The half that never arrives

There is one result in this subject where the component value visibly does not appear in the answer, and it is worth an essay for that reason alone. Charge a capacitor from a step of V through a resistance and the resistance dissipates exactly $\tfrac12CV^2$ — the same energy the capacitor ends up holding, so half of everything the source delivered is gone, and it is gone whatever it was lost in.

It is usually presented as a curiosity about resistors. It is not about resistors at all. It is a statement about the source, and once that is seen the result stops being a curiosity and becomes the reason a whole class of circuits is built the way it is.

The account, and how it is closed #

A microfarad, five volts, and a resistance. Nothing is assumed about where the energy goes: the network is marched forward in time, the current is rebuilt from the element law as the source voltage less the capacitor voltage over the resistance, and three integrals are taken.

$\int i^2R\,dt$ is what the resistor turned into heat. $\tfrac12Cv^2$ at the end is what the capacitor is holding. $\int v_s i\,dt$ is what the source delivered. The account has to close, and it does: the third is the sum of the first two to better than a part in $10^6$ for a step and a part in $10^{12}$ for every ramp, which is the check that the integration is right, since nothing in the computation reads a formula for any of the three.

For a step, the numbers are 12.5 µJ into the capacitor, 12.5 µJ into the resistor, 25 µJ out of the source. The measured ratio of the loss to $\tfrac12CV^2$ is 1.00000 — and it is 1.00000 through ten ohms, a hundred, a kilohm, ten kilohms and a hundred kilohms, which is the flat part of every curve in the figure.

Why the resistance is absent #

The reason is worth stating because it explains the whole rest of the essay.

The final state is fixed: the capacitor ends at V whatever the resistance is, so it ends up holding $\tfrac12CV^2$ whatever the resistance is. The charge delivered is also fixed — $Q = CV$ — and the source held V the entire time, so the source delivered $QV = CV^2$ whatever the resistance is. The difference between two quantities that do not depend on the resistance cannot depend on the resistance.

Both halves of that argument lean on the source being a step. The source delivered $CV^2$ because it was at V for the whole of the charge, including at the beginning, when the capacitor was at zero and the entire V was across the resistor. That is the moment the energy is lost: at the first instant the current is $V/R$ and the resistor is dropping the full supply, and every joule crossing it is a joule that never reaches the capacitor.

The resistance changes how long that moment lasts and how large the current in it is, in exactly compensating proportions.

What the ramp does #

So make the source not a step. Ramp it linearly to V over a time T and repeat the measurement.

The result has a closed form, derived rather than fitted: the current during the ramp is $(V/RT)(1 - e^{-t/\tau})$ and decays afterwards, and integrating $i^2R$ over both stretches gives

$$\frac{2\tau}{T}\left[1 - \frac{\tau}{T}\left(1 - e^{-T/\tau}\right)\right]$$

times $\tfrac12CV^2$. That expression never sees a netlist and the marcher never sees the expression; they agree over four decades of ramp duration to better than three parts in a thousand at the worst point and a part in a million over most of it.

At $T \ll \tau$ it goes to one, which is the step and the textbook answer. At $T \gg \tau$ it goes to $2\tau/T$ — the fitted exponent over the slowest three points is −0.994 — and there is no floor. Charge slowly enough and the dissipation goes to zero.

Where the joules actually go, instant by instant #

The total is the memorable part and the distribution is the part that explains it.

At the first instant the capacitor is at zero, the whole supply is across the resistor, and the instantaneous power into the resistor is $V^2/R$ while the power into the capacitor is nothing at all. At the end the capacitor is at V, the current is zero, and both powers are nothing. In between the split moves continuously from all-resistor to all-capacitor, and it crosses over when the capacitor is at half the supply — one time constant times $\ln 2$, which is 0.693 τ.

Integrate each half and both come to $\tfrac12CV^2$. That the two halves are equal is the theorem; that they are distributed so differently in time is why the ramp works. The loss is concentrated at the start, where the voltage across the resistor is large, and a ramp’s whole effect is to never let that voltage be large: with a ramp of twenty time constants the capacitor tracks the source about five per cent behind it, so the resistor never sees more than a twentieth of the supply and its dissipation is a four-hundredth of the peak the step produced, for twenty times as long.

