Depth

Ladders

A field says what an essay is about. A ladder says what else there is to say about it — the distinct arguments that stand against one idea, from the one that introduces it to the one that assumes all the others.
00.0500.10002.5057.5010time, in network time constantsoutput ÷ inputboth ratios: 0.1000the divider9M12.8 pF1Msolved, then checked — two ratios, one network0.0% out at 12.8 pF

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.

1 rung · instruments
-20020401001k10k100k1M10M100M1G10Gfrequency (hertz)gain (decibels)midband 43.2 dBsolved: 503 kHzMiller says 643 kHzsecond pole 336 MHzzero at g_m/C_μsolved, then checked — two networks, one measurementMiller is 22% optimistic

The frequency a device sets for itself

A common-emitter stage's bandwidth is decided by two picofarads between its collector and its base. The Miller approximation says how — lump it at the input, multiplied by one plus the gain — and predicts 643 kHz where the solved network gives 503 kHz. Twenty-two per cent optimistic, and it has no room at all for the second pole or for the zero in the right half-plane that the network also has.

1 rung · semiconductors
-101-0.100-0.05000.0500.100differential drive (volts)output, normalisedthe pair: odd, and it saturatesone exponentialharmonics, as a fraction of the fundamentalsingle, h218.876%pair, h21e-16single, h32.404%pair, h31.202%single, h40.231%pair, h46e-17single, h50.018%pair, h50.017%solved, then checked — evenness measured, not assumedthe pair's second harmonic is 1e-16

What a pair cancels, and what it only halves

A differential pair's transfer characteristic is an odd function, and an odd function driven symmetrically produces no even harmonics at all. Measured, the second harmonic comes out at 10⁻¹⁶ of the fundamental — the arithmetic's own floor, not a small physical residue. The third harmonic is a different story, and it comes out at exactly half the single stage's, which is a reduction and not a cancellation.

1 rung · semiconductors
the drive: a sinusoidthe current out, and a symmetric one for comparisonone cyclemeasured, against the Bessel ratioharmonic 29.61%harmonic 30.62%harmonic 40.03%harmonic 51.2e-5harmonic 63.7e-7agreement: 5e-10 relativesolved, then checked — a transform against a seriessecond harmonic 9.6% at 10.0 mV

The distortion a linear model cannot have

A small-signal model's output is a scaled copy of its input by construction, so it has no second harmonic and asking it for one is not a hard question but a meaningless one. Measured on the curve itself, an exponential produces one per cent of harmonic distortion at 1.03 mV of drive — seven times sooner than the 7.30 mV at which its gain is one per cent wrong.

1 rung · semiconductors
00.50011.50202468time, in propagation delaysvolts at the far endthe wave answer20 sectionsone delaysolved, then checked — a ladder against a latticeplateaus 0.08 V out at 20 sections

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.

1 rung · lines
02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ

The divider, and the thing it does not know about

A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.

1 rung · networks
1e-81e-71e-61e-51e-41m10m100m1amplitude (volts)Johnson noise of 1 kΩ: 501.6 nV1% distortion, exponential: 1.03 mV1% gain error, exponential: 7.30 mVlinear step, unity follower: 79.58 mV66.3 dB of rangesolved, then checked — a floor and a ceiling, both computed66 dB between them

A floor and a ceiling

Every other boundary on this site is a ceiling. This one puts a floor underneath and measures the distance between them — 4.00 nanovolts per root hertz at the bottom, one per cent of harmonic distortion at 1.03 millivolts at the top, and 66.3 decibels of range in a ten-kilohertz measurement. Both ends are computed, neither is on a datasheet, and the arrangement of the stages decides which one moves.

1 rung · noise
-90-60-3001001k10kfrequency (hertz)gain (decibels)ButterworthChebyshevBesselhalf power1.00 kHzthe passband, magnified-1-0.500000.2000.4000.6000.8001solved, then checked — three networks, 133 frequencies eachall normalised to a measured −3 dB at 1.00 kHz

Three families, one corner

Butterworth is flat, Chebyshev is steep, Bessel has good delay. None of those is a number, so the table they appear in cannot answer the question anybody has. Here each family's poles are computed from its definition, built as an actual network, and then measured — starting with the step every comparison skips.

