wuts the diff between these? why does 2 ohm have more wattage

ohm is a resistance on wattage. The lower the ohm, the higher the output
wattage.


"SeaN" wrote in message
. ca...
wuts the diff between these? why does 2 ohm have more wattage

mayhemkrew wrote:
ohm is a resistance on wattage. The lower the ohm, the higher the output
wattage.

That is...to say the least...misleading. You cannot say that the lower
the ohm (which should say "resistence" or more properly "impedence"
since we're talking about an AC current) the higher the output wattage.

First of all, you need to know the voltage the amplifier is putting out,
or at least assume that the voltage remains constant.

Secondly, you need to assume that the amplifier can handle the speaker.
Most amps can handle 2 ohm in two channel mode. A good number of amps
can handle 2 ohm when bridged, and will put out more wattage. But if you
go lower, the amp may shut down. Then your wattage is 0.

Thirdly, you need to keep in mind that the speakers impedence is the
lowest measurement of the speakers impedence, and usually occurs near
resonance. In general, the further away from the resonant frequency of a
speaker (as determined by the reactivity of the speakers voice coil, the
mechanical resonance of the speaker, and the enclosure the driver is
in), the higher the impedence.

Finally, don't assume that more wattage means more audio output. A two
ohm speaker at 400 watts will sound just as loud as a four ohm speaker
at 200 watts.

--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

On Thu, 17 Jul 2003 at 19:01 GMT, The Lizard wrote:

Finally, don't assume that more wattage means more audio output. A two
ohm speaker at 400 watts will sound just as loud as a four ohm speaker
at 200 watts.

Then I, Sam, wrote:

This is interesting. Why do people opt for a two ohm speaker at 400
watts over four owm speaker at 200 watts if there is no difference?

And finally Paul wrote:

Beacuse 400 watts is twice as much as 200 watts and most amps will make more
power at lower impedances. But 400 watts at 4 ohms on one amp is the same
as 400 watts at 2 ohms on another. All you should be concerned with is the
power output and if the amp can handle the load.

Paul,

What you said makes perfect sense. I cannot imagine where any one
would argue with you, but I am sure there are those that would.
What you stated and what I was questioning are two different things,
I think...

Let us pick one or two amp to discuss. I shall pick the two Z
Series Zapco amps (http://www.zapco.com).

Z300C2-SL Z600C2-SL
4 ohm - 2 x 75 watts 4 ohm - 2 x 150 watts
2 ohm - 2 x 150 watts 2 ohm - 2 x 300 watts
1 ohm - 2 x 300 watts 1 ohm - 2 x 600 watts

If I interpret you correctly, you are saying that a 2 ohm set of
speakers on the 300 will be just as loud as a 4 ohm set on the 600.
Correct? This makes perfect scene because in both cases you have 2
x 150 watts.

If I am reading what The Lizard said correctly, Taking just the 600,
it matters not if one uses a 2 ohm speaker or a 4 ohm speaker, the
audio output will be the exact same.

I then quoted what Boston Acoustics said about using 4 ohm compared
to 2 ohm speaker on the said amp. It is my interpretation that
Boston Acoustics is claming that there will be more audio output if
a 2 ohm speaker is used rather then a 4 ohm.

So the question is, who is right, The Lizard or Boston Acoustics?
And why?

Sam

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also, people get 2 ohm or 8 ohm speakers or dual voice coil subs to wire
differently (parallel and series) to acheive a different ohm load.

"Paul Vina" wrote in message
news:%hGRa.83553$ye4.61750@sccrnsc01...
Beacuse 400 watts is twice as much as 200 watts and most amps will make
more
power at lower impedances. But 400 watts at 4 ohms on one amp is the same
as 400 watts at 2 ohms on another. All you should be concerned with is
the
power output and if the amp can handle the load.

Paul Vina


This is interesting. Why do people opt for a two ohm speaker at 400
watts over four owm speaker at 200 watts if there is no difference?

basic ohms law
I = I X R
E- voltage
I - current
R- resistance (it is impedance when you have inductance and or capacitance
in the circuit)

I = E/R I = E/R
= 12v/4ohm = 12v/2ohms
= 3 amp = 6amps
that is simplified version of what happens but the therory is true. Most
amps will run between 11 and 14volts so substitute your own numbers. All
coils such as the one in the speaker produces inductance that is why
impedance is used and not resistance. All these numbers are constantly
changing (E,I,R) non is a constant, but basically the more amps that amp
sucks the more louder it sounds.

Hi,
Most car audio amp is MOSFET or Bi-polar power transistor driven.
Solid state device is current source device which means the lower the
load impedance the more power it can deliver(in a matched condition with
the ability of the power supply, the battery to supply extra demand for
juice) Under exact same condition (same speakers; same SPL and same
dispersion angles) with only difference in load impedance. Remember more
power will produce more acoustic energy(loudness). To be two times
louder, you need power of two times not straight two times power. So
100W is not much louder than 50W. Just it may have little more head
room. Sure 1000W is louder than 100W. Did I make any sense?
Tony

mayhemkrew wrote:

also, people get 2 ohm or 8 ohm speakers or dual voice coil subs to wire
differently (parallel and series) to acheive a different ohm load.

"Paul Vina" wrote in message
news:%hGRa.83553$ye4.61750@sccrnsc01...

Beacuse 400 watts is twice as much as 200 watts and most amps will make

more

power at lower impedances. But 400 watts at 4 ohms on one amp is the same
as 400 watts at 2 ohms on another. All you should be concerned with is

the

power output and if the amp can handle the load.

Paul Vina



This is interesting. Why do people opt for a two ohm speaker at 400
watts over four owm speaker at 200 watts if there is no difference?