That is where the $2\tau/T$ comes from and it is worth having in the arithmetic form. In the long-ramp limit the current settles to a constant $CV/T$ — the charge divided by the time — so the loss is $I^2RT = (CV/T)^2RT = C^2V^2R/T$, which is $\tfrac12CV^2 \times 2RC/T$. A constant current for a long time is the cheapest way to move a given charge through a given resistance, and the ramp is the drive that produces one.

A disagreement that turned out to be arithmetic #

The marched result and the closed form did not agree at first, and the size of the gap is worth recording because of how ordinary the cause was.

The first version asked for forty time steps per time constant, which is generous for the settling and is not what the shortest ramps need. At a ramp of three hundredths of a time constant that rule put slightly more than one step inside the ramp itself, so the marcher was integrating a drive it had barely sampled, and the two routes came apart by three parts in a thousand at that end of the sweep.

Three parts in a thousand is exactly the size of thing that reads as physics. It is small enough to look like a real second-order effect in the closed form’s derivation and large enough to be worth explaining, and a plausible explanation is available — the expansion assumes the ramp is slow compared with nothing in particular, so perhaps it fails when the ramp is fast. That explanation is wrong and it is the one that would have been written.

What settled it is that the disagreement moved with the step count rather than with the ramp duration. A departure that changes when the numerics change and not when the physics does is a numerical departure, and the repair is to make the step follow the ramp rather than the settling. With that done the two routes agree to a part in a million over the whole sweep except at the very shortest ramp, where the residue is a hundredth of what it was.

What the resistance actually decides #

Which brings back the slider. The plateau does not move with the resistance; the knee does.

Losing only a tenth of $\tfrac12CV^2$ takes about twenty time constants, which is 200 µs through ten ohms, 20 ms through a kilohm and two seconds through a hundred kilohms. So the resistance decides nothing about how much a fast charge costs and everything about how long a cheap charge takes — and in a circuit that has to repeat the charge at some rate, that is the constraint that actually binds.

The direction of the trade is the opposite of the intuitive one. A larger series resistance is usually thought of as the lossy choice. Here it makes no difference to a step’s loss at all, and it makes a slow charge harder to achieve, so the useful move is a smaller resistance and a slower drive.

The staircase, which is the same result in the useful form #

A ramp is difficult to make. A staircase is not, and it gives the same benefit in a form a switch can produce.

Charge the capacitor in N equal steps, allowing each to settle. Each step is a charge of $V/N$ through the same resistance from a source held constant, so each costs $\tfrac12C(V/N)^2$ by the argument above — and there are N of them, so the total is $\tfrac12CV^2/N$. Marched rather than argued, the ratios come out 1.00000, 0.500003, 0.250190, 0.125001 and 0.0625007 for one, two, four, eight and sixteen steps.

The loss falls in inverse proportion to the number of steps. That is the principle behind charge-recycling logic, behind resonant gate drivers, and behind the several-stage precharge sequences that appear in front of large capacitor banks. None of them is doing anything to the resistance.

What the model contains and what it does not #

No inductance. The network is a source, a resistance and a capacitance, so every result here is about a purely dissipative path. A real charging loop has inductance in it, which changes the shape of the current and — at low enough resistance — makes the charge oscillatory, at which point the energy argument above still holds for the total but the waveform it holds through is a different one.

An ideal source. The source holds its voltage at any current, including the $V/R$ that flows at the first instant. A real one does not, which is a series resistance of its own and lands back inside the same result.

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

Where the number matters #

A switched-capacitor circuit’s power budget. Every clock edge charges a sampling capacitor from one voltage to another, and the loss is $\tfrac12C\Delta V^2$ per edge regardless of the switch’s resistance. Multiply by the clock rate and it is the whole dynamic dissipation. Making the switch bigger buys nothing at all.

Driving a gate. A power MOSFET’s gate is a capacitance charged from a step, so the driver dissipates the same energy as the gate stores every time it turns on. A datasheet’s gate charge times the drive voltage times the switching frequency is that number, and it is why resonant drivers exist.

