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Old 1st February 2013, 09:29 AM   #1731
fas42 is offline fas42  Australia
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Quote:
Originally Posted by gootee View Post
For the second approach, we might put three or hopefully more film caps (probably using the largst value available in whatever case size is small-enough) as close to the device pins as we can possibly get them, with at least 5 uF but hopefully more, and hopefully with much less than the 100 nH of maximum inductance we calculated.
Poor, poor, a.wayne -- now look at what you've gone and done, Tom ... ,

As regards film caps, back when I did this fella, many years ago, it didn't make sense to add them into the mix, the benefits were not there from a theoretical POV. And the amp didn't seem to suffer from not including them. Which is not saying that there may not be audible impact if one were to use the very latest and greatest parts -- something to play with for someone ...

Overall, the biggest benefit of my approach was to give the chip amp the sense of being "unburstable", no restriction on dynamics. Which was due to the effectively huge size of the capacitor bank ...

Frank
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Old 1st February 2013, 09:39 AM   #1732
AndrewT is offline AndrewT  Scotland
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The capacitor charge/discharge is given by the relation
1Farad changes by 1Volt when 1Ampere is drawn for 1second.
From this you can examine any capacitor voltage change for any current over a specified time period.

eg
1uF 1V 1A 1us
or
100nF 1V 1A 100ns
or
100nF 1V 10A 10ns
Just to remind you what that means.
using a 100nF decoupling cap, you can draw 10A from it for a period of 10nanoseconds and the charge will drop by 1volt.
That would be your HF decoupling.

Now do the same for the MF decoupling and for the LF decoupling.

The difference between the HF & MF & LF decoupling is the Length (nanoHenries) of the route around the circuit
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Old 1st February 2013, 12:33 PM   #1733
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Quote:
Originally Posted by gootee View Post
Ripple reduction is secondary. It's more about being able to provide current accurately on demand.
perhaps this thread was hijacked, but I was restricting my comments to caps and ripple. You MIGHT have to go all the way back to the circuit breaker to "provide current accurately..". Maybe start a new thread?


Quote:
Originally Posted by gootee View Post
All of what you have mentioned has been discussed in ten different ways, here, already.
the pot calling the kettle black?
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Old 1st February 2013, 05:08 PM   #1734
gootee is offline gootee  United States
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Originally Posted by KMossman View Post
perhaps this thread was hijacked, but I was restricting my comments to caps and ripple. You MIGHT have to go all the way back to the circuit breaker to "provide current accurately..". Maybe start a new thread?




the pot calling the kettle black?
Please, here is a clue:

I have read this entire thread. I suggest that you do the same, so that your contributions can be relevant.
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Old 2nd February 2013, 12:10 AM   #1735
gootee is offline gootee  United States
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Hey Nico! Remember your VERY FIRST post, here? (Hint: It's Post #1.) You said:

"...one arrives at a reservoir capacitance of 80,000 uF."

Well looky there! It only took 1708 posts and about 6 1/2 months to get to "...giving C >= 83577 uF",

which is just a few screens back from here, at:

Power Supply Resevoir Size

I want to thank you, my friend, for putting up with me, in this thread. We have been through a lot, together, here, and have learned so much; all of us.

"The PSU CURRENT IS the audio signal, which almost all comes directly from the caps!"

Last edited by gootee; 2nd February 2013 at 12:13 AM.
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Old 2nd February 2013, 09:24 PM   #1736
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Quote:
Originally Posted by gootee View Post
Hey Nico! Remember your VERY FIRST post, here? (Hint: It's Post #1.) You said:

"...one arrives at a reservoir capacitance of 80,000 uF."

Well looky there! It only took 1708 posts and about 6 1/2 months to get to "...giving C >= 83577 uF",

which is just a few screens back from here, at:

Power Supply Resevoir Size

I want to thank you, my friend, for putting up with me, in this thread. We have been through a lot, together, here, and have learned so much; all of us.

"The PSU CURRENT IS the audio signal, which almost all comes directly from the caps!"
Well put .
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Old 3rd February 2013, 03:09 AM   #1737
gootee is offline gootee  United States
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Originally Posted by abraxalito View Post
Have a look at Rubycon ZL range - that's showing 330uF/50V in 10mm with a max 100kHz impedance of 28mohm. In my limited experience of extreme paralleling of caps, I found going above 330uF to be problematic in that the self-resonant freq comes down a bit too low for comfort. I doubt when paralleling 100 caps that the high frequency ESR is going to be limited by the caps' ESR, more likely the interconnections.
Still thinking about the caps needed for the 10x10 (or so) array.

