[ §1 · supply ]

Capacitance is the easy half

The microfarads are arithmetic and take an afternoon. What ends a bank is the loop it sits in and the volts it actually sees.

CommonGnd19 August 202642 min read

Everything the coil throws was sitting in a capacitor a moment before it left, and it sat in two of them: the bank behind the bridge and the tank in front of it. Picking the microfarads is the easy half of both. What ends banks is the loop they sit in and the volts they actually see.

What it is and how it works

A diode passes current one way and blocks it the other. Four of them in a bridge send both halves of an alternating supply the same way round, so what comes out never changes sign. Behind them sits a bank of capacitors, and the two together make a direct voltage out of an alternating one.

The part that is not obvious is when the diodes are actually working. The capacitor holds a voltage. The supply is a sine, so for most of every cycle it is below what the capacitor is already holding, and while that is true the diodes are reverse biased and off. Only near the top of each hump does the supply exceed the capacitor, and only then does any current flow through them at all.

The diodes conduct only in the shaded bursts, where the mains is above what the bank is already holding. Everywhere else the bank carries the load alone.

So the current through a rectifier is not a smooth trickle. It is a series of short, tall bursts, one per hump, and the average of those bursts is a small fraction of their height.

And that fraction can be computed. The diode opens when the sine climbs back above what the bank is holding, so the width of the window is set by how far the bank sagged since the last hump. For a droop of d below the peak the window is arccos(1-d) of a half cycle, and all the charge the load took over the whole half cycle has to go through in it:

droop between humps   window    current in the window, against the load current
       2 %             6.4 %                   15.7x
       5 %            10.1 %                    9.9x
      10 %            14.4 %                    7.0x
      20 %            20.5 %                    4.9x

Derived here from the geometry of a sine and from charge balance, not measured. The true peak inside the window is higher again, because the pulse is a shape and not a rectangle.

Which inverts the thing you were about to do. A bigger bank sags less, a bank that sags less gives a narrower window, and a narrower window means a taller pulse through the same four diodes. The capacitance you add to hold the top of a ramp up is paid for here rather than saved here. Sizing that bank is its own page on this page; what it leaves behind on this one is that the rectifier's job gets harder as the rest of the machine gets better.

What it does in a coil, and what you decide

Which arrangement

From 230 V mains:

A plain bridge gives about 325 V. Both half cycles, four diodes, nothing clever.
A Delon doubler gives about 650 V. Both half cycles again, and twice the voltage from the same wall socket.
A half wave gives about 325 V for half the period. Fewer parts, and worse in the way that matters: for the other half of the cycle nothing refills the bank, and residual voltage left on the bus is what makes the arc branch at the start of a ramp.

The reason to reach for the doubler is not the 650 V. It is what four times the stored energy does to the shape of a ramp, and that argument belongs with the bank, because the bank is the part it is an argument about.

What the wall socket is asked for

Two things, and the smaller one is the trap. The average power is the burst energy times the rate, and on the 406 J ramp the bank's page on this page works from:

burst rate      average draw   [derived, all four]
0.5 Hz     203 W
  1 Hz     406 W
  2 Hz     812 W
  5 Hz    2030 W

A 230 V 16 A socket is 3.7 kW, so even the top of that looks comfortable. It is not, because the current does not arrive smoothly. It arrives in the same narrow windows the diodes conduct in, and current made of narrow pulses has a poor power factor, which puts the RMS current well above what the wattage alone suggests:

5 Hz, 2030 W       power factor 0.45  ->  19.6 A RMS   [derived, all three]
                                0.5   ->  17.7
                                0.6   ->  14.7

The 0.45 is what a rectangular pulse in the five per cent row's window gives, sqrt(2 × 0.101) [derived], and it is the pessimistic end, because a real conduction pulse is a shape rather than a block. Kaizer's design guide puts the real figure a little above it and no higher: expect 0.5 to 0.6 from a rectifier with nothing done about it, 0.6 to 0.8 from a passive line choke, and 0.95 and above only from an active boost stage, which it calls complex and expensive. So the honest reading is fifteen to twenty amps, the bottom of it Kaizer's 0.6 and the top of it the pessimistic 0.45 rather than anything he says, and every one of those is most or all of a 16 A socket while the power budget still reads half used. Which is the part that survives whichever figure is right: the socket runs out before the wattage does, and it runs out on the shape of the current rather than the size of the load.

