Every essay — page 3
Frequency, which is the same solve
Reactance, phase, resonance and the corner frequency are four readings of one number. The models that fail here are the straight-line sketch every engineer draws, which is three decibels wrong exactly where it is read, and the capacitor, which is a capacitor only below a frequency its own leads decide.
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.
The reading a coil leaves nearly exact
On a series resonator three readings of Q agreed exactly and two did not. A parallel tank whose loss is in its coil has no exact readings at all, and the order is not the one its schematic suggests. Its poles are the series circuit's, but the coil's own zero puts the peak, the zero-phase point and the ring at three different frequencies, and all of them below 1/√(LC). At Q = 3 the phase slope reads 16.2 per cent low, the stored energy 5.72 low and the peak height 5.42 high, while the bandwidth reads only 0.307 low: the loss that a parallel resistance stands in for tilts across the band, and the tilt is nearly antisymmetric about the centre, so the reading that spans it cancels it. The instrument's own source then decides what is being read.
Measurement, which is a circuit on a circuit
An instrument is not an observer; it is an element, it goes in the netlist, and every reading is a reading of the circuit that includes it. A probe is a capacitance with a bandwidth. A divider with capacitance in it has two ratios and one equation that makes them agree. Two terminals measure the leads as well, and below ten ohms that is most of the answer.
The probe is part of the circuit
A one-to-one oscilloscope probe on a two-kilohm source gives a reading that is one per cent wrong at 6.8 kHz. Not because the instrument is inaccurate — it is reading correctly — but because the hundred and fifteen picofarads at the end of the cable are across the node, and above that frequency the trace on the screen is a picture of a circuit that only exists while the probe is attached.
A divider with two ratios
Put capacitance in a resistive divider and it divides by resistance at direct current and by capacitance at high frequency, and those are two different numbers unless one equation holds. The adjustable trimmer on every oscilloscope probe exists for that single equation, and the square wave on the instrument's front panel is a display of which of the two ratios is currently winning.
Two terminals measure the leads as well
Fifty milliohms in each lead makes a two-wire measurement one per cent high at ten ohms, ten per cent high at one ohm, and a hundred per cent high at a tenth. Not approximately — the reading is the resistance plus the leads, and below about ten ohms most of what is being reported is the wire between the instrument and the thing.
The instrument's own rise time
Rise times add in quadrature, so ten per cent of inflation needs an instrument 2.18 times faster than the edge. That constant contains no instrument. Measured on the solved network it is 2.79 for a one-pole front end, 3.97 for two, 4.87 for three and 5.62 for four — the rule is optimistic at every pole count, which is the wrong direction for a rule of thumb to err in.
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.
The corner that says nothing about an edge
An AC-coupled input is sold on a low-frequency corner, and a corner is a statement about steady sinusoids. What an instrument is usually shown is a pulse, and for a pulse the number is a sag: a hundred-millisecond pulse through a 0.159 hertz corner comes out 9.5 per cent shorter than it went in. A one per cent flat top needs a pulse rate 625 times the corner, which is a constant with no component in it — and above fifty per cent duty an AC-coupled pulse never reaches half its own height at all.
The rejection four resistors decide
Measuring a ten-millivolt sensor against a ground that somebody else's hundred milliamps is also using puts five hundred microvolts of their current into the reading. Subtracting the two ends of that conductor with a difference amplifier removes it — by 53.99 decibels with one-tenth-per-cent resistors, against a closed form of 53.98, which is a factor of five hundred and not a removal. The amplifier has nothing to do with it: the number is one plus the gain over four times the resistor tolerance.
The ammeter that is a resistor
Every direct measurement of a current is a measurement of a voltage across something the current was made to flow through, so the instrument has two errors pointing opposite ways: a larger shunt changes the current, a smaller one leaves less for the amplifier's offset to be compared with. Written in the burden voltage they are the burden over the supply and the amplifier's offset over the burden, and the best of them is at the geometric mean — 7.75 millivolts on a twelve-volt rail, with a worst-case error of 0.129 per cent. Neither number contains a resistance, and neither contains the current: ten milliamps and a hundred amps want the same 7.75 millivolts.
The four resistors that decide, and the two that do not
A difference amplifier's rejection is decided by four resistors and one-tenth-per-cent parts give 54 decibels. Putting a two-amplifier stage in front adds exactly twenty times the log of its gain — 94 decibels at a gain of a hundred — and the reason is not that the input stage rejects anything. It passes common mode at exactly unity, so the common-mode gain of the whole instrument is 1.998 millivolts per volt at every gain tried, and the improvement is entirely the differential signal arriving larger. The two resistors that set that gain may be ten per cent apart without moving the answer a hundredth of a decibel.
The ammeter that is not in the circuit
A shunt measures a current by putting a resistance in the circuit, and every objection to it follows from that. A current transformer puts nothing in the circuit at all — a thousand-turn secondary reflects twelve microhms into the primary — and charges for it in a different currency: no response at direct current, a ratio error that stops falling at one minus the coupling, and a phase error of half a degree at fifty hertz that costs eighteen per cent of a power reading at a power factor of 0.05.
The rejection the parts have
Two essays measured the architecture: four resistors decide an instrumentation amplifier's rejection, two do not, and the answer is the one-amplifier figure plus twenty log of the first stage's gain — exactly, with amplifiers of infinite rejection. Give each amplifier its own and something unobvious happens: two matched but individually mediocre parts cost nothing at all, because their error is a common-mode signal at the difference stage and is rejected there. What costs is the difference between them, and it sets a ceiling with no gain in it.
The corner the instrument has no part in
Three earlier essays 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.