Maybe terminology is getting mixed up or maybe I am under some misconception. The self-resonant frequency is usually a GOOD place to be. It's where the minimum impedance is. It's the frequency at which a cap can most-quickly supply the most current.

Basically, we want a very low impedance, everywhere, if possible. But if there's a band of lowest impedance, it seems like we would want that band to lie across the audio band, and at least across the bass portion of the audio band, if we couldn't have it all.

Attached is a plot of the impedance of Terry Given's 10x10 array of 1000 uF caps, on a 1.6mm-thick (instead of the desired 1mm, or less) PCB. The plot came from his HP 3577A analyzer (capable of 5 Hz to 200 MHz).

I would have thought that the cap array's impedance would be much lower than it is, at the lower frequencies. The 12.5 mF cap he compared it to looks lower, there. However, the 12.5mF cap doesn't include any connection wiring, whereas the cap array includes all of that already.

At any rate, the array's flat, lowest-impedance region appears to extend from somewhere between 2 and 5 kHz up to about 1 Mhz. (It does very well at the high frequencies.)

I think that, ideally, the lowest-impedance band should extend another two octaves lower, so that it includes down to somewhere in the 20-50 Hz region at its lowest-impedance level.

Would that mean using larger cap values or more total capacitance? Would it help to put caps on both sides of the pcb, to decrease the total PCB path length? (I realize that just getting 1/wC down to 0.02 at 20 Hz would require about 400000 uF. Is that what we would need?)

Then we see what Klaus got, here:

Network Analyser Measurements

Now I am scratching my head.

Suggestions?
Attached Images
File Type: png Terry_pcb_impedance.png (77.2 KB, 128 views)

Last edited by gootee; 3rd February 2013 at 03:12 AM.
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Old 3rd February 2013, 08:32 AM   #1738
fas42 is offline fas42  Australia
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Quote:
Originally Posted by gootee View Post
Basically, we want a very low impedance, everywhere, if possible. But if there's a band of lowest impedance, it seems like we would want that band to lie across the audio band, and at least across the bass portion of the audio band, if we couldn't have it all.

...

Would that mean using larger cap values or more total capacitance? Would it help to put caps on both sides of the pcb, to decrease the total PCB path length? (I realize that just getting 1/wC down to 0.02 at 20 Hz would require about 400000 uF. Is that what we would need?)
?
Ah, brings back memories ... how many times have I stared at curves like this ...

Easiest way of getting that nice, low impedance in the audio band is to use regulation, otherwise, bigger, meatier and rough and ready electro's will be needed. The latter can be as low quality as you like, what you want is capacity, capacity, capacity, the most you can get for your money. I would aim for better than 1 m.ohm over the whole audio band, and try to maintain that up to 100-200khz. This is where you use your best quality electro's, heavily paralleled, short paths to the point needed. Once past that frequency, heading towards the MHz it starts becoming a real struggle, absolutely everything has to be just right, otherwise you're just wasting your money. You worry about millimetres of conductor, fight to reduce every nH ... and it may not actually be worth it ...

Frank
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Old 4th February 2013, 04:20 AM   #1739
gootee is offline gootee  United States
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Thanks, Frank. I'm not really worried as much about the higher MHz range; only up to the reciprocal of Pi times the fastest rise-time.

But I just realized that I've not thought that through. For example, an amp that can do 40 V/us = 5 A/us could do 1 A in 1/5th μs, or 0.1 A in 1/50th μs, and so on. Where do I draw the line on that?! Wait, I might be getting it...

Z_target (max) = Δv_max / Δi , for any Δi, BUT, we pick a single Δv_max.

Before, I only looked at the worst-case full-range transient, like the 5A / μs, and made sure that it wouldn't generate more than a Δv_max glitch in the power rail voltage (i.e. the voltage across the decoupling cap).

But 0.1 A, with the same maximum di/dt (based on max slew rate), would be done a lot sooner, giving a much faster rise time (0.02 μs, for a 5A/μs max slew rate, for example), which would translate into a much higher equivalent frequency, 1/(π 0.02μ) = 15.9 MHz.