Which diodes

By peak current, not average, for the reason in the drawing above. A part sized on the average figure is sized on a number it never sees.

Look for I_FSM, the single half-cycle surge rating, and the repetitive peak alongside it. The average forward current is the last of the three to consult, not the first.

I_FSM is sized on the moment before any of the above. At switch-on the bank is empty, so there is nothing holding the diodes off and the first half cycle is limited by the mains impedance and the bank's own ESR and by nothing else. On 325 V into half an ohm of total resistance that is 650 A, and into a fifth of an ohm it is 1625 A [derived]. That is the event the single cycle rating exists for, and it is also the reason there is a resistor in the AC line with a relay to short it out a few seconds later, which is on the bank's page on this page. Size I_FSM for the start you get if that relay ever closes early, because it eventually will.

What the bank is for

The mains delivers in bursts, on the tops of the sine wave. The bridge draws whenever it is switched on. Between the two sits a bank of capacitors, and its job is to answer while the mains is not looking.

On a DRSSTC the bang is short and the bank barely notices it. On a QCW the ramp runs 6 to 25 milliseconds, which is long enough to pull the bus down while the arc is still growing. The last centimetres of an arc are the dearest ones to buy, so that is the worst possible moment to run short.

Everything the coil throws was sitting in this bank a moment before it left.

Whether to fit a doubler

The criterion is not volts. It is what fraction of the bank's stored energy one bang consumes.

E = C·V² / 2

The 406 J is V·I·t on the machine this site works from: about 615 V of average bus, close to 26 A while the ramp is running, over 25 ms. Put your own three numbers in and everything below follows from them rather than from ours.

Doubling the voltage quadruples the store at the same capacitance. One 406 J bang against a 9900 µF bank:

325 V, a plain bridge on 230 V mains: the bank holds 523 J and the bang eats 78 per cent of it.
424 V, an intermediate bus rather than a topology of its own: 890 J, and 46 per cent.
650 V, a Delon doubler: 2091 J, and 19 per cent.

All three are ½CV² on that bank against that bang [derived].

Below about a fifth the ramp holds its shape, and that is where the worked machine sits, at 19.4 per cent.

A fifth is comfortable rather than necessary, and the page should say where the good machines actually are. Two of the best documented QCW coils run at close to twice that fraction:

Gao's QCW 1.5 holds 10,000 µF at 385 to 397 V, which is 741 to 788 J, against a stated 275 J per pulse. That is 35 to 37 per cent of the store, and it throws seventy inches.
dr. kilovolt's SiC coil says its 800 V bus drops "to approx. 650 V" at 100 A peak. That is 18.8 per cent of the voltage and 34 per cent of the energy, and it throws two to two and a half metres.

So a third of the bank is where working machines live. Below a fifth is easy, a third is proven, and past that you are ahead of the published art rather than inside it. Where it stops being tolerable is not documented anywhere this corpus has read, so there is no ceiling quoted here.

Gao's page also says what that sag costs him. One logged run takes his bank from 397 V to 326 with 256 J out of it, a slightly lighter pulse than the 275 J above, and the three numbers agree with each other to a volt. His ramp tops out a little over 300 V, and the reason he gives is the bank running down, not the tank and not the bridge. So the top of the ramp and the sag of the bus are one limit seen from two sides. The buck cannot put out more than the bus has left at the moment the ramp peaks, which makes the ramp height a bank will hold up something you can work out on paper: take the ramp's energy out of the store and turn what is left back into volts.