The current the instrument draws
Every amplifier in this collection has had inputs that take no current, and that is not an idealisation of a small quantity — it is an idealisation of one whose size is decided by something outside the part. Fifty nanoamps is nothing until it flows in a megohm, and then it is fifty millivolts. The classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten, not a thousand, and it costs forty per cent of the noise density to get.
The current that does not reach the input
A teraohm across a board from a fifteen-volt rail is fourteen millivolts of error through a gigohm source, and a humid morning takes that resistance down two decades. A ring held at the input's own potential leaves a nanovolt — proportional to the signal rather than to the rail, so an offset has become a gain error of one part in a billion — and the same wire multiplies the input resistance by the loop gain, which makes it 10¹⁸ Ω at direct current and 10¹² Ω at a megahertz.
The probe that takes a tenth
A ten-to-one probe buys an order of bandwidth for a tenth of the signal, and on a two-kilohm source the bandwidth is exact: 6.79 kHz becomes 69.2 kHz. The tenth of the signal is not a tenth of the signal-to-noise ratio. Solved resistor by resistor, the noise referred to the tip goes from 1.782 µV to 55.78 µV — a factor of 31.3 — because the divider that does the attenuating is nine megohms and a megohm, and √(n(n−1)kT/C) on the cable's own capacitance has no source resistance in it at all.
The voltmeter four wires do not remove
A four-terminal measurement is described everywhere as removing the leads from the answer. It moves them. What is left is the voltmeter's own input resistance, and it grows with the resistance being measured rather than shrinking: with a ten-megohm voltmeter and fifty milliohms of lead, the four-wire reading is 0.0100 per cent low at a kilohm, where the two-wire reading is exactly right — 1.8 × 10⁻¹² — because its lead error and its loading error cancel. Above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.
The errors that arrive before the gain
Four earlier measurements each found a different owner of one instrument's common-mode rejection and each measured it alone. Solved together, the four are complex numbers that add — to 1.25 parts in ten thousand — and two of them carry a factor of the first stage's gain while two do not. The two that do cancel the two that do not at a gain of 776, and what is left there is 110.066 decibels, which is exactly the number the cable sets. Better-matched amplifiers move that gain from 93 to 3392 and do not move the ceiling by a hundredth of a decibel.
The cure that becomes a different circuit
The classical cure for an amplifier's input current is to make the two resistances its inputs look back into equal, and it reads as one instruction. Solved, it is two circuits meeting at 909 ohms — the feedback network's bottom resistor times (G−1)/G — and above that knee there is no resistor to add: the network is scaled to the source, which at a megohm means 11 megohms of feedback and at a gain of 1001 means 1001. Above the knee three different networks become one instrument to twelve figures, the noise penalty settles at 1.41420 against a √2 of 1.41421, and the benefit at 10.49 against two currents whose ratio is ten.
The sign of what the guard gives back
A guard ring is sold on two numbers and they are both about magnitudes: a teraohm of leakage multiplied to 10¹⁸ ohms, and a hundred picofarads of cable bootstrapped out of the way. The guard is also driving that capacitance with a copy of the input that lags it, and a capacitance driven by a lagging copy of its own voltage takes current out of phase with the voltage across it. What the guarded input presents is a negative conductance rising as the square of frequency — −15.9 megohms at ten kilohertz, −3.18 kilohms at a megahertz, flattening at the gain-bandwidth product times the capacitance — and a faster amplifier makes it worse.
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.
The rail that is an input error
The best burden voltage across a shunt is the geometric mean of the amplifier's offset and the supply, and moving the shunt to the high side does not change that form — it changes what the offset is. A hundred-decibel amplifier on a twelve-volt rail turns the rail into 120 µV of equivalent input error, twenty-four times its own five, so the optimum moves from 7.75 mV and 0.129 per cent to 38.73 mV and 0.644. The two are equal only at 128 dB, and with the common-mode term removed the optimum returns to the low-side value exactly.
The resistance that depends on the reading
Three of a shunt's errors are free of the current being measured, which is the whole content of the burden-voltage optimum. The fourth is not: the shunt dissipates, warms, and its resistance rises — so the divisor the reading uses is a function of the reading. Solved as a fixed point it agrees with R₀/(1 − αθI²R₀) to 2×10⁻¹⁶, and along the optimum, where the dissipation is I·u* rather than I²R, the error is the FIRST power of the current: 7.75 ppm at an ampere, 775 at a hundred, fitted exponent 1.0007.
Every tooth the same height
A shared return conductor's impedance rises a decade per decade above 79.6 kilohertz and a switching load's harmonics fall a decade per decade, so the two cancel exactly: every harmonic of a hundred-milliamp square wave puts 0.828 millivolts into a ten-millivolt reading, from the conductor's own corner up to the edge's. The comb is flat rather than falling, so the total is decided by how many teeth there are — and since they add in quadrature it grows as the square root of the edge rate, 39.7 millivolts at a nanosecond against 4.03 at a hundred.
One voltage added to two readings
A difference amplifier across a shared return rejects somebody else's current by 54 decibels, and the number is four resistors rather than the amplifier. Two sensors referenced to the same node do better by a different mechanism entirely: the interfering voltage is one voltage on one node, so it is added to both readings identically and the difference between them has none of it — 511 microvolts in each reading against two and a half picovolts in the difference, which is 166 decibels and is exact rather than good. What is left is not a tolerance but a mismatch between the two sensors' own return paths, ten nanovolts per per cent of it.
A star is half a millimetre long
The repair for a shared return is a star ground, and it is described as making the shared length zero. It does not: every return meets at a pad, the pad reaches the plane through a via, and a via of a quarter-milliohm and 0.35 nanohenries is worth half a millimetre of one-ounce track by its resistance and 0.35 by its inductance — two different lengths, so a via has a corner of its own at 114 kilohertz where a track's is at 79.6. What that buys is a change of law rather than a factor: a daisy chain's error grows as the square of the circuit count and a star's as the first power, so the advantage runs from sixty times at two circuits to three hundred and forty at thirty-two.