Knoll wrote:
basic ohms law
I = I X R
E- voltage
I - current
R- resistance (it is impedance when you have inductance and or capacitance
in the circuit)

I = E/R I = E/R
= 12v/4ohm = 12v/2ohms
= 3 amp = 6amps
that is simplified version of what happens but the therory is true. Most
amps will run between 11 and 14volts so substitute your own numbers. All
coils such as the one in the speaker produces inductance that is why
impedance is used and not resistance. All these numbers are constantly
changing (E,I,R) non is a constant, but basically the more amps that amp
sucks the more louder it sounds.

Well, you got ohms law right, but there is a basic flaw here. The output
voltage of the amp is all we should be concerned with. If you take it's
input power, you automatically disregard such issues as efficiency, or
consumption of power by parts of the amp that do not produce audio output.

--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

Finally, don't assume that more wattage means more audio output. A
two ohm speaker at 400 watts will sound just as loud as a four ohm
speaker at 200 watts.

This is interesting. Why do people opt for a two ohm speaker at 400
watts over four owm speaker at 200 watts if there is no difference?

Because Lizard made a mistake. His statement is demonstrably untrue.

Sam Carleton wrote:

This is interesting. Why do people opt for a two ohm speaker at 400
watts over four owm speaker at 200 watts if there is no difference?

I am still trying to figure out the whole ohms thing, but Boston
Acoustics has an interesting PDF on the subject.

source: http://www.bostonacoustics.com/Manuals/y2ohm.pdf

-------- snip --------
With a 2-ohm voice coil, the wire gauge is larger than that of a
4-ohm coil.

This is not necessarily true. You can also reduce the resistance of the
voice coil by shortening the length of the wire used to make it.

As a result, it can pass more current without the risk
of fusing (burning out).

No engineer would associate fusing - which implies the joining of two
materiels - with burning out - which implies the thermal destruction of
a materiel. At this point, it becomes clear that this was written by a
marketing type.

Additionally, the 2-ohm coil produces less
heat, as more energy is transferred directly into setting the cone
in motion.

A 2 ohm coil produces less heat because it has less resistance, not
because it more efficiently transfer pwoer into setting the cone in
motion. There are dozens of factors that affect the efficiency of a speaker.

This results in a lower voice coil temperature that
increases the power handling. This also leads to a more efficient
subwoofer. Given the same voltage, a 2-ohm driver will produce more
output than a 4-ohm driver since less of the amplifier's energy is
transformed into heat within the voice coil. Instead, the energy is
being used to create movement-which creates sound.
-------- snip --------

This is simply wrong. There is no more acoustic output as a result of
halfing the voice coils impedeance. There is more power output by the
amplifier, but this is a fudging of the numbers. Halving the resistance
at a given voltage results in double the flow of current.

This, however doesn't tell us about how the speakers voice coil is
converting that into mechanical energy. Suprise: it's the voltage that
determines the acoustical output of a speaker. That's what amplifiers do
- they amplify voltage.


It sounds good to me. Is this accurate or is Boston Acoustics full
of it?


I'm not trying to sell you a speaker ; )

--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

Mark Zarella wrote:

Because Lizard made a mistake. His statement is demonstrably untrue.


Start demonstrating, bucko...


--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

On Fri, 18 Jul 2003 at 07:37 GMT, The Lizard wrote:

It sounds good to me. Is this accurate or is Boston Acoustics
full of it?

I'm not trying to sell you a speaker ; )

But are you trying to sell me your ego? (That is a rhetorical
question)

--
See my links engine for a collection of sites that
might be of interest to you. Your additions will
make this engine more powerful global resource.
http://www.miltonstreet.com/scarleton/links/

Because Lizard made a mistake. His statement is demonstrably untrue.


Start demonstrating, bucko...

A speaker that dissipates 400 watts, when compared to a speaker that
dissipates 200 watts, with the only variable being impedance, will exhibit a
3dB increase in intensity. 3dB changes are well above known psychophysical
thresholds for virtually every part of the audible spectrum.

In article , "Mark Zarella" seesigfile wrote:
Because Lizard made a mistake. His statement is demonstrably untrue.


Start demonstrating, bucko...

A speaker that dissipates 400 watts, when compared to a speaker that
dissipates 200 watts, with the only variable being impedance, will exhibit a
3dB increase in intensity. 3dB changes are well above known psychophysical
thresholds for virtually every part of the audible spectrum.

Too bad almost everyone uses power to talk about SPL, when
it would be so much better to speak VOLTAGE.


greg

Mark Zarella wrote:

A speaker that dissipates 400 watts, when compared to a speaker that
dissipates 200 watts, with the only variable being impedance, will exhibit a
3dB increase in intensity. 3dB changes are well above known psychophysical
thresholds for virtually every part of the audible spectrum.

We're not talking about dissapation of wattage, we're talking about the
output current of the amp. We're not talking about psychophysical
thresholds, we're talking about real life, where the actual output of
this system is unchanged.

If the voltage remains constant (which in this model, it does), the
strength of the magnetic field remains the same. Current is a function
of field density not strength.

Why do you think amplifiers amplify voltage not current?

--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

On Fri, 18 Jul 2003 at 18:01 GMT, The Lizard wrote:
Mark Zarella wrote:

A speaker that dissipates 400 watts, when compared to a speaker
that dissipates 200 watts, with the only variable being
impedance, will exhibit a 3dB increase in intensity. 3dB changes
are well above known psychophysical thresholds for virtually
every part of the audible spectrum.

We're not talking about dissapation of wattage, we're talking
about the output current of the amp. We're not talking about
psychophysical thresholds, we're talking about real life, where
the actual output of this system is unchanged.

If the voltage remains constant (which in this model, it does),
the strength of the magnetic field remains the same. Current is a
function of field density not strength.

Ok, so if I understand you correctly, you are saying that it is the
increase in voltage that allows a speaker to play louder. What
exactly happen then when you replace a 4 ohm speaker with a 2 ohm
speaker. It is my impression that something from the amp is
doubled. Is that the voltage or the current?

