The band counted in rise times

The band a turns ratio holds over found that a transformer is the one common model that is wrong at both ends and right in the middle: its response falls away below a lower edge set by the magnetising inductance and above an upper edge set by the leakage, and the flat part between is the turns ratio times what the coupling and the winding resistances leave. What coupling buys, and where measured the two edges against the coupling and found the asymmetry that every winding engineer lives with: a tighter coupling moves the upper edge by a factor of 168 and the lower by 1.083.

Those are statements about sinusoids. A transformer is very often asked to carry pulses instead — a gate drive, a trigger, a data isolator, a pulse-forming network’s output — and the questions are then about time: how fast the edge rises, how far the top sags, what happens after the pulse ends. Each of the two edges has a time it becomes, and the time is not an approximation to the edge; for a band as wide as a decent transformer’s it is the edge, read in the other domain.

Two edges, two times #

The transformer is the band essays’ own: two 10 mH windings coupled at 0.99, half an ohm of copper in each, a 50 Ω source and a 50 Ω load, and 100 pF of winding capacitance across the output. Its band, bisected on the solved network, runs from 400 Hz to 81.3 kHz with a plateau of 0.4901 V per volt — the 1:1 ratio times what the coupling and the divider of source, copper and load leave.

The pulse response comes two ways. The netlist’s transfer function is recovered as a ratio of polynomials, its poles found, and the step response written as a sum of exponentials, one per pole; a pulse is a step followed by an equal and opposite step one pulse width later. And the same netlist is marched forward in time by the trapezoidal rule with the source switched on and off. The two agree to a third of a per cent of the plateau at every point of the hero figure, which is the march’s own error at that step size.

The pulse does three things, and each belongs to one part of the band. It rises — in 4.03 µs from 10 to 90 per cent — which is the upper edge: for a first-order corner the 10–90 per cent rise is $\ln 9/(2\pi f_H) = 0.35/f_H$, 4.31 µs here, and the leakage inductance against the loop’s resistance is a first-order corner at this load, as which picture sets the upper edge established. It droops — to 0.784 of its plateau after 100 µs — which is the lower edge: the magnetising inductance in parallel with the loop’s resistance is a time constant $\tau_L = 1/(2\pi f_L) = 398$ µs, and $e^{-100/398} = 0.778$. And when it ends, it swings below zero.

The droop is the lower edge #

The droop is the most useful number a pulse transformer’s data sheet can give, and it follows from the lower edge alone. A pulse of width $T$ ends with its top at $e^{-T/\tau_L}$ of the plateau, so it has drooped by $1 - e^{-T/\tau_L}$, which is $2\pi f_L T$ while that is small. Measured on the solved pulses at six widths from a twentieth of $\tau_L$ to twice it, the droop follows the law to within the few tenths of a per cent that the rising edge’s own tail leaves in the narrowest pulse; the slowest pole of the transfer function, which is what actually sets the droop, sits at $1/(2\pi f_L)$ of the bisected lower edge to two per cent. So a designer who wants no more than 10 per cent of droop on this transformer may send a pulse of 41.9 µs; for 1 per cent, 4.0 µs — which is the transformer’s own rise time, so that at a 1 per cent droop specification this transformer can carry nothing longer than its own edge.

The same arithmetic governs any coupling that blocks direct current, and the corner that says nothing about an edge found it in a coupling capacitor, where the low corner and the sag of a slow pulse are one fact read twice. The transformer’s magnetising inductance is that capacitor’s dual: it shunts the signal instead of blocking it in series, and its time constant is an inductance over a resistance rather than a resistance times a capacitance, but the pulse sees the same exponential.

The rise is the upper edge, and coupling buys only that #

The rise is proportional to $1 - k$, because the leakage inductance is: from 30.8 µs at a coupling of 0.9 to 4.03 µs at 0.99 and 422 ns at 0.999, a decade of rise for every decade of shortfall. Where the band is wide — above a ratio of about three hundred between its edges, at couplings of 0.995 and better — the rise is $0.35/f_H$ to within five per cent. Where the band is narrow, at 0.9 and 0.95, the solved rise reads shorter than the law, and that is an artefact worth understanding rather than a better transformer: with the two edges only twenty times apart the top has started to droop before the edge has finished rising, so the response never reaches its plateau and the 90 per cent point is reached early on a lower peak.

