The ring a blocking capacitor buys

Two coupling stages without a buffer found that a pair of resistor-and-capacitor coupling stages, however they load each other, can only spread their poles along the real axis: a network made of resistors and capacitors cannot ring. Its tail crosses zero once — the two zeros at the origin remove the waveform’s area and first moment, and one sign change is the least that balances both — and then returns to zero without oscillating. That essay closed with the arrangement where the guarantee fails, which is the one that matters most often in practice: a coupling capacitor in front of a transformer.

The capacitor is there for a reason that comes from the other end of this collection. The flux that walks drove a core with a square wave whose halves differed by one per cent and saturated it in 64 cycles, and the walk the core sees found a relaxation oscillator’s threshold noise doing the same thing slowly and at random. Both walks are the drive’s direct component accumulating in the core. A capacitor in series passes no direct current, so it removes the direct component and the walk with it. What it adds is a second reactance in a loop that already had one.

A high-pass with an inductance in it #

The transformer is reduced to what a pulse sees at its low-frequency end: its magnetising inductance, 1 mH, with the load reflected across it, 100 Ω. A coupling capacitor C sits in series between a driver of resistance $R_s$ and the transformer. Two elements block direct current — the capacitor in series and the inductance in shunt — so there are two zeros at the origin, exactly as for two resistive coupling stages, and the transfer function is

$$H(s) = \frac{s^2LC}{a\,LC\,s^2 + (CR_s + L/R_L)\,s + 1}, \qquad a = 1 + R_s/R_L.$$

The natural unit of time is $L/R_L$, ten microseconds, and the natural measure of the capacitor is $x = CR_L^2/L$, the ratio of the capacitor’s time constant against the load to the inductance’s: x = 1 is 0.1 µF. In those terms the damping ratio of the two poles is $(xR_s/R_L + 1)/(2\sqrt{ax})$, and nothing about it forces the poles to be real.

With a stiff driver and a 0.3 µF capacitor — x = 3 — the damping is 0.289 and the natural frequency 9.19 kHz. The march, by the backward difference the earlier coupling essays used, agrees with the closed form for complex poles to $10^{-3}$ of the pulse. The pulse’s top droops to 0.128 of the transformer’s high-frequency gain by its end, and after it the tail does what no resistive coupling ever did: it rings, crossing zero eight times with lobes larger than a thousandth of the pulse in the sixty units drawn.

The slider runs the capacitor from 0.03 µF to 1 µF. The smallest gives a damping of 0.913 and a tail with two crossings; 0.1 µF, damping 0.500 and five; 1 µF, damping 0.158, a slower ring of which only five crossings fit in the window. A larger capacitor rings more slowly and for longer, which is the first sign that the coupling capacitor’s usual sizing rule — make it large enough not to matter — is the wrong rule here.

Who decides whether it can ring #

The damping depends on the driver as much as on the capacitor, and the three cases a designer meets are different in kind.

Between equal resistances the damping is $(x + 1)/(2\sqrt{2x})$. It is least at x = 1, where it is exactly $1/\sqrt2$, and it is below one — the poles complex — only between $x = 3 - 2\sqrt2 = 0.172$ and $3 + 2\sqrt2 = 5.83$. A matched transformer can ring, but only mildly and only for a band of capacitors: at a damping of 0.707 each half cycle of the ring is $e^{-\pi} = 0.043$ of the one before, so after the crossing the two zeros force there is one shallow lobe a twentieth the size, and nothing a reader would see after it. Both ends of the window are the same physics: a very small capacitor makes the pair behave like a resistive coupling stage with the capacitor’s time constant, a very large one like the transformer’s own droop, and each is a first-order system with a slow second pole behind it.

From a driver of a tenth of the load resistance the floor drops to 0.302, and the window widens.

From a stiff driver the damping is $\sqrt{L/C}/(2R_L) = 1/(2\sqrt x)$ — 0.500 at x = 1, 0.158 at x = 10, 0.050 at x = 100 — and there is no floor at all. The driver’s resistance was what damped the series resonance of the capacitor and the inductance; a driver with none leaves only the load across the inductance, and the load damps less the larger the capacitor. Gate drivers, line drivers, the output stages of switching converters are all stiff drivers, and all of them feed transformers through blocking capacitors.

