errors go the same way and -78.8 dB if they oppose, a factor of 197 from the same tolerance on the same two parts. The slopes are 20.0 and 40.0 decibels per decade: first order in the error on the product LC, second order in the error on the impedance level.

What actually fills a null

A ten per cent error in the two components of a notch's arm leaves the null three hundred decibels deep — it moves it rather than filling it. What fills it is loss, at twenty decibels per decade of arm resistance exactly. And the depth at the frequency the notch was designed for splits into two orders depending on which way the two errors go: one per cent parts leave 79 decibels one way and 33 the other.

filters · Null depth

A follower's output impedance from 1 kΩ of source, bare and with 100 pF on it. computed by solving, not by drawing, on a small-signal follower at 2.0 mA with β = 150 and fT = 560 MHz. At 100 Hz the emitter presents 19.08 Ω against a textbook 1/gₘ + Rₛ/(β+1) of 19.55 Ω — the expression is an upper bound here and at every source resistance on the slider, 2.4% high at this one. What it cannot describe is the frequency axis: the β that divided the source resistance down is itself falling, so the impedance rises, and the reactance at 3 MHz is 4.3 Ω — an inductance of 0.229 µH against Rₛ/ωT = 0.284 µH. With 100 pF hung on the output that impedance peaks at 67.0 Ω at 29.3 MHz, 3.51 times its own low-frequency value: an inductive source and a capacitive load are a resonant circuit, and this one is inside a part whose output impedance is quoted as a single number.

The buffer that is not a buffer

An emitter follower is reached for when something has to be driven without being loaded: unity gain in, high impedance seen, low impedance presented. The last of those is a number with a range, and the range is narrow. At a kilohm of source the emitter presents 19.08 ohms at low frequency and 67 ohms at 29 megahertz, because the current gain that made it small is falling — and the peak is worst in the middle of the slider, so it cannot be avoided by making the source stiffer or softer.

semiconductors · Emitter follower

One Sallen-Key design at 10 kΩ, and the band of impedance levels it survives. computed by solving, not by drawing. A 10.0 kHz unity-gain Sallen-Key section realised at seven impedance levels three decades apart, with every resistance multiplied and every capacitance divided by the same factor. The design is exactly invariant: the solved magnitudes agree to 1.0e-15 of each other across all seven and every frequency in the passband, which is the last bits of a double rather than a good approximation. The realisation is not, because two of its quantities do not scale — 50 Ω of amplifier output resistance and 2 pF of stray capacitance to everywhere — and they bind at opposite ends. Inside 0.1 dB the band runs 31.6 Ω to 31.6 kΩ, with the least departure of 0.0133 dB at 1000 Ω; at this setting it is 0.036 dB at 20.0 kHz. The consideration that does not appear on either axis is noise: the resistors' density goes as √R, so across the six decades drawn the noise moves by 1000 times and points at the low end of the band.

The same filter a thousand times larger

Multiply every resistance by a thousand and divide every capacitance by a thousand and the response does not change — not approximately, but to a part in ten to the fifteenth, which is the last bits of a double. So a designer has a free parameter that the design says nothing about, and what decides it is the two quantities that refuse to scale: fifty ohms of amplifier output resistance at one end and two picofarads of stray at the other. Between them the realisation survives over three decades of impedance level and nowhere else.

filters · Impedance scaling

At the order a cascade runs out, a ladder still has 4.2 decades of level. computed by solving, not by drawing. How many decades of impedance level each structure can be built at while staying inside 0.1 dB of its own design, against order. Both carry two picofarads of stray at every node. The cascade also carries fifty ohms of amplifier output resistance, a fixed resistance, which binds it from below; its floor rises 8.5× over the four orders drawn, to 0.114 dB at order eight. The ladder's inductors carry a fixed resistance per henry instead — a fixed quality factor, which scales with the design and bounds nothing — so what limits it from below is a few milliohms of track, and at order nine it still has 4.20 decades with a floor of 0.0337 dB, 3.3× its own floor at order three against the cascade's 8.5×.