1 rung · filters
passband deviationdecibels, peak to trough below 0.8 f_cButterworth0.443 dBChebyshev0.500 dBBessel1.882 dBattenuation at three times the cornerdecibels downButterworth47.7 dBChebyshev64.0 dBBessel28.3 dBgroup-delay variation across the passbandper cent, slowest against fastestButterworth48.0%Chebyshev49.0%Bessel0.1%solved, then checked — nine measurements, three networksevery number here moves with the order

What a steep skirt costs

A filter's order buys attenuation at a known rate — twenty decibels per decade per pole, and no arrangement of components changes it. What varies between families is how quickly the slope is reached, and the currency it is paid for in is delay: the steepest of the three distorts delay eight hundred times more than the gentlest.

1 rung · filters
-6-4-201m10m100m1101001kresistance being measured (ohms)log₁₀ of the reading's errortwo wiresfour wiresone per centtwo wires are 1% out at 10 Ωsolved, then checked — two arrangements, not two formulastwo wires fail below 10 Ω

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.

1 rung · instruments
-60-40-200gain (decibels)the sketch: flat, then −20 dB/decade−3.01 dB-90-450101001k10k100kfrequency (hertz)phase (degrees)sketch within 0.1 dB below 152 Hzsolved, then checked — checked against a chain-matrix productthe sketch is 3.01 dB wrong at 995 Hz

One solve, read four ways

Reactance, phase, the corner frequency and the roll-off are not four ideas. They are four readings of one complex number, obtained from the same matrix that answers direct-current questions — and the straight-line sketch every engineer draws of them is itself a model, three decibels wrong exactly where it is read.

1 rung · frequency
010203040501101001k10k100k1Mfrequency (hertz)closed-loop gain (decibels)the ideal amplifier: two resistors, no frequencythe circuit+1% low at 1.35 kHz3 dB down at 10.0 kHzsolved, then checked — a nullor against a real devicethe ideal answer is 1% wrong above 1.35 kHz

The ideal amplifier, and where it stops being one

An ideal operational amplifier's closed-loop gain is set by two resistors and nothing else — a horizontal line at every frequency. The real one is already a tenth of a per cent low at direct current, one per cent low by 1.35 kHz, and above 10 kHz has no loop gain left, at which point the ideal answer is not an approximation to anything.

1 rung · feedback
00.5011.521001kfrequency (hertz)group delay (milliseconds)ButterworthChebyshevBesselthe corner, 1.00 kHzsolved, then checked — −dφ/dω on the unwrapped phaseflat magnitude is not flat delay

Flat magnitude, unflat delay

A filter that passes every frequency in its band at the right amplitude and the wrong time has not passed the signal. Group delay is the measurement that says so, it is absent from the classical comparison, and it varies by fifty per cent across the passband of the two families everybody uses.

1 rung · filters
±619.3 nV, the root mean square1200 of 192001 stepssampled, this seed621.2 nVsampled, 6 seeds619.3 nV ± 1.79%integrated from |H|²620.6 nV…ignoring the hold626.3 nVnoise bandwidth15.03 kHzthe −3 dB point10.00 kHzsolved, then checked — a sample against an integralagreeing to 0.21%

The floor a resistor sets

A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.

1 rung · noise
-40-200204060loop gain (decibels)unity loop gaincrossover 5.73 kHz46.1 dB of gain margin-180-135-901101001k10k100k1M10M100Mfrequency (hertz)loop phase (degrees)−180°34.9° of marginsolved, then checked — the loop cut and injected34.9° of phase margin at 5.73 kHz

What is left at crossover

A feedback loop is stable or not according to one number read at one frequency — how much phase remains before −180° at the point where the loop gain passes unity. The loop gain here is obtained the way it is obtained on a bench: cut the loop, drive one side of the cut, and measure what comes back to the other.

1 rung · feedback
1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz

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.

1 rung · limits
0501001500200400600closed-loop output (volts) for a 1 V stepthe 100× the divider asks for35.1% overphase margin, measured two waysfrom the loop gain34.9°from the overshoot35.0°apart by 0.1° — the relation assumes two poles and this loop has threesolved, then checked — margin against overshootthe second-order relation is 0.1° out here

Two measurements of one margin

A phase margin is computed from the loop gain in the frequency domain, without ever looking at a step. An overshoot is measured from the closed-loop step response in the time domain, without ever looking at a Bode plot. Inverting the standard relation on the second returns 34.9° against the first's 34.9°, and the residue is the third pole.