Set the two machines beside each other and the voltage side lands together as well. 800 V down to 650 is 18.8 per cent; 397 down to 326 is 17.9. The buses differ by a factor of two, both pairs of voltages are the builders' own and the percentages are ours, and that is a cheap plausibility check on a third machine. One caveat on reading it off a scope: the rectifier refills the bank while the ramp draws from it, so what you see is the net of the two, and it flatters the bank a little against the arithmetic.

And the ratio can be read off a scope without knowing the capacitance at all. Energy goes as the square of voltage, so the fraction of the bank one ramp consumes is 1 − (V_end/V_start)². Measure the bus at the start of the ramp and at the moment the interrupter drops, and that is the answer, whatever the cans say on the side.

Do not bring the DRSSTC rule of thumb here, and do not read its two versions as one rule. Both say twenty to fifty times. They say it about different denominators, and this page had them running together. Steve Ward means the energy drawn per bang and defines it in the same breath, as input power divided by bangs per second. That is the same quantity as the ramp energy above, so his rule can be brought here, and when it is the worked machine fails it: 406 J against a 2091 J bank is 5.2 times, not twenty [derived]. Twenty would want 8120 J, which at 650 V is about 38400 µF of effective bank and so 76800 a side once it is stacked [derived], and that is a bank nobody builds. Barnkob means something else: twenty to fifty times the energy in the MMC, the tank capacitor, which on a QCW is a fraction of a joule and which every bank on this page clears without trying. One rule is about what the coil draws, the other about what the tank holds. Same band, different question, and only Ward's is answering this page's.

One design looks like it goes past the top of that band, and the man who wrote the band is further past it than that. Jimmy Hynes's DRSSTC 2 lists sixteen 12000 µF 350 V capacitors in a voltage doubler, which is 48000 µF at 700 V and 11760 J of store, against his stated 200 J per bang: 59 times, eighteen per cent over the fifty [derived]. Read it as a design and not a measurement, because that is how he writes it: the power draw is estimated at about 20 kW and the sparks are what they should be. But the two estimates are consistent with each other, since 20 kW at 200 J a bang is 100 bangs a second, which is Ward's own definition of a bang applied to Hynes's own two numbers, so the multiple is not an artefact of mixing them. And on that same definition Ward's own DRSSTC-1 sits at 73 times, 880 J of storage against about a 12 J bang, both printed on the page the band comes from, and he notes in the same breath that it could really support much more [derived]. So 59 is not past anything. The pair is what the tolerance actually looks like once the denominator is the drawn energy. And a DRSSTC's bang is not the charge in its tank, which is worth saying because it is the tempting way to dissolve the conflict with this page. Hynes settles it himself: his MMC tops out at 2.55 µF, which even at the full 700 V bus holds 0.62 J [derived, ½CV²], about three hundred times under the 200 J he calls a bang. The energy comes out of the bank across the whole ontime, the same way a ramp draws it, and Hynes's bang and this page's ramp are a factor of two apart rather than a factor of two thousand.

A doubler also takes both half cycles, and it pulls the bus practically to zero every cycle, so no residual voltage is left to branch the arc at the start of the ramp.

And it costs twice the cans, for a reason that is not the voltage rating. The electrolytic ceiling is 550 V, which Barnkob's own electrolytic against film comparison sets against 1500 V for film DC link parts, so 650 V is two banks in series whatever else you do, and capacitance in series halves: the 9900 µF above is two stacks of 19800 µF. Twice the cans for the same effective microfarads, on top of the parts being dearer.

A series stack also does not divide itself. Leakage current differs from can to can, the one that leaks least takes the larger share of the voltage, and it goes on diverging until something gives. Balancing resistors across each section stop that, and they are usually the same resistors as the bleeder below, sized for whichever of the two jobs is the harder.

What it costs besides is higher-rated parts down the whole chain, and one that is easy to miss: the isolation rating on gate drive DC-DC modules. A module rated 400 V on a 650 V bus is a blocker nobody notices until something fails. How the three arrangements are built is on the rectifier's own page on this page.

