The energy two capacitors cannot share

The half that never arrives and the half a switch keeps measured the cost of charging a capacitor from a supply, and found it to be the mismatch between a source held at V and a capacitor that is not, dissipated somewhere along whatever path joins them. The swing a diode keeps found the one exception that path can make: an inductor and a diode, which stop the current at the top of its swing and leave the capacitor holding what a resistor would have burned.

All three had a supply in them. Take it away and the oldest puzzle in the subject is left: a charged capacitor, an empty one, and a wire. Charge is conserved, so the final voltage is fixed by arithmetic. Energy is not, and half of it goes missing even when the wire is given no resistance at all. This essay measures where it goes, what an inductor and a diode do to it, and why every switched-capacitor converter is that same missing energy charged for once a cycle.

The arithmetic, and the two ways the network is solved #

Charge first, because it is the one thing that cannot be argued with. A capacitor $C_1$ at voltage $V$ holds a charge $C_1V$. Connect it to an empty $C_2$ and wait. The charge has nowhere else to go, so it divides between them until both sit at one voltage, and that voltage is

$$V_f = \frac{C_1 V}{C_1 + C_2}.$$

The energy before is $\tfrac12C_1V^2$. The energy after is $\tfrac12(C_1+C_2)V_f^2$, which works out to $\tfrac12C_1V^2 \cdot C_1/(C_1+C_2)$. The difference, the energy that has gone somewhere, is

$$E_{\text{lost}} = \tfrac12 C_1 V^2 \cdot \frac{C_2}{C_1 + C_2} = \tfrac12\,\frac{C_1C_2}{C_1+C_2}\,V^2,$$

which is the energy of the two capacitors in series charged to the initial difference between them. Equal capacitors lose exactly half.

The measurement does not take that on trust. The circuit is built as a netlist — the charged capacitor represented by its standard equivalent, an empty capacitor in series with a step source of five volts — and solved from its own poles, exactly as the earlier charging measurements were. The current is the difference of the two solved node voltages across the resistance, divided by the resistance, and the heat is $\int i^2R\,dt$ integrated over sixteen thousand samples. Nothing in that route knows the formula; it has to arrive at it.

It does. At equal capacitors the two voltages meet at 2.500 volts and the resistance has taken 50.00 per cent of the initial energy.

The resistance is not in the answer #

The famous version of the puzzle asks what happens as the resistance goes to zero, since a perfect wire cannot dissipate anything and yet the arithmetic insists that half the energy leaves. The measurement can at least say what happens on the way there.

Across four decades of resistance the loss does not move. A tenfold smaller resistance makes the sharing ten times faster and the peak current ten times larger, so the instantaneous power is ten times larger for a tenth of the time, and the product is unchanged. This is the cancellation the supply-charging measurements found, with the supply removed: it was never a property of the source, only of a passive path joining two unequal voltages.

Putting an inductance in the path does not help either, and that is the result that dissolves the paradox rather than merely restating it. With 100 microhenries in series and 0.3 ohms of resistance the pair is an LC circuit with a quality factor near a hundred: the charge sloshes back and forth between the two capacitors for hundreds of cycles, each swing a little smaller than the last. When it finally settles, the resistance has taken 0.200, 0.500 or 0.800 of the energy, exactly as it did with no inductor. At zero resistance it would never settle at all. The energy that “goes missing” would oscillate between the capacitors and the inductor’s field for ever, and the textbook’s final state — both capacitors at $V_f$ — would never be reached.

So the puzzle’s premise is the problem. A circuit with no resistance and no inductance is not a circuit with zero loss; it is a circuit whose equations have no solution, because a finite charge would have to move in zero time. Give the path any physical property at all — resistance, inductance, or the radiation a real loop of wire has — and the energy has somewhere to go. Make the path a pure inductance and it goes nowhere, but then the capacitors never come to rest at the shared voltage. Either way, the final state the arithmetic describes is reached only by dissipating the difference.

The size ratio is the whole of it #

Since the resistance and the inductance have both left the answer, the only parameter remaining is the ratio of the two capacitances, and it decides everything.

The two curves cross at equal capacitances and swap their roles on either side. Emptying a small charged capacitor into a large empty one throws away almost everything: a capacitor ten times the size of the charged one ends up holding 9.1 per cent of the energy, at a tenth of the voltage. Emptying a large charged one into a small empty one costs almost nothing, because the voltage barely falls.

That asymmetry is the practical form of the result, and it is better stated in terms of voltages than sizes. The loss is half the series capacitance times the square of the initial voltage difference. So what is cheap is sharing between two capacitors whose voltages are already close, and what is expensive is sharing across a large difference — whatever the sizes. Topping up a reservoir that sits a tenth of a volt below a flying capacitor costs very little, and that observation, read the right way round, is the whole design rule for a charge pump.

