The swing a diode keeps

The half that never arrives measured the energy account of charging a capacitor from a step: the supply gives $CV^2$, the capacitor keeps $\tfrac12CV^2$, and the other half is heat in the resistance — the same 12.50 microjoules through ten ohms as through a hundred kilohms. The half a switch keeps then took every lever a designer holds on a switch, its on-resistance and the shape of its gate edge, and found that none of them moved the half either. The free variable, it concluded, is the voltage of the source while the charge is moving, and only that.

That conclusion has a gap in it that the switch result points straight at. Every circuit measured so far put a resistance between the source and the capacitor and nothing else. Real charging paths contain inductance, sometimes by accident and sometimes on purpose, and an inductance does something no resistance can: it holds energy while current flows, and gives it back. Whether that changes the half, and under what condition, is what this essay measures.

The answer has two parts and they point in opposite directions. An inductor alone changes nothing at all. An inductor with a diode in series changes nearly everything, and the reason is a single instant in the waveform.

The circuit, and the two ways it is solved #

The apparatus is the one the two earlier measurements used, with one element added: a one-microfarad capacitor, a five-volt step, a series resistance, and now a one-millihenry inductor between them. The characteristic impedance $\sqrt{L/C}$ is 31.6 ohms and the resonant period is 199 microseconds, and the resistance is chosen to set the circuit’s quality factor $Q = \sqrt{L/C}/R$, which is the one dimensionless number this network has.

The capacitor voltage is solved from the netlist’s own poles: the network is assembled, its determinant rooted, and the step response built from the residues. The current is not computed from a formula either. It is the difference between the source voltage and the solved voltage at the far end of the resistance, divided by the resistance — two node voltages the solver returned, subtracted. From those two traces three energies are rebuilt from the element laws: what the supply delivered, $V\int i\,dt$; what the capacitor holds, $\tfrac12Cv^2$; and what the resistance burned, $\int i^2R\,dt$, integrated by the trapezoidal rule over sixteen thousand samples.

The three must close, and that closure is the check on everything below. A diode is modelled in the only way an ideal diode acts on this circuit: the solution is cut at the first instant the current reaches zero, and from then on nothing moves. The zero is interpolated between samples rather than taken at the next grid point, so the stopping voltage is not a property of the sampling.

An inductor alone changes nothing #

Start with the inductor and no diode, since it is the case most likely to be guessed wrong. The inductor slows the current’s rise, so the peak current is lower; it lets the capacitor overshoot, so the voltage swings above the supply and back; and a low-loss circuit rings for many cycles before it settles. Any of those could plausibly change the account.

None of it does. Across four decades of inductance and more than two decades of Q, every settled charge loses $\tfrac12CV^2$ to within 1.4 parts in a hundred million, which is the integration’s own residue. The reason is the same accounting argument that made the resistance disappear, with one more term in it. Once the circuit has settled, the current is zero, so the inductor holds nothing: its energy $\tfrac12Li^2$ starts at zero and ends at zero. The capacitor ends at V and holds $\tfrac12CV^2$. The supply sat at V the whole time and pushed a total charge $CV$ through it, so it delivered $CV^2$. What is left over is $\tfrac12CV^2$, and nothing in that argument knows what the path contained.

What the inductor does change is the route to that answer. In the overdamped runs it spreads the current out, which lowers the peak power in the resistance and lengthens the time it is dissipated for, in exactly the proportion that leaves the product alone — the same cancellation the half a switch keeps watched happen inside a single transient. In the underdamped runs it does something stranger: on each swing above five volts the current reverses, and charge flows back into the supply. The supply has to absorb energy on those half-cycles, then push it out again on the next. The resistance takes its toll on every crossing, forward and back, and the toll adds up to the same half. A Q of a hundred is a hundred cycles of paying a small fee, and a Q of 0.3 is one large one.

This is worth stating carefully because it is the reason the half is so robust. The loss is not “in” the resistance in any way that depends on what the resistance is. It is in the mismatch between a source that sits at V and a capacitor that does not, and a passive path that eventually brings them together has to dissipate that mismatch somewhere, however long it takes.

Where the energy is while it moves #

Before adding the diode it helps to watch one swing in detail, because the diode acts on a single instant and the instant only makes sense once the energy’s whereabouts are visible.

On the way up, the current rises as the supply pushes against a capacitor that is still nearly empty, and most of what the supply delivers in that first quarter-period goes into the inductor’s field. Half-way up the swing, at 49.7 microseconds, the inductor holds 0.86 of $\tfrac12CV^2$ — very nearly as much energy as a plain charge to five volts would leave in the capacitor. In the second quarter-period the capacitor’s voltage passes the supply’s, the current starts to fall, and the inductor empties itself into the capacitor, driving it well past five volts.

At the top of the swing the current is zero. That is the instant that matters. At that instant the inductor holds nothing — every joule it took on the way up has gone on into the capacitor — and the capacitor holds more than it ever will again, because without a diode the current now reverses and the capacitor starts handing charge back to the supply.

