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Old September 15th 19, 10:49 PM posted to rec.audio.high-end
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Default Introducing a New Horse to the Stable

On Saturday, September 14, 2019 at 9:58:44 AM UTC-4, Trevor Wilson wrote:
> So, a little Ohm's Law should tell you if you are demanding more current
> than the output devices are capable of delivering. 14 Amps is, by high
> end audio standards, a relatively modest current ability for a (say) 100
> Watt @ 8 Ohms amplifier. Provided the driver impedance is relatively
> benign, you should be OK.


Hmm, that's not what a little Ohm's law tells me.

100 Watts into 8 Ohms is a tad over 3.5 amps. Let's say it's a VERY
robust 100 watt amplifier, delivering 200 Watts into 4 Ohms requires
about 7 amps, and, let's pretend it has essentially ZERO output
impedance and an effectively limitless power supply, you're not reaching
14 amps until you're driving 400 watts into 2 ohms.

So, a couple of questions that Mr. Ohm may ask; what kind of loudspeaker
presents a broadband 2 ohm impedance or, conversely, what kind of
musical content would generate that kind of power requirement over
the pretty narrow band of frequencies where a loudspeaker has the
kind of pathological impedance curve that would dip to as low as
2 ohms.

(Yes, there exist SOME rare examples of loudspeakers with
2 ohm impedance, but such are confined to a VERY narrow
band of frequencies)

Okay, let's pretend we have real examples of the above. Let's assume
such a speaker has a moderately low efficiency, say the equivalent
of, oh, 86 dB SPL/1W/1M. We're blowing in 400 watts that means the speaker
is putting out 112 dB 1 meter way, a stereo pair, assuming the two
channels are uncorrelated, that's 115 dB. Really? This is a serious
requirement?

But wait, you explicitly stated:

"Provided the driver impedance is relatively benign"

and you specified 8 Ohms. So let's assume it's a nominal 8 ohm
impedance 3-way speaker using at least a 2nd-order crossover
network. The impedance will be around 6.5 ohms below system
cutoff (DC resistance of woofer voice coil), will rise to
perhaps 30 Ohms at and around system cutoff, then drop down
to perhaps 15% above the DC resistance above there, start
rising again until the woofer-midrange crossover starts working,
maybe betting to 10-12 ohms, then dip to perhaps 60% of the rated
impedance, so about 4.8 ohms, rise again to about 12 ohms or so
at the mid-tweeter crossover point, drop down to about 10-15% above
the tweeter DC resistance (which, for the purpose of argument, we'll
take to be a nominal 4 Ohm tweeter, so about 4.5 Ohms, after which
it starts rising again.

So, minimum impedance of about 4.8 ohms will occur over perhaps
a 2-octave band around 1 kHz, then about 4.5 ohms around 5 kHz.

Let's take your 14 amps, produce a musical signal where ALL the energy
is concentrated from about 500-2000 Hz and from about 4000-8000 Hz
ALONE, and see what 14 Amps does.

Well, since

P = I^R

And we'll assume the impedance at these points is largely resistive,
which is is, then:

P = 14^2 * 4.5

882 watts. And to do that, the amplifier must be capable of outputting

E = I R

E = 14 * 4.5

63 volts RMS.

Really?

Oh, wait! Everyone knows that under transient conditions, the loudspeaker
impedance can actually go well below the lowest impedance of the
speaker for brief moments due to back EMF, Otala said so.

Oh, wait! Everyone who knows that is wrong and has yet to advance any
confirmed data sowing this to be the case and, by the way, Otala DID
NOT say so: he basically said that the peak current requirement under
actual transient conditions is exactly what is expected from the actual
measure steady state impedance, and the only thing he really said
that's even remotely like this is that the peak current requirements
are greater than predicted by the "nominal" impedance of the loudspeaker.

Give me a shovel, Mr. Ohm wants to go back to sleep.
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