It also explains a family of effects elsewhere in this subject that look unrelated. The noise a clock does not make and the sampled-data measurements around it lean on charge redistribution between a sampling capacitor and its load; the charge injected by a switch, which the capacitance a third switch moves measures, is charge shared between a channel and a hold capacitor. In each case the voltage that results is fixed by conservation of charge, and the energy is something the circuit pays without being asked.

An inductor and a diode make it an exchange #

The inductor that rang for hundreds of cycles did not reduce the loss, because it was allowed to ring. The swing a diode keeps showed what happens when it is not: a diode in series stops the current at its first zero, and the energy that was in flight is kept rather than returned.

In the two-capacitor circuit the first half-swing of current carries the charge from the full capacitor to the empty one and then past the shared voltage, because the inductor’s stored energy keeps pushing after the voltages have met. At the top of that swing the current is zero, and with a diode in series it stays zero. With no resistance the second capacitor ends at $2C_1V/(C_1+C_2)$ and the first at $V(C_1-C_2)/(C_1+C_2)$, and no energy has been lost at all — it has only moved.

Equal capacitors are the striking case. The first ends at zero and the second at the full five volts: the charge has been handed over entirely, not shared. At a loop Q of fifty the measured exchange leaves the first capacitor at 0.08 volts and the second at 4.92, with 96.9 per cent of the energy moved. A resistor alone could never move more than a quarter of the energy into the second capacitor, and only at equal sizes; the inductor and diode move all of it at equal sizes and still most of it at a ratio of four, where the transfer falls to 64 per cent in the ideal case and 62.0 per cent at Q = 50.

This is the principle of the bucket-brigade and of every resonant switched-capacitor converter: the charge is moved between capacitors through an inductance timed by the circuit’s own resonance, and the switch opens at the current’s zero. What it buys is the difference between a quarter and almost all.

A charge pump is the sharing loss, billed every cycle #

The most common circuit made of shared charge has no inductor at all. A charge-pump doubler charges a flying capacitor to the supply voltage in one phase, then stacks it on top of the supply and connects it to an output reservoir in the next. Every phase is a sharing event between two capacitors at unequal voltages, and the loss measured above is paid on every one of them.

The march here is not the residue route: the switches change the circuit twice a cycle, so the two phases are marched as two coupled equations with the trapezoidal rule, every joule counted as it leaves the supply, enters the load or heats a switch. The pump is run until a whole cycle moves its output by less than a part in a hundred billion, and the energy account is then taken over forty cycles.

The result is an identity that holds to four parts in ten million at every load and both clocks: the efficiency is the output voltage divided by twice the input. The reason is charge conservation, once more. For every unit of charge the load takes, the supply delivers exactly two — one to fill the flying capacitor and one more while that capacitor is stacked on top of it — and it delivers both at the full input voltage. So the input energy is $2V_{\text{in}}q$ and the output energy is $V_{\text{out}}q$, and their ratio is $V_{\text{out}}/2V_{\text{in}}$ however the losses are distributed.

That makes a charge pump, in the one respect that matters for efficiency, a linear regulator: whatever voltage it drops below its ideal output, it burns. A doubler delivering 7 volts from a 5-volt supply is 70 per cent efficient, exactly as a series-pass regulator from a 10-volt rail would be. The only question left is how much it drops, and that is where the clock comes in.

The clock decides the drop, until the switches do #

At a slow clock each phase runs to completion, so each phase is a full charge-sharing event, and the energy it loses is half the series capacitance times the voltage difference squared. Per cycle the flying capacitor carries a charge $q = I/f$, its voltage moves by $q/C$, and the loss works out to make the pump look like a resistance of $1/(fC)$ in series with an ideal doubler.

The measurement follows $1/(fC)$ to four figures at a kilohertz, which is the same law a resistor made of a clock uses to make a resistance out of a capacitor and two switches. It is not a coincidence: a switched capacitor between two voltages is a sharing event per cycle, and its average current is $fC$ times the voltage difference, which is a resistance by definition. The charge pump’s output resistance is the same element, seen from its output.

Raise the clock and the phases stop completing. The flying capacitor’s voltage then moves very little in each phase, the sharing loss falls with it, and what is left is the resistance of the switches the charge has to pass through. In this doubler each phase carries the whole cycle’s charge through two switches, at a current of about $2q/T$, so the heat is $4q^2 \cdot 2R_{\text{sw}}/T$ per cycle and the output sees $8R_{\text{sw}}$ — 0.80 ohms for 0.1-ohm switches. The measured plateau is 0.80 ohms. The two limits cross near 1.25 megahertz, and above that a faster clock buys nothing but switching loss of a kind this model does not include.