The account at that instant is the whole argument of this essay. The supply has delivered 3.709 units of $\tfrac12CV^2$. The capacitor holds 3.439. The resistance has burned 0.270. The capacitor is charged to almost twice the supply, and the charge has cost about seven per cent of what it put there.

A diode keeps the swing #

A diode in series does one thing to this circuit: it refuses the reverse current. So at the top of the first swing, where the current falls to zero and would reverse, it simply stops. The capacitor is left at its peak, and the circuit has no way to give anything back.

The peak is a quantity the step response already has a name for. A second-order circuit’s step overshoots by $d = e^{-\pi\zeta/\sqrt{1-\zeta^2}}$, where the damping ratio $\zeta$ is $1/2Q$, so the capacitor is left at $V(1 + d)$. At a Q of ten $d$ is 0.8545 and the diode holds the capacitor at 9.272 volts, which is what the solver returned to the fourth figure. The supply delivered a charge of $CV(1+d)$ at voltage V, the capacitor stores $\tfrac12CV^2(1+d)^2$, and the difference is the loss:

$$E_{\text{lost}} = CV^2(1+d) - \tfrac12CV^2(1+d)^2 = \tfrac12CV^2\,(1 - d^2).$$

That expression is a third route to the number, sharing nothing with the other two — it never sees a netlist, and the solved waveform never sees it.

The two agree at every Q from 0.6 to 300 to within the integration’s tolerance. And the curve has the shape that makes resonant charging worth building: at high Q, $d \approx 1 - \pi/2Q$, so the loss as a fraction of what is kept falls as $(1-d)/(1+d) \approx \pi/4Q$. Ten times the quality factor is a tenth of the loss. Unlike a resistor alone, the loss here has no floor except the resistance itself, and the resistance can be made small.

There is a subtlety in reading the vertical axis. The loss is quoted in units of $\tfrac12CV^2$ — what a plain charge to the supply voltage stores — but the diode-stopped capacitor holds nearly four times that. So the fair comparison is not loss against loss but loss against energy kept, and it is that ratio which falls as $\pi/4Q$: 7.85 per cent at a Q of ten, against one hundred per cent for a resistor at any Q at all.

The same comparison at the same final voltage #

The diode-stopped charge ends at nearly twice the supply, which makes a direct comparison with a resistor charge slightly unfair to the resistor: they are not charging the capacitor to the same place. The comparison that a design actually faces is the other way round — a capacitor that must reach five volts, and a choice of how to get it there.

Choose the supply as $V/(1+d)$, so that the swing lands exactly on five volts. At a Q of ten that is a 2.70-volt supply, the capacitor arrives at 5.000 volts after half a resonant period, and the loss is 7.85 per cent of what is stored. A resistor from a five-volt supply loses one hundred per cent of what it stores, at any resistance. The resonant path does in a single half-period what the half a switch keeps needed a thirty-two-step staircase to do, and it does it without any staircase: the inductor builds the gentle drive out of a fixed source.

It is also the same physics that makes a rectifier’s leakage inductance dangerous rather than useful. The inductance that limits, and lifts found a rectifier whose reservoir sits 29 per cent above the peak of its own supply, permanently: that is this diode-kept swing, arriving uninvited, with the rectifier’s own diodes doing the stopping.

The supply voltages are themselves informative. As Q rises the required supply falls toward half the target — 2.52 volts at a Q of a hundred for a five-volt charge. A lossless resonant charge from a supply of V reaches exactly 2V, so to reach V it wants exactly V/2. This is the voltage doubling that flash-lamp chargers, pulse-forming networks and the charging chokes of radar modulators have relied on for most of a century, and it is the reason a “resonant charging” stage is quoted with a voltage gain of two.

The half-sine is nearly a ramp #

That leaves the question of why the resonant charge is so cheap, and the answer connects it to the ramp the half that never arrives measured. There the loss of a ramped source of duration T fell as $2\tau/T$, because a slow ramp keeps the capacitor only slightly behind its source and the resistance never sees more than a sliver of the supply. Resonant charging also takes a definite time — half a resonant period, $\pi\sqrt{LC}$ — and the comparison worth making is between the two at the same duration.

They are close, and the ratio between them settles to a constant: $\pi^2/8$, or 1.2337. The measured ratio at 314 time constants is 1.2376, still drifting down toward it as the duration grows. The constant has a clean origin. A ramped source charging a capacitor drives an almost constant current once the start-up transient has passed, and for a fixed charge in a fixed time a constant current is the waveform with the least $\int i^2\,dt$. A resonant charge drives a half-sine of current instead. Carrying the same charge in the same time, a half-sine has an $\int i^2\,dt$ larger than the constant’s by exactly $\pi^2/8$ — the square of the form factor that what a meter multiplies by uses to translate an average into a root-mean-square.