The resistor that moves the lag
A guarded input driven by a one-megahertz follower presents a negative resistance of −1/(ωₜC), −1.59 kΩ behind 100 pF of cable. The standard remedy is a resistor between the follower and the ring. It weakens the negative resistance, to −6.5 kΩ at 1 kΩ and −403 kΩ at 100 kΩ, but it does not remove it: the resistor and the ring's own capacitance to ground make a lag of their own. With a large enough resistor the worst value is −R(C + Cₛ)²/(C·Cₛ) for a drive resistor R — minus four times the resistor when the ring is as capacitive to ground as to the input — and the same at every length of one cable. The length moves somewhere else instead. The frequency where the bootstrap is given up, 1/(2πR(C + Cₛ)) for a drive resistor R, falls with every metre.
The source that rings against its guard
A bootstrapped cable is a capacitance in series with a negative resistance that falls as the square of frequency, −1/(ω²·C/ωₜ). Drive it from a source with resistance R and inductance L and the loop's reactances cancel at 1/√(LC). Its resistances cancel where R = L·ωₜ, so the input oscillates once the source's own time constant L/R passes the follower's 1/ωₜ. The poles of the whole netlist say 17.6 µH for a 100 Ω source behind a 1 MHz follower, against 15.9 from the expression. The cable's length moves that by at most a factor of two and sets only the frequency it rings at. The same source without the guard is stable. A resistor in the guard drive raises the limit, to 164 µH through 1 kΩ, and pays for it in bootstrap bandwidth.
The pulse that ends before the heat
A shunt's self-heating error is a fixed point in the steady state — 775 ppm at 100 A for the shunt the burden optimum picks, through 20 K/W. A pulse never reaches it. Heat leaves the element through a ladder network — the element's own 5 mJ/K, its terminations, the board — and a pulse short against the element's millisecond heats it adiabatically, by P·t over its heat capacity alone: 0.737 ppm for 100 µs at 100 A, a thousandth of the held figure, 10.7 ppm for 10 ms. Along the burden optimum the error is linear in the current, so the current a pulse may carry for 100 ppm is exact: 12.9 A held, 936 A for 10 ms, 13.6 kA for 100 µs. A reading averaged over the pulse carries half the error of one taken at its end while the heating is adiabatic, and the section-by-section thermal model a data sheet quotes is 10% wrong at 7 ms.
The rejection two capacitors lose at their corner
AC-couple both inputs of a differential instrument and the two coupling networks become part of its common-mode path. Two 1 µF capacitors one per cent apart turn common mode into difference at exactly half their mismatch at their shared corner — 46.0 dB of rejection at 0.159 Hz, from a pair that is perfect at direct current and far above. Behind a 90 dB amplifier they decide the instrument's rejection over five decades around that corner. A 10 V common-mode step arrives as a 36.8 mV differential pulse one time constant later. Four hundred pairs of 20 per cent electrolytics reject a median 25 dB. Coupling the difference instead of each side removes the dip and hands the common mode to the amplifier.
The second capacitor, and the tail that changes sign
One coupling stage turns a pulse into a sagging top and an undershoot that returns to zero from below. Add a second stage with the same one-second time constant and a half-second pulse sags to 0.303 of itself instead of 0.607, undershoots to −0.697, crosses zero 2.27 s after it began and overshoots positive before settling. The step response becomes (1 − t/τ)·exp(−t/τ), through zero at exactly one time constant and down to −e⁻² at two. The reason is arithmetic: one capacitor removes the waveform's area, two remove its area and its first moment, and a pulse with a one-signed tail cannot have both. Built to the same stated corner, two stages still sag 1.29 times as fast, and a 1% flat top needs a pulse rate 806 times the corner rather than 625.
The second pole the ringing guard sees
A guarded input driven through an inductive source oscillates once L/R passes 1/ωₜ — for a follower with one pole. Give the follower the second pole every real amplifier has and the rule stops being a bound in either direction. A 9.49 Ω source that rang above 1.66 µH now needs 6.85 µH with a second pole at 3 MHz, four times safer; a 949 Ω source that rang above 177 µH now rings above 141, a fifth worse. The dividing line is the frequency the input rings at: above the geometric mean of the follower's crossover and its second pole, the pole takes gain from the bootstrap and helps; below it the pole adds phase and hurts. On three poles a decade apart the change of sign falls within 2.4 per cent of √(fₜ·f₂).
Two coupling stages without a buffer
Hang a second coupling stage directly on the first, with equal parts, and the pair's denominator gains a third time constant — the first resistor carrying the second capacitor's current. The poles spread from a double pole to −0.382 and −2.618 per time constant, a half-second pulse's top sags to 0.175 instead of the buffered pair's 0.303, and the sag rate is exactly 2 + 1/k per time constant for a second stage k times the first's impedance — the sum of the two short-circuit time constants. The tail still crosses zero, because the two zeros at the origin are untouched, and a step's lowest point is still at exactly twice its zero crossing. Then a reversal: specified by the overall corner rather than by the parts, the loaded pair sags only 1.12 times as fast as one stage and undershoots 7.6% rather than 13.5% — the buffer the cascade essay assumed makes the specified input worse.
