The collection

Every essay

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence.

Networks, and how a solve is checked

A circuit has one answer and a matrix finds it. What matters is not that the answer exists but that it can be checked: the branch currents are rebuilt from the element laws and summed at every node, and the energy is counted twice. Networks with no answer are refused by name rather than returned as a large plausible number.

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.

6 figures
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.

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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.

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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.

-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.

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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.

4 figures
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.

4 figures
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.

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Before the steady state

Everything on the frequency axis assumes a settled circuit. These are the figures about getting there — and about the one limit no transfer function contains, where a step becomes large enough that the response stops being a scaled copy of a smaller one.

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.

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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.

4 figures
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.

4 figures

Filters, measured not tabulated

Butterworth is flat, Chebyshev is steep, Bessel has good delay. None of those is a number. Each family's poles are computed from its definition, built as a network, and then the ripple, the skirt, the delay variation and the ringing are measured on the network that results.

-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.

4 figures
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.

5 figures
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.

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Feedback, and the margin

The site's strongest construction, and the subject supplies it free: a phase margin computed from the loop gain in the frequency domain, an overshoot measured from the step response in the time domain, one circuit, and a required agreement between two numbers that share no arithmetic.

-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.

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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.

4 figures
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.

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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.

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.

5 figures
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.

4 figures
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.

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