posted April 17, 2000 11:55 PM            

Jeff

posted April 18, 2000 10:10 AM         

I have a 50 amp Jacobs Accuvolt in my system and it seems to work. My voltage to the amps stays at 14.4 when the engine is off.

What is the REAL story on the JA?

posted April 18, 2000 10:46 AM            

Say you have some induced noise in your main power line feeding back to a couple of caps. From the caps it goes into an amp that is driving some mids. Basically you have a little AC ripple, whether it be coming from the alternator or what ever.

Don't caps help filter some high frequency noise when ran in parallel?

Just a thought for another use of the cap other than voltage stabalization...

"Hey man we're not looking for any trouble... Well maybe trouble look for you!"

------------------
(Art S.)

posted April 18, 2000 11:50 AM            

Not to complain or anything, but what would really drive the argument home would be an exact replica of the experiment you did on "the cap" (15F) but with a low ESR, 1F cap and a "high" ESR version of the same if if such caps are even available. I believe you have proved that "the cap" is in reality, a load on the alternator if anything, and that it is a grossly inadequate current/power storage device, and so is of no conceivable use in an audio system other than to get points.

posted April 18, 2000 01:14 PM            

thank you Richard for the great electronics lessons, and the test. the test is really what helps us see.

but there are a few points of interest;

first being that Cade was actually right, in that ESL of the cap actually dominates the impedance of the capacitor. the ESR was 0.01 ohms, and the ESL contributed about 0.7 ohms if i remember correctly. how much ESL do the typical 1F caps have?

second, i agree that a comparison with the regular caps is extremely crucial to closing this argument. if the reason for the CAP not working is due to the high impedance, then the logical thing to do to disprove this is to show how a cap that has low impedance DOES work. right now, that seems like a little unfinished business. since you used about 2200 watts of power, 2F would be a good test, i would think.

thanks, Richard!

posted April 18, 2000 02:28 PM            

Golf clap for BlackMax88! "clap clap"

Let us review.

Quote from Richard Clark..
"Now I have been told that the ESL value of the giant cap is 0.2 mh. Somebody check my math but I think this would put the reactance of the cap near .063 ohms at 50 Hz. This means that if we wanted to refresh our amps at a rate of 50 Hz (seems reasonable if we were playing bass real loud) our ESL of .07 ohm would be in series with our .017 ohm ESR for a total value of
.08 ohms. "
Ok so the facts in this is: (15 Farad cap)
1) ESR = .017 ohms.
2) ESL = .2 mh with a reactance of the cap near .063 ohms at 50hz
3) ESL of .07 ohm would be in series with our .017 ohm ESR for a total value of
.08 ohms

Now let us review the 1 farad cap.
1) ESR = .0034 (hmmm nice and low)
2) ESL = 10 mh

posted April 18, 2000 02:51 PM            

It would be a much higher frequency than 50hz. Most switching power supplies in amps switch and draw current at many thousand times a second, most at 20,000 hz or above. The inductance would be a much more important part of the equation than the ESR, according to an old article I read that Richard wrote in A2TB

posted April 18, 2000 07:33 PM            

I've scanned the article and put it online at
http://www.damnloud.com/ass_cap1.jpg
http://www.damnloud.com/ass_cap2.jpg
http://www.damnloud.com/ass_cap3.jpg

These images are fairly large: 132K, 149K and 96K respectively.

Focal

------------------
1998 Acura EL
IASCA Amateur 1-150
http://www.damnloud.com
COM0008583

posted April 18, 2000 08:17 PM            

------------------
OBJECTS IN THE MIRROR ARE MORE STABLE THAN THEY APPEAR
My Install
Learn How to Make Fiberglass Kickpanels:
Nasty's Kickpanel Fabrication Page
Lot of step-by-step pictures and links for ideas and help!!!

posted April 18, 2000 09:10 PM            

As I understand it, that's the reason for needing 1/0, 2, and 4 etc. guage power cables. It's not the DC resistance of the cables that makes such large diams. necessary, it's the inductive reactance.