Sam

--
See my links engine for a collection of sites that
might be of interest to you. Your additions will
make this engine more powerful global resource.
http://www.miltonstreet.com/scarleton/links/

Sam Carleton wrote:

Ok, so if I understand you correctly, you are saying that it is the
increase in voltage that allows a speaker to play louder. What
exactly happen then when you replace a 4 ohm speaker with a 2 ohm
speaker. It is my impression that something from the amp is
doubled. Is that the voltage or the current?

Yes, the current is double, and therefore the total wattage is doubled.
But the actual amount of work getting done by the woofer hasn't changed.
Think of it like this: Ice has less resistance than concrete, but it's
hard to walk on ice than on concrete. Think of it in terms of running
down hill. It is harder to run down hill than uphill, even though there
is less resistance going downhill.

The backstory on this, just so you can get some perspective into the
story of this 2 ohm cult, is this:

When you add a second speaker in parallel with another, the actual
acoustic output goes up by ~3db. That is because you have added a second
moving surface, not because the load - which has now dropped to 2 ohms -
is causing the amp to put out more wattage. Competitors soon demanded
high current amps - amps that could handle putting out into lower loads,
or more precisely, that could put out more current without frying up.
Car audio companies published the specs for these amps at 2 ohm, 1 ohm,
even 1/4 ohm loads. Lo and behold, joe consumer sees higher wattages on
these amps, and assumes that means his system will thump harder. Speaker
companies have now seized on this to produce 2 ohm voice coils, and joe
consumer is snapping them up because their amps will put out more wattage!

Usually you only see this kind of BS marketing hype form interconnect
makers, and b-market audio companies. I'm suprised Boston is putting out
this kind of junk science. In the old days, guys like myself, Eddie
Runner, Jay B. Haider, and Andrew C Ohnstad used to pester manufacturer
reps on this NG about making bogus unscientific claims. We ran the BS
ones off...(Shameless plug)...DEI is one of the few manufacturers that
could justify their marketing.

Now, I'm going to confuse you. There is some truth to the idea of 2 ohm
speakers increasing efficiency in audio systems. Ever wonder why we use
4 ohm speakers in car audio, and 8 ohm in the home? In the car, you have
a limited amount of power you can draw on. An amplifier can only ever
get 12-14 volts in the average car, but it can draw as much current as
the car can produce for it. An amplifiers DC to DC power supply converts
that current into voltage, and the output stage regulates the amount of
voltage sent to the speaker. The higher the resistance of the speaker,
the more voltage you need to produce. However, in the examples we've
talked about, dropping the resistance has not resulted in an increase in
voltage, but in current. You can't create voltage by reducing the
resistance of the speaker - ohms law doesn't work that way. In fact, if
you halve the resistance, and try to maintain the same current, you'll
actually halve the voltage (rather counterintuitve eh?).

Do the math yourself: http://www.teamrocs.com/technical/pages/ohmscalc.htm



--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

On Fri, 18 Jul 2003 at 19:34 GMT, The Lizard wrote:
Sam Carleton wrote:

Ok, so if I understand you correctly, you are saying that it is
the increase in voltage that allows a speaker to play louder.
What exactly happen then when you replace a 4 ohm speaker with a
2 ohm speaker. It is my impression that something from the amp
is doubled. Is that the voltage or the current?

Yes, the current is double, and therefore the total wattage is
doubled. But the actual amount of work getting done by the woofer
hasn't changed. Think of it like this: Ice has less resistance
than concrete, but it's hard to walk on ice than on concrete.
Think of it in terms of running down hill. It is harder to run
down hill than uphill, even though there is less resistance going
downhill.

The backstory on this, just so you can get some perspective into
the story of this 2 ohm cult, is this:

When you add a second speaker in parallel with another, the actual
acoustic output goes up by ~3db. That is because you have added a
second moving surface, not because the load - which has now
dropped to 2 ohms - is causing the amp to put out more wattage.
Competitors soon demanded high current amps - amps that could
handle putting out into lower loads, or more precisely, that could
put out more current without frying up.

Ok, I am with you, FINALLY Sorry it took so long for me to
understand. If I am reading you correctly, when only using one sub,
there is absolutly no reason to opt for a 2 ohm sub.

Along the same line, it is my impression that two 2-ohm speakers in
series would be a better solution then two 4-ohm speakers in
parallel given the same amp because the two 2-ohm would draw half
the current. Correct?

I know that if you take four speakers, break them into two sets that
are in series and hook those two sets up in parallel that the
impendence will be that of one of the speakers. Considering adding
one speaker will give a ~3db gain, I assume adding two speakers to
that would give another ~3db gain for a total of ~6db. I might be
off, but the point is that I assume there is more acoustic output.

I understand that you cannot gain something for nothing, which if
you look at ohms law, it appears that you are gaining acoustic
output without any price other then space. But I have a hard time
believing that it is that simple.

I have a feeling that there are forces at play of which I don't
know. The only think I can come up with is that there is more
energy then the one speaker can utilize. But that would fall apart
at some point, say with 9 speakers, if not 9, maybe 16, or 25, or
36, but at some point the theory falls apart.

It could be like CPU's. Two is faster then one, four is faster then
two, but as some point adding CPU's slows the system down because of
the overhead of maintaining all the CPU's. But I have a feeling
that things are simpler then that

Can you explain what is going on?

Sam

Sam Carleton wrote:

Ok, I am with you, FINALLY Sorry it took so long for me to
understand. If I am reading you correctly, when only using one sub,
there is absolutly no reason to opt for a 2 ohm sub.

Well, the idea to adopt a 2 ohm sub would depend on the situation. For a
single sub set up, it probably would'nt make a whole lot of sense, it
would just be unnecessary strain on the amplifier. As part of a dual sub
setup, you could put them in series so that the amp would have a single
four ohm load - the optimum load for most car audio amplifiers. But
then, people have been running dual speaker setups in parallel,
resulting in two ohm loads, for many years now without many problems.

