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Old 23rd June 2012, 04:03 PM   #6281
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Quote:
Originally Posted by nigel pearson View Post
40 year ago and I probably remember incorrectly ?
Good memory, i'd say.

(overhere it's easy, volt/ohm/ampère sound incredibly strange when pronounced with an added s. One advantage the French lack )
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Old 23rd June 2012, 04:15 PM   #6282
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I think in French I would say my amp is not very sensible ( needs more gain ) . I think Dvv would say so in English also ?
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Old 23rd June 2012, 04:19 PM   #6283
DF96 is offline DF96  England
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Quote:
Originally Posted by nigel pearson
I remember at college the derived units were always single
Never come across that, except when the number happened to be 1: 1 watt, 2 watts. I imagine it might depend on how the language formed plurals.

Last edited by DF96; 23rd June 2012 at 04:21 PM. Reason: add sentence on plurals
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Old 23rd June 2012, 04:23 PM   #6284
SY is offline SY  United States
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OTOH, one says, "A 47 ohm resistor," but, "The resistance of that lead is 47 ohms."

Hobgoblins, small minds, that sort of thing.
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Old 23rd June 2012, 04:26 PM   #6285
Pano is offline Pano  United States
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Quote:
Originally Posted by dvv View Post
Briefly, the say we will need 1-2 Joules of energy per every 10 Watts of dissipated power, depending on how easy or hard a load is.
Thanks DVV, at least that's a bit of info and a start. And provieded by someone who has built a few amps, Motorola.
Much better than:
Quote:
Originally Posted by john curl View Post
And it is usually 'relative' rather than absolute, so you can not just compute the ideal solution. Usually, more is better, unless you go to extreme excess.
Which tells me, basically, nothing. How can one learn to design from that?

Certainly I don't expect to get a rigid formula, but knowing what you might need to overcome ripple is a starting point. After that we might be talking about what the extra energy can do for sound quality, either objective or subjective.

Many of the amps I've liked over the years have stored massive amounts of energy compared to the power dissipated into the load. That might be a low power amp with a disproportionately large PSU, or an amp designed for much higher powers than it's used at. (High power amp used on efficient speakers at home).

It would be interesting to know from amp builders who have experimented with this if they've found a point of vanishing returns. Or a point of "good enough." Beyond ripple reduction (which DF9 has outlined) what are the advantages of more stored energy? Or are there any advantages?
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Old 23rd June 2012, 04:29 PM   #6286
Pano is offline Pano  United States
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Quote:
Originally Posted by nigel pearson View Post
I think in French I would say my amp is not very sensible ( needs more gain )
Yes, sensible in French has to do with what we call sensitivity in English. The more "sensible" an amp or speaker, the less voltage it needs to reach a certain output.
The adjective is the same for people, too.
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Old 23rd June 2012, 04:45 PM   #6287
popilin is offline popilin  Argentina
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Quote:
Originally Posted by DF96 View Post
Cap size is set by things like required ripple voltage, transformer winding resistance, peak current draw, PSRR etc. The biggest demand on the caps' stored energy would come from a full level 50/60Hz square wave, as if it is aligned to the supply phase then all the energy has to come from the cap. For all other situations things are easier, as either the cap gets recharged partway through (lower frequencies, or not phase aligned) or the current is only drawn for part of the recharge cycle (higher frequencies).

Peak power out = V x I. Suppose we want V to droop by no more than 5%. Then 0.05V = I x 0.01 / C (assuming 50Hz mains and full-wave rectification). Rearrange this to get CV = I/5 . Now substitute I=P/V and get CV = P/5V. Finally 0.5CV^2 = P/10. So for 5% max droop on a phase-aligned 50Hz square wave you need 1 joule of stored energy for every 10 watts of peak output power (or 5 watts of mean sine wave power). I guess a calculation like this is where Motorola got their rule of thumb from.
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Something like that is what I was looking for.
Thanks.

Quote:
Originally Posted by DF96 View Post
Having done the calculation, I have no idea how relevant it is to real amplifiers handling real music.
Don't worry it doesn't matter, in this thread is a minor detail.
We have almost one hundred post talking about ... Economy.

Quote:
Originally Posted by SY View Post
Shouldn't that be "joules"?
There is no place for philosophy of language, here the units are €, £, US$, etc.
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Old 23rd June 2012, 04:53 PM   #6288
dvv is offline dvv  Serbia
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I found the app note - Motorola AN1308, featuring a 100 WPC and a 200 WPC power amp, based on their MJL 3281/1302 power transistors.

The said formula is in the "Designing the power supply" section, and yes, it does require a division by 2.

I stand corrected.

When you see how it's all typeset, you'll also see why it is so easy to leave something out. Good app note, awful typesetting, formulas should never be included in a line as regular text, they should always be set apart.

Sorry, Pano and Popilin, my bad.

Last edited by dvv; 23rd June 2012 at 04:55 PM.
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Old 23rd June 2012, 05:11 PM   #6289
dvv is offline dvv  Serbia
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Quote:
Originally Posted by Pano View Post
Thanks DVV, at least that's a bit of info and a start. And provieded by someone who has built a few amps, Motorola.
Much better than:

Which tells me, basically, nothing. How can one learn to design from that?

Certainly I don't expect to get a rigid formula, but knowing what you might need to overcome ripple is a starting point. After that we might be talking about what the extra energy can do for sound quality, either objective or subjective.

Many of the amps I've liked over the years have stored massive amounts of energy compared to the power dissipated into the load. That might be a low power amp with a disproportionately large PSU, or an amp designed for much higher powers than it's used at. (High power amp used on efficient speakers at home).

It would be interesting to know from amp builders who have experimented with this if they've found a point of vanishing returns. Or a point of "good enough." Beyond ripple reduction (which DF9 has outlined) what are the advantages of more stored energy? Or are there any advantages?
I know John is neither blind nor mute, but I think you're being a little too hard on him.

I asked him specifically, a few pages back, what would he consider to be the right amount of capacitance for a nominally 100 WPC/8 Ohm amp, and he very specifically said 22,000 uF should be good. I assumed he meant per channel, although that was not expressly mentioned.

Point is, a very specific question gets you a very specific answer.

But then, John already knew I was in for 4 pairs of 200W output transistors (which arrived a few days ago), so he had a frame of reference. I think THIS is what he refers to - it's one thing to have a PSU capable of great things, but that in itself is not enough to make sure all of that capability actually gets to the speaker, if need be.

Ultimately, designing a power amp is really designing a system, with several subsystems. Inflating any one of them makes no sense, rather we are looking for an even handed approach to it all, remembering that the weakest link in any chain is the weakest link of the whole system.

This is WHY I asked John that. In my particular case, I'm pretty sure I could get away with less, given that my speakers never go below 6.5 Ohms, their worst case phase shift is -25 degrees and they are relatively efficient at 92 dB/2.83V/1m - in my 14 sq. m room. Sheesh, I could get away with a tube amp rated at 10 WRMS easy. By 3 WRMS, my window panes start shaking.

But that's a minimum setup approach, and I don't do minimum setup, EVER. I'm more with Wayne on this, as he puts it "doubling down" (more or less, it can never be quite exact), and I don't build power amps every day.
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Old 23rd June 2012, 05:14 PM   #6290
popilin is offline popilin  Argentina
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Quote:
Originally Posted by dvv View Post
Sorry, Pano and Popilin, my bad.
No man, you were right, you were talking about dual polarity PSU, the factor 1/2 disappears

E = (1/2) C V^2 = (1/2) (C/2) (2V)^2 = (1/2) (C/2) 4 V^2 = C V^2
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