How much capacitance

The criterion is the sag over one ramp, and the current that goes into it is the one the ramp draws, not the average over the minute. Those two differ by more than a hundred times, and using the wrong one is the easiest mistake on this page to make.

The right one falls out of the section above. Taking 406 J out of 9900 µF at 650 V leaves the bank at 584 V, so:

dV = 650 - sqrt(650² - 2E/C)
I  = C·dV / t_ramp  =  0.66 C in 25 ms  =  about 26 A

Twenty six amps while the ramp runs. The same machine at half a hertz draws about a third of an amp from the wall, because the ramp is brief and the gap between ramps is not, and it is the first of those two that the capacitance has to answer.

capacitance    sag over one 406 J ramp   [derived, all four]
   470 µF      drained before the ramp ends
  2200 µF      419 V,  64 %
  4700 µF      150 V,  23 %
  9900 µF       66 V,  10 %

So the target is not a volt or two. No capacitance reaches that, and a page that asks for it is asking for something that does not exist. The target is a percentage, and it is the same criterion as the doubler section above: 10 per cent of the voltage is 19 per cent of the energy, which is the case that section calls holding its shape.

What it costs is ripple current. The bank delivers in bursts and refills in bursts, and both heat it through its ESR. On electrolytics that is a rated parameter and it is not one to exceed, so the part that satisfies the sag has still to satisfy the heating.

Put a number on it, because otherwise it cannot be acted on. Twenty six amps for 25 ms at five hertz is 12.5 per cent duty, so about 9 A RMS on the discharge alone, and the refill from the mains adds to it. At half a hertz it is nearer 3 A. Divide that by the ripple rating on the datasheet of the can you are actually buying, and what comes out is how many cans, not how many microfarads. This is the answer to a bank that meets its capacitance and still runs hot.

And it costs something at the other end, which is easy to miss because it is on somebody else's page. A bigger bank sags less between mains humps, and a bank that sags less takes its refill in a narrower window, which makes the pulse through the rectifier taller. At five per cent of droop the diodes carry about ten times the load current inside that window, and at two per cent about sixteen [derived, from the conduction angle arccos(1 − d) against the half cycle; the same two figures are worked on the rectifier's own page]. So the capacitance that holds the top of the ramp up is bought partly at the rectifier on this page, and the two choices cannot be made one after the other as though they were separate.

Whether the source behind it is stiff

The criterion is how fast the bank refills between bangs, which is not the same question as whether the supply can carry the average.

The mains through a doubler refills willingly: the mains impedance is a fraction of an ohm. A transformer's leakage inductance limits the rate. On average heating a transformer is comfortable, because a QCW's average power is low, and that is exactly what makes the trap: the number that looks fine is not the number that matters.

How the bank gets charged in the first place

A bank of thousands of microfarads at switch-on is a short circuit across the mains. A resistor in the AC line before the rectifier, shorted out by a relay after a few seconds.

Averaged over time, not over the bang

Same peak, and the bars under them are the heating, which goes as the square of the current across the whole bang. Sizing a bank on the peak reads both of these as the same duty.

I_rms(average) = I_rms(during the bang) · sqrt(duty)
duty = bang length × bangs per second

And the shape of the envelope inside the bang moves the answer by a factor of the square root of five:

a flat top gives I_pk / sqrt(2), so 0.707 of the peak;
a linear rise gives I_pk / sqrt(6), 0.408;
a quadratic rise, which is what a QCW does, gives I_pk / sqrt(10), 0.316.

Dissipation is I_rms^2 · ESR. The catalogue current is quoted for a stated temperature rise, and the figure to look for is ten degrees on most MKP film parts, which is davekni's reading of the catalogues and which he adds is often not printed on the sheet at all. Where it is printed it is not always the ten: of the two TDK sheets cited further down, the B3265* one states 20 °C and the B3264*B one 15, and the 2015 edition of that second sheet stated 10. Whatever the stated rise turns out to be, it gives you the thermal resistance for free and lets you work out the real rise rather than guessing at it. This page used to give twenty degrees as the general figure with nothing behind it; the twenty that survives further down is one sheet's own footnote and holds for that sheet's parts, not as a rule.