Along the same line, it is my impression that two 2-ohm speakers in
series would be a better solution then two 4-ohm speakers in
parallel given the same amp because the two 2-ohm would draw half
the current. Correct?

It would draw half the current, provided the voltage remained unchanged,
yes. However, it is really no more or less a "better" solution, since
most amps can handle a 2 ohm load such as that from a pair of 4 ohm
speakers in parallel. The amp would run cooler with the first setup
though, so if heating and cooling are an issue, then the 2 ohm sub is a
good idea. However, you achieve the same effect with two 4 ohm dual
voice coil speakers. Wire each speaker up so that their voice coils are
in parallel, then wire the speakers in series, and blammo! A single 4
ohm load for the amplifier.

snip: see above I might be
off, but the point is that I assume there is more acoustic output.

Yup. Each additional moving surface gives you ~3db more of output. It's
a small change, but if you have plenty of room for woofers and no desire
to see out your rearview mirror, you can always expand. However, when
space is a premium, there's no substitute for higher excursion, more
power, and a good enclosure.


I understand that you cannot gain something for nothing, which if
you look at ohms law, it appears that you are gaining acoustic
output without any price other then space. But I have a hard time
believing that it is that simple.

It has nothing to do with ohms law. You're increasing output because
you're adding another moving surface, i.e., not because you're lowering
the resistence of teh speaker. In fact, two identical speakers will have
the same output power regardless of the coils impedence (for the most
part, there are details).

It could be like CPU's. Two is faster then one, four is faster then
two, but as some point adding CPU's slows the system down because of
the overhead of maintaining all the CPU's. But I have a feeling
that things are simpler then that

Yeah, it's simpler. The impedence of a woofer? That resistance? We're
taught to think of resistence as like a pinch in a water hose, but it's
sometimes better to think of it as a measure of work getting done, a
sort of factor by how much applied power is turned into actual work.
(But don't forget the pinched hose analogy. It is quite useful in many
cases.) Let's say you're on an excercise bike, and you are doing as much
as you can do. If you lower the setting (think of it as resistance),
you'll be able to peddle hella faster. If you increase the setting
(i.e., ramp up the resistance), you'll peddle slower but you'll have to
use more strength. Is more work being done either way? Of course not!
Either way you were already doing as much as you can! Alot of excercise
bikes actually have digital displays that will show your output either
as BTU's or Watts. You can verify this, and get your cardio done at the
same time.

--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

Hi,
Voltage? Why alternator and battery can't keep up? SS device is current
driven by nature. Tube is voltage driven. You have vacuum tube amp?
Now I made you confused!
Tony

GregS wrote:
In article , "Mark Zarella" seesigfile wrote:

Because Lizard made a mistake. His statement is demonstrably untrue.


Start demonstrating, bucko...

A speaker that dissipates 400 watts, when compared to a speaker that
dissipates 200 watts, with the only variable being impedance, will exhibit a
3dB increase in intensity. 3dB changes are well above known psychophysical
thresholds for virtually every part of the audible spectrum.


Too bad almost everyone uses power to talk about SPL, when
it would be so much better to speak VOLTAGE.


greg

In article ,
Tony Hwang wrote:

Hi,
Voltage? Why alternator and battery can't keep up? SS device is current
driven by nature. Tube is voltage driven. You have vacuum tube amp?
Now I made you confused!
Tony

GregS wrote:
In article , "Mark Zarella" seesigfile
wrote:

Because Lizard made a mistake. His statement is demonstrably untrue.


Start demonstrating, bucko...

A speaker that dissipates 400 watts, when compared to a speaker that
dissipates 200 watts, with the only variable being impedance, will exhibit a
3dB increase in intensity. 3dB changes are well above known psychophysical
thresholds for virtually every part of the audible spectrum.


Too bad almost everyone uses power to talk about SPL, when
it would be so much better to speak VOLTAGE.


greg


Did everyone here visit a party together? I think the punch was spiked.

Tony Hwang wrote:
Hi,
Voltage? Why alternator and battery can't keep up? SS device is current
driven by nature. Tube is voltage driven. You have vacuum tube amp?
Now I made you confused!
Tony



Tony,

We're talking about speakers. Speakers are not solid state devices.

Secondly, not all solid state devices are current driven.

FETs are voltage driven. An applied voltage at the gate controls the
current that flows between source and drain.

Transistors are current driven. A current applied at the base controls
the flow of current between the emitter and the collector.

Nobody even mentioned tubes, and the type of amplifier is really not
relevent to the discussion anyway.

--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

Tony Hwang wrote:
Hi,
Better go back to text book. SS device is low impedance device which is
current driven. Vacuum tube is hi impedance device which is voltage
driven. Just look at what happens at N-P junction(between base and
collector/emitter) MOSFET is little better acting like tube.
Tony


Tony...pay real close attention...A SPEAKER IS NOT A SOLID STATE DEVICE!

You need to go to a text book...an english one.


--
Lizard

teamROCS #007 / Technical Director / Founding Member *res derelicta*
http://www.teamrocs.com/
Save Farscape http://www.watchfarscape.com

A speaker that dissipates 400 watts, when compared to a speaker that
dissipates 200 watts, with the only variable being impedance, will exhibit a
3dB increase in intensity. 3dB changes are well above known psychophysical
thresholds for virtually every part of the audible spectrum.

We're not talking about dissapation of wattage, we're talking about the
output current of the amp.

No, we're talking about power dissipation. You said: "A two ohm
speaker at 400 watts will sound just as loud as a four ohm speaker at
200 watts." That clearly states that we're talking about two speakers
of different impedance "at" 400 and 200 watts respectively. Did you
mean something else?