Does the PWM pwr. supp. switch at a variable frequency, dependent on the current demands of the output stage?

posted April 19, 2000 12:35 AM            

A capacitor can filter high freqs as well as low freqs depending on whether it's wired in parallel or series with a speaker.

An inductor also can work the same. The reason I know for not usually doing it this way is that the amp "sees" a short across its leads when the first component in the passive xover network is in parallel...

"The word of the day is SHUNT."

------------------
(Art S.)

posted April 19, 2000 04:15 AM            

The author measures 17 milliohms ESR (I will keep the values for ESR, ESL and inductive reactance for the sake of simplicity), so there will be a loss inversely proportional to the rate of current flow, even if there is no ESL. However there is ESL, 0.2 millihenries, which equates to 63 milliohms of inductive reactance to be added to the ESR if we assume that the PWM power supply is switching at 50 Hz to deliver even the nominal 100A of current.

By the way, what would the switching freq. be if the amp was drawing more current? (less I suppose) So what effect would a lower switching freq. do to the inductive reactance value? Would it stay the same, or would it vary in a frequency dependent manner?

In the Test Results/"Real World intrudes" section (green inset): the cap draws 560 amps for some period of time. (I can't read the scope shots in the scanned/posted article) Can the battery produce current if there is a voltage drop across its posts to <10V? I don't believe that it can (not according to Richard's argument, and anyway, the amp will go into protection mode), so there will be some limit in the amount of current that can be drawn by the cap from an automotive battery, therefore lengthening the time required to fully charge it.

The author does not account for these factors.

It looks like his test method (to model ESL) uses a power supply that is able to maintain its voltage, unlike a battery. Could that be because it is plugged into a wall/110-220VAC outlet and or that it has voltage regulation circuitry/capabilities? He also claims that the inductance of "the cap" is .2 microhenries, which is differs by a factor of 100 from the value used by Richard of 0.2 millihenries. Even so, the ESR alone is enough to show that they don't work.

I hope someone will email Mr. Lewis so he can respond to our forum.

posted April 19, 2000 12:52 PM            

Most batteries have an internal resistance of between .5 and 1.5 ohms, depending on the size of the cells and the lead content.

[B]By the way, what would the switching freq. be if the amp was drawing more current? (less I suppose)

The same in most amps. The rails stay at the same voltage no matter what the output is (in theory,they do sag). Class D amps do change the frequency and pulse width of their power supply with the input signal.

[B]It looks like his test method (to model ESL) uses a power supply that is able to maintain its voltage, unlike a battery. Could that be because it is plugged into a wall/110-220VAC outlet and or that it has voltage regulation circuitry/capabilities? He also claims that the inductance of "the cap" is .2 microhenries, which is differs by a factor of 100 from the value used by Richard of 0.2 millihenries.

If the power supply was stable enough to hold voltage at that current, you would see the sag in voltage without the CAP. Whatever test he did, real world or not, the CAP helped.

[B]Even so, the ESR alone is enough to show that they don't work.

Not necessarily. The reactive inductance is probably much more important. What do you think the reactive inductane of the battery/power wire is in most systems?

posted April 19, 2000 01:03 PM            

quote:

---

Originally posted by STACKHOUSE1:
Just some thoughts here..

Say you have some induced noise in your main power line feeding back to a couple of caps. From the caps it goes into an amp that is driving some mids. Basically you have a little AC ripple, whether it be coming from the alternator or what ever.

Don't caps help filter some high frequency noise when ran in parallel?

Just a thought for another use of the cap other than voltage stabalization...

"Hey man we're not looking for any trouble... Well maybe trouble look for you!"

---

posted April 19, 2000 01:09 PM            

quote:

---

Originally posted by GoneGator:
StackHouse,

Capacitors are usually used as high pass filters. Inductance coils are usually used to filter high frequency noise.

