The DRSSTC bus capacitor bank: loop, ripple and the volts it sees

Everything the coil throws was sitting in a capacitor a moment before it left, and it sat in two of them. This page is the first: the bank, thousands of microfarads behind the bridge, and the rectifier that fills it. The second is the tank, nanofarads in the primary, and it is sized by current rather than by capacitance. They share a name in conversation and nothing else, not even a ripple rating. Picking the microfarads here is the easy half. What ends a bank is the loop it sits in and the volts it actually sees.

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.

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.

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 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.

And that fraction, the average against the peak, 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 below, under How much capacitance; what it leaves behind here is that the rectifier's job gets harder as the rest of the machine gets better.

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 below, under How the bank gets charged in the first place. 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.

Which is why a second, far smaller capacitor sits at the bridge, and it is not there for its capacitance. The bank is electrolytic, and an electrolytic has series inductance of its own before the busbar adds any more. Above some frequency it stops behaving as a capacitor at all and behaves as an inductor, so it cannot answer a demand that changes faster than that. The bridge's demand does exactly that: current steps at every switching edge, whose spectrum runs far above anything the bank can follow. The film capacitor bolted across the module is physically small, has little inductance, and sits centimetres from the die, so it supplies the fast part locally while the bank supplies the slow part from across the chassis.

So the two are not a big one and a spare. They divide the job by frequency, and neither substitutes for the other: more bank does nothing for the edges, and more film does nothing for a twenty five millisecond ramp. The ordinary Chinese market bridge shows the split as built, 5600 µF or more of bank behind a 2 µF 1200 V snubber across the module, which is with the bridge devices. The reasoning is ours; the parts list is theirs.

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 whether the gap between bangs is long enough to put it back is a question about rates, not about time constants. It gets asked directly on a 160 mm DRSSTC running 10 ms bursts: is 10 ms enough to power the bus caps back up? Hydron's answer is that at 100 bangs a second, off 50 Hz mains through a full wave rectifier, "the bus caps will be charged (by 50Hz mains) at approx the same rate as the burst rate" (HVF 117). One hump of the rectified mains per bang. So the starting voltage is repeatable not because the bank refills quickly but because the two rates are locked together, and he hedges it himself: "so I'd hope (but cannot guarantee) that it's a reasonably constant starting bus voltage".

Read that as the condition rather than the number, because it is what breaks when either rate moves. Run at a repetition rate that is not a whole fraction of the rectified mains and successive bangs start from different points on the refill, which is a bang to bang variation with nothing wrong with the bank at all. And on a half wave supply the humps come at half the rate, so the same 100 bangs a second get one hump between every other pair, which is the mechanism behind the residual bus voltage named at the top of this page.

A doubler also takes both half cycles, so it leaves no residual voltage to branch the arc at the start of a ramp.

The alternative that gets rejected, and why. Raise the AC with an autotransformer to about 300 V and a plain bridge gives 424 V without a doubler at all. It has one real advantage: a gate DC-DC rated for 400 V of isolation is over by only 1.06 rather than by a lot [derived, 424 over 400]. But the bank's energy falls by 2.35 times [derived, the square of 650 over 424], so the same bang eats 46 per cent of it instead of 19, which is precisely the thing the exercise was for. And it wants a couple of kilowatts of variac.

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 above, under Which arrangement.

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 eighty times at half a hertz, which is just the reciprocal of the duty, 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)  =  66 V
Q  = C·dV                     =  0.66 coulombs
I  = Q / t_ramp               =  0.66 coulombs in 25 ms  =  about 26 A

Twenty six amps while the ramp runs. The same machine at half a hertz averages about a third of an amp out of the bank, 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. What the wall sees is a different number again, and it is above.

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. Both are floors: they take the ramp's draw as flat, and it is not. 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 lands in a different section. 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 under Which diodes above]. So the capacitance that holds the top of the ramp up is bought back at the rectifier, and the two choices cannot be made one after the other as though they were separate.

The second criterion, which is not the binding one here

Everything above sizes the bank against one burst. General purpose guides size it against the mains instead, on the ripple between rectifier conduction intervals, and Barnkob gives that method with the intervals it needs (his rectifier chapter):

V_ripple = I · t / C          t = conduction interval, seconds at 50 Hz
  1 phase, half wave      0.02
  1 phase, full wave      0.01
  3 phase, half wave      0.0067
  3 phase, full wave      0.0033

His own conclusion from it is the argument for three phase in one line: to hold ten per cent of ripple at 20 A takes 6000 µF on single phase full wave and 2000 µF on three phase, a third of the bank for the same job. The arithmetic reproduces both, 6154 and 2031 µF [derived, on his intervals].

Run it on the machine this page works from and it says almost nothing, which is the point of putting it here. That bank averages about a third of an amp between ramps, so on single phase full wave the mains ripple across 9900 µF is 0.33 V [derived]. Against the 66 V one 406 J ramp takes out of the same bank, the mains criterion is two hundred times slacker.

So the two criteria are not alternatives and the larger wins. On a coil the burst wins by a margin that makes the ripple figure decorative, and Barnkob says as much in the same paragraph, that the ripple calculation "does not make it alone in choosing the filtering capacitance for the DC link" and that the tank may demand more. A builder who sizes only by the ripple rule arrives two orders of magnitude short.

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.

And the reason to build it rather than reach for a variac is stated plainly by a builder who publishes the wiring: 山猫 adds soft start "为了使用特斯拉线圈更加 简便，脱离调压器。接好地线，插电即用！", to make the coil simpler to use and to be free of the variac, connect the earth, plug it in and go (hvdiy 35139). His implementation is the ordinary one and his sizing rule is the useful half of it: charge the main filter capacitor through the limiting resistor, then let a time relay drive a contactor that shorts the resistor out, and "延时时间略大于充电时间即可", the delay only needs to be slightly longer than the charging time.

Which is where the two time constants below earn their keep. "Slightly longer than the charging time" is not a number until you know whether you are charging one bank or two in series behind a doubler, and those differ by four.

The other bank, which is not this one

Everything above is the bus bank: thousands of microfarads behind the bridge, charged from the mains and sized on droop. The tank is the other thing that gets called a bank, nanofarads of film in series with the primary, and it is sized on a different quantity entirely, its ripple current at the ring frequency rather than its capacitance.

That used to be argued out here, which put two parts with two unrelated ratings under one heading and left both of the machine's tank sections empty. It is now the MMC is sized by current, including the ripple ratings and their conditions, the two editions of the TDK sheet that disagree about the same part, why smaller capacitors carry more current per nanofarad, and why parallel strings do not share the load evenly.

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.

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.
• 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 9 A RMS at five hertz, 3 A at half a hertz, both floors, which set 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.

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.

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

What goes wrong

• 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.
• A diode fails early with everything apparently within rating. It was sized on the average forward current. The peak inside the conduction window is seven to sixteen times it, and the table under Which diodes says which.
• The arc branches in the first few milliseconds of a ramp. Residual bus voltage from the previous bang, which a half wave supply leaves and a full wave one does not. Either arrangement that takes both half cycles fixes it, and the bank does not empty between bangs in any of them.
• 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.
• 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.

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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.