We're not talking about psychophysical
thresholds, we're talking about real life,

What's the difference?

where the actual output of
this system is unchanged.

If the voltage remains constant (which in this model, it does), the
strength of the magnetic field remains the same. Current is a function
of field density not strength.

Current is a function of voltage and impedance. If voltage is indeed
the same as you suggest, and the impedance is halved, then ohm's law
tells us that the current will be doubled. Joule's law of heating in
turn tells us then that the power is also doubled.

Ok, I am with you, FINALLY

There;s no reason to be with him. He's still wrong. And I can't
understand why because I know that he knows this stuff. My guess is
that he's misinterpreting your question.

Sam, it's simply ohm's law. Halve the resistance and double the
current. Double the current (while maintaining the same voltage) and
you double power dissipation. Double power dissipation and (assuming
linearity, which is a somewhat safe assumption) you double acoustical
output.


Sorry it took so long for me to
understand. If I am reading you correctly, when only using one sub,
there is absolutly no reason to opt for a 2 ohm sub.

If your amp delivers more power into 2 ohms than 4, that's an awfully
good reason to do it.



Along the same line, it is my impression that two 2-ohm speakers in
series would be a better solution then two 4-ohm speakers in
parallel given the same amp because the two 2-ohm would draw half
the current. Correct?

No. Two 2 ohm speakers in series will not be as loud as two 4 ohm
speakers in parallel. Why? Less amplifier output.



I know that if you take four speakers, break them into two sets that
are in series and hook those two sets up in parallel that the
impendence will be that of one of the speakers. Considering adding
one speaker will give a ~3db gain, I assume adding two speakers to
that would give another ~3db gain for a total of ~6db. I might be
off, but the point is that I assume there is more acoustic output.

If you add another speaker yet halve the amount of power that each
speaker is getting, there's no net gain (assuming negligible power
compression and acoustical factors - not a safe assumption).

I hope this clears it up for you.

On Sun, 20 Jul 2003 at 02:12 GMT, Mark Zarella wrote:

Sam, it's simply ohm's law. Halve the resistance and double the
current. Double the current (while maintaining the same voltage)
and you double power dissipation. Double power dissipation and
(assuming linearity, which is a somewhat safe assumption) you
double acoustical output.

Ok, as most everyone has figrued out I am anything but an expert in
electronics or car audio. But I do know a bit about photography.
Back in the old days of motor drives being seperate from the camera,
some folks made after market kits to all one to hook up more
batteries to the motor drive. The additional batteries where added
in series so it upped the voltage, not current. The purpose was to
increase the SPEED of the motor drive. On the other hand, there are
aftermarket big batteries for flashes. This unit up the current
going to the flash to speed up the recycle time.

My take, which can be completely wrong, is that voltage equals speed
and current equals power. The motor drive needs more voltage to go
faster, the flash (filling up capacitors) need more current (power)
to recycle faster.

With this basic idea in mind, it makes perfect sence to me that for
a speaker to be louder it needs more speed, not more power. Granted
those really big subs need quite a bit of power to operate, but more
power will only result in more heat, not faster moving voice coils.

If, and that is a big if because I don't know... But IF when
putting a 2 ohm load on an amp, it doubles the current, not the
voltage, I can easily see where a 4-ohm speaker will be just as loud
as a 2-ohm speaker on the same amp. The current can double and
leave the volage alone. This of cource is assume I am correct in
saying that it is voltage that makes this louder and cutting the
load to the amp only ups the current. If I am not mistakin, this is
exactly what Lizard is trying to say.

Please, oh, please let me know where I am wrong, if I am. I find
this very interesting and would like to get to the truth of the
matter

Sam

Sam, it's simply ohm's law. Halve the resistance and double the
current. Double the current (while maintaining the same voltage)
and you double power dissipation. Double power dissipation and
(assuming linearity, which is a somewhat safe assumption) you
double acoustical output.

Ok, as most everyone has figrued out I am anything but an expert in
electronics or car audio. But I do know a bit about photography.
Back in the old days of motor drives being seperate from the camera,
some folks made after market kits to all one to hook up more
batteries to the motor drive. The additional batteries where added
in series so it upped the voltage, not current. The purpose was to
increase the SPEED of the motor drive. On the other hand, there are
aftermarket big batteries for flashes. This unit up the current
going to the flash to speed up the recycle time.

My take, which can be completely wrong, is that voltage equals speed
and current equals power. The motor drive needs more voltage to go
faster, the flash (filling up capacitors) need more current (power)
to recycle faster.

I unfortunately don't know anything about cameras so I won't pretend
that I do. But yes it seems perfectly reasonable to assume you're
correct since it is indeed true that a capacitor will charge faster if
the source (in this case the flash battery) has a lower output
impedance - which is basically what a "high current battery" is.

With this basic idea in mind, it makes perfect sence to me that for
a speaker to be louder it needs more speed, not more power. Granted
those really big subs need quite a bit of power to operate, but more
power will only result in more heat, not faster moving voice coils.

Not quite. "Speed", in the case of loudspeakers, is a function of
frequency and amplitude. That is, the speed of the cone increases as
frequency increases and amplitude increases. The former does not
translate into increased acoustical output though. It simply
increases the frequency of the air vibration. Increasing sound
intensity is accomplished *only* by increasing the amplitude of the
cone's motion. Cone motion is increased by a number of different
factors: increased speaker sensitivity to the frequency you're
playing, air pressure effects (enclosure, etc), and increased current.
According to ohm's law, current is dictated by two things: voltage
and impedance. In this discussion, we're assuming that the amplifier
is delivering a certain voltage waveform in both the 2 and 4 ohm case
(it does not, but that's a different topic altogether). So the only
variable here is impedance. If we're assuming 2 ohms vs. 4 ohms,
twice the current will flow in the 2 ohm case (I = V/R). In other
words, more power (P = VI).