---

posted April 19, 2000 04:50 PM            

What occurs is that as more and more current is drawn from the alternator, the output impedance of the alternator produces a voltage drop proportional to the current drawn. Therefore, as the current draw increases, the voltage at the terminal of the amplifier decreases. This is where the capacitor comes into play. If a capacitor is placed in parallel, the voltage is smoothed due to the fact that the capacitor discharges "through the ESR" to the lower potential at the power terminal of the amplifier. Well, not really. What happens is that the alternator and capacitor form a balance so that the highest constant voltage appears at the power terminal. Therefore, it won't necessarily draw all the current possible from the alternator AND THEN from the capacitor afterwards. Because technically, the maximum amount of current to be drawn from the alternator is greater than the rated current and can be found by grounding the alternator to the chassis. hehehe don't try this at home. It would be the alternator's voltage divided by the shunt resistance...until of course the device failed.

quote:

---

Assume that when a bass note hits it forces the amplifier to demand current of 100 amps (100 coulombs per second) and further assume that the electrical system (Alternator) can only supply 90 amps (90 coulombs per second) Then a capacitor must make up the difference. It must release current AT THE RATE OF 10 amperes (10 coulombs per second). If this is done over 0.02 seconds, the number of coulombs released is (10 coulombs per second) x (0.02 seconds) = 0.2 coulombs.

For more on electricity go to Read & Learn:
Electricity http://encarta.msn.com/find/Concise.asp?z=1&pg=2&ti=03AEF000

[/B]

---

posted April 19, 2000 04:59 PM            

As I understand it, that's the reason for needing 1/0, 2, and 4 etc. guage power cables. It's not the DC resistance of the cables that makes such large diams. necessary, it's the inductive reactance.

Large wire, especially high strand count, has extremely high inductance. The less the strands, the lower the inductance. Smaller wire has less inductance, but of course it doesnt have the current capacity that the larger wire has. That is why we use caps.

posted April 20, 2000 12:01 AM            

I was thinking early on while reading this thread that this was going to be fun seeing Richard prove Cade so blatently wrong, I mean did this guy really think that those caps were raising the system voltage as he described? And I thought that this could be somewhat damaging to a company such as Phoenix Gold's (or Alumapro's) sales of such a product. Not to mention their credibility in a technical market such as car audio. But PG is a well respected company with many years of outstanding car audio products. So this is leading me to this conclusion.

Cade's title (per his signature) is a Technical Specialist for Phoenix Gold.
Understand that I am certainly no expert at this stuff, and have absolutely no clue whatsoever as to Cade's expertise or qaulifications. But when a "Technical Specialist" for any company would seem to have such a blatant misunderstanding of a product that they are marketing, such as the PowerCore, it makes me wonder if this is more Sales and Rep talking than technical talking on his part. If he were to come out and say at the end of all this, "Yeah, I guess you were right, our EE's and myself were so wrong and misundertanding of this capacitor all together, I better let our marketing department know!", I would be very surprised. I will have to give Cade the benefit of the doubt on this subject and say that this is the case. Being a Tech Spec for any company in the field of electronics would necessitate you to have as least the basic electronic theory shown here.
I feel that his thinking has been clouded and is being caused by his defense of his product rather than ignorance (not that anyone called him such). As Richard has made several points in this thread, sometimes our minds are tricked into thinking things are better, based on time or money spent. But there is also the pride factor.

Anyway, sorry to ramble. Just some thoughts and a little understanding and (defense?) of Cade's position.

