Two capacitors that are one

A reservoir capacitor is often two capacitors. The value a supply needs is not always a value somebody stocks, the can that holds it may be too tall for the enclosure, and two smaller parts spread the heat and the ripple current over more surface. A designer who does it usually has a vague sense of a second benefit as well: paralleling halves the series resistance, and series resistance is bad.

The resistance that lowers the ripple marched a thousand-microfarad reservoir with its own series resistance and found that “bad” needs qualifying. The resistance lowers the ripple a little up to seventeen milliohms and raises it after, and it lowers the peak diode current throughout. The resistance that belongs in the winding moved the same resistance to the transformer’s side and found the peak current indifferent to where it sits and the ripple very much not.

A pair of capacitors changes the capacitor’s resistance and nothing on the winding side, so it is the capacitor-side question asked again. What the pair adds is a second question the single part could not raise: when two capacitors share one job, which one does the work, and which one gets hot.

A pair is one part, exactly #

Start with the claim that has to be true before any comparison means anything. Two identical capacitors, each a capacitance $C$ behind a resistance $r$, connected in parallel between the same two nodes. Each branch has the impedance $r + 1/j\omega C$ at every frequency, the two are equal, and two equal impedances in parallel are one of half the value:

$$\frac{1}{2}\left(r + \frac{1}{j\omega C}\right) = \frac{r}{2} + \frac{1}{j\omega\,(2C)}.$$

That is a capacitance of $2C$ behind a resistance of $r/2$, at every frequency and therefore for every waveform, including the rectifier’s pulses. Nothing about the argument needs the circuit around the pair to be linear: the diodes see only the terminal behaviour of whatever is across the output, and the terminal behaviour is identical.

The march does not take this on trust. The rectifier was marched with a bank of two 500-microfarad parts at 60 milliohms each, every part a capacitance and a resistance of its own in the netlist, and with one 1000-microfarad part at 30 milliohms. The ripple is 1.330303 volts both times, the crest factor 12.4903 both times, and the root-mean-square current in each half of the pair is 0.2187 amperes — and 0.4374 amperes in the single part, exactly twice, which is the exact equivalence showing up in the branches as well as at the terminals. The figure checks the agreement to a microvolt.

This is worth checking because the netlist for the pair is genuinely different — two more nodes, two more unknowns at every step, a different matrix — and a march that discretises a capacitor with a companion model could have broken the symmetry in principle. It does not, and so everything that follows can be said about one part.

One curve, read at two places #

So the question “is a pair better than one part of the same total capacitance?” is not a comparison of two circuits. It is one circuit — a capacitance of $2C$ with some resistance — read at two resistances. The pair of $C$ parts at $r$ each sits at $r/2$. The single $2C$ part sits wherever its own resistance puts it.

The figure’s two curves are therefore the same curve, the pair’s drawn at half the abscissa. Put them on one axis of effective resistance and the ten marches this essay made collapse to a single table for two thousand microfarads:

The rows at 5, 15, 50, 150 and 500 milliohms are pairs; the others are single parts. They interleave into one smooth column in every quantity, which is the equivalence again, from the other direction.

Read down it and the shape is the one the thousand-microfarad sweep found, shifted. The ripple has a flat bottom around ten milliohms, 704 millivolts to within a millivolt from five to thirty, and rises past that. The crest factor falls monotonically, from 17.4 at five milliohms to 6.6 at an ohm.

What halving the resistance buys #

Now the pair against the single part is a step of a factor of two to the left along that table, and what the step does depends entirely on where it starts.

Past the minimum it lowers the ripple. At a hundred milliohms a part the single part is at 728 mV and the pair at 709 — 2.7 per cent better. At three hundred it is 867 against 758, and at an ohm 1.334 volts against 1.013. The further out the parts are, the more the halving is worth.

Near the minimum it does nothing. At thirty milliohms a part the single part gives 705.2 mV and the pair 704.2. At ten milliohms a part the two are 704.1 and 704.2, and the pair is fractionally worse. A pair of good low-resistance parts has no ripple advantage over one part of the same series at all.

At every value it raises the peak current. The crest factor at a hundred milliohms is 12.61 for the single part and 14.41 for the pair, so the pair draws 14 per cent more peak current from the transformer. At thirty milliohms it is 15.48 against 16.51, and at ten, 16.93 against 17.40. Halving the resistance never lowers the peak, because the peak is the one quantity the resistance acts on with a single sign.

