Do caps stay "fully formed" even under minimal load? (SS)


Theoretical question. If my ss amp can easily handle 120wpc into 8 ohms, but most of the time I'm only using less than 1 watt, will the "unused headroom" (if there is any?) of the power supply caps degrade? Suppose one day I get very hard to drive speakers and all of a sudden I'm pushing 15+ watts. Will the caps be "out of shape"? 
lostark
Nope. Caps are the ultimate couch potatoes: less exercise, better performance! Tubes are even better. Tubes can just lay around for years, people actually pay more, they are so well rested you can't believe!
Filter caps won’t see a difference if the speaker draws one watt or fifteen watts since the voltage across it is constant from no load to rated power draw. The headroom is in the mains transformer and power transistors.
@gs5556  - that's not really true. The caps are charging when the rectified voltage from the transformer is higher than their current stored voltage, and discharging otherwise.

With 60hz power and assuming a full bridge rectifier, the caps will get a charging pulse every 120th of a second. Once the caps reach the steady-state voltage after power-up, if the load on the supply is fairly light, the rectified voltage will only be high enough for current to flow into the cap for a small percentage of the time, but you'll still always be charging or discharging the caps since there is always at least a small current draw from the amp.

The amount the voltage changes on the capacitors every cycle will depend on the capacitance and the current draw. Assuming a low bias class A/B amp, this can be significantly different between low-power operation and high power.

With a class A amp, the amp is always drawing a lot of current, so the difference between idle and fairly loud playback will be much less, but the amount of current flowing into and out of the caps will be considerably higher.

Lots of current flowing into and out of the cap puts a lot of stress on it and will shorten it's life, so the power supply caps will last longer if the amps current draw is low.

I suspect that any operation of the caps will keep them working optimally, but while I believe strongly that capacitors that pass audio signals change noticeably during break-in, I'm not sure that this applies to power supply caps. It's quite possible that power supply caps can respond quicker to changes in current requirements after break-in, which would significantly change the sound of an amp. 
DBS is a patent for a reason. The sharp also apply it inside the DAC, preamp, etc.... Some of those monster botique caps never form in the circuits they get put in......
worst of all is pulse units, like class d amps. Someof the capacitors and other components are stressed to the maximum, all the time.

Imagine a 400 hp car. with a 7k rpm redline.

Imagine said car running at 400hp and at the 7k red line, all day long, every time it is turned on.

How long do you think it is going to last?

answer: not very long.

Ladies and gentlemen, I rest my case. Class d and pulse drive systems of similar nature (pulse power supplies) do not last very long, due to maximum stressing of certain components therein.

then we buy..based on appearance and price, etc, knowing nothing of all those non moving parts or what they do.

these problems can be addressed and they are, at lets say... the aerospace satellite construction level. But the average consumer won’t pay for it, so the vast majority of these efficient pulse designs have short lifespans, specifically due to component over stressing.

Eg, $2 led light bulbs with an advertised 50,000 hour lifespan, but a 20 cent 5 year (if that) lifespan power supply built in to it. right, right.....

No heat, no moving parts.... but the electromagnetic fields are going nightmare level haywire in there.

Hint: this stressing is tied to the achieved efficiencies. These efficiencies.. come at a cost to the environment (due to high failure rates) that is so high, re the idea of short lives, that the world would be better off if you stayed the hell away from almost all of them.

the manufacturers, the mainstream ones, they all bought into it, like the two of you are good friends, sharing kisses, saving the world.... but really, back in the real world.... they get to sell sell sell and sell again and again, all while filling landfills, and you get your monkey fix with the new stuff all the time.
I stand corrected. I should have said the voltage across the filter cap is constant, plus or minus the 50mv ripple voltage.
"...like class d amps. Someof the capacitors and other components are stressed to the maximum, all the time."


Output Mosfets connect output to positive and negative supply with duty cycle dependent on desired output voltage.  When duty cycle is 50% average output voltage and current are zero.  At this point Mosfets' current would be 100%, switching load between positive and negative voltage, if not for the fact that load is not resistive (speaker).  Not only that speaker presents very high impedance at the carrier frequency (about 500kHz), but there is also low-pass filter between Mosfets and the speakers (Zobel network), that filters out 99% of the switching pulses.  Voltage on the speaker becomes zero (with very small switching noise) and current thru both output Mosfets is close to zero.  It is very easy to verify, just by placing hand on them, or measuring amp's supply current.  My Rowland model 102 small class D amp total supply power was 10W with no signal (no stand-by) to 500W at full output power.

Switching devices in any electronics switch all the time, often heavy currents, and last for decades, including every computer where switching regulators often switch tens of amperes.  SCR's in spot welding machines switch hundreds of amperes and also work for very long time.   We don't have to look far - diodes in linear power supply switch large currents constantly 60 times a second.

To sum this:  No, switching elements in class D amplifier are not "stressed" all the time, but even if they were it doesn't mean at all that they have to fail.