The floor has a one-line reason. With equal resistances the two damping terms in the denominator’s middle coefficient are $CR$ and $L/R$ — the capacitor’s time constant against the driver and the inductance’s against the load — and the damping is their sum over twice the geometric mean of the capacitor’s and inductance’s product, scaled by $\sqrt a$. A sum of two positive numbers is never less than twice their geometric mean, so the damping is never less than $1/\sqrt a = 1/\sqrt2$, and it reaches that only when the two terms are equal, at x = 1. The driver’s resistance is doing half of the damping’s work at every capacitance, and taking it away removes exactly the term that was holding the damping up as the capacitor grew.

The droop is the capacitor’s #

The reason designers make the capacitor large is the droop: a pulse through a transformer sags as the magnetising current builds, and the obvious expectation is that a series capacitor adds its own sag to the transformer’s. From a stiff driver it does something stranger.

At the instant of a step the inductance carries no current — its current can only grow — so all of the load’s current flows through the capacitor, which charges at $1/(R_LC)$, and the output falls by exactly that rate. The initial droop from a stiff driver is the capacitor’s, $1/(R_L C)$, and not the transformer’s $R_L/L$; the inductance enters the response only in its second term, through the curvature. The march measures 3.30, 1.000, 0.334 and 0.100 per unit of $L/R_L$ at x = 0.3, 1, 3 and 10, against 1/x. With source resistance the inductance does contribute at first order — at x = 1 from a matched driver the rate is 1.000 and at x = 10 it is 0.550, the extra being the source resistance carrying the inductance’s current.

So the designer’s lever against droop, from a stiff driver, is the capacitor alone: ten times the capacitance gives a tenth of the initial droop. And the same tenfold capacitance cuts the damping by $\sqrt{10}$, to 0.158 from 0.500. The two requirements pull the one part in opposite directions, and nothing about the inductance changes the trade.

The inductance does return, in the second term, and for long pulses it is the larger. The step’s curvature at its start is $(1/R_LC)^2 - 1/LC$, which in units of $L/R_L$ is $1/x^2 - 1/x$: for any capacitor larger than x = 1 the inductance makes the top bend downwards faster as the pulse goes on. Over a pulse of width w the droop is about $w/x + (1/x - 1/x^2)\,w^2/2$, and the second term overtakes the first once w exceeds $2x/(x-1)$ — 2.22 units at x = 10, and a little over two for any large capacitor. So for a pulse longer than about twice $L/R_L$, a large blocking capacitor has handed most of the droop back to the transformer, which is where it was before the capacitor was added, and kept the ring. Making the capacitor larger still buys almost nothing for such a pulse: its own term shrinks, the transformer’s does not.

Every half cycle, another crossing #

Two resistive coupling stages crossed zero once. The coupled transformer crosses as often as its ring stays above whatever level a reader can see.

Counted until the ring has fallen below a thousandth of the pulse, over a window of twenty-five of each capacitor’s own ring periods — a larger capacitor rings more slowly, so a fixed window would undercount it — the tail crosses zero once at x = 0.1, where the damping is 1.58 and the poles are real, and then 2, 5, 9, 14, 22 and 35 times at x = 100. The march, over the same windows at x = 1 and 10, finds the same counts to within the one lobe that peaks on the threshold.

The first crossing is owed to the two zeros at the origin and would be there with any damping; every further one is a half cycle of the inductance and capacitor ringing. Their number grows roughly as one over the damping, because each half cycle shrinks by a factor of $e^{-\pi\zeta/\sqrt{1-\zeta^2}}$, and a lightly damped pair needs many of them to fall by a thousand.

The contrast with the resistive pair is the whole of the inductance’s contribution. Both networks have two zeros at the origin and both remove the waveform’s area and first moment. The resistive pair satisfies those two constraints with the least possible oscillation; the coupled transformer satisfies them too, and then rings on top, because a series resonance with too little damping has its own business that the constraints do not forbid.

Reading the ring on a bench #

The ring is also a measurement, and a convenient one. Its frequency is the series resonance of the blocking capacitor and the magnetising inductance, $1/(2\pi\sqrt{aLC})$ — 9.19 kHz for 0.3 µF and 1 mH from a stiff driver, 5.03 kHz for 1 µF — and its decay per cycle gives the damping. A designer who sees a few-kilohertz oscillation on the tail of every burst from a MOSFET’s gate-drive transformer has, in that oscillation, the magnetising inductance (if the capacitor is known) and the effective damping resistance, without disconnecting anything. The tail’s first crossing, the one the two zeros force, comes first and is the same whatever the damping; the crossings after it count the damping.