So why is power the ultimate determinant in this situation anyway?
Well, with increased power dissipation by a 2 ohm (or 4 ohm) speaker,
more current is required (again, assuming voltage remains constant).
How does a speaker move? The current passing through the voice coil
creates a magnetic field (Faraday's law). This magnetic field
interacts with the magnetic field provided by the big magnet on the
back of your speaker. This results in either an attractive or
repulsive force. That moves the speaker coil/cone assembly with
respect to the speaker basket/magnet assembly. According to Faraday's
law, more current means more force.


If, and that is a big if because I don't know... But IF when
putting a 2 ohm load on an amp, it doubles the current, not the
voltage, I can easily see where a 4-ohm speaker will be just as loud
as a 2-ohm speaker on the same amp. The current can double and
leave the volage alone. This of cource is assume I am correct in
saying that it is voltage that makes this louder and cutting the
load to the amp only ups the current. If I am not mistakin, this is
exactly what Lizard is trying to say.

I don't think Lizard was trying to say that, because I know that he
knows better. My guess is that he misspoke or I misinterpreted what
he was trying to say. I can't quite figure out what he could have
been saying though. But the simple answer to your question is: more
power = more sound intensity, as I described above.

On Mon, 21 Jul 2003 at 20:18 GMT, Mark Zarella wrote:

Not quite. "Speed", in the case of loudspeakers, is a function of
frequency and amplitude. That is, the speed of the cone increases as
frequency increases and amplitude increases. The former does not
translate into increased acoustical output though. It simply
increases the frequency of the air vibration. Increasing sound
intensity is accomplished *only* by increasing the amplitude of the
cone's motion. Cone motion is increased by a number of different
factors: increased speaker sensitivity to the frequency you're
playing, air pressure effects (enclosure, etc), and increased current.
According to ohm's law, current is dictated by two things: voltage
and impedance. In this discussion, we're assuming that the amplifier
is delivering a certain voltage waveform in both the 2 and 4 ohm case
(it does not, but that's a different topic altogether). So the only
variable here is impedance. If we're assuming 2 ohms vs. 4 ohms,
twice the current will flow in the 2 ohm case (I = V/R). In other
words, more power (P = VI).

So why is power the ultimate determinant in this situation anyway?
Well, with increased power dissipation by a 2 ohm (or 4 ohm) speaker,
more current is required (again, assuming voltage remains constant).
How does a speaker move? The current passing through the voice coil
creates a magnetic field (Faraday's law). This magnetic field
interacts with the magnetic field provided by the big magnet on the
back of your speaker. This results in either an attractive or
repulsive force. That moves the speaker coil/cone assembly with
respect to the speaker basket/magnet assembly. According to Faraday's
law, more current means more force.

Man is it hard to argue when folks start quoting laws But then I
was, well am not out to argue in the first place.

Now if you would allow me to iterate what you said: Thanks to
Faraday's law, we all know that it is current, not voltage, that
makes a electro magnetic stronger. When the impendance is cut in
have, it is current, not voltage that goes up. Because of this the
voice coil moves more resulting is loud music. Correct?

If, and that is a big if because I don't know... But IF when
putting a 2 ohm load on an amp, it doubles the current, not the
voltage, I can easily see where a 4-ohm speaker will be just as loud
as a 2-ohm speaker on the same amp. The current can double and
leave the volage alone. This of cource is assume I am correct in
saying that it is voltage that makes this louder and cutting the
load to the amp only ups the current. If I am not mistakin, this is
exactly what Lizard is trying to say.

I don't think Lizard was trying to say that, because I know that he
knows better. My guess is that he misspoke or I misinterpreted what
he was trying to say. I can't quite figure out what he could have
been saying though. But the simple answer to your question is: more
power = more sound intensity, as I described above.

I very much look forward to what Lizard has to say on this *new*
twist

Sam

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In article , Sam Carleton wrote:

batteries to the motor drive. The additional batteries where added
in series so it upped the voltage, not current. The purpose was to
increase the SPEED of the motor drive.

Increased voltage will increase the speed of a DC motor. This could be
short-lived if the motor can't handle the higher voltage.

On the other hand, there are
aftermarket big batteries for flashes. This unit up the current
going to the flash to speed up the recycle time.

This, too, makes sense. More on this below:

My take, which can be completely wrong, is that voltage equals speed
and current equals power. The motor drive needs more voltage to go
faster, the flash (filling up capacitors) need more current (power)
to recycle faster.

What it really is is like a garden hose. Current = the amount of water
flowing out. Voltage = pressure. You can have a skinny hose at high
pressure and it puts out the same amount of water as a big hose at low
pressure. Volts times current = power - volts times amps = watts.
10A of 12V = 120W
20A of 6V = 120W

Same amount of power. One has twice the current (and will need a thicker
wire to carry it) but both provide the same amount of power. Your home
stereo might only draw 3A from the wall socket, but 3A * 120V = 360W. To
get 360W in your car with a 12V feed you need 360 / 12 = 30A. So you need
ten times the *flow*, because you have one tenth the *pressure*.

The above in simplified. If anybody starts yapping about laminar hole
flow in solid vs. stranded wire I'm going to ignore it. 8-)

If, and that is a big if because I don't know... But IF when
putting a 2 ohm load on an amp, it doubles the current, not the
voltage, I can easily see where a 4-ohm speaker will be just as loud
as a 2-ohm speaker on the same amp.

Now we're talking about impedence matching, and this is a whole 'nother
ball game from the one we talked about above. The most relevant equation is
P = I^2 R2, where R2 is the input impedance. I won't bore you with the
math, but suffice it to say that the maximum power flows when the input and
output impedance is the *same*. This is true for DC circuits, not always
true for AC circuits. If you're using 4 ohm speakers then use a 4 ohm amp,
and vice versa.

The current can double and
leave the volage alone.

