The half a switch keeps

The half that never arrives measured the energy account of a capacitor charge and found two things. Charging from a step loses exactly $\tfrac12CV^2$ in the resistance, the same 12.50 microjoules through ten ohms and through a hundred kilohms; and driving the same network with a ramp instead takes the loss down as $2\tau/T$, with a fitted exponent of −0.994 and no floor beneath it at all.

It also recorded what it had not done. The result is a statement about the source, and the machinery it was measured on takes an arbitrary drive — which had only ever been given a step and a ramp. A switch is not a drive. The supply on the far side of a switch sits at five volts the whole time and what moves is the path: a channel conductance climbing from nothing to one over the on-resistance across the gate’s edge. Whether the charge-loss result survives that was left open.

It does, completely, and the completeness is the finding.

The calibration, which is the rung below unchanged #

The apparatus is one microfarad, five volts and a series resistance, marched forward in time with the current rebuilt from the element law and three energies integrated: $\int i^2R\,dt$ into the resistance, $\tfrac12Cv^2$ into the capacitor, $\int v_s i\,dt$ out of the supply. Nothing reads a formula, so the account closing is the whole check on the integration.

Everything below is quoted against that. When a new drive is tried, the first thing asked of it is that it reproduce the step, and every number here comes with the account gap that says whether it may be believed — the same discipline one step, computed twice applies to a waveform rather than to an energy.

The instrument, and the two routes it is checked by #

A conductance that changes in time is the one thing the network assembly cannot hold. Every element it stamps is a constant, and the whole reason the poles of a circuit can be recovered from its own matrix is that the matrix is affine in s — which a channel resistance moving during the transient is not. So the switch is marched as a scalar equation instead, $C\,dv/dt = g(t)(v_s(t) - v)$, with both $g$ and $v_s$ taken as functions and the three energies integrated from the element laws exactly as before.

That is a second implementation of the same physics, and a second implementation is only worth having if it is made to agree with the first where both apply. Held at a constant conductance it must reproduce the network solved as a matrix, and there is no arithmetic in common beyond the trapezoidal rule itself: one inverts a five-by-five system once per step, the other divides two scalars. On a step they give 1.00000002 and 1.00000009 of ½CV² — 7.4 parts in a hundred million apart. On a ramp they agree to eight figures at one, ten and a hundred time constants of duration, and both agree with the closed form $2(\tau/T)[1 - (\tau/T)(1 - e^{-T/\tau})]$, which sees neither a netlist nor a marcher: 0.735758884 against 0.735758882 at one time constant, 0.0198000000 against 0.0198000000 at a hundred.

Two routes that agree establish the assembly. Three, where the third is an expression derived on paper, establish the marching as well — and the switch result below is a number no third route exists for, which is precisely why the two that do exist are pinned to one that is not negotiable. It is the same arrangement what a network answers makes for a solved circuit, moved from a steady state to an integral over a transient.

The staircase, computed rather than argued #

Charge the capacitor in N equal risers instead of one, letting each settle. The usual argument is two lines: each riser is a charge of $V/N$ from a source held constant, so each costs $\tfrac12C(V/N)^2$, and there are N of them, so the total is $\tfrac12CV^2/N$.

The argument is right and it is worth having the measurement anyway, because the argument’s hidden premise is the whole of the next section. Marched, with sixteen time constants of dwell, N times the loss comes out 1.000004, 1.000002, 1.000001, 1.000001, 1.000000, 1.000000 for one, two, four, eight, sixteen and thirty-two risers. That is a law rather than a trend: the fitted exponent over the six is −1.00000, and the residual four parts in a million is the trapezoidal rule’s, not the circuit’s.

The premise is the word settle. Each riser costs $\tfrac12C(\Delta V)^2$ only if the capacitor actually reaches the level it was raised to before the next riser arrives. It never quite does, and how nearly it does is the only free parameter in the whole scheme.

What the dwell has to be #

A riser held for d time constants leaves the capacitor short of its level by $e^{-d/\tau}$ of the riser. The next riser therefore has to move that much further, and a charge transfer costs the square of the distance it moves, so the excess is twice $e^{-d/\tau}$ to first order. The first riser starts from a capacitor that is genuinely empty and the last dwell is a settling rather than a dwell, so N − 1 of the N carry it:

$$\frac{\text{loss}}{\tfrac12CV^2/N} - 1 \;\approx\; \frac{N-1}{N}\,2e^{-d/\tau}$$

That expression never sees a netlist and the marching never sees the expression, and they agree to better than two per cent from six time constants of dwell to twelve. Which puts a number on the word settle: the 1/N law is a hundredth below the truth at a dwell of 5.272 time constants, bisected on the marched runs rather than read off the sweep, against $\ln(2\cdot\tfrac{31}{32}/0.01) = 5.267$ from the expression.