Meanwhile the droop’s time constant hardly moves: 414 µs at 0.9, 396 at 0.999. The magnetising inductance is $k^2 L$, which changes by twenty per cent over the whole range, and the loop’s resistance does not change at all. In time, the asymmetry what coupling buys, and where found in frequency is complete: an hour spent on the winding buys a faster edge and not a single microsecond of flat top.

The figure of merit is the band ratio #

A pulse transformer is good if it can carry a pulse that is long compared with its edge. The natural measure is the longest pulse whose top droops no more than some tolerance, divided by the rise time — how many rise times of flat top the transformer offers.

Written out, the longest pulse at 10 per cent droop is $\tau_L \ln(1/0.9) = \ln(1/0.9)/(2\pi f_L)$, and the rise is $0.35/f_H$, so their ratio is

$$\frac{\ln(1/0.9)}{2\pi \times 0.35}\cdot\frac{f_H}{f_L} = 0.048\,\frac{f_H}{f_L}.$$

The figure of merit is the band ratio, scaled by a constant that depends only on the droop allowed. Measured on the solved pulses it is 10.4 at a coupling of 0.99 — ten rise times of flat top — and 98.8 at 0.999, on the law to five per cent wherever the band is wide. At 0.9 the transformer manages 1.4: a 10 per cent droop allows a pulse barely longer than its own edge, which is to say it cannot carry a pulse at all in any useful sense.

Read as a specification, the law says how good a transformer has to be before anybody winds it. A transformer that drives a power transistor’s gate and must pass a pulse twenty times as long as its own edge, drooping no more than 10 per cent, needs a band ratio of $20/0.048 \approx 420$ between its edges — on this winding, a coupling of about 0.995. One that must hold a 1 per cent droop over the same pulse needs ten times the ratio, because the allowed droop enters through its logarithm, $\ln(1/0.99) = 0.010$ against $\ln(1/0.9) = 0.105$: a tighter flat top is bought almost linearly with band, and band on a transformer is bought with coupling.

Since the upper edge goes as $1/(1 - k)$ and the lower edge is fixed, the figure of merit goes as $1/(1 - k)$ too. Every factor of ten in the coupling’s shortfall is a factor of ten in how many rise times a pulse can last. The inductance that is a shape found the leakage set by the window’s geometry — by interleaving and by insulation thickness — so in a pulse transformer those are the variables that decide how long a pulse it can carry.

Reading a band off a pulse #

The correspondence runs both ways, which makes it a measurement method. A bench without a network analyser can find both edges of a transformer from one pulse and an oscilloscope. The rise gives the upper edge, $f_H = 0.35/t_r$; the droop over a known width gives the lower, $f_L = -\ln(1 - \text{droop})/(2\pi T)$. On the hero figure’s pulse those readings return 86.8 kHz and 387 Hz against the bisected 81.3 kHz and 400 Hz — the upper reading 7 per cent high and the lower 3 per cent low, and both for the same reason: the two edges are only two hundred times apart, so the top is already drooping while the edge is still rising, which shortens the measured rise and leaves the droop to be measured from a plateau the pulse never quite reaches.

Two cautions come with the method. The oscilloscope’s own rise adds to the transformer’s in quadrature, and the instrument’s own rise time found the quadrature rule optimistic for every instrument with more than one pole, so the edge measured on a 4 µs transformer with a 1 µs instrument is not quite the transformer’s. And the droop has to be read on a pulse long enough to droop measurably and short enough not to approach saturation; on this transformer a 100 µs pulse droops 22 per cent, which is easy to read, and the core’s volt-second limit is far away at a volt. On a power transformer driven at its rated voltage the same pulse might not be.

Why the constant is 0.35, and when it is not #

The number 0.35 is $\ln 9/(2\pi)$: a single-pole step reaches 10 per cent at $\ln(10/9)$ time constants and 90 per cent at $\ln 10$, and the difference is $\ln 9 = 2.197$ time constants, which divided by $2\pi$ turns a time constant into a frequency. It is exact for one pole and only for one pole. One step, computed twice put a step through networks of known poles and read their rise against their band. A transformer with its winding capacitance is a three-pole network, but at 50 Ω the third pole, the capacitance against the load, is two decades above the upper edge and slows the rise by a fraction of a per cent; what moves the solved 4.03 µs away from the law’s 4.31 is the other end of the band, the droop that has already begun before the edge finishes. Where the edges are a thousand times apart, at a coupling of 0.998, the two agree to two per cent, and the narrower the band the more the edge and the droop share the same microseconds.