And it raises the mean output a little. 15.954 against 15.926 volts at a hundred milliohms a part, which is the conduction ending nearer the crest of the sinusoid, the same mechanism run backwards.

So the pair is a trade and the terms are fixed by the table. Where the parts are well past the minimum — ordinary general-purpose electrolytics at a few hundred milliohms each — a pair buys real ripple and a little mean output for a higher peak current. Where they are near it, the pair buys nothing in ripple and still costs the peak current, which sizes the transformer.

The fair comparison is not at equal resistance #

The table compares a pair and a single part of the same resistance per part, and that is not the comparison a designer actually makes. A 2000-microfarad can in a given series is physically larger than a 1000-microfarad can in the same series, and its series resistance is usually lower — how much lower is on its data sheet, and it is seldom exactly half.

That is where the one-curve view pays. There is no need to march the specific pair against the specific part. Look up both resistances, halve the pair’s, and read both off the same column. If the larger part’s resistance is below half the smaller part’s, the single part sits to the left of the pair, and it has the lower ripple past the minimum and the higher peak current everywhere. If it is above half, the pair does. The comparison is decided by one inequality and the direction of each consequence is fixed.

What the curve cannot supply is the part a designer is really paying for when choosing between them, which is heat and life. Those depend on how the current divides between the parts, and for identical parts the division is trivial: half each. The interesting case is when the parts are not identical.

A mismatched pair divides by capacitance #

Take two thousand-microfarad parts and let one of them have twice the other’s resistance: thirty and sixty milliohms. It is not an exotic case. Electrolytic series resistance is specified as a maximum, two parts from different batches or different ages routinely differ by that much, and the resistance of a part rises as it ages.

The intuition from resistive dividers is that current prefers the lower resistance, so the good part carries more. Marched, the two parts carry 0.2607 and 0.2560 amperes root-mean-square. The good part carries 1.8 per cent more current, not twice as much.

A current divider between two branches splits by admittance, and each branch’s impedance is its resistance plus its reactance. The branch the other resistance decides made the point that a current divider’s output is set by the branch the current does not take, and here the relevant branch quantity is dominated by the capacitor, not the resistor. At a hundred hertz — the ripple’s fundamental in a full-wave rectifier — a thousand microfarads has a reactance of 1.59 ohms, twenty-six times the worse part’s sixty milliohms. At a kilohertz, where the charging pulse still has real content, the reactance is 0.159 ohms, still between two and a half and five times the resistance. Two equal capacitances divide the current equally, and a resistance that is a small fraction of the reactance barely moves the split.

The heat is a different matter, because heat is the current squared times the resistance. With the currents nearly equal, the dissipation follows the resistance nearly exactly: 2.038 milliwatts in the good part and 3.931 in the worse, a ratio of 1.93 against a resistance ratio of two.

Across the sweep the pattern holds and slowly weakens:

At five times the resistance the worse part’s current has fallen by eight and a half per cent, which is the resistance becoming comparable to the reactance at the charging pulse’s higher harmonics, and the heat ratio is 4.18 rather than five. But across any realistic spread of part-to-part resistance, the part with the higher resistance dissipates in proportion to its resistance, because the parts share the current as though the resistances were not there.

The tolerance that decides the split #

Everything above is the resistance mismatch, and resistance turns out to be a weak lever on the split. The strong lever is the one a resistive intuition overlooks entirely: if the current divides by capacitance, then a difference in capacitance moves it directly.

An aluminium electrolytic’s capacitance is commonly specified to twenty per cent, so two parts with the same marking can reasonably be 800 and 1200 microfarads — and the marking is itself one reading of a quantity that the capacitance that is not one number shows depending on how it is measured. Marched at thirty milliohms each, that pair carries 0.2094 and 0.3122 amperes — a ratio of 1.491 against a capacitance ratio of 1.500 — and dissipates 1.316 and 2.925 milliwatts, a ratio of 2.22 against the square of the capacitance ratio’s 2.25. The larger part does half as much work again and heats more than twice as much, and it does so with identical resistances.

Set that beside the resistance table. Doubling one part’s resistance moved the current by 1.8 per cent and the heat by a factor of 1.93. A capacitance split well inside an ordinary tolerance moves the current by half and the heat by more than a factor of two. In a pair of electrolytics, the capacitance tolerance is the larger cause of unequal heating — and, unlike the resistance, it puts the extra heat in the part with the higher capacitance, which is usually thought of as the better one.