The current is restricted by the wiring (because the fuse is sized
according to the wire). The voltage can be choked by the wiring, because
resistance causes a voltage drop, which causes heat and loss of power, but
it's much much easier to raise voltage in a DC system than it is to raise
current.

This of cource is assume I am correct in
saying that it is voltage that makes this louder and cutting the
load to the amp only ups the current.

The real kicker is power. Volts and amps can be raised and lowered
without changing the amount of actual power. You know those static electric
shocks you get when you get out of your car, or touch a doorknob after
shuffling your feet on carpeting or whatever? Those shocks can run up to a
couple of thousand volts. They have screw-all for power because the
amperage is screw-all, so they don't hurt you.

Oh yeah, regarding that flash, the flash is powered by a relatively large
capacitor which has to charge up from the batteries. The capacitor is a
temporary storage location for power. It's like having a tank on your
well-pump so that you can occasionally use water faster than the well can
supply it. The larger battery delivers more current at the same voltage,
and this results in more power. The hose is filling the bucket faster, you
see. When the bucket is filled up you can dump it into the flashbulb really
quickly and 'pop', nice bright high-amperage flash.

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In article , Mark Zarella wrote:

and impedance. In this discussion, we're assuming that the amplifier
is delivering a certain voltage waveform in both the 2 and 4 ohm case
(it does not, but that's a different topic altogether). So the only
variable here is impedance. If we're assuming 2 ohms vs. 4 ohms,
twice the current will flow in the 2 ohm case (I = V/R). In other
words, more power (P = VI).

I'm afraid it doesn't work that 'cleanly'. Impedance matching is a prick,
and I didn't cover it until second year university physics. If you want the
math, then here it is:

R1 = output resistance of the amp (impedance)
R2 = input resistance of the speaker (impedance)
P = I^2 R2
V = IR, so I = V/R, but we have two resistances in the completed circuit so:
I = V/(R1 + R2)

So:

P = I^2 R2 = (V^2 R2) / (R1 + R2)^2

So far it's looking pretty clean, but how to maximize it? Divide by
R1/R1 to get:

P = (V^2 (R2/R1)) / (R1 (1 + (R2/R1)^2)

Now, take the derivative of P with respect to R2, and set it to zero.
This gives us a maximum:

0 = dP/dR2 = (V^2 (1 - R2/R1)) / (R1^2 (1 + R2/R1)^3)

Now, the only way to make the equation equal zero is for the numerator
(V^2 (1 - R2/R1)) to equal zero. Since we can't make the voltage zero and
have a functioning circuit then (1 - R2/R1) must equal zero. So R2/R1 must
equal 1, so that 1-1=0. For R2/R1 to be 1, R2 = R1.

So, maximum power is transmitted in a DC circuit if the impedance is
matched, input to output. Note, I'm talking about maximum power, not
necessarily maximum *current*. Maximum power.

For this reason many audio amps have a transformer in it tapped to provide
output at various impedances, so you can match the impedance.

So why is power the ultimate determinant in this situation anyway?

The above. Really.

Well, with increased power dissipation by a 2 ohm (or 4 ohm) speaker,
more current is required (again, assuming voltage remains constant).
How does a speaker move? The current passing through the voice coil
creates a magnetic field (Faraday's law).

This is taking me back! I'm with you so far...

This magnetic field
interacts with the magnetic field provided by the big magnet on the
back of your speaker. This results in either an attractive or
repulsive force. That moves the speaker coil/cone assembly with
respect to the speaker basket/magnet assembly. According to Faraday's
law, more current means more force.

You just lost me. More current means more force? Hmm... From what I
remember, emf is proportional to the rate of change of the magnetic flux.
Now, if the magnetic flux is defined as the integral of the magnetic field
over an open surface, or basically the integral of the magnetic flux. In
fact, Faraday's Law of Induction is simply that the emf equals the negative
of the rate of change of the magnetic flux through the circuit. So while
I'm not arguing with the statement that more current means more force, I
don't see how you got it from Faraday's Law. Now, Ampere's law might help
us, here. The coil in a speaker is just a solenoid, so Ampere's law tells
us that the power of a given solenoid is directly proportional to the
current flow.

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In article , Sam Carleton wrote:

Man is it hard to argue when folks start quoting laws

Not if you know them.

But then I
was, well am not out to argue in the first place.

Fair enough. We're all here to learn and share, right?

Now if you would allow me to iterate what you said: Thanks to
Faraday's law, we all know that it is current, not voltage, that
makes a electro magnetic stronger.

Well, certainly according to Ampere's Law we know that the strength of the
magnetic field formed by a solenoid is directly proportional to the current.
Double the current and you double the strength of the field. Basically. The
law only perfectly applies to solenoids of infinite length, which makes it a
bit of a prick, but the reality of the situation is that it is close enough
for jazz with solenoids of any length.

When the impendance is cut in
have, it is current, not voltage that goes up.

If you reduce the resistance then you increase the current, all other
things remaining equal. That's the kicker. Equal. It doesn't work quite
that nicely in real world situations. In the text book with perfect
conductors and whatnot you can say this and be marked as correct. Given
sufficiently large conductors and sufficiently clean connections then given
a certain unchangeable voltage in the feed, the amperage will increase with
a lowered resistance (impedance) in the powered circuit.

Because of this the
voice coil moves more resulting is loud music. Correct?

Not necessarily, no. It depends on the impedance of the driving circuit.
If you use a given amp and hook up 4 ohm speakers to the 4 ohm outputs,
then you will get such-and-such power out of it. If you hook up 2 ohm
speakers to the 2 ohm outputs then you will get just a shade more power.
NOT twice as much.
None of this works cleanly, math-wise, because the simple equations don't
properly apply. Let's look at some of the ones that have been used in this
thread:
V=IR Volts = amperage times resistance
P=IV. Power = amperage times volts.