The band that does not close

A cascade of active sections can be built at three decades of impedance level at second order, one at sixth, and none at all at eighth — the band shuts by half a decade per order because two fixed quantities bind it from opposite ends. A doubly terminated ladder has only one of those quantities, because an inductor's loss is a fixed quality factor rather than a fixed resistance and therefore scales with the design. At order nine it still has four decades.

filters · Impedance scaling

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

95 dB of instrument, 290 Hz corner — and the corner belongs to the source. computed by solving, not by drawing. The common-mode rejection of the same three-amplifier instrument the rungs below measured, with 1 kΩ of imbalance between the two source resistances and 10 pF at each input. The instrument's own curve is drawn beside it. Below 290 Hz the two agree; above it the measurement falls at twenty decibels a decade while the instrument does not, reaching 84.0 dB at a kilohertz against the instrument's 95.0. What converts common mode into differential is the difference of the two input time constants — 10.0 ns here — and once it is differential no rejection repairs it.

The corner the instrument has no part in

Three rungs of this argument measured a three-amplifier instrumentation amplifier's rejection at direct current and found 95 dB, of which the resistors' matching decides one part and the amplifiers' own mismatch another. Connect it to a source with a kilohm of imbalance and ten picofarads at each input and the rejection has a corner at 290 Hz and falls twenty decibels a decade after it — reaching 84 dB at a kilohertz on an instrument that is still doing 95. What converts common mode to differential is the difference of two time constants, and the cure is a capacitor on the quiet input.

instruments · Common-mode rejection $The floor rises as the 1.06 power of the order, and the arithmetic gives out first. computed by solving, not by drawing. The least departure a doubly terminated ladder can achieve at any impedance level, against order — the floor, which is what is left when the tolerance the band is drawn at is taken away. It rises from 0.0087 dB at order three to 0.0389 dB at order 13, as the 1.06 power of the order, so the order at which it reaches the 0.1 dB the band is drawn at is near 32. Drawn over it are the same networks with one imperfection removed at a time, and they do not add: the winding loss alone leaves a larger departure than the complete realisation does, because the track resistance is in series with the load and lifts the passband exactly where the winding loss droops it. The stray capacitance contributes nothing at all, since the level that minimises the departure is a few ohms. What ends the sweep is neither: the continued-fraction synthesis loses its leading coefficients to cancellation and stalls at order 9 for a Butterworth and 14 for a Chebyshev.$

The floor that outlives the arithmetic

A band of impedance levels is what a tolerance allows; a floor is what a structure has. Measured at six orders, a doubly terminated ladder's floor rises as the 1.06 power of the order and would not reach a tenth of a decibel until order thirty-two — an order nobody builds. What stops the sweep is neither the structure nor the parasitics: the continued fraction that turns a reflection polynomial into element values loses its leading coefficients to cancellation and stalls at order nine for a Butterworth and fourteen for a Chebyshev.

filters · Impedance scaling

A coil of 500 nF and a capacitor of 100 µH give the loop three features, not one. computed by solving, not by drawing. The current round the loop for a volt across it, with the coil carrying 500 nanofarads of its own capacitance and the capacitor 100 microhenries of its own inductance. Three features rather than one: the resonance the two nameplate values set, at 4.02 kHz; the coil's own self-resonance at 7.12 kHz, which is a parallel tank and so a null in a series loop, 5.67e+3 times below the peak beside it; and a second series resonance at 28.2 kHz that belongs to neither component. Above the null the coil is a capacitance, that capacitance is in series with the tuning capacitor, and the capacitor's own inductance resonates with the pair — which is why removing either parasitic removes this peak and neither alone can produce it.