I am very looking forward to the conclusive lesson, and hope, as mentioned, there is some comparison to the standard 1-2 farad caps that are mainstream.

posted April 20, 2000 12:34 AM            

quote:

---

Originally posted by Richard Clark:
guys you all have some interesting things to say---shows a lot of good thought patterns-- we have sure come a long way since two weeks ago where we were challenging the basics of ohms law! I will cover most of that stuff when I wrap things up but I would like to give Chris another day to comment---I will try to call him tomorrow---however let me say a couple of things. Figuring the output impedance of an alternator is not always as easy as loading it's output and measuring the drop. The problem is that the regulator controls the field pulse width. Loading the output causes the field voltage to drop causing the pulse to widen. Eventually the field pulses cannot widen any more therefore the ability of the rotor to generate porportional current is diminished. This then causes the output to drop which causes the field voltage to drop more which causes the output to drop........and on and on and on. There is a complex relationship due to this and the best way to determine an alternators output is to measure it and develop a matrix for calculations. And while doing this don't forget to include speed and temperature---very complex model!! And while you are all speculating about whats really happening with this amp/battery/cap/alternator thing never forget that in a circuit where there are multiple voltage sources the current will only flow from the source that is at the HIGHEST voltage potential. Only when the voltge of the highest potential is pulled down till it is equal to the next highest does current start to flow from the second source. At that time they start to share based on their relative esr and storage capacity-----think about it!!!!!!!!!

---

posted April 20, 2000 03:16 AM            

There must be a clue in Richard's last post. The final answer, although there are probably some more concepts to consider, is going to rely on knowing that current will always flow from the highest to the lowest potential, as he has said.

So when the automobile is running, the current must be flowing away from the alternator, to either charge the battery, the cap, or power the amp, since at 14.4V or so, it is at the highest potential. In no circumstance can current flow toward the alt. because of the diodes.

When the amp "asks" for higher current than can be supplied by the alt. then the battery kicks in, and during this time the battery and the alt. are in equilibrium i.e. at equal potential, so the current coming from the alternator is divided between the amp, the cap and the battery, with the greatest portion going to the lowest impedance. I would say that during musical peaks, the current is almost exclusively going to the amp and during the other times, the current is more likely to be going to the battery (to reverse the chemical rxn.), because the ESR & ESL of the cap probably make it have the highest impedance of the three devices.

We know that the ESR and the ESL of the cap are high, so it's not too likely that current is either going into it or coming out of it. In order for current to be able to come out of the cap, it must be at a higher potential than the alt./batt. subsystem. Even if the cap is at some time at a higher potential than the alt./batt. subsys., it (the current) must overcome the ESR and ESL and therefore it's potential will be lowered just by passing thru the impedance. So when what puny amount of energy there is in it has finally gotten out, the voltage across it's terminals will be lower by some amount, and therefore current from the alt./batt. will try to flow back into it?.

At this point I think its obvious that I have run out of ideas, so what I would next ask, is does current just bypass the cap or does the cap become yet another load for the alt./batt. subsys.? Possibly, the cap alternates between these two conds.?

posted April 20, 2000 12:11 PM            

quote:

---

Originally posted by OverKill:
The thing that is occurring to me as I am thinking about this is that why doesn't "the cap" with it's high ESR, & ESL show itself as a detriment to the automotive power subsystem? The tests that CarSound did showed virtually identical results both with and without it! There must be some condition under which this (the cap would hinder performance) would be the case.

There must be a clue in Richard's last post. The final answer, although there are probably some more concepts to consider, is going to rely on knowing that current will always flow from the highest to the lowest potential, as he has said.

So when the automobile is running, the current must be flowing away from the alternator, to either charge the battery, the cap, or power the amp, since at 14.4V or so, it is at the highest potential. In no circumstance can current flow toward the alt. because of the diodes.

When the amp "asks" for higher current than can be supplied by the alt. then the battery kicks in, and during this time the battery and the alt. are in equilibrium i.e. at equal potential, so the current coming from the alternator is divided between the amp, the cap and the battery, with the greatest portion going to the lowest impedance. I would say that during musical peaks, the current is almost exclusively going to the amp and during the other times, the current is more likely to be going to the battery (to reverse the chemical rxn.), because the ESR & ESL of the cap probably make it have the highest impedance of the three devices.

