God (god at almighty.com) wrote:
: >> Consider an amplifier with an 8 ohm output impedance feeding 8 volts into
: >
: >> a speaker with an 8 ohm impedance. The speaker will be dissipating 8
: >> watts of power and will generate a specific sound pressure level.
: >>
: >> Next, consider the same amplifier putting 8 volts into a 16 ohm load.
: >> Since P = E^2 / R, the power being pumped into the load is 8^2 / 16, or 4
: >
: >> watts, which will yield a lower sound pressure level.
: >
: >You're totally missing the point You are confusing voltage level with
: >power levels.
: >
: >You are putting LESS POWER into the 16 ohm speaker which results in a lower
: >sound level. This has absolutely nothing do do with efficiency which is
: >function of the physical engineering design of the speaker and the
: >enclosure.
: >>
: >> The assumption, of course, is that both speakers have the same efficiency
: >
: >> and are, in fact, identical except for their impedances.
: >
: >If thet are of identical impedances, puting 1 watt of power into either
: >speaker will result the EXACT SAME sound level!
: >>
: >> John Fields
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I believe the original poster asked whether impedance had anything to do
with loudness. It surely does, as proven by my argument.
Your problems with the use of the written language seem to have caused
you to misunderstand my post, so let's see if I can make it easier for you.
Let's assume you have a nine volt battery, a nine ohm ten watt resistor
and an eighteen ohm ten watt resistor. Let's also say that the
efficiency of the resistors is 100%, just to level the playing field.
Next, connect the battery across the nine ohm resistor. The resistor
will dissipate nine watts, will exhibit a rise in temperature, and will
eventually come to thermal equilibrium at some temperature.
Now, disconnect the battery from the nine ohm resistor and connect it
across the sixteen ohm resistor. The resistor will dissipate 4.5 watts,
will exhibit a rise in temperature, and will eventually come to thermal
equilibrium at a temperature _lower_ than the nine ohm resistor used in
the previous example.
The point is that if you have a constant voltage source, the terminal
temperature (which is analogous to sound pressure level) will vary as a
function of the load impedance, which is the correct answer to the
question originally posed. By the use of DC In the cases above, the
imaginary portion of the impedance has been reduced to zero in order to
allow you to, perhaps, grasp the concept.
John Fields
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: John Fields is absolutely correct in this respect and hit it right on
: the money. If you consider sensitivity to mean SPL as a function of
: voltage, then the original theory is correct: the higher impedance
: speaker would produce lower SPL at a given voltage. However,
: sensitivity generally refers to SPL as a function of power. As John
: Fields said, if the efficiency (dB @ 1W, 1m) of both drivers is the
: same, they will produce the same SPL at a given power level.
: As an aside, the argument is not begging the question (petitio
: principii) although it does seem to be. After all, he did prove his
: point that the higher impedance speaker would produce lower SPL at a
: given voltage. However, when he translated this to lower SPL at a
: given wattage, he committed the fallacy of four terms, a non sequitur
: fallacy of ambiguity.
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An answer to this critique will soon be posted.
John Fields
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