And precharging anything large. A capacitor bank charged straight from a supply through a contactor loses as much as it stores, in the contactor. The staircase result is the standard answer, and its 1/N is the reason precharge circuits have the number of stages they have.

Worth putting numbers on the last one, because the quantity is not small and the arithmetic is the essay’s. A bank of ten millifarads charged to four hundred volts holds 800 joules and therefore dumps 800 joules into whatever charged it, in one event. Through a one-ohm precharge resistor the time constant is ten milliseconds and the whole of it is over in a tenth of a second, so the resistor is absorbing eight kilowatts on average over that tenth of a second — which is why a precharge resistor is a physically large object and why its rating is an energy rather than a power. Halving the resistance does not halve the energy. It halves the time and doubles the power, and the resistor is no better off.

Where the lost half is paid #

Half the energy of every charging step, independent of the resistance it is lost in, is a result three other essays depend on. The first cycle, which no steady state contains is where the whole of it arrives at once, in a reservoir being charged from nothing. The heat a recovery leaves behind is where it becomes one of the two positive feedback paths on a thermal node. A resistor made of a clock pays it thousands of times a second by construction, and The energy is in the gap is the magnetic version of the same accounting — where the energy is stored rather than where it is lost. One step, computed twice is the machinery all of it is measured on.

The gate #

The step’s loss is asserted to be $\tfrac12CV^2$ to three parts in $10^5$, at every resistance on the slider, which is the claim the essay is named for and the one that has to be exact rather than close.

The account is asserted to close — delivered equals lost plus stored — to a part in $10^6$, and the check on the ramps is a part in $10^8$. That is the only thing standing between a marched integral and a number that looks plausible.

The closed form is asserted against the marched result, over four decades of ramp duration, and the tolerance is three parts in a thousand because that is what the shortest ramp’s discretisation costs rather than what either route is worth.

And the asymptote is fitted rather than asserted: −0.994 over the slowest three ramps, which is the statement that there is no floor under the loss.

A loss with no resistance in it, twice #

Charging a capacitor from a step losing exactly as much energy as it stores, with the resistance absent from the answer — the same 12.5 microjoules through ten ohms and through a hundred kilohms — is one of two results in this collection where a resistance cancels out of a total, and the pair is worth reading together because the mechanism is the same.

The total that has no resistor in it is the other: a resistor’s noise density goes as $\sqrt R$ and the noise bandwidth of what it charges goes as $1/R$, so five decades of resistance give one total — 63.2762 microvolts, the square root of $kT/C$. Here the current goes as $1/R$ and the dissipation as $I^2R$, so the power goes as $1/R$ and the duration as $R$, and the product is a constant.

In both cases the resistance sets how fast and not how much, and in both cases the answer is a property of the capacitor and the excitation. Which is why the ramp result on this page is the interesting half rather than the step result: driving the network with something whose duration is a free parameter is the one way to change a quantity that no component value moves, and the loss then falls as two time constants over the ramp with no floor beneath it.

That has a reader in the digital field. Every logic transition is this circuit, and the absence of a floor is why a slower edge costs less energy — bounded, in practice, by the millimetre that becomes common mode and the rest of the timing budget rather than by anything in this measurement.

The other reader is the power field, where the same energy appears as a loss to be limited rather than as a cost per transition. The first cycle, which no steady state contains is this circuit at scale — a reservoir capacitor charged from a mains peak through whatever resistance is in the way — and its 32.4 amperes is what the arithmetic on this page produces when the capacitance is a thousand microfarads and the step is a hundred and seventy volts. That the resistance does not appear in the energy is why the inductance that limits, and lifts finds adding resistance limiting the peak and not reducing the energy at all, and why the inductive repair — which changes the shape of the drive rather than the resistance in it — is the one that reduces both.

Which is this essay’s result read as a design instruction rather than as a curiosity. The energy lost charging a capacitor is decided by the drive and not by the path, so every repair that acts on the path — a larger conductor, a lower resistance, a better switch — moves where the heat appears and not how much there is. Only a repair that acts on the drive reduces it. Which is the whole content of the ramp measurement: two time constants over the ramp, no floor beneath it, and the design parameter is a duration rather than a component.