And the rise you measure on the first day is not the rise the part settles at. One ceramic tank capacitor has been run to destruction with the heating logged the whole way. davekni took a Samsung CL31C222JHHNNNE, 2.2 nF at 630 Vdc in a 1206 package with no AC rating at all, and characterised it at 450 kHz: 8 to 10 °C of rise at 250 Vrms, 18 to 20 at 340. Carried forward on the square of the voltage, as dielectric loss should go, the 340 V pair predicts 34 to 38 °C at the 466 Vrms he then pushed it to [derived, 18 to 20 times (466/340)²]. He measured 36. The two measured points do not quite follow the square between themselves, an exponent of 2.3 to 2.6 [derived], so the 250 V pair on its own would have said 28 to 35 and the prediction that lands is the nearer one. So far the method on this page works, and the part is characterised rather than guessed: 2.9 Arms into its 160.8 Ω of reactance is 466 V, which is the voltage he states [derived]. Then it kept climbing. Over four weeks of almost continuous running the rise went from that 36 °C to 127 °C, a case at 147 with the room at 20, and the part failed at thirty five days. Nothing in the drive changed. The loss did, by 3.5 times [derived]. So a catalogue current at a stated rise describes a part in its original condition, and this is the only published measurement of what a tank's duty does to that condition. Size for the rise you measure, then go back and measure it again.

Check which edition of the datasheet you are reading

TDK's B32642B0333J is not the part it was on paper. EPCOS's edition of May 2015, mirrored by DigiKey, lists the part at VR = 1000 V DC against VRMS (f ≤1 kHz) = 600 V AC and draws its permissible-voltage curves for "Self-heating TA ≤10 °C". The June 2018 edition lists the same part at 500 V AC and redraws those curves for "ΔT ≤15 °C": a larger rise allowed, and a lower rating anyway. Read the 33 nF curve at 400 kHz off each and it goes from about 68 V to about 34. Half, on the same part number.

Those two voltages are read off graphs, not quoted. Neither edition prints a number at 400 kHz, both give curves, and 68 and 34 are what the 33 nF curve measures on the published plots at that frequency. This page used to print the pair as 70 and 35 with nothing under either. The reading checks against a figure that is printed: TDK's current edition, June 2026, carries the 2018 plot unchanged, plot number KMK1821-E, and adds a column the older sheets did not have, B32642B0333 at 2.9 A RMS at 85 °C and 100 kHz. If that ampere still holds at 400 kHz it is 35 V across 33 nF [derived, 2.9 / (2π · 400 kHz · 33 nF)], against the 34 the curve reads there. So the halving is not an artefact of one edition's draughtsmanship, and it is still in force.

The public capacitor tables everybody links to were built on the older numbers.

Small capacitors carry more current per nanofarad

This is the counterintuitive one, and it is why "I will fit bigger ones to be safe" is backwards. At a fixed bank capacitance, smaller parts give you more current. From TDK's B3265* sheet of June 2026, the B32652 rows at 15 mm lead spacing and VR, DC = 1000 V DC, the amperes being that sheet's IRMS at 85 °C and 100 kHz "for a T ≤ 20 °C":

10 nF: 1.3 A each, 130 mA per nF. A 12.2 nF bank, the one on the machine this site works from, as 9 series by 11 parallel is 99 parts and 14.3 A.
22 nF: 1.9 A, 86 mA/nF. 9 by 5, 45 parts, 9.5 A.
100 nF: 5.4 A, 54 mA/nF. 9 by 1, nine parts, 11.1 nF, 5.4 A.