Two parasitics, and the resonance neither of them has

A resonator's quality factor is supposed to sit below the worst of its components, because reciprocals add. Give the capacitor half a millihenry of its own inductance and the loop measures 92.21 against a coil that allows 64.55 and a reciprocal sum that predicts 62.47 — the ceiling passed exactly where ESL/L crosses ESR/DCR, at 5.00 per cent. Add the coil's own capacitance beside it and the loop grows a second resonance at 28.2 kHz that neither part has alone, taller than the first by 40.4, and the half-power level is then crossed four times.

frequency · Resonance

A photodiode's own capacitance sets the bandwidth, as its -0.50 power. computed by solving, not by drawing. The bandwidth of a 1.0 MΩ transimpedance stage against the capacitance of the diode driving it, with the feedback capacitor at each point bisected to give exactly forty-five degrees of phase margin on the solved loop. The classical expression √(GBW/2π·rf·cd) is drawn over it: the right shape, and conservative by about a fifth at every capacitance. The bandwidth falls as the -0.497 power of the capacitance — a square-root law, so a diode of four times the area costs half the bandwidth rather than three quarters of it. At 30 pF the compensation is 0.598 pF against the expression's 0.725, and the bandwidth 295 kHz against 220.

Where the trouble is at the input

Every other arrangement in this field has its difficulty at the output — a capacitive load, an isolation resistor, a load that draws current. A photodiode amplifier has it at the input, and the capacitance causing it is not a parasitic: it is the diode's junction, which is the price of its area, and area is what a photodiode is bought for. The feedback resistor's own noise is 127 nV/√Hz against the amplifier's 4, and the amplifier is still ninety-five per cent of the noise at a large diode.

feedback · Transimpedance

A hold capacitor's band closes at 6.43 MHz, where a resistor's does. computed by solving, not by drawing. The switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it, driving a capacitor. The lower edge is the smallest capacitance onto which the open switch feeds through no more than 1%: 495 pF far above 318 Hz, rising as the reciprocal of frequency below it because the leakage charges the capacitor. The upper edge is the largest capacitance the closed switch tracks to 1%, counted as a vector. The two meet at 6.43 MHz; a resistor on the same switch closes at 6.43 MHz, and counted as a magnitude the capacitor's band closes at 91.6 MHz. At the closure the closed switch's phase is 0.57 degrees.

The width no load can change

A switch's band was drawn against a load resistor and closed at 6.43 megahertz. Put a hold capacitor where the resistor was and the band changes axis and shape — a diagonal below 318 hertz, a floor of 495 picofarads above it — and stays exactly as wide: 1.8083 decades at 100 kilohertz for both loads, closing at 6.43 megahertz for both. Count the capacitor's error as a magnitude, as the resistor's always was, and the band appears to stay open to 91.6 megahertz. That extra room is 8.11 degrees of lag the count cannot see.

limits · Ideal switch

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

Where 8 channels leak to: 91.0 MHz from buffered sources, 901 kHz from 50 Ω. computed by solving, not by drawing. The frequency at which the open channels of a multiplexer built from the 0.5 Ω, 100 MΩ, 5 pF switch leak 1% of the signal onto the shared output, against the impedance of the source driving the selected channel, into 1 MΩ. 2 channels: 637 MHz buffered, 6.30 MHz from 50 Ω; 8 channels: 91.0 MHz buffered, 901 kHz from 50 Ω; 16 channels: 42.4 MHz buffered, 420 kHz from 50 Ω. Each falls as the reciprocal of the source impedance plus the on-resistance, and the single switch's own band closes at 6.43 MHz.

Where an open switch leaks to

A lone switch has a band whose width no load can change, because its open state leaks into the load. In a multiplexer the seven open channels leak into a node the selected channel holds, so the load leaves the answer and the source takes its place: one per cent of leak at 91.0 megahertz from buffered sources and 901 kilohertz from fifty ohms, moving as the first power of the tolerance rather than the second. Adding the channels' capacitance into one forty-picofarad switch puts it at 804 kilohertz, near the fifty-ohm figure by coincidence and a hundred and thirteen times low for a buffered one.

limits · Ideal switch

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 $From 50 Ω into 50 Ω: a T isolates to 643 MHz, a changeover to 6.34 MHz. computed by solving, not by drawing. The fraction of the drive that arrives with the path open, against frequency, from a 50 Ω source into 50 Ω, for the 0.5 Ω, 100 MΩ, 5 pF switch used three ways. A T reaches 1% at 643 MHz; a changeover reaches 1% at 6.34 MHz; one switch reaches 1% at 3.18 MHz.$

