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Glad they’re people like you who can grind things like this out. Also reassure me that supporting STEM is worthwhile.
I’m kindda at the “is there an app for that” level.
fpitas
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Heavy Load: How Loudspeakers Torture Amplifiers | Stereophile.com
Why, in loudspeaker reviews, is impedance measured (assuming that the magazine in question bothers to measure anything)? Generally, for one principal reason only: to establish whether the speaker presents an "easy" or a "difficult" load to its partnering amplifier. In the design context, much...
All this was clear from Eric Benjamin's aforementioned AES article, but the message seems not to have filtered through in the 13 years since (footnote 5). Loudspeaker impedance continues to be assessed by considering modulus and phase separately when, as Benjamin showed, there is a much better way to reflect the load's severity from the amplifier's viewpoint. What he did—as Douglas Self did later (footnote 6), though by the less elegant means of SPICE circuit simulation rather than mathematical analysis—was to plot peak output-stage power dissipation vs frequency with respect to a stated resistive load. Benjamin chose 4 ohms as his reference, while Self preferred 8 ohms. Such curves can be calculated analytically from conventional modulus and phase data—no additional measurements are necessary.
Alice of Old Vincennes
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Magnepan emphasizes high current amplifiers.
Alice of Old Vincennes
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What does that mean? I'm confused.
Alice of Old Vincennes
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Just use 12 gauge.
valerianf
Addicted to Fun and Learning
Let us take an example of a class D amp power dissipation with a load of 4 Ohms.
For 100W at the output the class D amplifier will dissipated a little more than 17W (blue curve).
To that we need to add the dissipation (11.7W) of the power supply itself that can have an efficiency of 90% if well designed.
In total the class D amplifier+power supply will dissipate 28.7W for 100W delivered on 4 Ohms.
Some nice heat-sink will be needed to reach a good reliability at high sound level.
Here is the link to the reference paper from TI (page 4).
https://www.ti.com/lit/an/slaa996a/...32974&ref_url=https%3A%2F%2Fduckduckgo.com%2F
For 100W at the output the class D amplifier will dissipated a little more than 17W (blue curve).
To that we need to add the dissipation (11.7W) of the power supply itself that can have an efficiency of 90% if well designed.
In total the class D amplifier+power supply will dissipate 28.7W for 100W delivered on 4 Ohms.
Some nice heat-sink will be needed to reach a good reliability at high sound level.
Here is the link to the reference paper from TI (page 4).
https://www.ti.com/lit/an/slaa996a/...32974&ref_url=https%3A%2F%2Fduckduckgo.com%2F
Nice post! A far cry from impression of super efficiency that has been promoted/marketed for Class D.
- How representative is this specific example for Class D overall?
- What happens at 8 ohms?
- And at lower wattages?
I recently measured the FOSI V3:
Magnepan quotes that a high current amp is required to drive Maggies. My simple test to see if a particular amp is ‘high current’ is whether the amps output doubles for a 4 ohm load…ie. If an amps output is 100w into an 8 ohm load and it doubles to 200w into a 4 ohm load it’s considered ‘high current’I recently measured the FOSI V3:
The NAD you mentioned states 160w into 8 ohm and 300w into 4 ohm.. ok not quite double but close enough
I use a Mark Levinson No. 383 integrated to drive my Maggie MG.7’s. Quoted specs for the 383 are: 100w into 8 ohms and 200w (minimum) into 4 ohms
Even so the volume control has to be turned up quite high to drive the Maggies. I’ve just added a DWM bass panel to the mix which adds additional load…
Maggies simply soak up the power of any amp!
It makes little difference because the output devices aren't operating in the linear mode.
Simplistic explanation :
This is all about the energy stored in the equivalent capacitance/inductance of the load which results in the large phase shift. This energy must be supplied by the amp while being stored, and absorbed by the amp when being released. With a Class A or AB amp, that extra energy is absorbed by the output devices as extra dissipation.
With a class D amp, the energy is passed back through the ouput stage into the PSU. The only consequence for class D is that the PSU must have enough capacitance to absorb the energy. There is little extra power loss (only the losses resulting from the current passing through the on state resistance of the output devices) Effectively for class D the reactive energy is passed back and forth between the load reactance, and the PSU capacitance.
