charging audio system using a stator

I wouldn't say you are a broken record, but you may be going around in circles. I want to straighten you out (even though I get along better with bent people!). If your "stator" isn't an alternator, what is it - a generator, or a motor? If your "stator" is "constant current", does it keep increasing the voltage until the load accepts the the current? I think you will find your charging system is an alternator. (Alternators have a stator and a rotor as do generators and motors). Not that (I think) it makes any major difference - it outputs whatever power at whatever RPM. I also think you will find it is NOT a constant current system, but capable of a certain (max) current at a certain speed/RPM. We could do the complex theory - like a X,000W RMS amp at 1/2 volume hence 1/4 power hence (X,000/4)W RMS hence (X/4 x 1000W/12V) Amps required and will the "stator" keep the battery charged if an average of Y RPM? What about the "frack that crap" approach? Instead, take your pick - is my system up to it, or is my system STILL up to it? (IE - it was big enough, but are my alternator brushes deteriorating; is that new amp to big' etc?)    The solution - a voltmeter. A voltmeter (at the battery!) tells you everything. EG - above ~13.6V - we are charging. - above ~14.4 we are overcharging. - 12.7-13.6V - we are charging, but not at full voltage. (Ok, so our load is bogging down the alternator, or the alternator isn't big enough, but...) ~12.6-12.8V - battery is fully charged. ~11.6-11.8V with no load - we have a FLATTERY (Flat Battery - a term AFAIK coined by a Darwinian (in Aus) known as "Paddles") 10.6V - the general "worst case" voltage for a battery assuming a load. I am now talking the bottom line. It doesn't matter what know-it-all R-Soles like OldFart think your system can do (based on your supplied average RPM with average Watts-volume output etc)..... So your 1,000W rated alternator happens to put out 10,000W at 3RPM, or your 250A alternator only puts out 50A at 2,000 RPM. Is it enough?    Voltmeter. The ultimate confirmation. Does that make sense? Do you require further clarification? Mind you, IMO, I love redundancy - and that's for cars etc. In a boat, I would DEFINITELY want a second battery. (Unless you have a back-up rope-pull, or good enough fins/flippers to push start?) BTW - here in Aus being "down under", we merely pull out our sea-anchor. We then start falling down hill and eventually gain enough speed to start our outboard/inboard engine. It's equivalent to push starting a car (downhill). I've been told that you can't do that in the northern hemisphere. Apparently water there is "flat". Is that really true - or just more "I read it on the internet" bullsh? But back to serious doodiee... I just purchased a pre-loved 12V refrigerator that I am currently (no pun) testing. I'm running it off a 12V battery with no charger. So why is its voltage higher now than when I started? Is my DVM (meter) battery flat, my DVM stuffed, or is it AC being "reflected" by the fridge's inverter?    (Damn - I should have bought that Tue-RMS multimeter!) Alas, a few of the complexities of electricity for us peasants with inadequate primitive tools. But we make do. Do you have a multimeter? Or a voltmeter that you can connect across the batteries? If so, you should be able to get a feel for what RPM for what "volume" output (with/without lights etc). 40-45A means about 500-650W max output.    The capacitor has negligible effect on reserve time. I connect my 2nd battery directly to the alternator (across the 1st battery) ONLY whist the alternator is charging. The batteries are otherwise isolated. Hence I have my untouched 1st battery for engine cranking etc irrespective of whatever the load (fridge, amplifier) does to my 2nd battery. That relates to my "redundancy" desire mentioned above. And I hear the birds chirping. I think it's time for bed! Thanks too for you compliment! ZZZzz... Peter.