The capacitance a third switch moves

A changeover's open channel leaks into the source of the channel that is closed, so its isolation into fifty ohms falls from 643 megahertz with a buffered source to 6.34 megahertz with a fifty-ohm one. Put a third switch to ground between two series switches and the leak lands on half an ohm of closed switch instead: 814, 643 and 2,240 megahertz from sources of nothing, fifty ohms and a kilohm, rising forty decibels a decade where a changeover's rises twenty. The price is the shunt switch's own capacitance, moved under the closed path, which makes the T one per cent wrong as a waveform at 6.59 megahertz beside the changeover's 6.61.

limits · Ideal switch

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 $One switch is never better than 70.71 ppm; a T of three reaches 10 ppb only into 200 MΩ. computed by solving, not by drawing, at direct current. The worse of a switch's two errors — closed, the fraction the load fails to receive; open, the fraction it receives anyway — against the load, for one 0.5 Ω, 100 MΩ switch and for a T of three, from a buffered source. The lone switch is best at 7.07 kΩ, the geometric mean of its two resistances, where both errors are 70.71 ppm, 13.79 bits: no load does better. The T has no best load. Its worse error falls with the load towards Rₒₙ/(Rₒₙ + Rₒff) = 5 ppb, the square of the lone switch's resistance ratio rather than its root; it is within twice that from 200 MΩ, it passes the lone switch's floor only above 14.1 kΩ, and into 7.07 kΩ it is 141.4 ppm, worse than one switch. Solved on the network up to 1000 MΩ and continued, dashed, from the closed form it matches.$

The floor below any load

A switch of half an ohm closed and a hundred megohms open is within one per cent of ideal for loads between two edges, and the edges close on each other as the tolerance tightens. At direct current they meet at 70.71 parts per million, into 7.07 kilohms: no load makes that switch better, which is 13.79 bits and a boundary with no frequency in it. A T of three such switches has no best load at all. Its error falls with the load towards five parts per billion — the square of the lone switch's resistance ratio rather than its root — and reaches ten only into two hundred megohms. Into the 7.07 kilohms that suited one switch, the T is worse than one switch.

limits · Ideal switch

A core loss of 100 kΩ gives the resonance a floor at −66 dB, and by 10.0 GHz the choke is worth 0.02 dB. computed by solving, not by drawing. The common mode at the load, in decibels against the same circuit with no choke in it, with 5 pF across each winding and a core loss of 100 kΩ across each, drawn over the lossless curve. At the winding's resonance, 1.59 MHz, the inductance and the capacitance cancel and what is left is the loss, so the deepest point is 66.02 dB — 20·log(1 + (Rp/2)/(Zs + Zl)), with no inductance and no capacitance in it. Above the resonance the part is a capacitor across the path it was fitted to block, and by 10.0 GHz it attenuates 0.02 dB.

The depth a resonance does not have

The inductor one mode cannot see printed its choke's best attenuation as 102 decibels at 1.59 megahertz. Swept with 1,999, 2,000 and 2,001 samples the same lossless part reports 110.0, 96.3 and 103.6 decibels, because a resonance with nothing to dissipate has no bottom and a sweep reports how near its nearest sample fell. Give each winding a core loss and the depth is 20·log(1 + (Rp/2)/(Zs + Zl)) — 66.02 decibels at a hundred kilohms, twenty more per decade of loss, with no inductance and no capacitance in it, and half of it belonging to the circuit the choke sits in.

lines · Common-mode choke

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

What is warm in a capacitor, by its two loss models. computed by solving, not by drawing. One 100 nF capacitor of loss tangent 0.02, written as a 3.183 Ω resistance in series with it and as a 7.958 kΩ resistance across it — the pair that converts exactly at 10.0 kHz and nowhere else. The noise at the terminals is 4kT times the real part of the impedance, so the two models give the same density at 10.0 kHz and are 33.0 dB apart at 100 Hz and 40.0 dB apart at a megahertz. The dots are the same quantity computed the other way — the resistor split out of the netlist, a source put in its place and the network re-solved — agreeing to 3.3e-16. The reactance itself contributes nothing at either end: a lossless capacitor has no real part and is not warm.