The ring a blocking capacitor buys
A series capacitor in front of a transformer stops its flux from walking: a 55%-duty drive that walks a bare core by 0.40 units of volt-time a period settles within a few dozen periods once the capacitor takes the drive's average. The price is a second-order high-pass with the magnetising inductance, and an inductance lets its poles go complex where two resistive coupling stages never could. Between equal resistances the damping cannot fall below 1/√2, and the poles are complex only for a capacitor between 0.17 and 5.8 times L/R². From a stiff driver the damping is √(L/C)/2R and has no floor: a capacitor large enough to make the droop small — the droop is 1/RC, the capacitor's and not the transformer's — makes the tail ring, from one zero crossing at a tenth of L/R² to thirty-five at a hundred.
The time constant one trimmer cannot reach
A ten-to-one divider whose lower arm is 100 pF of instrument input and 15 pF more reached through a kilohm reads flat on a 1 kHz calibrator with its trimmer at 12.8 pF — the same setting as a lower arm of 115 pF lumped — and overshoots 12.8% on a 1 ns edge, settling within 1% only after 35.5 ns. No setting of the one trimmer is flat: the best judged on a fast edge is 6.1%, at 11.9 pF, and a trimmer set by eye so the fast edge lands flat leaves the calibrator's square wave 9.9% low with a 114 µs slope on its top. Flat at every time needs the upper arm to be a scaled copy of the lower one — a second branch of 9 kΩ and 1.7 pF — and then a 1 ns edge is flat to 10⁻¹⁰, with a window of −18 to +24% on that resistor and ±6% on its capacitor.
The current a capacitor adds up
An amplifier's input current in a resistor is an offset; in a capacitor it is a rate, and time is the gain. An integrator of 100 kΩ and 1 µF with its input grounded ramps at 50.0 mV/s from 50 nA, and balancing the inputs leaves the 5.00 mV/s of the offset current, still unbounded. A leak resistor bounds it at the current times the leak and costs the lowest frequency the circuit integrates. A hold capacitor's droop times its acquisition time is fixed, so 50 nA through a 100 Ω switch needs 655 nF and 0.77 ms for half a 16-bit step. A charge amplifier's offset times its low corner is Ib/2πCf, whatever its bleed.
The cure that needs one current
Balancing the resistances an amplifier's inputs look into removes whatever its two input currents have in common and leaves their difference. On a plain bipolar input the two are one current at two pins, correlated at 0.991, and the cure is worth ten. On an input whose bias current has been cancelled they are two residues of either sign, correlated at 0.10, and over 400 seeded parts the cure makes the error 38 per cent worse. It is worth 1/√(2(1 − ρ)) for a zero-mean current, nothing at ρ = ½, and in general exactly the input current over the offset current.
The voltage four wires carry in
Four wires remove a resistance measurement's leads by keeping the sense loop free of current, and that loop is exactly where a thermoelectric voltage lives. Ten milliamps through a milliohm is ten microvolts, so every microvolt where a copper lead meets a brass terminal moves a single-polarity reading by a tenth. Reversing the current removes any EMF that holds still and leaves the change over one interval; the + − + delta leaves the second difference; four readings leave the third — 501 ppm, 4.14 and less at 10 mA while a disturbed junction settles. Against the shunt's own heating each scheme has a best current, and the least error rises only as the cube root of the shunt's temperature coefficient: the reading method moves the error by orders of magnitude, the alloy by a cube root.
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.
The source the hook hides behind
A compensated divider with a late branch in its lower arm overshoots 13.2 per cent on an ideal step, and every measurement of that hook has been made from a source with no resistance. Through a real source the input capacitance charges first, and the hook and the source's lag act on the same edge. The overshoot is 10.5 per cent from 50 Ω, 4.13 from 300 Ω and gone by a kilohm, where the source's lag equals the hook's. The excess has not gone away: it is in the edge, which the screen shows 15 per cent faster than the source and the probe's capacitance allow. And the second branch that made the divider exact only makes its ratio exact, so from a kilohm the exact divider settles in 65.6 ns and the one with the hook in 44.
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.
Where the models stop
The boundaries as the subject rather than as a mark on something else. Four of them are frequencies and one is an amplitude; the last is the frequency at which Kirchhoff's laws themselves become an approximation, and it is set by nothing but the size of the board.
Every model has an edge
Four assumptions the circuits in this field run on, with the frequency at which each stops being true, on one axis. The ordering is not the one most readers would guess — an ordinary amplifier circuit runs out of model at 1.42 kHz, three thousand times sooner than a ten-centimetre circuit board does.
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.
How small is small signal
Linearising an exponential replaces a curve by its tangent, which is exact at a point and progressively wrong away from it. The amplitude at which it is one per cent wrong is 7.3 millivolts at room temperature — 28 per cent of the thermal voltage, not a small fraction of it, and a good deal smaller than "small signal" suggests.
The one current a constant is right at
Seven-tenths of a volt is the true forward drop at 5.748 milliamperes and at no other current, and every circuit built on it crosses zero error there — four different resistors at four different supplies, all exact at the same current. What decides whether the model is any good is not the diode at all; it is how much of the supply the diode is taking.
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.
The edge that is a region
Every boundary drawn so far in this subject is a number on one axis, and the figure that gathers four of them admits in its own caption that the fifth is an amplitude and cannot go there. Drawn on both axes at once, the ideal amplifier's one per cent boundary is a region with three sides and a corner — and the corner sits at 38.5 kilohertz where the two numbers a data sheet quotes cross at 48.8, because the two mechanisms are lags on the same waveform and add as magnitudes rather than in quadrature.
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.
The constant that is a window
A diode's ideality factor is quoted as a number and defined as a derivative, which means it has a value at every current and no value anywhere. Eight one-decade fits to one curve return factors from 1.23 to 1.98, two of them straight to a few parts in a thousand — so the residual gives no warning at all. Asked for the forward voltage at a milliamp, the window containing it is right to a third of a millivolt and the worst is out by 186, which is a current a hundredth of the truth.