So the inductor is not a trick that sidesteps the ramp’s law. It is a way of building a ramp — or something within twenty-four per cent of one — out of a supply that only knows how to sit at a fixed voltage. A switch cannot do that, which is the whole of the half a switch keeps: a switch in front of a fixed supply presents the capacitor with a step however its edge is shaped. An inductor in front of the same switch converts the step into a current that rises and falls smoothly, and the diode decides when it ends.

A real diode, and the voltage scale it brings #

Everything so far used an ideal diode, which has no voltage across it while it conducts. That was not an innocent choice. The circuit above has no voltage scale of its own — every loss is a fixed fraction of $CV^2$ whatever V is — and a real junction brings one.

While a real diode conducts it is close to a constant voltage $V_f$ in the loop. That lowers the effective step to $V - V_f$, and it dissipates $V_f$ times the charge that passes, which here is the capacitor’s whole final charge. With an ideal diode the efficiency at a Q of twenty is 96.2 per cent at every supply voltage, because nothing in the loop has a voltage scale. A 0.7-volt silicon junction at a five-volt supply takes 81 per cent of all the loss for itself and brings the efficiency down to 82.8 per cent — worse, in fact, than the resistance it is protecting the capacitor from. A 0.3-volt Schottky drop does better, and both fade toward the ideal only once the supply reaches tens of volts.

This is the same kind of boundary how many straight lines a diode is draws around a junction’s own model: the ideal diode is a perfectly good description above some voltage and a misleading one below it. Resonant charging has historically lived at hundreds or thousands of volts, where a junction’s drop is a rounding error. At logic-level supplies it is not, and the diode is replaced by a synchronous switch timed to open at the current’s zero — which brings its own on-resistance back into the loop, by an amount the half a switch keeps measured for a channel.

What the model leaves out #

The inductor here has no resistance of its own and no capacitance between its turns. The first is easy to include, because it simply adds to R and lowers the circuit’s Q: a real millihenry with an ohm of winding resistance in this circuit is a Q of about thirty, and the curve above says what that costs. The second matters more than it looks. The inductor that is a capacitor measures the self-resonance a winding’s own capacitance produces, and an inductor used near its self-resonance stops being one. A resonant charger’s inductance is chosen for a resonance at the charging period, far below the coil’s own; a coil chosen carelessly for its inductance alone can put the two closer than they should be.

The diode is ideal in a second way that is easy to miss: it switches off instantly at the current’s zero. A real junction stores charge while it conducts and conducts backwards briefly before it recovers, which is the subject of the diode that conducts backwards. In this circuit the reverse current flows out of the capacitor it has just charged, so it costs energy twice: once as the charge lost and again as heat in the junction while it recovers. At high Q and a slow resonance the recovery is a small fraction of the period; at a fast resonance it is not.

Finally, the supply is ideal. A real supply has its own impedance, which sits in series with R and lowers Q — and at switch-on it has a history the steady state forgets, as the first cycle, which no steady state contains measures for a rectifier — and it may not tolerate the reverse current at all in the no-diode case. That is the practical reason a resonant charger always has its diode: without one, the swings above the supply push current back into a source that may be a rectifier and cannot take it, which is a different failure from the energy account above but the same waveform.

The number to keep #

An inductor in a charging path changes nothing about the half by itself: twenty-four settled charges across four decades of inductance lose $\tfrac12CV^2$ to 1.4 parts in a hundred million. Stop the current at its first zero with a diode and the capacitor keeps the swing, ending at $V(1+d)$, and the loss becomes $\tfrac12CV^2(1-d^2)$ — a fraction $\pi/4Q$ of what is kept at high Q, 7.85 per cent at a Q of ten. Measured at equal durations, a resonant charge costs $\pi^2/8$ of a ramp’s loss, because a half-sine carries a charge with that much more heat than a constant current does. A real diode’s drop is the boundary: at five volts a silicon junction burns more than the resistance does.

Still open: the switch that opens at the zero, the energy that comes back, and the second capacitor #

A synchronous switch in place of the diode. A transistor opened at the current’s zero has no forward drop but has an on-resistance, and the timing of its opening is a control problem: open it early and current is cut while the inductor still holds energy, which appears as a voltage spike; open it late and the reverse current flows. The cost of a timing error of a given fraction of the period is computable on this same circuit, and it is the number that decides how good the timing must be.

Recovering the energy instead of keeping it. The no-diode circuit pushes energy back into the supply on every swing above V. A supply that can absorb it — a battery, or a large capacitor — makes the ringing a lossy oscillation between two stores rather than a charge, and adiabatic logic is built on doing that deliberately: charge a node through an inductor and later recover most of the energy rather than dumping it. How much comes back at a given Q is the companion number to the one measured here.

The second capacitor. A charged capacitor emptied into an empty one through a resistance loses energy by the same mismatch argument, and through an inductor and a diode it need not. That is a circuit with no supply in it at all, and it is the basis of every charge pump and of the switched capacitor that a resistor made of a clock treats as a resistance, so the energy of a charge shared between two capacitors is the natural next measurement.