Only the real part is warm

Johnson's 4kTR is the special case of a statement about impedances: the noise across any passive two-terminal in equilibrium is 4kT·Re{Z}, so a reactance contributes nothing however large it is. That turns a modelling convenience into a noise figure. A 100 nF capacitor of loss tangent 0.02 written as 3.183 Ω in series and as 7.958 kΩ across it — the pair that converts exactly at 10 kHz — gives 0.226 and 10.10 nV/√Hz at 100 Hz, 33 dB apart, and 40 dB apart the other way at a megahertz.

noise · Johnson noise

1 pF across a 50 Ω line: a dip of 0.320 V and an area of 25 ps. computed by solving, not by drawing as a cascade of two-ports, with a raised-cosine edge of 59 ps sent into it. The incident edge is the faint curve; what comes back is the shaded dip and what goes on is the third. The dip reaches -0.3202 V and its area is 25 ps, which is Z₀C/2 to a part in ten thousand. Driven by an edge fifty times faster the same cascade returns the single exponential the closed form gives, to 9.3e-6 of a volt. The transmitted edge leaves at 81.1 ps, against 59 ps arriving.

The dip whose area is fixed

A picofarad across a 50 Ω line makes a dip in what comes back. Its depth is 0.833 volts to a six-picosecond edge and 0.0196 volts to a nanosecond one, forty-two times less; its area is 25 picoseconds to both, to six parts in a hundred thousand, because the area is Z₀C/2 and contains nothing about the edge. Two half-picofarad discontinuities too close to tell apart read as exactly one picofarad, and so do two far enough apart to be separate — the area is additive where the depth is not. And what a reflectometer calls the capacitance of an impedance step is the step's real excess capacitance times 1 + Z/Z₀.

lines · Reflections

The optimum that hands back a bandwidth

The best burden voltage across a shunt is 7.75 mV and contains neither the current nor the resistance, which is what made it worth having. A shunt has two nanohenries whatever it is made of, so holding the burden fixed fixes the resistance at u*/I — and the corner R/2πL then falls in exact proportion to the current: 6.16 MHz at a tenth of an ampere, 616 kHz at one, 6.16 kHz at a hundred. A resistor and a 258.2 pF capacitor across it, found by search on the solved response, flatten the reading to 8×10⁻⁵ per cent across six decades.

instruments · Current sensing

The best shunt switch for a T is 3.8×, 0.33×, 0.082× a series switch from sources of 0 Ω, 50 Ω, 1 kΩ. computed by solving, not by drawing. The frequency at which the band of a T closes — where no load leaves it within 1% of ideal in both states, the closed state counted as a shortfall in amplitude — against the size of its shunt switch, as a multiple of the 0.5 Ω, 100 MΩ, 5 pF series switches, with every conductance and the capacitance scaled together. From a 0 Ω source the band closes latest with a shunt 3.775 times the series switch, at 2.40 GHz, against 702 MHz with three identical switches — 3.42 times later. From a 50 Ω source the band closes latest with a shunt 0.325 times the series switch, at 243 MHz, against 89.8 MHz with three identical switches — 2.71 times later. From a 1 kΩ source the band closes latest with a shunt 0.082 times the series switch, at 59.4 MHz, against 4.53 MHz with three identical switches — 13.12 times later. A larger shunt holds the open node harder and hangs more capacitance on the closed path, and the source decides where the two meet.

The shunt switch the source sizes

A T is two series switches and a third to ground, and it is always drawn with three of the same part. The shunt switch pulls its own size two ways: larger, it holds the open node harder; larger, it hangs more capacitance on the closed path. The size at which the band closes latest is a balance of the two — the fourth root of 2/ε times √(Rₒₙ/(Rₛ + Rₒₙ)) — when the closed state is counted as an amplitude — 3.8 times a series switch from a buffered source, a third of one from fifty ohms, a twelfth from a kilohm — and it buys a band 3.4, 2.7 and 13 times wider. Counted as a waveform the root of ε goes, and from a buffered source the best shunt is exactly the series switch.

limits · Ideal switch

A via's area changes sign at 44.7 Ω, and two impedances give its 0.5 pF and 1 nH back. computed by solving, not by drawing, as a cascade of two-ports: a via of 0.25 pF, 1 nH and 0.25 pF, met by an edge of 59 ps from reference lines of 20 to 150 Ω. The area under the reflection is −Z₀C/2 + L/2Z₀ at every impedance: a bump below 44.7 Ω, where the inductance's term is the larger, nothing at it, and a dip above. From 50 Ω the area is 2.5 ps of dip, which a single-capacitance reading calls 0.100 pF. From 50 and 75 Ω together the two areas give 0.5000 pF and 1.0000 nH. An error of 50 fs on each area moves them by up to 0.8% and 1.5%.