The volt-seconds come back #

The last feature of the pulse is the one that most often surprises: after the pulse ends the output does not return to zero, it goes negative.

When the source switches off, the output falls by the full plateau — the transformer passes the falling edge as faithfully as the rising one — from a top that has already drooped. So it lands below zero by exactly the amount the top lost: 0.1900 V after a 200 µs pulse whose top fell from 0.4901 to 0.2984. Then it decays back towards zero with the same 398 µs time constant.

The depth of the backswing is the droop, so the two specifications are one: a transformer allowed 10 per cent of droop will also put a 10 per cent negative excursion on its output after every pulse, and a circuit downstream that treats any negative voltage as a fault — a gate driver’s undervoltage lockout, a logic input’s clamp diode — has to be designed for it. The excursion’s duration is the droop’s time constant, 398 µs here, which is long against the pulse and so outlasts it by several widths.

That is not a fault of this transformer. It is a theorem: a transformer has no gain at direct current, so the integral of its output over all time is the direct-current gain times the integral of its input, which is zero. Every volt-second the pulse delivers above zero is taken back below it afterwards. Measured over three milliseconds the two areas are 76.90 and 76.84 V·µs, equal to eight parts in ten thousand — what remains of the tail beyond the window. Physically, the magnetising inductance has been charged by the pulse, and the current it carries at the end of the pulse must go somewhere: it drives the load the other way until the field it stored is gone.

The backswing matters wherever pulses repeat. A train of pulses at a duty cycle $D$ settles to a direct-current level such that the areas balance cycle by cycle, so the tops ride at $1 - D$ of the plateau’s height above the baseline and the gaps sit at $-D$ — a transformer-coupled pulse train has no fixed zero, which is why such circuits need a clamp or a restorer when the absolute level matters. It is the same balance a boundary in volt-seconds required of the core’s flux: the core will not tolerate a net volt-second per cycle, and neither will the output.

When the upper edge is a resonance #

All of this used a 50 Ω load, at which the upper edge is the first-order leakage corner. Where the band goes entirely found that past a few hundred ohms of load there is no band at all: the leakage resonates with the winding capacitance, lightly damped, and a well-coupled transformer peaks at 3.03 times its turns ratio. In time that is a ringing edge — an overshoot and a decaying oscillation at the leakage resonance on every transition — and the rise time is then the quarter period of the ring rather than $0.35/f_H$. The figure of merit stops being the band ratio because there is no band ratio; what replaces it is the ring’s decay against the pulse length. The ring a blocking capacitor buys met a ring of the same family from the other side of the band, where a capacitor in series with a transformer’s drive stops its flux from walking and resonates with the magnetising inductance instead.

The numbers #

The k = 0.99 transformer with a band of 400 Hz to 81.3 kHz carries a pulse with a 4.03 µs rise — $0.35/f_H$ is 4.31 — and a droop of time constant 398 µs = $1/(2\pi f_L)$, so a 10 per cent droop allows 41.9 µs and a 1 per cent droop 4.0, its own rise time. Coupling moves the rise from 30.8 µs at 0.9 to 422 ns at 0.999 and the droop’s time constant from 414 to 396 µs. The longest pulse at 10 per cent droop over the rise is 0.048·$f_H/f_L$, 10.4 at 0.99 and 98.8 at 0.999, growing as 1/(1 − k). After a 200 µs pulse the output swings to −0.190 V, and the areas above and below zero are equal to 0.08 per cent.

Still open: a pulse that saturates, a train with a duty cycle, and a ringing edge priced #

A pulse long enough to saturate. The droop here is linear: the magnetising current rises steadily under the pulse. A long pulse carries the core towards saturation, where the magnetising inductance falls and the droop accelerates; the pulse width at which droop stops being exponential is a volt-second boundary read in time, and the hysteretic core of the area a curve cannot have can march it.

A train at a duty cycle. The backswing makes a repetitive pulse train settle to a baseline set by its duty cycle. How long it takes to settle, how the droop of each pulse and the baseline shift combine, and what that does to a threshold detector on the far side, is the calculation a data isolator’s designer needs.

The resonant edge. At a light load the edge rings at the leakage resonance. The damping resistor that makes that ring critically damped, and what it costs in rise time and in droop, is the time-domain version of the load at which the band goes, and it has a closed form nobody has drawn here.