Ah - now I know the system - it's a permanent magnet rotor. (Sorry - those birds must've been chirping loud this morning!) The stator is a stator like any other. But instead of its rotor being a variable current (controlled by a regulator) and hence varying the stator output, it is a magnet. You cannot therefore vary its output (other than via RPM). Instead, you dump the excess power. The regulator is probably 650+ Watt Zenor diode. Hence why it is water cooled. Many old motorbikes use that system. I had a bike with a similar 120W system but that used SCRs - Silicon Controlled Rectifiers - that shunted/shorted the output to ground when voltage exceeded (eg) 14.4V. (Or did it only pass its output when batt voltage was lower than whatever voltage?) (SCRs are diodes that can be turned on using a relatively small current. They are like a solid-state relay, though once turned on, they latch on until the current through them drops to zero.) Your engine looses about 1HP (~745W) when producing its 650W whether it is used or not. Apart from what happens when the regulator blows, it will look the same as any charging system. If the regulator is a "pass mode" system, hopefully it fails in open mode and doesn't pass current. (Diodes have about a 50/50 chance of failing as short circuit or open circuit; but I'm not sure with SCRs etc.) If the regulator is a dump system, the same failure modes occur though a short is likely to burn open - especially if the battery can blow it. These days components like power FETs could be used instead of SCRs and Zenors. Although I suspect if they were "advanced" enough for that, they would use some brushless induction rotor system for "normal" alternator control, and up to 1HP more output (LOL). But I digress. If that regulator fails so it passes or doesn't dump, then the alternator's (no-load) voltage will be proportional to its speed. That could blow downstream equipment - but that's why that equipment has fuses and spike protection etc. The battery(s) will also act at a big capacitor and should limit the rate of voltage rise. Interesting system to contend with.... the issue being how bad can any overvoltage be (and it won't be as long as demand matches or exceeds its output capability). A lot depends on what regulator circuit/components they are using. Now who was that brilliant person that suggested further reasoning could solve this issue? And who was the clever person that investigated and provided more info? Yep - iteration. Investigate, discuss, brainstorm etc. (Edwards Demming would be proud!) NEXT! (Thanks Mr Big! Alas I'd better go now... I hear another "bird" calling, and I'm not big enough to ignore that one!)

Jason - I apologise for my rude opening line in my first reply above. I confused you and this Topic with another lengthy one. Sorry! And in my last reply I got so fixated with what the system is - and I was probably feeling guilty for the second time because I was wrong - ie, it was not a regular alternator, but a permanent magnet system (like a magneto)... a "stator" system as you called it. (Apols again - I'm not up with industry jargon, though I see lots of ambiguous terms like "analog" CDMA phones; or and ignition systems that are Hall (Effect) or "magnetic" - ie, reluctor - when both are magnetic; etc.) So, reconsidering.... No matter how you power your audio or whatever, I assume you never want to have a flattery (flat battery)? (Unless rope-pull or push-start is a viable fall-back?) So why not a second battery from which you power your audio? If you flatten that battery, you still have the first battery as per normal. The 2nd battery is otherwise a spare in case your 1st battery dies. The assumption is that the two batteries are ONLY connected together whilst they are charging. Hence a typical auxiliary battery situation. Whilst I use the alternator's charge lamp circuit to energise a relay to connect auxiliary batteries, you can't do that (ie - no charge-lamp circuit; it's a permanent magnet stator system). So instead, an overvoltage sensor that connects the 2nd battery when the 1st battery or the stator output is above (say) 13-13.5V. There are some considerations like switching hysteresis and desirable delays (ie, a short dip below 13V should not disconnect the 2nd battery, etc). And although elsewhere I recently mentioned a kit with an 80A latching relay, I think I'd prefer a normal relay. (The few Watts needed to drive a relay is negligible in a 650W system.) How does that sound? You have battery redundancy. (And extended battery reserve time if the engine is a no-go.) You get extra AC (noise) filtering with the 2nd battery. The cost is the voltage-sensing relay, and the second battery (whatever size - either the best bang-for-bucks, or identical to the first battery, or whatever is needed to suit enclosures, audio systems etc). (Then there are optional voltmeters or under-voltage alarms or cut-outs etc.) If your stator system is insufficient, you could upgrade it later - or add a wind or water generator if efficient enough etc. Any other suggestions or thoughts? Regards, Peter.