Voltage rating on filter caps

[Malcolm Irving]

To get a perfect equilibrium you just need the DC supply to be constant. If that could be achieved, there would eventually be current through the resistors but no current through either capacitor (except leakage current). In practice, unless you use a battery supply, there will always be some ripple, etc.

For AC and during any transients the capacitors 'win' the battle to establish the voltage sharing (assuming they are quite large values), while in the longer term, the more things settle down to DC, the resistors take over.

EDIT: by 'perfect equilibrium' I just mean all voltages and currents are constant.

[Malcolm Irving]

Leakage current through the capacitor can just be regarded as a resistance in parallel with the capacitor, so this resistance is in parallel with the balancing resistor.

For the balancing resistors to work well, the current they draw must be large compared with the capacitor leakage current.

[jmaf]

There is not zero current at equilibrium.
We are dealing with a rectified AC voltage and the 1st storage cap.

Pulsated DC, not AC.

There is charge current flowing into the capacitors from the rectifier whenever instantaneous rectifier voltage is above the capacitor voltage and load current from the rectifier being delivered at this time too. When instantaneous rectifier voltage is below the capacitor voltage then load current is sourced from the capacitors. This is an ongoing thing which happens every half cycle of the High voltage AC winding. It never becomes static.

You mention it just ahead, it's called ripple, no need to school us on that. But when there is no load this current tends to zero and you should design your circuit for the no load voltages + a margin, unless you're the kind that forgets that once a while an amplifier runs without power tubes - because that's when you blow the capacitors up.

That RIPPLE CURRENT is important cause capacitor lifetime is dependent upon its self heating and the self heating power is the ripple current squared x the ESR (Equivalent Series Resistance).

You're throwing more concepts into the conversation and steering way off, that's a very complex subject you've just branched into (capacitor failure modes). Capacitors fail for numerous reasons. RIP Bob Pease.

Not going to argue about it anymore, just will point out that my day job is Electronic Design Engineer (30 years experience with 10 years before that as an Electronics Technician) and I do this stuff for a living.

Here in Brazil we had shortages of components in the 1980's and we had to do with whatever we had and it was common to use different capacitors and that's how I know, because I've actually built tens of amplifiers. When I open any Brazilian tube amp for maintenance it's always a bunch of such situations, dissimilar components, modern and old resistors and caps mixed up.

In practice when you build 2 different capacitors in series in DC and leave them there, they always conspire against you and blow up.

I also have a degree hanging on the wall but that's not part of my argument and it's not really an argument IMO. Even Richard Feynman was wrong on occasion and you're no Feynman.

Been thru' this theory and the calcs 100's of times. TRUST ME

Do you have some SPICE available near you right now? Try something simple like this.

Let’s say we have two capacitors in series, initially uncharged. Energize them from a constant DC voltage. (No resistors in there yet.) The current through the two capacitors must be equal at all times. Charge is the time integral of current, and obviously time is the same for both capacitors, so the charge in each capacitor must be the same. The definition of capacitance gives V = Q / C so the voltage is shared in inverse proportion to the capacitance (as Gingertube says).

When it's a DC totem it's electrostatic (for circuit design) until you demand load from it. Picture this there is a plate from the smaller capacitor, that plate holds charge Q. It is physically impossible for more than Q charge to flow onto the next capacitor in series, therefore you're charging up the biggest capacitor based on the smallest one in the chain because the only electrostatic charge you've got is from the smallest capacitor.

That's where the logic in previous arguments is missing it: for this calculation you do not use the nominal C of the bigger capacitor, you use the series capacitance to determine how much charge there is on the series capacitance. Charge ends up being the same on all caps and the bigger capacitors are only used up to the smallest value in the chain, so the 220uF cap becomes 30something in this specific case here, it never reaches 220uF capacitance. Only charge proportional to 30something uF will be built on those plates.

Surely such a perfect equilibrium seems unfeasible, as (given component value tolerance to nominal) the cap ratio is unlikely to match the resistor ratio; so current flow between the nodes seems inevitable. And any imbalance in the cap's leakage resistance, which seems likely, would compound this?
Sorry if I'm in over my head

You are correct, there is no perfect equilibrium but assuming that the voltages will tend to exceed the 100v nominal cap without the resistors is the secure theoretical reasoning for this circuit.

That Fender schematic is descriptive, not prescriptive - it's not good practice and it shouldn't be taken as the correct approach for a theory and design forum discussion IMO.

On real circuits here although there is no perfect balance like that, we still always plan for the empty load + a margin. Every totem should handle at least 30% (that's just my margin as mentioned earlier post, not based on any standards that I know) more nominal tension than the empty circuit tension.

Designing a circuit with a 100V cap in series with a 400v one and counting on this DC proportion to hold is timebomb design. The resistors are the only guarantee in this particular circuit, which is my argument for Gingertube. Take those out and eventually it'll blow up during some specific situation, and it normally happens live based on Murphy.

[Malcolm Irving]

…
Picture this there is a plate from the smaller capacitor, that plate holds charge Q. It is physically impossible for more than Q charge to flow onto the next capacitor in series, therefore you're charging up the biggest capacitor based on the smallest one in the chain because the only electrostatic charge you've got is from the smallest capacitor.
… .

When a capacitor is charging up, exactly the same current is always flowing into the +ve terminal as flows out of the -ve terminal.

Electrons flow into the -ve end, building up a negative charge on that plate, at the same time electrons flow out of the +ve end at exactly the same rate, building up a positive charge on that plate.