We know that the ESR and the ESL of the cap are high, so it's not too likely that current is either going into it or coming out of it. In order for current to be able to come out of the cap, it must be at a higher potential than the alt./batt. subsystem. Even if the cap is at some time at a higher potential than the alt./batt. subsys., it (the current) must overcome the ESR and ESL and therefore it's potential will be lowered just by passing thru the impedance. So when what puny amount of energy there is in it has finally gotten out, the voltage across it's terminals will be lower by some amount, and therefore current from the alt./batt. will try to flow back into it?.

At this point I think its obvious that I have run out of ideas, so what I would next ask, is does current just bypass the cap or does the cap become yet another load for the alt./batt. subsys.? Possibly, the cap alternates between these two conds.?

Perhaps it is the case that since there is such a pitifully small amount of energy in the cap (compared to the battery and the production of the alternator) and because in any case half of it is lost to ESR and ESL, that it makes no discernible difference either way.

---

posted April 21, 2000 04:20 AM            

But since this whole discussion is about ESR and ESL.....

How do you determine the ESR of a cap by applying a voltage to it and looking at the initial charge current.....

Doesn't this current come from other caps, slow batteries or other sources that have their own lag, ESR and ESL ? And isn't this what this post is all about ?

Surprised no one ever attacked this point before.

Also, where can I find info on ESL's of caps, because if they are as big a people here claim they are, even caps in regular homeamps become almost questionable, especially the practise to have 10mF caps paralleled with 100uF and 1uF, both for filtering (would still work) and fast peakcurrents (is bull according to the here found measurements).

I'll follow the rest yet to come...

posted April 21, 2000 04:58 PM            

Richard, thank you spending so much time with your audience. The lessons
you gave are of course flawless. (and much needed)

The cap review was difficult. Coming up with real uses is tough.
I still think the best use is in a retail display to extend the power
supply.

I agree that the response time of the alternator is what causes lights
to blink. I tried to use an alternator as an amplifer in 1982 but the
frequency response would be limted to about 20 Hz.

The cap really does behave like a good 20F cap with 0.017 ohms ESR.
The nature of the construction does not lend itself to significant ESL.
0.2 MicroHenry is an approximation to the upper limit of ESL value,
judging from the rise time of the 500 ampere inrush.

Your examples are accurate. But, I would like to see a more realistic
system where the alternator is pre-loaded to 60% capacity without the
stereo. The transient demands of even a 100 watt stereo (20-50 amperes)
should cause fast drops to the 12.5 battery discharge voltage. This is
where caps cause a visible improvement.

second battery? my car plays longer before dying, and that is a real
benefit.

Hmm, midnight afer a long day. did my repsonse help?

BTW, for the guy that asked.

no energy is you, standing on the ground.

Potential energy is you standing on the roof after expending chemical
energy to climb the stairs against the force of gravity.W=FxD=PE.

As you fell PE was converted to kinetic energy KE.

posted April 21, 2000 09:59 PM            

your example is of "gravitational" potential energy defined by U = -GMm/R (G is the gravitational constant, R is the distance, and M and m are the mass of the two objects between which the gravitational force occurs).

there are many different types of potential energy, some being gravitational potential energy, some being elastic potential energy, and yes, electric potential energy. a capacitor is one form of getting electric potential energy.

and to REALLY nitpick (sorry Richard) , well, the kinetic energy is not what killed you, the impulse from the collision is what killed you. and the momentum is not transferred in a lossy function at all, the momentum of the center of mass should be conserved to a pretty high degree. however, from the inelastic collision, kinetic energy is lost, not total momentum. if the momentum was transferred in a very lossy fashion, the impulse would be pretty small and you would survive.

posted April 21, 2000 10:58 PM            

I think Richard calculated 63 ohms impedance or so due to the ESL. I don't know if it is the frequency of the tone produced by the speaker (I would think not) or if it's the frequency that the power supply switches at that is pertinent to the analysis. Even still, it still appears that the ESR is enough to take a large chunk out of any potential benefit that the cap might produce.

posted April 22, 2000 12:16 AM            

[This message has been edited by Richard Clark (edited 04-22-2000).]