BUT, it's only 0.1 A and yet it's still allowed to produce the same Δv_max as 5 Amps was. So the Z_target at 15.9 MHz is allowed to be Δv_max/0.1, which is 50 TIMES higher than the Δv_max/5 that it had to be for the 5 Amp rise in 1 μs, which corresponded to 1/(π 1μ) = 318 kHz.

Cool. Now I have a way to get the high-frequency part of the Z_target impedance curve.

I was "assuming", without really thinking about it, that Z_target was a constant from say 20 Hz to 318 kHz, and then didn't matter, except that we would want it as low as possible, and with no impedance peaks, so that no HF resonances would get too excited.

So now I'm wondering, above what frequency DOES the z_target curve cease to matter?

Does the stuff above really mean that if the PSU impedance at 15.9 MHz is too high, a 0.1 Amp or smaller transient cannot be produced at the maximum slew rate without causing a rail disturbance that is greater than the Δv_max that dictated that impedance at that frequency?

Well, never mind that, for now. The amplifier will be limited in its frequency response, probably well before that point. We just want to make sure that it's not unable to correct whatever needs to be corrected, which it does eventually become unable to do, above some frequency.

Anyay, as I implied, I am thinking more about the lower frequencies, right now. What is a reasonable impedance to aim for, down to 20 Hz?

Frank, you mention 1 mOhm; 0.001 Ohm. For the example with 5A/μs slew rate and 0-5A range, that would mean a worst-case Δv_max = 0.005 Volt.

Ignoring ESR, which, at very low frequencies, would be swamped by the capacitance when paralleling lots of electrolytic caps, anyway, to get 0.001 Ohm at 20 Hz we would need

C = 1 / (2∙π∙20∙0.001) = 7.96 million μF; almost 8 Farads!

I'm just not going there. But the question is, how far WILL I go? And, WHY?

With 100000 μF (from the 10x10 1000 μF), we would have Z = 80 mOhms, at 20 Hz. That means that 0.080 = Δv_max / Δi , which for our example 100-Watt amp with the 5-Amp swing, would give Δv_max = 5(0.080) = 0.4 Volt, at 20 Hz, which would be "1% ripple" at the maximum rated output power.

PSRR is usually very good at lower frequencies, though. So as long as we have enough capacitance to produce the demanded current, then how much ripple we create shouldn't matter much, within reason.

I guess we would need a plot of PSRR vs frequency, to be able to create a plot of acceptable ripple amplitude vs frequency, for a chosen maximum distortion level (or something like that).

But we also know that above some relatively-low frequency, the impedance will be almost identical to the ESR divided by the number of caps, which will be some small fraction of 1 mOhm, for 100 large-value caps in parallel, with unbroken copper power and ground planes. So we only have to worry about the low and low-mid frequencies, maybe, where the capacitance's impedance starts to dominate over the ESR.

Xc = 1 / (2∙π∙f∙C) .

For C = 100000 μF, that becomes Xc = 1.59 / f .

For C = 220000 μF, we get Xc = 0.723 / f .

At 100 Hz, we'd have 15.9 mOhms with 100kμF, or 7.23 mOhms with 220kμF, giving 0.0795 V worst-case ripple amplitude for a 100 Hz 100-Watt sine with 100kμF, or 0.0362 V worst-case ripple for a 100 Hz 100-Watt sine with 220kμF, for example.

We would just need to know what effect it might have, and how low the ripple needs to be at each frequency, and then we could decide how much capacitance was actually needed. Again, the PSRR might tell us those answers. I'll have to go investigate PSRR and ripple-induced distortion, now, I suppose. DF96, Frank, anyone? Any ideas about that?

-------------------------------

I happened to run across the following paper which some might find interesting or enlightening:

http://www.illinoiscapacitor.com/pdf...or_fiction.pdf

- Tom

Last edited by gootee; 4th February 2013 at 04:38 AM.
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Old 4th February 2013, 07:13 AM   #1740
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Hey Nico! Remember your VERY FIRST post, here? (Hint: It's Post #1.) You said:

"...one arrives at a reservoir capacitance of 80,000 uF."
What do you know, and that was sucking my left thumb only

Tom, your work on this thread remains exemplary.
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