Of course, that usually just means the 8Ω power is underrated, for instance the 100W/200W ohm probably clips at like 133W/200W.* And as we see in this informative thread
https://www.audiosciencereview.com/...ing-a-complex-load-by-power-amplifiers.43900/
started by @pma there's much more to amps than driving resistors. However at least if the manufacturer rates at 4Ω and better yet 2Ω it shows some confidence in ability to drive those low impedance loads, and it's the only information we usually have.
I'd always thought it was Martin Logan that wanted high current, due to the vaguely capacitor nature of their electrostatic panels. The Magnepan LRS for instance has a pretty benign impedance. I suppose this varies by model, and if I were to be cynical I'd say that the thought process of the company pushing towards high current amps might be "the high current amps tend to be the expensive "good stuff" so if we say that's necessary customers will pair our speakers with really good amps instead of some weak thing."
*I've only seen like 3 amps that actually doubled power at clipping...I think a CH Precision is one? And this:
https://www.stereophile.com/content/ps-audio-stellar-m700-monoblock-power-amplifier-measurements
I think you are inadvertently conflating something else. Having worked many years in auto engineering, that discussion is usually about what wire gauge is needed from the battery to the amp not from amp to speaker. I was about to say you don't have to worry about it on the speaker side but let's see...
- Pumping 1000W continuous into a subwoofer, a pretty extreme case, into 4Ω would mean
--> P=I^2 * R (power = current squared times resistance)
--> 1000 = I^2 * 4
--> I = sqrt (4000) = 20 amperes if I did that right.
[as @antcollinet points out, I didn't all, OMG, see below. Hmm no time stamp on the posts, I wonder how late it was...]
https://www.powerstream.com/Wire_Size.htm would say you need about AWG 9, and AWG12 could get you like 400W @4Ω...but then this chart quite differs, maybe not really showing quite the same thing
https://www.electricaltechnology.or...gauge-awg-chart-wire-size-ampacity-table.html
it seems the AWG 12 would be OK at the 1000W. I'd think you would not really be hitting 1000W continuous even into a loud sub but a clipped one quite possibly (???)
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Agreed, and how many of those 3 tested for any longer than seconds, or just a couple minutes at best, when loaded with a 4 ohm resistor? I don't have use for "continuous" literally meaning indefinite, just making the point that this double down talks are myth talks, it is not a good guideline for assessing amp's so called "high current" capability, without the details of how it is measured.
You didn't
You have to divide both sides by 4 to get I^2 on it's own.
So I^2 = 1000/4 (Not * 4)
So I = sqrt (250) = 15.8A
(I'm also not sure where you get sqrt of 4000 = 20, it would have been 63.2A.
The AWG of a wire (or mm squared in EU) determine how much current a wire can carry rather than how much power. The voltage multiplied by the current then determines the power.
In car audio applications the supply is normally only 12V - which means for a given power you need a much higher current, which is why consideration of wire size is more important in car applications.
D
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So this is why EPDR isn't relevant for class D amplifiers, right?
Pretty much.
That's right, but he wasn't right to say "pumping 1000 W" into a subwoofer to begin with either. Subwoofer is not a pure resistance load! It is unfortunate that people started using watts for speaker loads from the beginning. Using volatge and current would have been a little less confusing, and technically more correct.You didn't
You have to divide both sides by 4 to get I^2 on it's own.
So I^2 = 1000/4 (Not * 4)
So I = sqrt (250) = 15.8A
(I'm also not sure where you get sqrt of 4000 = 20, it would have been 63.2A.
Yeah, amperes probably would have been more useful. I guess power is an ingrained concept. And no the subs aren't a pure resistance load by a mile! However to calculate wire current in the absence of impedance data then nominal resistance will have to do.
Once upon a time, Matti Ottala presented an Audio Engineering Society paper showing that an "8 ohm" speaker could eat up to "1 ohm" of current.* Later someone published a pushback, saying that Ottala's paper was based entirely on simulation and that an empirical study had shown no evidence for current draws greater than the D.C. resistance of the voice coil would indicate.
I don't know if there was ever further study made. I wanted to do that with some kind of simultaneous current and voltage readings into a data logger but have never gotten around to it. Another way would be a long term trace on a storage scope and looks at the trace slopes. Someday...
*note this is different than EPDR, since no dissipation is factored in, it's just equivalent resistance.
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