The edges that move with the room
Every boundary in this collection is quoted at one temperature and most of them are functions of it. The small-signal edge is proportional to the thermal voltage, so it runs from 5.67 millivolts at −40 degrees to 9.69 at +125 — a factor of 1.71, the ratio of the absolute temperatures exactly. A realised Q is 1.54 per cent high at one end of that range and 2.58 at the other. The numbers are right; the condition attached to them was left off, and it is the same condition every time.
A boundary is a model and a tolerance
Every edge in this collection is computed from a fraction of error nobody states, and the four on its opening axis use three different ones. Swept over two decades, each boundary moves as a power of that fraction — Kirchhoff's laws exactly as the first power, the amplifier and the capacitor and the small-signal model as its square root to within three per cent, and a full-power bandwidth not at all. The exponent identifies the mechanism, and it re-orders the axis twice: the board fails before the capacitor below 0.424 per cent, and the output before the amplifier above 21.7.
Where the mechanisms are one mechanism
An amplifier is said to run out of three separate things — bandwidth, slew rate and rails — and errors from separate mechanisms add in quadrature while errors from one mechanism add as magnitudes. Measured as waveforms rather than as numbers, two of the three sit 18.4349 degrees apart, which is exactly the angle between a sinusoid and its own cube, and the third sits at ninety: its cosine against both of the others stays under 0.0025 wherever it exists. So the arithmetic is neither of the two anybody reaches for, and a budget built the right way is within 2.9 per cent where quadrature is 17 per cent low and a straight sum 37 per cent high.
Two currents with one name
A junction's saturation current is two currents with one name. The one in the forward law doubles every 4.49 kelvin; the one a reverse-biased junction actually conducts is generated in its depletion region, doubles every 8.98, and on this model junction is 3.06 × 10⁵ times larger at a volt of reverse bias. Doubles every ten kelvin is the second current's rule, and applied to the first it turns the forward drop's −1.81 millivolts per kelvin into +0.39.
The resistance a slow curve cannot see
A diode's series resistance is read off the top of its forward curve, and a bench curve is a slow one: each point is held until the junction has warmed to it. Through 350 kelvin per watt the held curve sits 39.7 millivolts below the pulsed one at 100 milliamps, and the three-parameter fit that reads 0.580 ohms from the pulsed curve reads 0.184 from the held one — with a smaller residual. The fitted resistance reaches zero at 531.7 kelvin per watt, and the resistance it hid is what keeps the junction from folding back: with 0.05 ohms instead of 0.6 the held curve turns over at 87.6 milliamps.
The logarithm is in the collector
A diode is the textbook logarithm, and a real junction holds one to within one per cent over only 2.40 decades — from 50 nanoamps to 12.6 microamps, at an ideality of 1.98 — because two mechanisms and a series resistance share its terminals. The same junction read at a transistor's collector, with its recombination current supplied from the base, holds 8.50 decades at an ideality of 1.0001. How far the logarithm reaches is decided by which terminal the current is taken from, and at the bottom of the range by a leakage current a millivolt is enough to switch on.
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.
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.
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.
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.
The straight line between two models
Between a constant seven-tenths of a volt and an exponential sits the model a designer actually reaches for: a drop plus a resistance. Fitted so that its worst error over a decade of current is as small as it can be, it is out by ±8.00 millivolts where the best constant is out by ±29.76 — and both numbers are the same over every decade, because a decade of a logarithm is the same shape wherever it is taken. On a real two-mechanism junction the line does best in the top decade, ±8.71 against a constant's ±58.53, because a series resistance is exactly the term the line has and the constant has none. And the resistance the fit returns is the part's plus 701 milliohms that is not there.
Three amplitudes, all of them one per cent
The amplitude at which linearising an exponential is one per cent wrong is 7.30 mV, and that is a statement about the gain. Two other quantities are also one per cent wrong somewhere: the second harmonic reaches one per cent at 1.03 mV and the shift in the operating point the model was linearised about reaches it at 5.17 — which is √2 below the gain boundary exactly, because the mean goes as a²/4 and the fundamental as a²/8. And the spacing is not a property of the device: tighten the criterion to a part in ten thousand and the same three spread to 1 : 50 : 70.7.
Two boundaries removed, and one moved
A differential pair's transfer is odd, so its mean and every even harmonic are zero — −3.2×10⁻¹⁷ and 9.1×10⁻¹⁷ at a drive of two thermal voltages, absent rather than small. Two of the three one-per-cent boundaries therefore do not exist for a pair at all. The third, the loss of gain, moves by √2 in the limit and by 1.42610 at the one per cent usually quoted: 10.42 mV against 7.304. So a pair reached for as headroom has bought forty per cent of it, and twenty millivolts of imbalance brings the even orders straight back.
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.
The resistance that bends the signal
A switch of half an ohm and a hundred megohms has a floor of 70.71 parts per million, 13.79 bits, because its on-resistance and its off-resistance cannot both be small beside one load. Most of that floor is a gain error, and a gain error calibrates away. Give the on-resistance a realistic ten per cent of movement across the signal range and the uncalibrated floor slips to 74.16 parts per million, while the part no calibration can touch — the curvature — balances the leak at 13.89 parts per million, 16.14 bits, into 1.39 kilohms. The floor was never set by the on-resistance. It is set by how much the on-resistance moves, as its square root.
The leak no switch can hold
A T of three switches reaches five parts per billion because its shunt switch holds the node a leak has to cross. A junction leakage does not cross anything: it flows out of the outer switch's terminal straight into the load. With 100 picoamperes of it at 25 °C, doubling every ten kelvin, the T's floor is 9.998 parts per million rather than five parts per billion — 16.61 bits, not 27.6 — and it has a best load again, at 100 kilohms. A lone switch loses nothing at room temperature. Above 91.4 °C, where the junction current reaches the off-resistance's conductance at one volt, the T is worse than one switch.
