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

AC coupling — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A 100 ms pulse through a 0.159 Hz corner, 9.52% shorter by the end of it. computed by solving, not by drawing. Marched. The dashed line is the pulse that was sent. The solid line is what a 1 MΩ input with 1 µF in front of it receives: the top decays as exp(−t/RC) for the whole 100 ms, ending 9.515% down, and the trailing edge undershoots by exactly the same amount. The closed form for the same two components gives 9.516%. The input's specification is a corner at 0.159 Hz; a top flat to one per cent needs a pulse shorter than 10.1 ms, which is a rate 625.2 times the corner — a constant with no component in it, and the reason a low-frequency specification says nothing useful about an edge.

    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.

    part 1 · instruments
  2. Two coupling capacitors 1% apart reject common mode by only 46.0 dB at their corner, 0.159 Hz — and perfectly at direct current and far above it. computed by solving, not by drawing at 281 frequencies: one common-mode source driving two coupling networks, each 1 µF into 1 MΩ, with the capacitors 1% apart. The difference between the two outputs, per volt of common mode, agrees with the difference of the two first-order high-passes to 6.9e-14. It is zero at direct current, where both block, and zero far above the corner, where both pass; between, it peaks at the corner, 0.159 Hz, at 5.000e-3 — half the mismatch — which is 46.0 dB of rejection. Behind a 90 dB amplifier (grey), the coupling network decides the rejection from 0.000503 Hz to 50.3 Hz.

    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.

    part 2 · instruments
  3. A second coupling stage with the same time constant: the top sags to 0.303 instead of 0.607, and the tail crosses zero at 2.27 s and overshoots to +0.025. computed by solving, not by drawing, marched and checked against the closed form to 3.0e-4: a half-second pulse through a 1 µF, 1 MΩ coupling stage (dashed), and through that stage followed by a buffered second one with a time constant of 1 s (solid). One stage sags to 0.6065 by the pulse's end and its undershoot decays monotonically to zero. Two stages sag to 0.3033, undershoot to −0.6966, cross zero 2.271 s after the pulse began and overshoot positive — the second capacitor removes the first moment of the waveform as well as its area, so the tail has to change sign.

    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.

    part 3 · instruments
  4. Without the buffer, a second stage of 1 times the first's impedance: the top sags to 0.175 instead of 0.303, and the tail crosses at 2.00 s and overshoots to +0.010. computed by solving, not by drawing, marched and checked against the closed form to 4.5e-4: a half-second pulse through two one-second coupling stages, buffered (dashed) and with the second hung directly on the first (solid), its resistor 1 times the first's and its capacitor 1/1 of it, so that each stage alone keeps its one-second time constant. Buffered, the top sags to 0.3033 and the tail crosses zero 2.27 s after the pulse began. Unbuffered, the denominator gains the cross term R1·C2 and its poles move to −0.382 and −2.618 per second; the top sags to 0.1751, undershoots to −0.8245, crosses zero at 2.002 s and overshoots to +0.0103. The slider is the second stage's impedance relative to the first's.

    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.

    part 4 · instruments
  5. With a 0.30 µF coupling capacitor and a stiff driver the transformer's damping is 0.289: the pulse's tail crosses zero 8 times in sixty units of L/Rₗ. computed by solving, not by drawing, marched and checked against the closed form to 2.3e-4: a pulse two L/Rₗ long through a series coupling capacitor into a 1 mH magnetising inductance with 100 Ω reflected across it, from a stiff driver. The capacitor is 0.30 µF, x = CRₗ²/L = 3. The poles are those of 3.000e-10·s² + 1.000e-5·s + 1, a damping of 0.2887 and a natural frequency of 9.19 kHz. The top ends at 0.128 of the high-frequency gain, and after the pulse the tail crosses zero 8 times, with lobes above a thousandth of the pulse, in the sixty units drawn. The slider is the capacitor.

    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.

    part 5 · instruments

All series