So the design rule reads straight off the figure. A charge pump’s efficiency is set by its drop, its drop is set by whichever is larger of $1/(fC)$ and the switches’ resistance, and a designer chooses the clock to sit near the corner — fast enough that sharing is cheap, no faster than the switches justify.

A pump sized from the two laws #

The two measured laws are enough to size a pump without any further simulation, and doing it once shows how little freedom there actually is. Suppose a circuit needs 9 volts at 20 milliamps from a 5-volt rail. An ideal doubler gives 10, so the pump may drop one volt, which at 20 milliamps is an output resistance of 50 ohms at most.

The efficiency is fixed before any part is chosen: 9 volts out of an ideal 10 is 90 per cent, by the identity measured above, and no choice of capacitor, switch or clock will change it. A designer who wants better than 90 per cent from this specification does not want a better pump; they want a pump with a different ideal ratio, or an inductor.

The resistance budget then divides between the two terms. With 0.1-ohm switches the plateau is 0.80 ohms, which is negligible against 50, so the whole budget can go to the sharing term: $1/(fC) \le 50$ ohms, or $fC \ge 0.02$. A one-microfarad flying capacitor needs a 20-kilohertz clock; a 100-nanofarad one needs 200 kilohertz. Both sit far below the 1.25-megahertz corner, so both are in the region where the clock buys efficiency directly, and either is a reasonable design — the choice between them is a choice between a larger capacitor and more switching events.

The output reservoir is decided separately, by ripple. During the phase in which the flying capacitor is being recharged the output capacitor carries the load alone, so its voltage falls by about $I/(2fC_o)$ each cycle — five millivolts for a 100-microfarad reservoir at 20 kilohertz. That is a sawtooth on a direct voltage of exactly the kind the direct voltage that is a sawtooth measured on a rectifier, with the clock in place of the mains, and a reservoir’s own series resistance adds a step to it in the way two capacitors that are one found. The difference is that a pump’s clock is a free choice, so its ripple is a design parameter rather than a consequence of the mains frequency.

What the model leaves out #

Every switch here is a resistance that turns on and off instantly, and that leaves out the cost that dominates a real pump at high frequency: the energy to charge each switch’s own gate electrode and the parasitic capacitance at every switched node, paid once a cycle and so rising in proportion to the clock. That is a charge-from-a-supply loss of exactly the kind the half that never arrives measured, and it is why a real pump’s efficiency peaks at some clock and then falls, rather than flattening at the plateau drawn here.

The capacitors are ideal. A real flying capacitor has series resistance, which adds to the switches’ and raises the plateau, and a ceramic one loses capacitance under bias, which raises $1/(fC)$ at exactly the voltage it is being asked to hold. The second effect is the subject of the capacitance that is not one number.

And the two-capacitor exchange used an ideal diode. The voltage a real junction drops was measured for the supply-charging case in the swing a diode keeps, and it matters more here, because there is no supply voltage to make it small against: two capacitors at a few volts exchanging charge through a silicon junction lose most of the advantage to the junction itself.

The number to keep #

Two capacitors joined by any passive path lose $\tfrac12 \cdot C_1C_2/(C_1+C_2)$ times the square of their initial difference, which is $C_2/(C_1+C_2)$ of what the charged one held — through a resistance of 1 ohm or 10 kilohms, with or without an inductor ringing between them. An inductor and a diode turn the sharing into an exchange, and equal capacitors then swap 96.9 per cent of their energy at a loop Q of fifty. A charge pump is the sharing loss billed once a cycle, so its efficiency is $V_{\text{out}}/2V_{\text{in}}$ exactly and its output resistance is $1/(fC)$ until the switches’ own $8R_{\text{sw}}$ takes over.

Still open: the gate electrode charge, the ladder network of capacitors, and the resonant pump #

The gate electrode charge as a fourth term. Adding each switch’s gate capacitance as a charge drawn from the supply every cycle would put a rising line under the plateau drawn above, and the clock at which the two cross is the optimum a real pump is designed for. It is computable on the same marched circuit.

A pump of several stages. A Dickson or series-parallel converter shares charge through a ladder network of flying capacitors, and the slow-switching resistance becomes a sum over the stages weighted by the square of the charge each carries. Whether the measured output resistance matches that weighting, and which topology wins at a given ratio, is a measurement rather than an argument.

The resonant pump. The exchange measured above is what a resonant switched-capacitor converter does deliberately: a small inductance in each charge path, the switches opened at the current’s zero. Its output resistance should fall below the plateau a resistive pump reaches, and how far it falls at a given Q is the number that decides whether the inductor is worth its size.