At one time constant of dwell the real loss is 2.106 times what 1/N predicts. That is not a correction to a law, it is a different law, and the reason is that a staircase whose risers do not finish is not a staircase at all — it is a coarse ramp, and it obeys the ramp’s arithmetic instead.

Where the two laws cross, which is the design number #

Put the total time on a budget instead — a switched-capacitor stage has a clock period and a precharge sequence has a contactor rating, and neither has an unlimited number of stages — and the two arithmetics meet.

Inside a fixed T, the staircase promises $\tfrac12CV^2/N$ and the ramp of the same duration delivers $\tfrac12CV^2 \cdot 2\tau/T$. Those are equal when the dwell $T/N$ is two time constants, and the measurement lands at 1.980, which is two less the ramp’s own end correction of one part in the hundred time constants budgeted. Below fifty steps in that budget the staircase is doing better than the ramp and the extra step is worth building; above it the staircase has become the ramp and every further step buys nothing, arriving at 0.020056 against the ramp’s 0.0198000 — a per cent and a quarter apart, entirely.

The shape of the family is worth reading as well as its crossing. Every curve in it starts on the 1/N line and leaves it downward-flattening, and none of them ever goes below the ramp: no arrangement of risers inside a fixed time beats the drive that fills the same time smoothly, which is asserted at every step count drawn. A staircase is an approximation to a ramp made out of parts a switch can produce, and the arithmetic says it is a good approximation exactly while its risers finish and a redundant one afterwards.

That is the answer to “how many stages”, and it is the sort of number that is normally chosen by custom. A precharge sequence has three stages because three is a common number of stages. The arithmetic says the useful count is the total time divided by two time constants, and past it the right move is to stop adding switches and lengthen the drive.

The switch, which is the other half of the same product #

Now the question the rung below left open. Every drive above moves the source voltage. A switch does not: the supply is at V before the gate moves, during, and after, and the thing that ramps is the conductance of the path.

Both are factors of the same current — $i = g(t)\,(v_s(t) - v)$ — so it is not obvious that they should behave differently at all, and the whole essay turns on the fact that they do.

A switch’s edge buys nothing whatever. Not a little; nothing measurable at nine figures across five decades of edge rate. Against a source ramped over the identical thousand time constants through the identical kilohm, which costs 1.998 × 10⁻³ of ½CV², that is a factor of 500.5 for the same transition time — and the transition time is the only thing the two drives have in common. And the on-resistance does not appear in the switch’s answer either: with an edge of a hundred time constants the loss is 1.000000000 of ½CV² through ten ohms, through a kilohm and through a hundred kilohms, with the energy account closing to seven parts in $10^{14}$.

The reason is two sentences and neither of them mentions the path. The capacitor ends at V, so it holds $\tfrac12CV^2$ however it got there. The charge delivered is $CV$ and the source was at V for every instant of the delivery, so the source gave $CV^2$ — and the difference of two quantities that do not depend on $g(t)$ cannot depend on $g(t)$. A time-varying conductance changes the shape of the current and the duration of the transfer, and neither is in the answer.

It is worth watching the same thing instant by instant, because the total is the memorable part and the distribution is what explains it. At the first moment the capacitor is empty and the whole supply stands across the switch, whatever the channel is doing. The power there is $g V^2$, so a channel that has barely opened dissipates very little — and takes correspondingly longer to move the charge, in exactly the proportion that leaves the product alone. That is the same cancellation the rung below found across four decades of fixed resistance, arriving now within a single transient as the conductance sweeps through those decades on its way up. A ramped source has no such cancellation available to it, because it never lets the voltage across the resistance be large in the first place: with a drive of twenty time constants the capacitor tracks the source about five per cent behind, so the resistance sees a twentieth of the supply and dissipates a four-hundredth of the step’s peak power for twenty times as long.

So the free parameter is the source voltage and only the source voltage. Slowing a gate edge, widening a switch, choosing a better part, adding a series resistor: all of these move where the heat appears and how long it takes to appear, and none of them changes how much there is. Only a drive that is somewhere other than V while the charge is moving does that — which is what a ramp is, and what a staircase is, and what a switch by itself can never be.

The on-resistance, which comes back the moment the drive is finite #

There is a reading of the rung below that says the resistance is simply absent from this subject, and it is wrong in a way that only shows up when the time axis stops being measured in time constants.