1 rung · feedback
00.2000.4000.6000.8000.50011.50frequency ÷ the frequency it was cut forreflection |Γ|the bare junction: 0.600|Γ| = 0.1, or 20.0 dB return lossexact here only17.1% of a bandsolved, then checked — the band scanned, not approximatedexact at one frequency, 17% of band

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.

1 rung · lines
101001k10k100k1M10M100M1G10Gfrequency (hertz)the ideal operational amplifier1.42 kHz — a gain of 100 from a 1 MHz part is 1% low herea 10 V output at full amplitude7.96 kHz — above this the output cannot move fast enoughthe ideal 100 nF capacitor4.69 MHz — 1.2 nH of lead makes it 10% wrong hereKirchhoff's laws on 10.0 cm3.97 MHz — the board is one degree long hereeach bar is where the model may be used; the rule at its end is the numbersolved, then checked — each boundary from its own modeland one that is not a frequency: 7.3 mV

Every model has an edge

Four assumptions this collection runs 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.

1 rung · limits
a bridge, which no series-parallel reduction reachesnode a7.5566 Vnode b4.7993 Vcurrent law, rebuilt from the element laws2.71e-16 of the largest branch currentpower delivered against power dissipated4.33e-16 apart · 48.07 mWsolved, then checked — 6 elementsa linear network has no edge: this one is exact

What a network answers, and how the answer is checked

A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number.

1 rung · networks
00.500101234frequency ÷ the −3 dB point|H|², the power that gets throughhalf the power: the −3 dB pointthe brick wall: 1.5706× the corner1.571×solved, then checked — the area integrated, not tabulated1.571× the corner, not 1×

The bandwidth noise sees

A single pole passes π/2 times as much noise power as a brick wall at its own corner frequency, so a noise voltage computed with the −3 dB point is twenty-one per cent low. Measured by integrating the solved response rather than taken from the table it usually comes from, the ratio is 1.5706 and π/2 is 1.5708. A five-pole Chebyshev's is 0.964 — less than one.

1 rung · noise
02.5057.5000.2000.4000.6000.800voltage across the diode (volts)current (milliamperes)load line: (5 V − v)/1.0 kΩ123450.6925 V, 4.307 mAthe circuit5 V1.0ksolved, then checked — two Newtons, no shared arithmeticthe drop moves 60 mV per decade

A bias point is a solution, not a choice

The phrase "the diode drops 0.7 volts" is a constant standing in for the root of a transcendental equation. Solved properly, from a five-volt supply through a kilohm, it drops 0.692544 V — and from forty-eight volts through the same kilohm it drops 0.754459 V, because the drop moves about sixty millivolts for every decade of current through it.

1 rung · semiconductors
realimaginaryacross Racross Lacross Cthe source, 1 Vmagnitudes|v_R| = 1.000 V|v_L| = 2.128 V|v_C| = 2.128 Vsum 5.255 Vvector sum 1.000 Vsolved, then checked — one solve at 1.59 kHzsteady state only: 3 cycles to settle

Three voltages that close on one, and the steady state they assume

Kirchhoff's voltage law drawn as a polygon in the complex plane. The three element voltages of a series circuit add head to tail to the source exactly — while their magnitudes add to five times it. And the whole picture is a statement about a settled circuit, which takes a computable number of cycles to arrive.

1 rung · frequency
the step this produces00.50011.5001234σζ = 0.3000ω₀ = 1592 Hzsolved, then checked — poles by rooting the determinantnatural frequency recovered to 6 digits

Where the behaviour is written down

Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.

1 rung · transients
00.250.500.75110100load resistance (ohms)power factorno capacitorsized here: pf 1.000000leading above 21 Ωthe 0.95 an installation is usually required to holdsolved, then checked — 61 loads, one capacitorexact at 20 Ω, 0.23 elsewhere

The capacitor that was right once

Cancelling a load's reactive power needs one division and no iteration, and the answer is exact. It is exact for the load it was computed from, at the frequency it was computed at, and the figure here is about what it does everywhere else — which includes making the installation worse than it was before anything was fitted.