Nearly a factor of three in current for the same capacitance, and the price is count. Across the two ends of that list it is eleven times the parts for 2.6 times the current [derived, 99 against 9 and 14.3 against 5.4], because every extra parallel string needs its own series stack. That is faster than the square of the current gain, which is what this page used to say it was: 2.6 squared is 7.0 against the 11 the rows actually give.

Nor is a bigger case a guarantee. In the same sheet the B32653 at 22.5 mm pitch and 33 nF, ordering code B32653A0333, gives 1.6 A, less than the B32652 at 15 mm pitch, B32652A0333, and 2.5 A. Both are 1000 V DC rows and both amperes are on the same 85 °C, 100 kHz, 20 °C basis. Read the table, not the outline drawing.

And on a QCW, small in the other sense too

A QCW at the size most people build wants a small bank outright. Gao Guangyan summarises the machine as needing a high impedance primary, "coupling of >=0.3, many turns for primary and with a small tank cap of around 8 - 15nF", and on the forum Hydron told a builder who was planning fifty that "50nF is very large for a QCW coil - the required MMC normally works out to be under 20nF". The bank on the machine this site works from is 12.2 nF and sits inside that.

The reason is the tank's characteristic impedance. At a fixed frequency L and C are tied together, and:

Z = sqrt(L/C),   primary current goes as C,   power goes as C

while the voltage across the bank does not depend on C at all. So the fix for too much current is more primary turns and less capacitance, and the fix for too little is the reverse. Landon Kageler built his first QCW on a 12.35 nF primary capacitor, eight series by three parallel of the same B32642B0333J the edition change above is about, was disappointed by what it made, and then "simply increased the current by increasing the primary capacitance and lowering the inductance" to 30 nF, at which point he was reaching one and a half feet.

And the current does not divide the way you think

Between parallel strings, the current divides in inverse proportion to ESR and path resistance, not to capacitance. At the frequencies a tank runs at, capacitance is not what decides it.

One idea worth disposing of, and the disposal is ours rather than anybody's we can cite: small capacitors do not make a poor bank. A network's ESR does not depend on the size of its unit, because R_bank = R_unit·n/m and tan δ/(2πf·C_bank) are the same number with the unit capacitance cancelling out [derived], and dividing the heat between more parts improves the cooling. This page carried the opposite claim in its first version and it was wrong. What limits it in practice is the number of joints and the symmetry of the paths, not the physics.

One more thing that follows from where the heat goes: the leads are a thermal path and not only a connection. davekni takes his film parts' case rise "at hottest point at center between leads", and notes that in his fixture the "Leads provide heatsinking to copper foil". Cool the leads, and space the parts so air can pass between them.

How the heat divides between the leads and the case is not established on this page. It used to print about two thirds of it leaving along the axis and one third radially. No source for that split was found, and the maker's own model does not work that way: TDK's general technical information sizes the heat a film capacitor can shed as α·A·ΔT with A the surface area of the whole part, not as a fixed share out of the ends. So the split is withdrawn rather than defended. The instruction above does not depend on it.

The rig on the DRSSTC department page draws the bank as a matrix rather than one symbol, because the number of parts is the point. Take the tank capacitor out on the QCW page and you have a ramped SSTC, which is a different machine with a different set of problems.

The two capacitors that were measured here, the Samsung ceramic and davekni's tank, are linked to the threads they were measured in, and so now are Gao Guangyan's range, Hydron's ceiling and Landon Kageler's rebuild. So are the catalogue rows, which used to carry no source at all: the 1000 V family and the B32652 against B32653 comparison are TDK's B3265* sheet of June 2026, and the edition change is TDK's own two editions of the B3264*B sheet, the older of them reachable only through a web archive. The one thing in that comparison that is not quoted is the pair of voltages at 400 kHz, which are measured off the published curves and are marked as such where they appear. The arithmetic on all of them is ours.

What will get you

Balancing and bleeder resistors are still wanted across those sections, but for the spread in electrolytic leakage current and for discharging the bank afterwards. Not for dividing anything.

