Electronic – What happens when a Cap fails

Short: Add a 1 ohm resistor in series with the transformer :-).

Longer:

A "perfect" transformer and 'perfect" capacitor will have infinite current spikes, as I know you realise.

While real world results will vary with transformer maker's 'ethos and philosophy', the real world experience is that you wil usually get superior results by adding a small "conduction angle spreading resistor" in series with the transformer winding feed to the capacitors. This is counter intuitive to what you may expect from an efficiency point of view and is often not done in practice. Theoretical calculation of the effect of such a resistor is surprisingly annoying but simulation will show the effects instantly.

Given that the mean DC level under load is 0.7071 ( = sqrt(2) ) of V peak, you have quite a lot of headroom to work with and can afford a modest amount of drop in the series resistance. There are several scondary effects which may be useful depending on environment. Spreading the conduction angle improves the power factor of the otherwise very peaked load - but probably not enough to make a difference in meeting or failing formal power factor requirements. Sometimes more importantly, spreading the conduction angle greatly reduces peak loads on the diodes and reduces EMC issues (ie less radiated electromagnetic noise) - probably not an intuitive effect of adding a few ohms of series resistance.

Lets have a play with some figures:

You have 15 VAC secondary voltage and are aiming at 12VDC at 2A.
Assume for now that about 15VDC minimum on the filter caps is acceptable 9giving the regulator 3V headroom minimum).
Vpeak is 15 x 1.414 = 21.2 V
Load power is VI = 12 x 2 = 24 Watts.
If you managed to filter this well enough to achieve say about 20VDC on the cap you would dissipate Vdrop x I = (20-12)x 2 = 16 Watts in the regulator and "as a bonus" achieve massive ripple CURRENT in the caps but little ripple VOLTAGE. This does not seem like a marvellous idea :-).

If you can manage to spread conduction over 25% of the voltage cycle you will get mean current during conduction down to 4 x Iavg = 8A.

Assuming 21V peak, 25% conduction occurs at about 19V transformer output, and a very useful 50% conduction happens at just under 15V. See graph below.

This suggests that inserting even one ohm series resistance is going to have a substantial effect. If the 8A mean that is required for 25% conduction is dropped across 1 ohm the 8 volt voltage drop is going to ensure that the 8A does not happen (as 21-8 = 13V which is lower than the 15V DC target this was based on).

If 50% conduction occurs then mean current during this period will be 4A and mean drop across 1 ohm would be 4V so this may be "about right" as if the filter cap was at about 15V you'd get (21-15)/1 = 6A peak at waveform peak - and as the cap will have "rippled up" in voltage by then you'll get less than 6A). And so on.
Yes, you can analytically work out what happens. But, just put 1 ohm in the simulator and see what happens.

This has the effect of putting MORE ripple voltage on the capacitor(s), LESS ripple current, less regulator losses and less transformer losses, less diode EMI.

The series resistnce could be in the transformer but then addes to heat generatoion inside a relatively costly component where you'd rather be trying to optimise power transfer rather than heat loss. A 5 Watt 1 ohm resistor will probably work OK here. 10W would be safer due to peaks. eg 4A at 50% = I^2R x 50% = 15=6W x 0.4 = 8W BUT waveform is complex so actual heating needs to be calculated.

Note that in many cases the ripple current rating of two capacitors is superior to that of a single capacitor of equal total capacitance.

Use 105C (or better) caps as a matter of course in this sort of application. 2000 hours+ a good idea. Cap life ~~~ 2^((Trated-tactual)/10) x Rated_life

As ever, a full circuit diagram would be invaluable - even if to show that there is nothing much more present than has been stated.

VAC = 220V so Vpeak = 220*1.414 =~ 310V. 180V DC/310 =~ 0.58 This is the sine of thge angle when the rectifiers start (or end ) conducting + 35 degrees. For 35/90 of the cycle the voltage in is below Vdc so the cap MUST provide the motor current. If you do not have any energy storage in inductors then the cap is seeing a ripple current of in the order of the motor current and peak currents will very likely be higher (depending on transformer and wiring resistsance and more.)

As dissipation will be in the order of proportional to current squared you probably have about 10 x rated dissiation due to excess ripple current.

Nichicon are a well respected brand. Chances are the actual ripple current capacity on a genuine Nichicon meets or exceeds specifications. But it is unlikely to exceed it by enough to save you here IF the circuit is as it seems. It is possible that the cap is a counterfeit. This definitely happens and Nichicon are a well enough known brand that people MAY counterfeit them, although I have no specific knowledge of this happening in this case.

UUCAP I know not.
It is not unusual for little known Asian components to not come close to spec sheet claims.
In this case it appears that they exceed the specs handsomely !!!!
I'd not complain!
But do look at the actual ripple current.
A small sense resistor in the cap ground lead will allow a scope to be used with due care (or in the "hot" side with an isolation device AND if you know what you are doing. Or a Hall clamp / proximity meter or ... .

Note that cap lifetime ~+ Rated hours x 2 ^ [(Trated-Trun) / 10 ]
It is usual to run a cap at WELL below rated temperature.
30C below = 2 ^ (30/10) = 8 x rated lifetime.
So a 2000 hour rated cap would last about 2000 x 8 = 16000 hours ~= 2 years.
The larger margin the better.

Note that an Al electrolytic cap with NO applied voltage, held at high temperature will die faster than when voltage is applied !