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Amplifier Output Impedance (Damping Factor) and Speakers

xnor

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@xnor, Tnx for the confirmation.

Again from post #239:
Actually, an amp with low DF should sound like less low frequency power regarding the frequency response measurement. But exactly the opposite happens!

What explanation do you have for this observation?
How can you thank me but still not get it and still have it backwards?

A low DF causes more voltage drop overall, but that is irrelevant.* What's important is that a low DF causes relatively less voltage drop at a speaker's impedance peaks, effectively resulting in a bass boost in a passive speaker with a woofer.

*) overall level differences are typically eliminated by level-matching different amps/settings during a comparison


It's really simple, and people have told you several times that you've got it backwards. Not sure what you're still struggling with.
 

pogo

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A low DF causes more voltage drop overall
This seems to be the case for a current drive amp design, but we mainly use voltage drive amp designs, don't we? This is also shown in the measurements of the T+A A200.
Loudspeakers are typically designed to work with ordinary voltage controlled amplifiers!

Here is an example of my loudspeakers.
fg_1139550.png


I would rather expect the opposite of a relative bass boost here, see also Benchmark examples at 119Hz.

Thank you for your efforts to try enhancing my understanding.
 

KSTR

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A low DF causes more voltage drop overall, but that is irrelevant.* What's important is that a low DF causes relatively less voltage drop at a speaker's impedance peaks, effectively resulting in a bass boost in a passive speaker with a woofer.
This!
 

pogo

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resulting in a bass boost in a passive speaker with a woofer.

I have calculated the conditions on my impedance curve at significant points and I do not see overall any bass boost at DF=60. But you are welcome to correct me.

8ohms/60 = 0.133ohms (amp output resistance)

Averaged assumption of 5ohms in the low frequency range:
5/(5+0.133) = 0.974
20*Log(0.974) = -0.23dB <-- loss of bass

25ohms at 1.3kHz:
25/(25+0.133) = 0.995
20*Log(0.995) = -0.04dB

Averaged assumption of 8ohms in the high frequency range:
8/(8+0.133) = 0.984
20*Log(0.984) = -0.14dB
 

kchap

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I have calculated the conditions on my impedance curve at significant points and I do not see overall any bass boost at DF=60. But you are welcome to correct me.

8ohms/60 = 0.133ohms (amp output resistance)

Averaged assumption of 5ohms in the low frequency range:
5/(5+0.133) = 0.974
20*Log(0.974) = -0.23dB <-- loss of bass

25ohms at 1.3kHz:
25/(25+0.133) = 0.995
20*Log(0.995) = -0.04dB

Averaged assumption of 8ohms in the high frequency range:
8/(8+0.133) = 0.984
20*Log(0.984) = -0.14dB
You be be comparing 2 different damping factors. Assume a DF of 30 and redo your calculations. While the difference is trivial an amp with a DF of 30 will give you less bass compared to an amp with a DF of 60.
 

Geert

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Averaged assumption of 5ohms in the low frequency range:
5/(5+0.133) = 0.974
20*Log(0.974) = -0.23dB <-- loss of bass

25ohms at 1.3kHz:
25/(25+0.133) = 0.995
20*Log(0.995) = -0.04dB <- Less loss at higher driver impedance (-0.04 vs -0.23), like at woofer resonance frequency

Why didn't you make the calculation for the woofers resonance frequency, where impedance is often the highest?
 
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xnor

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This seems to be the case for a current drive amp design, but we mainly use voltage drive amp designs, don't we?

Do you understand the difference between an (ideal) voltage source and current source? Ohms law?
I = V/R
An ideal voltage source outputs V regardless of R. If R is high (such as at a resonance peak where the speaker is more efficient) then naturally I drops. If R is lowered then I increases. If R is close to infinity (like an open circuit) then I goes towards 0. If R is close to 0 (like a short) then I goes towards infinity. V stays the same.

In practice, R is not just the speaker's impedance but includes everything outside the amp's feedback loop, like resistance of the speaker terminals, the wires etc. and the voltage across those components splits accordingly.

V = R I
A current source outputs I regardless of R. If R is high (such as at the resonance peak) then V is high. If R is close to infinity (like an open circuit) then V goes towards infinity as well. If R is close to 0 (like a short) then V goes towards 0. I stays the same.

As you increase output resistance (and lower DF towards 0), you effectively turn a voltage source more and more into a current source.
It's very simple to do this with a power/headphone amp, just add a resistor to each channel. But at some point, it becomes impractical. For example, with a 1 kohm output resistor (yielding a DF of 0.008 with 8 ohm load), you'd need your amp to output 126 V unloaded to get just 1V across the 8 ohm load. (Btw, if the load drops to 2 ohms then voltage drops to 0.25 V, which is like a 12 dB difference).

--

Analog audio signals are typically transmitted as voltages. Power amps may be current sources in integrated and special designs, but make little sense in general use cases. Most off-the-shelf power amps are voltage sources. Most passive speakers are designed to be driven by those.
Headphone outs historically had like 120 ohm output impedance, but dedicated headphone amps nowadays are typically very low output impedance as well.
 

xnor

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Here is an example of my loudspeakers.
View attachment 212137

I would rather expect the opposite of a relative bass boost here, see also Benchmark examples at 119Hz.

Thank you for your efforts to try enhancing my understanding.