The via two lines can weigh

The area under a reflection is a property of the discontinuity rather than of the edge, and for a via it is one number made of two: −Z₀C/2 from its pads and +L/2Z₀ from its barrel, with opposite signs. A via of half a picofarad and a nanohenry, seen from fifty ohms, leaves 2.5 picoseconds of dip — which a reading that assumes a capacitor calls 0.100 pF, a fifth of what is there. From fifty and seventy-five ohms together the two areas give 0.5000 pF and 1.0000 nH back, and fifty femtoseconds of error on each costs 0.8 per cent of the capacitance and 1.5 of the inductance. A second line at fifty-five ohms costs five times as much.

lines · Reflections

At 44.7 Ω the via's area is zero; its reflection is a doublet falling as the 1.99 power of the edge, not the first. computed by solving, not by drawing, as a cascade. A via of 0.25 pF, 1 nH and 0.25 pF on a line of √(L/C) = 44.72 Ω, met by an edge of 59 ps, against the same capacitance alone on the same line. The via's reflection is a doublet — a dip and a bump of equal area — whose largest excursion is 24.4 mV against 166 mV for the capacitance alone. Over edges from 5.9 ps to 295 ps the capacitance's reflection falls as the −0.97 power of the edge and the via's as the −1.99 power: 24.4 mV at 59 ps, 994 µV at 295 ps. The via delays the edge going past it by 22.4 ps, against √(LC) = 22.4 ps.

The via that is a piece of line

A via of half a picofarad of pad and a nanohenry of barrel puts no area under its reflection from a line of √(L/C) = 44.72 ohms, from any edge. It has not vanished. It delays the edge going past it by 22.4 picoseconds, which is √(LC) exactly, and it still reflects: a doublet whose largest excursion falls as the −1.99 power of the edge where a lone capacitance's falls as the −0.97 — 24.4 millivolts for a 59-picosecond edge against 166 for the pads alone, and 994 microvolts for a 295-picosecond one against 35. The balanced via is a short piece of line, and the reflection it leaves is the reflection of its length rather than of its size.

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

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

The same imbalance at five winding resistances, and the fixed point three of them reach. computed by solving, not by drawing. The peak flux density in each cycle, against the cycle, under a square drive whose positive half carries 1 per cent more volt-seconds than its negative one, for winding resistances of 0, 0.05, 0.15, 0.4, 1.5 ohms. With none the flux walks to 0.35 T in 15 cycles, which is the boundary this ladder's second rung measured. With 1.5 Ω it settles at an offset of 5.67 millitesla and stays there, because the offset draws a direct magnetising current and that current's drop across the winding opposes the imbalance. The resistance at which the two outcomes change places is 0.1014 ohms, bisected on whether saturation is reached at all. Every curve carries the identical drive; the resistance is the only difference between them.

The walk that stops

A drive whose two half-cycles differ in volt-seconds walks the flux to saturation in a count of cycles, and the rung that measured it concluded that no amplitude puts the design inside a limit. That model has a stiff source and a winding of no resistance. With resistance in the loop the walk has a fixed point, held by an identity the material is not in — the mean magnetising current is the drive's direct component divided by the resistance, to seven parts in 10¹³ — and the boundary becomes a resistance rather than a time: 0.1014 ohms bisected, at a one per cent imbalance and half the volt-second limit.

magnetics · Saturation

Named alongside it

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

Model range Design tradeoff Loading Verification Lumped-element Loop area Off isolation On-resistance Return current Self-resonance Measurement condition Characteristic impedance

All concepts