The terminals do not see this either. The ripple is 704 millivolts at every split from 1000 and 1000 microfarads to 700 and 1300, to three figures: the rectifier sees two thousand microfarads, and how they are distributed between two cans is visible only to the cans.

Why that matters more than the ripple #

An aluminium electrolytic’s life is set by its core temperature, and its core temperature is its ambient plus the dissipation times its thermal resistance. So the result has a direction that the terminal behaviour never shows: in a pair whose resistances differ, the worse part runs hotter by the ratio of the resistances, and the hotter part is the one that ages faster.

Nothing at the output says so. The mismatched pair of thirty and sixty milliohms, marched, gives 704.8 millivolts of ripple, a crest factor of 15.87 and a mean of 15.967 volts. A matched pair of forty-five milliohms each — the average of the two — gives 704.6 millivolts, 15.96 and 15.967. A designer measuring the supply sees a healthy two-thousand-microfarad reservoir with an ordinary resistance, and the heat distribution inside it is invisible from the terminals, and lopsided.

It is worth noticing which matched pair it resembles, because the resistive intuition picks the wrong one. Thirty and sixty milliohms in parallel are twenty, and a matched pair of forty milliohms each — the pair whose resistances combine to twenty — gives a crest factor of 16.14, further from the mismatched pair’s than the forty-five-milliohm pair is. When the capacitances dominate each branch, the two resistances add to the bank’s impedance in their sum over four, not in parallel: the branches are paralleled as capacitors, and the resistances ride along. That is 22.5 milliohms rather than 20, which is why the average is the better stand-in.

The pair that is worse than either found a different failure of the intuition that two capacitors in parallel are simply more capacitor: a bulk part and a ceramic that together present six times the impedance of either, at the frequency where one’s inductance resonates with the other’s capacitance. That is a failure at megahertz and it needs two unlike parts. The one here needs only two like parts that are not quite alike, and it happens at a hundred hertz. At this frequency the capacitor that is an inductor is still entirely a capacitor, which is exactly why the current divides the way it does.

What the pair is for #

Putting the pieces together, a reservoir made of two parts rather than one does these things, and each has a number:

• It is one part of the summed capacitance and half the resistance, exactly, whatever the surrounding circuit does.
• Against one part of the same capacitance and the same resistance per part, it lowers the ripple by an amount that is zero near the ripple’s minimum and grows past it — 2.7 per cent at a hundred milliohms a part, 12.6 per cent at three hundred.
• It raises the peak current at every resistance — 14 per cent at a hundred milliohms a part — and that is the current the transformer, the diodes and the fuse are sized for.
• It divides its ripple current by capacitance and its heat by resistance, so the worse part of a mismatched pair runs hotter by nearly the resistance ratio — and the larger of two capacitances inside an ordinary tolerance carries half as much current again and more than twice the heat.

The honest reasons to fit two parts are therefore mechanical and thermal — height, footprint, surface area — and the electrical consequences are small and mostly unfavourable unless the parts are well past the ripple minimum. The textbook ripple expression sees none of this, because it has no resistance in it, and it sees the pair and the single part as the same capacitance, which is the one statement here that is exactly right.

Still open #

A bank on a board, where the copper between the parts is a resistance in one branch. Two parts placed side by side share the rectifier’s node; four parts in a row do not, because the far part is reached through a few milliohms of track the near one is not. By the argument here that track barely moves the current split — it is a resistance added to a branch whose impedance the capacitance dominates — and so it should mostly move the heat. But the charging pulse’s harmonics reach the kilohertz range where the reactance and the resistances become comparable, and the part nearest the rectifier may carry a disproportionate share of the pulse even while the ripple current divides evenly. Marching a row of four with the track in the netlist would say how much.

Ageing, where the two mechanisms push against each other. As an electrolytic dries out its resistance rises and its capacitance falls. The rising resistance makes it hotter at the same current, which accelerates the drying; the falling capacitance takes current away from it, which cools it. Which one wins decides whether a pair’s worse part runs away or settles, and it is a march over a slowly changing netlist whose rate is set by a temperature — the kind of fixed-point question the ripple that is a temperature asks of a filter section.

The cold start, which these essays have been deferring. At low temperature an electrolytic’s resistance is several times its room-temperature value, so a cold supply sits further right on the one curve above — a lower peak current and, past the minimum, a higher ripple — and a pair that was near the minimum when warm may be well past it when cold. The pair’s advantage, small warm, is largest exactly when the supply is coldest.