The boundary that is a rise time
Kirchhoff's own frequency is 3.97 megahertz for ten centimetres of track, and a one-kilohertz clock is four thousand times below it — which is the reading that concludes wrongly. Marched as the exact staircase of reflections it is, the same track leaves forty per cent of overshoot on a one-nanosecond edge and 0.40 per cent on a hundred-nanosecond one, reaching one per cent at 40.0 nanoseconds. The boundary is a rise time, it is twenty-nine times the conductor's round trip, and the repetition rate is not in it anywhere.
How many straight lines a diode is
One drop-plus-resistance line fitted minimax over a decade of a diode's current is out by ±8.00 millivolts, and it is the same over every decade. Chain such segments over six decades, a nanoamp to a milliamp, and the error of each is a closed form in its current ratio alone. Minimax segments meet at their joins without being asked to, so the chain is a continuous piecewise-linear model for free. The error falls as the inverse square of the segment count: one segment is out by 132 mV, six by 8.00, eighteen by 0.94. A millivolt takes 18 segments and a tenth of one takes 56. A real junction needs more, 23 and 72, because recombination gives it an ideality near two over most of the range and each segment narrows by the square root of the local ideality.
Symmetry and a resistor together
A pair linearises by symmetry and an emitter resistor by feedback, and the two are normally used together. Solved together, the pair's one-per-cent gain boundary is 0.4·Vₜ times (1 + gₘRₑ) to the three-halves power up to gₘRₑ = 3: 10.4 mV bare, 29.2 mV at gₘRₑ = 1, 80.9 at 3. The √2 by which it outlasted a single transistor does not survive: at gₘRₑ = 1 the two are within three per cent, near ½ the single transistor outlasts the pair fourfold, because its cubic vanishes there and a pair's never can, and with heavy degeneration the ratio climbs to 1.95 at gₘRₑ = 50 — two, for a reason that has nothing to do with junctions: a single transistor cuts off when the drive reaches Ic·Re, and a pair has steered its whole tail at 2·Ic·Re, its two emitter resistors in series. What symmetry alone buys is the even orders, at every degeneration.
The offset that brings the even orders back
A balanced pair has no second harmonic. An offset gives it one, and the closed form is exact: the second harmonic over the fundamental is (u/2)·tanh(Vos/2Vₜ), with u the drive over 2Vₜ — a single transistor's second harmonic times the fraction of the tail the offset steers aside, 1.93% of it at a millivolt, not the twentieth usually estimated. It is first order in the offset up to 9.0 mV. It does not grow with the drive for ever: it peaks as the pair saturates and falls again, so below 1.36 mV of offset it never reaches one per cent at any drive, and above 3.14 mV it reaches one per cent before the pair's own third harmonic does, at 17.9 mV. Read the other way, a matching specification is a distortion specification: −60 dB of second harmonic at 10 mV of drive allows 0.535 mV of offset.
The load that rings with the line
Put 10 pF on the far end of ten centimetres of 50 Ω track and the reflection stops being a number. A one-nanosecond edge from 10 Ω into 200 Ω overshoots by 47.5 per cent instead of 40.0, and rings near 174 MHz — the track's own 35 nH against its 14 pF and the load's 10 together. The one-per-cent boundary, a rise time of 39.9 ns on a bare track, becomes 50.4 ns with 10 pF and 83.4 with 50: a rise time and a capacitance together. As the load outweighs the track the lumped model comes back, missing the line's waveform by 55 per cent of the settled value with no load capacitance and by 2.2 with 200 pF. And a source matched to the line turns every capacitance's ring into a delay of about 2.2·Z₀·C.
How long a track may be
The lumped-element boundary read the way a layout is designed: given the edge, how long may the track be? For 50 Ω track driven from 10 Ω into 200 Ω, a 1 ns edge stays inside one per cent of overshoot for 2.51 mm, and every nanosecond of edge buys 2.52 mm more — proportional to within half a per cent from 0.1 ns to 30. The mismatch decides the rest. From 45 Ω the same edge may cross 23.5 mm; from 50 Ω, or anything above it into this load, no length overshoots at all; and into an open end, a CMOS input, 1.80 mm. The common rule — a track is short while its round trip is under the edge — allows 49 per cent of overshoot from a 5 Ω driver, and a tenth of its length still allows 3.3.
Two decibels of matching for one of gain
An offset gives a differential pair back the second harmonic its symmetry removed. Put resistors in the emitters and that second harmonic falls as the cube of one plus gₘRₑ at a fixed input and as its square at a fixed output — measured slopes of −3.00 and −2.00 — so each decibel of gain given up buys two decibels of distortion, or two decibels more offset at the same distortion. A matching specification that allowed 0.535 mV for −60 dB at 10 mV of output allows 2.14 mV at gₘRₑ = 1. The worst case over all drives relaxes more slowly: the offset below which the second harmonic can never reach one per cent goes from 1.36 mV to 4.39 mV as gₘRₑ goes to three, because the peak it limits sits where the pair saturates, and degeneration only moves that further out.
The heavier input that allows a longer track
Put a capacitance across a CMOS input and the longest track a 1 ns edge from a 10 Ω driver may cross inside one per cent of overshoot does not shrink steadily. It falls from the bare open end's 1.80 mm to 0.86 mm at 1.5 pF, and then it rises — 2.71 mm at 20 pF, 10.9 at 100 — nearly in proportion to the capacitance. A heavy load turns the track into the inductor of a series RLC whose damping, (Rₛ/2)·√(C/L), the driver's resistance supplies, and at every heavy load the one-per-cent length sits at the same damping, about 0.8. So the limit scales as Rₛ²·C: a slower edge buys a 20 pF input only the 0.4 power of its length, and a stronger driver — 2 Ω instead of 10 — cuts the track it may drive from 2.71 mm to 0.20. Into a heavy input, a weak driver is the clean one.