For indication, a bright LED with a dropper resistor, which bleeds the bank slowly at the same time, and a neon as backup. Mind which neon voltage you are thinking of: it strikes at about 90 V but does not go out until nearer 60, so a dark neon means under sixty rather than under ninety, and it can still be glowing at a voltage where it would not restrike if you switched it in fresh. Sixty volts across thousands of microfarads will still hurt you badly, so it cannot be the only indicator.

And a bleeder wants a size, not just a presence. It is an exponential, so choose the wait you will tolerate and work back. From 650 V down to 50 is a factor of thirteen, which is 2.6 time constants:

2.4 kΩ across 9900 µF    RC 24 s     50 V in 1 minute       176 W continuous
6.0 kΩ                   RC 59 s     50 V in 2.5 minutes     70 W
                                     [derived, both rows]

Both are serious resistors in their own right, which is why the value ends up picked by what you can afford to dissipate rather than by how long you would like to wait. Two and a half minutes is a long time to stand in front of a machine you want to adjust, and that wait is the reason the indicator is not optional.

A hundred watts on a resistor is a continuous rating. It does not mean the part will swallow half a kilojoule in a second and a half, which is what a precharge start asks of it: a plain rectifier filling 9900 µF to 325 V leaves 523 J in the bank and puts of the order of the same again through the resistor, over about two and a half time constants of the 0.59 s below [derived, ½CV² and R·C]. Look for a pulse energy or surge withstand line, and if the datasheet has not got one, use the massive cement-bodied ceramics, which take the pulse in their heat capacity.

The mechanism is that the wire element takes the whole pulse before the body has heard about it, and only afterwards does the heat spread out into the ceramic for a far smaller rise. Repeated jumps fatigue the joint between the two, so leave a minute between starts. How that splits is not established here. This page used to give it as about three grams jumping three hundred kelvin into a fifty gram body for a ten degree rise, and those four numbers had no source and no working behind them. The pulse energy above is knowable; the division of it needs the datasheet for the part in your hand.

And the big bank on its own will not feed the bridge. The inductance of the power cable will not let it deliver quickly, so the bridge needs capacitors of its own beside it. The same goes for driver supplies: a local electrolytic plus a fast ceramic bypass right at the pins.

Which is why that run is flat copper and not wire. The problem is not resistance. A metre of heavy cable is a few milliohms and the loss in it is nothing worth arguing about. The problem is inductance, and inductance is set by the area of the loop that the outgoing and returning conductors enclose between them. Two round wires held apart enclose a large loop. Two flat straps face to face, separated by nothing thicker than their insulation, enclose almost none, and the wider they are the less is left. That is the entire argument for busbar: the same copper, arranged so the loop is small.

It matters in one of the two runs and not the other, and the two are worth separating because the wrong one gets the attention. Between the rectifier and the bank the current is large but it arrives at a hundred hertz, so that run is a heating question and ordinary heavy cable answers it. Between the local capacitors and the bridge the current reverses at the tank's own frequency, hundreds of kilohertz, and there V = L·dI/dt turns a small inductance into a voltage across the devices that nothing on the schematic accounts for. Steve Ward's rule for that run is bus traces an inch and a half wide or wider, both sides of the board paralleled.

The numbers in one place

Plain bridge, 230 V mains: about 325 V.
Delon doubler, 230 V mains: about 650 V.
Half wave, 230 V mains: about 325 V, half the period.
The rating to size on: I_FSM first, then the repetitive peak, then the average.
Window current against load current: about 10× at five per cent of droop, about 16× at two.

Two machines for scale, both stated by their builders. CJ's is a 420 V bus, "rectified and filtered by the mechanical voltage regulator", which is his whole description of the arrangement. The regulator is in front of it, so his 420 V says nothing about which of the three is behind it and you cannot read one out of the voltage. Gao's QCW 1 is 340 V maximum on the bridge from 240 V rectified, behind 6800 µF, and he does say which. The same page has him start on a doubler, find the 3400 µF that two cans in series left him too little, and swap it for a single 6800 µF on full wave rectification, which is what the 340 V off 240 V is. Note what he gave it up for: capacitance, not volts. Neither machine is a recommendation.