A ramp of duration T costs $\tfrac12CV^2 \cdot 2RC/T$ once T is long, and R is right there in the numerator. The rung below’s sweep held T in units of $\tau$, so a change of resistance moved the axis by exactly as much as it moved the answer and the dependence cancelled out of the picture. Held in seconds it does not cancel: at one ohm the millisecond drive costs 1.998 × 10⁻³ of ½CV², at a hundred ohms 0.180, and the proportionality holds with a fitted exponent of 0.9961 across the decades below the knee. Both edges are bisected — 100.0 Ω where the proportionality is a tenth out, 33.08 kΩ where the answer is within a hundredth of the step’s — so the band the resistance is visible in is two and a half decades wide and has an edge at each end, in the way a band rather than an edge collects.

Which reconciles the two halves. The resistance is absent from a step’s loss and from a switch’s, because in both cases the source is at V throughout. It is present, linearly, in the loss of any drive whose duration is fixed independently of the circuit — and a real design’s durations are fixed by a clock, so that is the usual case rather than the exception.

What the model does not contain #

A switch that is only a conductance. The channel here goes from zero to $1/R_\text{on}$ and has nothing else in it. A real one has an off-capacitance across it that conducts when the channel does not, and an off-resistance that does the same more slowly.

A gate that is a voltage rather than a schedule. The conductance here is ramped linearly in time because that is the simplest statement of a finite edge. A real channel’s conductance is a function of the gate voltage, which is itself the output of a driver charging a gate capacitance — this same circuit, one level down — so the true $g(t)$ is neither linear nor independent of the load. None of that matters to the result, which is the useful part: the argument never uses the shape of $g$, only that it is positive and that the charge finishes. A linear ramp is drawn because something has to be drawn.

No overlap loss. The energy measured here is what the capacitor’s charge costs. A switch carrying a load current while its voltage falls dissipates that as well, and that loss does grow with the edge — so a slower edge is free in the accounting on this page and expensive in the accounting next to it. The two are added, not traded, and the whole practical content of “buys nothing” is that the first term does not fall to pay for the second rising. The heat a recovery leaves behind is the same sum taken over a diode’s turn-off and finds nine tenths of it in the transistor rather than the diode, which is the general lesson: the place a switching loss is named after is often not the place it is dissipated.

No inductance. The path is purely dissipative, so every result here belongs to a charge that crosses a resistance rather than one that resonates through an inductor. A resonant transfer moves the energy into a magnetic field and back, and what is left in the resistance is then set by the quality factor — a different mechanism with a different answer, and the one the energy is in the gap accounts for from the storage side.

And a linear capacitor. A ceramic’s capacitance falls with the voltage across it, so $\tfrac12CV^2$ is not what one holds; the honest quantity is $\int v\,dq$ over the actual curve, and the ratio of loss to stored is no longer exactly one.

Where these numbers are spent #

A switched-capacitor stage. Every clock edge charges a sampling capacitor from one voltage to another and pays $\tfrac12C\Delta V^2$, whatever the switch is made of — which is why a resistor made of a clock has a dynamic dissipation set by the clock rate and the capacitor ratio and by nothing in the switch at all. The measurement above says a bigger switch buys nothing there and a slower one buys nothing either.

A gate driver. A power device’s gate is a capacitance charged from a step, so the driver dissipates as much as the gate stores on every turn-on. The result on this page says that the fashion for slowing the gate edge — done for other and good reasons, among them the emission that a millimetre and a half of skew turns into common mode — recovers none of it, and that the schemes which do recover it all move the drive rather than the switch: a staircase of intermediate supply rails, or a resonant transfer.

And a precharge. A capacitor bank charged through a contactor loses as much as it stores, in the contactor, and the first cycle, which no steady state contains is that event at scale — a reservoir charged from a mains peak, with 32.4 amperes in the first conduction against a repetitive 1.23. Staging it is the standard answer and this page prices the staging: one Nth for N stages while each stage settles, and nothing further once the dwell is under two time constants. That is also why the inductance that limits, and lifts finds series resistance limiting the peak without reducing the energy, while the inductive repair reduces both — it changes the drive rather than the path.

The pattern across all three is the one the total that has no resistor in it draws from the noise side. A resistance that sets how fast and not how much cancels out of every total taken over the whole event, and it comes back the moment something else fixes the duration.

The number worth carrying #

One Nth for N risers, exactly, while each is held longer than 5.272 time constants; the ramp’s $2\tau/T$ below that; and a switch’s edge worth 1.00000000 of ½CV² however slowly it is made to move.

The habit that goes with it is a question to ask of any energy result with a component missing from it, and it is the question the half that never arrives asked about a resistor and did not ask about a switch. Ask which factor of the current the component is in. The loss of a charge is decided by the voltage the source held while the charge crossed, and a resistance — or a conductance, or a switch, or a wire — is the other factor. Changing it changes how long and how hot and never how much. The only repair that reduces the energy is one that puts the source somewhere other than its final value while the charge is moving, and the whole of the design space is the shape of that drive.