The load spread along the track
A hundred picofarads at the end of a track ring with its inductance as a series circuit, and a 1 ns edge from a 10 Ω driver may cross only 10.9 mm inside 1% of overshoot. The same hundred picofarads spread over 32 inputs along the track allow 29.5 mm, and 300 pF spread allow 74.2. Spread capacitance is not a load at the end; it lowers the line's own impedance to Z₀ divided by the square root of one plus the ratio of the added to the line's own capacitance, and slows it, and once the loaded line's impedance falls below the driver's — 8.2 Ω for 150 pF over 30 mm — every reflection at the source returns with the same sign and the far end climbs a staircase that cannot overshoot. The valley the lumped load has at a picofarad or two survives, shallower: 1.14 mm at its lowest, against 0.88. Two taps are no better than one; from eight the track is a loaded line.
The track that forks
Two 2 pF inputs on one track from a 10 Ω driver with 1 ns edges may sit only 1.05 mm away inside 1% overshoot if both hang from its end. Fork the track at the driver and each branch may run 2.33 mm — more than twice as far — because two fifty-ohm branches in parallel are a 25 Ω line, half as mismatched to the driver. Every millimetre of trunk before the fork costs more than a millimetre of branch: with a millimetre of shared trunk the branches may run only 0.11 mm. Equal branches are the only good ones: a 2.2 mm branch overshoots 0.91% beside an equal partner and 1.7% beside a short one. And a star reaches further the more inputs it serves — 2.33 mm for two, 11.85 for four, where four branches in parallel make 12.5 Ω — while a daisy chain of the same inputs reaches 2.54.
The edge of a chain
A system built from stages that are each inside their own edge is not inside its own: the stages share one error budget. Four ×10 stages from a 1 MHz amplifier are 1% from ideal at 7.06 kHz, not at the single stage's 14.2 kHz — the edge moves as N to the −0.502. A signal path through four ten-centimetre boards is a degree long at 993 kHz, a quarter of one board's edge, as N to the −1. The chain's exponent is minus the edge's own exponent in its tolerance, measured on the line y = −x. And the small-signal edge of a chain with gain belongs to its last stage: four junction stages of ten are 1% off at 7.26 µV, a thousandth of one stage's 7.30 mV.
The table that has to be refitted
Eighteen straight segments model a real diode from 100 nA to 100 mA to within 1.886 mV at 300 K. Carried to 350 K unchanged they are 149 mV out; moved by one common shift, 38.3 mV, because the drops at the bottom of the table fall 3.31 mV a kelvin and those at the top 1.54; moved by a shift each, 3.84 mV. That last correction holds within twice the table's own error over fifty kelvin either way and fails at 400 K, where the bottom of the table has dropped below the junction's own recombination current and is no longer an exponential at all. Refitted on the same breakpoints the table holds 1.57 to 2.38 mV at every temperature: the breakpoints never need to move, and the lines always do.
The delay a chain adds
A chain of one-pole stages leaves its flat band as the square root of the number of stages, and it leaves its settling time faster: four stages take 10.0 time constants to settle within 1 per cent against one stage's 4.61, sixteen take 26.7 — 5.81 times one stage, where the frequency rule allows 4.02. The difference is a delay. The chain's step is the sum of N exponential lags, so it is an edge of mean N time constants and spread √N; the magnitude response sees only the spread. At a thousand stages the 1 per cent time is 1.074 of the delay. A chain budgeted to stay flat to 1 per cent settles to 1 per cent 45 per cent late.
The resistor mismatch that is only an offset
Two emitter resistors that differ by δ unbalance a degenerated pair, and with an ideal tail they do it in the simplest way possible: the arm currents always sum to the tail, so the mismatch adds the constant (Iₜ/2)·R·δ = Vₜ·gₘRₑ·δ to the input and nothing else. Solved arm by arm, a 1% mismatch at gₘRₑ = 3 is the balanced pair moved by 0.776 mV to 4 × 10⁻¹⁶ of the tail, and a trim of that size leaves no second harmonic at any drive. What separates it from a junction's offset is that it grows with the degeneration meant to divide it: at a fixed output its second harmonic falls as gₘRₑ/(1 + gₘRₑ)², one decibel per decibel of gain rather than two, and it takes over from 0.5 mV of base–emitter mismatch at gₘRₑ = 1.93. For −80 dB at 10 mV of output the resistors must match to 0.823% at gₘRₑ = 1 even with perfect transistors.
The stage that overshoots a little
A stage with a little overshoot settles sooner than a one-pole stage of the same bandwidth, 3.53 time constants against 4.61 within one per cent, and a chain of them should do better still, since a stage at the right damping adds almost no spread. It does not. The saving is 17 per cent at two stages, 11 at four and 2 at eight, and at sixteen the chain is 5 per cent slower than the one-pole chain, because a stage's extra delay adds as the number of stages and the spread it removes adds only as the root. The best damping stays near 0.83 at every length, just short of letting the overshoot out of the band. In frequency the same stages are five times flatter at their best, at a different damping, and so the two edges of a chain ask for two different stages.
Lines, where a wire has a length
On the far side of the frequency at which Kirchhoff's laws give out. What a source drives into is decided by geometry before the load has any say; what comes back one delay later decides the rest. The wave picture is checked against a lumped ladder that has never heard of a wave — and the useful result is how badly the ladder does.
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.