1 rung · power
-6-4-201101001k10k100k1M10M100Mfrequency (hertz)log₁₀ of the error in the reading1× probe, 115.0 pF at the tip10× probe, 11.5 pFone per cent1× is 1% out at 6.79 kHz10× at 69.2 kHzsolved, then checked — the node with and without the probe1% wrong at 6.79 kHz with a 1× probe

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.

1 rung · instruments
0500500100010001500150020002000real power (watts)reactive power (volt-amperes reactive)1636 W1285 var2080 VAreal power1636 Wreactive power1285 varapparent power2080 VApower factor0.7864angle38.15°line current9.04 A…doing work7.11 Athe load230 V20 Ω50 mHsolved, then checked — three routes to Qonly 0.79 of the current works

The current that does no work

A solved network has been reporting its own real power on every page of this collection, as the second of the two checks each answer passes before it is drawn. What that check discards is the imaginary half — the power that flows out to a reactance and back again, does nothing, and is still carried by the cable, still heats the transformer, and is still on the bill.

1 rung · power
10m1.0e+2m1101001k10k10k100k1M10M100M1Gfrequency (hertz)impedance magnitude (ohms)1/(ωC), the symbol's promise10% off above 4.69 MHzinductive above 14.5 MHz30 mΩ — the floor the resistance setssolved, then checked — the part as three elementsa capacitor below 14.5 MHz, an inductor above

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.

1 rung · frequency
00.50011.5002468time, in propagation delaysvoltsthe divider: 1.0000 V0.833 V: the source against the linethe far endthe near endfirst arrival, 4.83 nssolved, then checked — a series against a dividernothing at the far end before 4.83 ns

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.

1 rung · lines
00.200.400.600.8011001k10kfrequency (hertz)fraction of the source across the resistorhalf the power198.9 Hz measuredresonance 1.59 kHzsolved, then checked — half-power points by bisectionf₀/Q predicts 198.9 Hz — exactly

Resonance, and the bandwidth it sets exactly

The half-power bandwidth of a resonant circuit is f₀/Q — not approximately, but to every digit the arithmetic has, which is rare enough to be worth checking. What is not exact, and is drawn as though it were, is the idea that the band sits centred on the resonance. At a quality factor of one its middle is twelve per cent above.

1 rung · frequency
00.2500.5000.750105101520time (microseconds)output, divided by the size of its own step20 mV step1.0e+2 mV step5.0e+2 mV step2 V step8 V steplinear below 79.6 mVsolved, then checked — integrated with the rate limitscaling fails above a 79.6 mV step

The step that is too big

A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.

1 rung · transients
1.0m10m1.0e+2m110100100m110100drive amplitude (millivolts)how much the linear model understates the gain (per cent)1% understated10% understated1% at 7.3 mVV_T = 25.9 mVsolved, then checked — the Bessel ratio from its seriesthe tangent is 1% wrong above 7.3 mV

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.

1 rung · limits
024681010m100m110current drawn from the source (amperes)terminal voltage, solvedthe ideal source: 9 V at any current1% low at 180 mAthe model9 Vrloadsolved, then checked — the load swept over four decadesthe ideal source is 1% wrong above 180 mA

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.

1 rung · networks
00.50011.50024output (volts), for a 1 V step inthe final valuesolid: from the poles · dashed: stepped forwardtime (milliseconds) above · the same span as a fraction, belowgap between the two routes (volts)1e-71e-61e-51e-41.0m10m1.0e+2m1solved, then checked — residues against 500 trapezoidal stepsthe numerical route is out by 1.7e-3 V

One step, computed twice

A step response from the poles is exact. The same step walked forward in time is not, and the difference between them is the trapezoidal rule's own error rather than anything about the circuit. It falls by a factor of four every time the step is halved, which is a claim about a method and can be watched.

1 rung · transients
the three line currents, and their sum2.65 A in the neutral00.2000.40000.2000.4000.6000.8001imbalance in one phaseneutral current ÷ line currenta tenth of a line current11.1%solved, then checked — neutral by two routesa tenth of a line at 11.1% imbalance

Three phases, and the wire that carries nothing

Three sources a third of a cycle apart, feeding three equal loads, return a current of 5×10⁻¹⁵ amperes down the wire between the star points. That is zero, and the whole of three-phase distribution rests on it. What is worth measuring is how fast it stops being zero, and the answer is that an eleven per cent imbalance in one phase puts a tenth of a line current down a conductor often sized on the assumption that it carries none.

1 rung · power

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