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

Stray capacitance — where it appears

The capacitance between a node and everything else, contributed by the board and the package rather than by any component in the design. Two picofarads at every node is invisible in a passband at a few ohms and is the whole of what fills a null in a stopband.

Named by 4 essays across 3 fields — each of them below, with the objects they name alongside it.

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

The same filter a thousand times larger

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

filters · Impedance scaling An order-6 cascade at 10 kΩ: a band 10× wide. computed by solving, not by drawing. A 10.0 kHz unity-gain Butterworth of order 6, 3 Sallen-Key sections in cascade, realised at seven impedance levels three decades apart, with every resistance multiplied and every capacitance divided by the same factor. The design is exactly invariant: the solved magnitudes agree to 4.0e-15 of each other across all seven and every frequency in the passband, which is the last bits of a double rather than a good approximation. The realisation is not, because two of its quantities do not scale — 50 Ω of amplifier output resistance and 2 pF of stray capacitance to everywhere — and they bind at opposite ends. Inside 0.1 dB the band runs 316 Ω to 3.16 kΩ, with the least departure of 0.0763 dB at 1000 Ω; at this setting it is 0.182 dB at 12.9 kHz. The consideration that does not appear on either axis is noise: the resistors' density goes as √R, so across the six decades drawn the noise moves by 1000 times and points at the low end of the band.

The band that closes with the order

One Sallen-Key section is inside a tenth of a decibel of its own design over three decades of impedance level, bounded below by fifty ohms of amplifier output resistance and above by two picofarads of stray. Give it three more sections and the band is one decade; give it four and there is no impedance level at all that meets a tenth of a decibel. Every section brings three more nodes each carrying their own stray and one more amplifier carrying its own output resistance, so the floor rises with the order until it crosses the tolerance — a boundary in the order rather than in the impedance.

filters · Impedance scaling A 10 ps stub under a balanced via, and the pad it has to come out of. computed by solving, not by drawing, as a cascade. The reflection from a via of 0.5 pF of pad and 1 nH of barrel on its balancing line of 44.72 Ω, met by an edge of 59 ps, four ways: with no stub, whose largest excursion is 24.4 mV; with a 10 ps open stub below its bottom pad, 90.5 mV and an area of 5 ps; with the stub's 0.224 pF taken out of both pads, 56.1 mV; and taken out of the bottom pad alone, 24.6 mV. Both trims balance the area again. Only the second restores the doublet.

The pad the stub comes out of

A via used to a middle layer leaves barrel hanging below the connection: an open stub. Under a via balanced on its √(L/C) line, a 10 ps stub is 0.224 pF of extra capacitance, 5 ps of area, and a reflection of 90.5 mV where the balanced via gave 24.4. Taking that capacitance out of the pads balances the area again whichever pad it comes from. Only one restores the doublet: out of both pads the via still reflects 56.1 mV, out of the bottom pad the stub hangs from, 24.6. The trim works for every stub whose capacitance fits inside that pad, delays up to Z·C/2 = 11.2 ps, whose quarter wave is above 22.4 GHz. A longer stub has nothing left to come out of. At 40 ps it is a notch in the transmission at 6.25 GHz.

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

The delay an inductance does not add

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

transients · Value theorems

Named alongside it

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

Impedance scalingModel rangeOutput impedanceParasiticsRealisationSallen-keyBiquad cascadeCharacteristic impedanceDamping ratioFilter orderInitial-value theoremJohnson noise

All concepts