A ladder is not a line
A transmission line is usually introduced as the limit of a chain of inductors and capacitors as the number of sections goes to infinity. That is true, and it gives entirely the wrong impression of how close a finite chain gets. Forty sections still ring through every plateau by five per cent, and extrapolating the fitted convergence, reaching one per cent would need about nine hundred and sixty.
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.
Several sections, and the band they buy
A quarter-wave transformer is exact at one frequency, and a 4:1 transformation holds |Γ| under 0.1 over 17.1% of it. Several sections whose reflections cancel over a band take that to 47.7, 67.9, 82.1 and 92.6% — which is filter design with the reflection as the shaped quantity. An equal-ripple design found by minimax search buys a further 35 to 46% at the same worst reflection, and its ripples come out level to four decimals.
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.
The far end that cancels
Two mechanisms couple two parallel tracks: a mutual capacitance injecting a current and a mutual inductance injecting a voltage. They add at the near end of the quiet track and subtract at the far end, with the same constant in front of both — measured here as 1.048 times ten to the minus two either way, across a slider that moves their ratio by five times. So the far end is not a smaller effect: when the two couplings are equal it is 3.5 times ten to the minus nineteen of the drive, which is zero to the last bits of a double.
The mismatch that the cable hides
A lossless line carries a reflection back unchanged, so the standing-wave ratio at the instrument is the standing-wave ratio at the load. A real line does not, and the departure is exact: ten decibels of one-way loss improves any mismatch by twenty. A four-to-one load at the end of twenty metres of ordinary coaxial cable measures 1.13 at the near end, a return loss of 24 decibels, and passes an acceptance test the load could never pass. The boundary is a loss rather than a length, which makes it a frequency: 30.2 metres at 100 megahertz, 9.5 at a gigahertz, 3.0 at ten.
The resistor at the wrong end
A lattice diagram is read at the two ends of a line, and that is where the two respectable terminations look identical: a clean step, one delay late, at the receiver. Anyone standing halfway along a series-terminated net sees half the swing held for a full round trip, which for a logic input is not a level at all. The interval is 2(1−x) delays exactly, it is zero only at the far end, and the scheme that never has it draws sixty milliamperes for as long as the level is held.
The receiver that is a branch
A lattice diagram has two ends, and an interior receiver is not a point on a net — it is a short piece of track leading off it to a pin, open at the far end. Three lines of equal impedance meeting at a junction present each other with half the impedance, so a wave arriving is reflected by exactly minus a third and two thirds goes on: the far end receives two thirds of the swing and sits there for twice the stub's own delay, whatever the net is terminated with and wherever on it the branch is.
The delay that is not one number
Nine essays in this field quote a delay: a length divided by a velocity, the same for every frequency, and the edge that comes out is the edge that went in. A real trace has a series resistance, and below the frequency where the reactance overtakes it — 910 kilohertz for ordinary copper — the line is a diffusion rather than a wave, with a velocity proportional to √f. What survives is that the arrival is still exactly linear in the length. What does not is the rise time, which grows as the square of it.
The millimetre that becomes common mode
A pair carries two modes rather than two signals, and a length mismatch between its halves converts one into the other. A millimetre and a half of skew is ten picoseconds, a tenth of the signal is common mode by three gigahertz, and the differential signal has lost five thousand parts per million of itself getting there — so the error is not missing from the signal, which is why a pair can pass its own eye and fail an emissions test. The product in the answer is ωΔτ, which is what the instruments field's rejection corner is one over.
The inductor one mode cannot see
Two windings on one core present a millihenry to a current that goes the same way round both and a microhenry to one that goes opposite ways, so the same component has a corner at 3.98 kHz and another at 7.86 MHz — a ratio of two thousand, which is 2/(1−k) and contains no inductance at all. It is bought to remove the conversion the previous essay measured, and five picofarads across each winding turn it over at 1.59 MHz and leave it worth 0.02 decibels by ten gigahertz.
The number that was wrong
An earlier figure printed 97.3 per cent of band for a five-section transformer where the answer is 134, said in its own text that the figure was wrong, and blamed a search that had converged to eight digits. The search was fine. The objective was the worst of a grid rather than the worst of a band, the band was then measured by bisecting a function that crosses its threshold five times, and the check guarding all of it passed — because 97.3 is still more than 92.6.
Terminated at both ends
A series resistor at the driver and a parallel one at the receiver cost exactly half the swing — 1.650 volts of 3.300 — and no reflection ever gives it back, because there is no reflection. What the half buys is measured rather than assumed: a driver thirty per cent off its assumed impedance rings a series-terminated net by 16.3 per cent and a doubly terminated one by nothing at all, and an interior receiver on a series-terminated net sits in the undefined band for every far-end resistance above 125 ohms, which is Z₀/(1−2b) and contains no length, no driver and no frequency.
How wide a null is
A far end at 3.5×10⁻¹⁹ of the drive is a statement about arithmetic until somebody asks how far the two couplings may differ before it comes back. The answer has no flat bottom in it: the far end divided by the near end is |1−r|/(1+r) exactly, so the null is a V and a ratio one per cent off returns 4.98×10⁻³ of the near end. On the axis a board is built to that is a difference of 0.0081 between the two modes' effective permittivities, out of 3.99 — two parts in a thousand, and 0.675 picoseconds of mode skew over a hundred millimetres.
The cable that hides two things
A length of cable improves a return-loss reading by twice its loss and raises a noise figure by once it, so the rule of thumb is two decibels of apparent match per decibel of floor. Both halves are owned here and neither essay had the other. Put together they say what an acceptance limit costs: making a 4.0:1 load read 1.50:1 spends 6.53 decibels of noise figure, against a mismatch that was itself costing 1.938 — and the exchange rate is not two but 2(1−Γ²)/(1+Γ²), which is 0.94 where a pad is actually short.
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.
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.