This is very simple: as you increase output impedance, you will get a small peak at 70 Hz and a large one at 1-2 kHz with that specific speaker.

"Large" here is relative as it may still just be a fraction of a dB in absolute terms, depending on the actual output impedance.

In fact, you can imagine that the impedance plot is a bit like the frequency response curve of an EQ. The strength of the equalization increases with increasing output impedance.


Btw, the general statements about a bass boost are based on impedance plots of a woofer like this:
spkr-zplot.gif


Obviously, crossovers and other drivers can mess up the impedance curve. You can get very nasty dips and peaks that result in a really messed up frequency response with high output impedance. That's another reason why designers of passive speakers typically assume and design for voltage drive...
 
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pogo

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@xnor
So you wouldn't expect a bass boost with my specific speakers, would you?
 

Geert

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So you wouldn't expect a bass boost with my specific speakers, would you?

Why didn't you make the calculation for the woofers resonance frequency, where impedance is often the highest?

Your woofers don't have an impedance peak at resonance frequency?
 

Geert

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See my post #262

Well yes, why not let me figure it out.. :mad: So there is a rather small impedance peak in the low end around 70Hz. Let's reuse your own calculations to assess the impact, as that impedance is close to the numbers you used there:

Averaged assumption of 5ohms in the low frequency range:
5/(5+0.133) = 0.974

Averaged assumption of 8ohms in the high frequency range at woofer resonance :
8/(8+0.133) = 0.984

That's only a minor difference (bass peak) with an amplifier with DF = 60, which is what we've been telling from the start. (And most solid state amplifiers will have a higher DF).
 

pogo

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But even with this small bass peak, my overall low frequency range is attenuated, isn't it?
 

xnor

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But even with this small bass peak, my overall low frequency range is attenuated, isn't it?
Useless question. Everything is attenuated, the question is relative to what. I always assume relative to the nominal impedance.
In the specific impedance plot you posted, you get exactly what I said above. Basically a peak at 1-2 kHz, everything else is probably negligible.

Just out of curiosity, I assume your amps are connected to that M6000 sub and from there to the M200 bookshelf speakers?
Seems like an outdated design to me.
 

pogo

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Just out of curiosity, I assume your amps are connected to that M6000 sub and from there to the M200 bookshelf speakers?
No, the speakers are connected in bi-wiring to a NAD M33 and yes they are 30 years old ;)

And now I'm back to the same point I posted earlier:
The simple math for it do not correlate at all with the sound representation in reality with my speakers.
With a higher amp DF it sounds like less bass power.
And this is probably due to the fact that with a music signal, proper braking cannot be applied here when a low amp DF is present.
 

solderdude

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I don't really get all the fuss.

What's important is that a low DF causes relatively less voltage drop at a speaker's impedance peaks.

The impedance plot of the speaker in question has the largest peak at 1.5kHz (X-over or correction circuit ?)
A DF of 60 (8 ohm) is not what I would call problematically low. In fact I would not call it low at all.

The calculations pogo made are correct.
Effectively there would be a 0.2dB 'boost' around 1.5kHz.
Eyeballing the speaker is about -1dB too low in output (@ 1m, anechoic, on axis) that 0.2dB boost will not make it worse nor a very audible change.

Now... if this were a current amplifier this speaker would sound radically different. But lets face it... 0.13ohm output R is not really problematic.
 

Geert

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I don't really get all the fuss.

Indeed you don't :), but can't blame you. The last page was only an interlude. Pogo's last post is what's is all about:

The simple math for it do not correlate at all with the sound representation in reality with my speakers.
With a higher amp DF it sounds like less bass power.
And this is probably due to the fact that with a music signal, proper braking cannot be applied here when a low amp DF is present
.
 

solderdude

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The bolded part of course is nonsense. Pogo makes the usual (incorrect) assumption here.

People often assume an amp can 'grip' drivers and that damping currents are the reason.
A woofer with 8ohm DC resistance acts as a 'generator' with a 8ohm internal resistance.
Whether or not that generator is loaded with 0ohm or 0.13ohm doesn't make any difference.

The damping current is determined by Rsource (8ohm driver) + Rseries.
1V in 8ohm = 125mA
1V in 8.13ohm = 123mA
0.14dB difference in damping current is negligible.

Any real effects heard are caused by voltage division.
 

Geert

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Whether or not that generator is loaded with 0ohm or 0.13ohm doesn't make any difference.

The damping current is determined by Rsource (8ohm driver) + Rseries.

For Pogo to do the math; according to Small: Qec = 2 pi Fc Mmc (Re+Rs) / B l

Re: voice coil resistance
Rs: amplifier output resistance

Use Qec to calculate Qts, which defines the frequency and step response of a speaker system. The step response gives an indication of decay or ringing,.

A well designed speaker (low QTC) is a tuned resonator that's sufficiently damped by itself. With an amplifier with a DF of about >50 the amplifiers DF won't have any impact on the speakers decay time anymore. (That DF number will be a bit higher for low impedance drivers).
 

pogo

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This is actually the same but described in different words, as it is written in the manual of the A200 (except for the height of the decisive DF):

A high damping factor tends to produce a more clearly defined, very precise and analytical sound image, whereas a reduced damping factor produces a more warm and softer sound image.
Since this setting depends very much on the connected speakers, no universal recommendation can be given here. Choose the setting after a listening test and your personal listening preferences.
 
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