The same reading tells a designer which side of the trade a circuit is on. A tail that crosses once and returns is a capacitor small enough, or a driver resistive enough, that the pulse’s droop is the cost; a tail that rings for a dozen crossings is a capacitor sized for flatness from a stiff driver, and the ring is the cost. Neither is wrong. What the crossings figure adds is that the count can be predicted from two numbers on the schematic, before the circuit is built.

What the capacitor is for, measured #

None of this is an argument against the capacitor. It is there because a transformer driven with any direct component walks its flux to saturation, and the capacitor stops that.

A square drive with 55 per cent duty has a direct component of a tenth of its amplitude. Into the bare transformer from a stiff driver it walks the flux by 0.40 units of volt-time a period — four microvolt-seconds at one volt — and keeps walking, exactly as the magnetics essays found for a fixed imbalance from a stiff source, until, as the walk that stops found, a winding’s resistance halts it. With a 1 µF capacitor in series the flux settles: the capacitor charges towards the drive’s average, the voltage across the inductance loses its direct component, and over the last twenty periods the period-end flux varies by less than three thousandths of a unit. How it settles is the series resonance again, ringing with a damping of 0.158 from the first edge until the capacitor has caught up.

So the blocking capacitor converts an unbounded walk into a bounded ring. That is a good exchange, and it has terms. The ring’s amplitude after a change in the drive’s duty is set by the size of the change and the damping; a drive whose duty is modulated — a regulator’s pulse width moving with the load — keeps exciting it. And the flux during the ring overshoots the level it settles at, by the same factor a step response overshoots its final value, so the core’s headroom has to cover the ring as well as the settled offset. The spread a winding resistance holds found a winding’s resistance bounding the walk at a stationary spread; the capacitor bounds it at zero, and buys the ring with the difference.

Sizing the capacitor, with both constraints in view #

The two figures that constrain the capacitor from opposite sides give a designer a range rather than a rule. From a stiff driver the initial droop is 1/x per unit of $L/R_L$, so a pulse of width w droops by about $w/(xL/R_L)$, and a droop of five per cent over a two-unit pulse needs x of at least 40. At x = 40 the damping is 0.079, and the tail rings for a few dozen crossings. Halving the droop to two and a half per cent doubles x and cuts the damping by $\sqrt2$, to 0.056.

Two ways out of the trade are visible in the damping figure. A driver with some resistance raises the floor: at a tenth of the load resistance the damping cannot fall below 0.302 at any capacitance, and the cost is the driver’s own dissipation and a smaller pulse at the load. Or the damping can be added where it does no harm to the pulse — a resistor across the capacitor, which passes the direct component slowly and lets the walk return in proportion, or a snubber across the transformer, which costs power at every edge. Each is the capacitor’s ring being paid for in some other currency, and the figures above say how much of it there is to pay for.

What it leaves out #

The leakage inductance and winding capacitance. The transformer here is its magnetising inductance and nothing else, which is right for the slow end of a pulse — the droop and the tail — and wrong for its edges, where leakage inductance and winding capacitance make a second, much faster resonance. The two rings are separated by the ratio of magnetising to leakage inductance, usually a few hundred, and do not interact.

A core that saturates. The inductance is linear. A core driven near its limit has an inductance that falls with the flux, so the series resonance’s frequency rises and its damping changes during the ring; whether the ring then helps or hinders the settling is a march with the core’s own curve in it.

Still open: the digital high-pass, a duty cycle that moves, and the capacitor that is also a sensor #

The same arithmetic in software. A recording system that removes drift with a recursive digital high-pass puts zeros at the origin without any inductance, and a second-order digital filter can be given complex poles deliberately, for a steeper corner. Whether that choice puts a ring into every event the system records — the coupled transformer’s behaviour reproduced in arithmetic — is the question this essay leaves for a filter with no parts at all.

A drive whose duty is modulated. A regulator’s pulse width follows its load, so the drive’s average moves continuously and the blocking capacitor’s ring is excited continuously. The flux then carries a component at the load’s frequency, amplified near the series resonance, and a load that happens to vary near that frequency drives the core harder than any fixed imbalance would.

The capacitor’s voltage as a measurement. The blocking capacitor settles at the drive’s average, so its voltage is a direct reading of the drive’s imbalance — the quantity the hysteresis essays measured as a walk. A circuit that watches that voltage and trims the duty to null it is a loop around the ring, and its stability is the series resonance’s damping, read as a margin.