Bus from 230 V mains: about 325 V through a plain bridge, about 650 through a Delon doubler.
Current while the ramp runs: about 26 A, against about a third of an amp averaged over the minute. Do not put the second one in a sag formula.
Sag to aim at over a ramp: a percentage, not a fixed number of volts. 9900 µF gives 10 per cent of voltage and 19 of energy, which is comfortable. 4700 gives 23 and 41, which is past the two documented machines but in the same country as them. 2200 and 470 are out.
Bang energy as a fraction of the bank: a fifth is easy, a third is what the two best documented coils above actually run, and nothing published says where it stops working. The mapping between energy and voltage is arithmetic, since 19 per cent of energy is 10 per cent of voltage; where the useful limit falls is not, and nothing here measured it.
Reading the ratio without a capacitance value: 1 − (V_end/V_start)² off the scope.
Ripple current: about 10 A RMS at five hertz, 3 A at half a hertz, which sets the number of cans rather than the capacitance.
A 650 V bank is two in series: 19800 µF a side for 9900 effective, with balancing resistors, because leakage spread will not divide it for you.
Precharge current: 5.4 A through 60 Ω at the mains peak.
Precharge time constant behind a doubler: 2.38 s against 0.59 s for a plain bridge, since the series stack doubles the capacitance and half-cycle feeding doubles it again. Three seconds of relay delay leaves it 72 per cent charged.
Precharge element: the element takes the pulse and the body takes it afterwards. The split is not established here; the pulse is about 523 J on a plain rectifier at 325 V.
Bleeder: 6 kΩ takes 2.5 minutes to reach 50 V and dissipates 70 W; 2.4 kΩ takes a minute and dissipates 176.
Neon: strikes near 90 V, extinguishes near 60. A dark neon means under sixty, which is not a safe bank.

Figures here are from the component notes this site is built on, and the arithmetic on them is ours.

What goes wrong

A diode fails early with everything apparently within rating. It was sized on the average forward current. The peak at the top of the sine wave is several times that.
The arc branches in the first few milliseconds of a ramp. Residual bus voltage from the previous bang. A half wave supply leaves it; a doubler pulls the bus practically to zero every cycle.
The two halves of a doubler read different voltages. They are not meant to divide; check the bleeders and the leakage spread, not the capacitance matching.
One diode of a parallel pair fails and its twin looks untouched. Paralleling to raise the current rating does not divide the current. A silicon diode's forward drop falls as it warms, so the hotter of the pair drops less, takes more of the current, heats further and takes more still. It runs away quietly and the survivor is the evidence. Series resistance in each leg, or one bigger part, or thermal coupling tight enough that they cannot diverge.

The top of the ramp is flat, or collapsing. Bus sag. It looks exactly like a tuning problem and it is a supply problem, which is why it is at the top of this list. The height it flattens at is the volts the bank has left by then, so it can be worked out before it is discovered.
The arc branches in the first milliseconds of the ramp. Residual bus voltage left from the previous bang. A doubler pulls the bus to zero every cycle; a half wave supply does not.
A thump of inrush when the precharge relay closes. The delay was set for a plain rectifier and there is a doubler behind it. Size it on the doubled time constant.
A precharge resistor fails well inside its rating. The rating was continuous, and a start puts hundreds of joules through the element in a second or two.
The bank runs hot with nothing else wrong. Ripple current exceeded. It heats on the delivery and on the refill both.
The bus does not recover between bangs on a transformer supply. Leakage inductance. The average heating is fine and the refill rate is not.
Gate drive dies as soon as the bus is raised. The isolation rating on the DC-DC modules.

The bank and the tank on the QCW diagram are the two capacitors everything the coil throws was sitting in a moment before it left, and the rectifier is what fills the first of them.

buscapacitorsrectifier