Since V=IR
P=I(IR)
P=I^2 R power in watts = current squared times resistance

or

P=V(V/R)
P=V^2 / R power in watts = voltage squared divided by resistance

These are lovely equations, but they require values for voltage and
current that we can't be sure of. When you change one side of the balance
the other side changes to compensate. The amplifier is rated to a certain
number of watts or peak watts. Let's pretend for a moment that these are
*real* watts. So say it's a 50Wx4 amplifier. We have 50 watts per channel.
Fine.
P = I^2 R
I = Sqrt(P/R)
I = Sqrt(50/2) = 5A for 2 ohms
I = Sqrt(50/4) = 3.535A for 4 ohms.

So to get the same power requires more amps with the lower resistance.
Why's that? Because power is a measure of doing work, and it's easier to
move through 2 ohms than 4 ohms. The resistance is what causes work to be
performed. It's *easier* for the amp to run 2 ohm speakers than 4 ohm
speakers if the impedance is matched. But what would happen to the voltage?

P = V^2 / R
V = Sqrt(PR)
V = Sqrt(50 x 2) = 10V for 2 ohms
V = Sqrt(50 x 4) = 14V for 4 ohms

Do you see how these can't apply? Every time you set something as
'static' to calculate something else, it doesn't work. Not on a system like
this. The supplied voltage can vary, and the current can vary. These
simple equations just *can't* tell you what you need to know. They
give information that just isn't terribly useful. To get 50W in a 12V
system (if it really *was* 12V, which it really isnt) you'd need

P = V^2 / R
50 = 12^2 / R
50R = 144
R = 144 / 50
R = 2.88 ohms.

Anything other than 2.88 ohms isn't going to work out in the equations
with 50W being the power supplied at 12V. 2.88 ohms is close to 3 which is
mid-way between 2 and 4 ohms, which is why the voltages above are
symmetrical around the 12V mark.

The bottom line is that if you match the impedance of the amp to the
impedance of the speakers then you will get the MAXIMUM amount of power
flowing, given the existing wiring, connections and power source. Using a
lower impedance in both the amp and speakers (matched) gives you a slight
boost in efficiency, but not enough to get really excited about unless
that's the kind of thing you enjoy being excited about.

It's been well over a decade since I had to use this stuff, so I could be
wrong somewhere along the line. I'm sure someone will point it out if I am.

--
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|
http://www.metalmangler.com| Caledon, Ontario, Canada

and impedance. In this discussion, we're assuming that the amplifier
is delivering a certain voltage waveform in both the 2 and 4 ohm case
(it does not, but that's a different topic altogether). So the only
variable here is impedance. If we're assuming 2 ohms vs. 4 ohms,
twice the current will flow in the 2 ohm case (I = V/R). In other
words, more power (P = VI).

I'm afraid it doesn't work that 'cleanly'.

I kept it simple to provide an understandable explanation to Sam.
Output impedances and the like are not worth bringing up considering
the point of the conversation. The ultimate "argument" was whether or
not sound intensity increased upon decreasing the load from 4 to 2
ohms. The short answer is that it does. Lizard disagreed apparently.
But yes, as you suggest, it doesn't double - primarily because of
power compression and amplifier output impedance effects, as well as
the reactive component of loudspeakers.

snip
So, maximum power is transmitted in a DC circuit if the impedance is
matched, input to output. Note, I'm talking about maximum power, not
necessarily maximum *current*. Maximum power.

"Maximum power" is not very relevant, considering that load impedances
approaching amplifier output impedances are unfeasible.


This magnetic field
interacts with the magnetic field provided by the big magnet on the
back of your speaker. This results in either an attractive or
repulsive force. That moves the speaker coil/cone assembly with
respect to the speaker basket/magnet assembly. According to Faraday's
law, more current means more force.

You just lost me. More current means more force? Hmm... From what I
remember, emf is proportional to the rate of change of the magnetic flux.

Induced emf from the typical "moving coil in a magnetic field"
explanation (or stationary coil in a pulsing mag field; or combo of
the two - choose your poison) is different from what I described.
However, it is relevant in loudspeaker operation, but not really worth
getting into in this conversation (unless you want me to).

And yes, I meant Ampere's law.

In article , Mark Zarella wrote:

I kept it simple to provide an understandable explanation to Sam.

Fair enough.

Output impedances and the like are not worth bringing up considering
the point of the conversation.

I didn't see the root message of the thread - just Sam's post about what
*he* figured was the case.

The ultimate "argument" was whether or
not sound intensity increased upon decreasing the load from 4 to 2
ohms. The short answer is that it does.

Yes. If the impedances are matched. (otherwise I wouldn't count on it).

"Maximum power" is not very relevant, considering that load impedances
approaching amplifier output impedances are unfeasible.

Really? So what is an example of a common amplifier output impedance,
then?
Maximum power is always relevant, given a particular voltage.

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In article , Mark Zarella wrote:

Well under an ohm, typically.

Hmm... got a reference for that?

But maximum power in the case of today's SS audio amplifiers occurs at the
lowest impedance presented before the amp shuts off.

If the output impedance of the amp is really under an ohm, then that makes
sense, because the lowest impedance presented will be the closest to
matching the impedance of the amp.


By the way, in your header it says "seesigfile" but you don't seem to have
a .sig file posted in your messages.

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at the risk of going down in flames...

I beleive that Lizard's statements were based on the idea that speakers
are voltage dependent devices (ie double the voltage = double the sound
output of the speaker) which would validate his 400W@2ohm = 200W@4ohm
example.

I beleive that Mark Zarella's statements were based on speakers being
current dependent (ie double the current = double the sound output of
the speaker).

So I guess the $10,000 question is "Is a typical car speaker's sound
output voltage or current dependent?" I am inclined to beleive that
sound output of a speaker is a function of current through the coil.

With regards to impedence matching, I would think that the 2 ohm/4 ohm
setting on the amp would be inclusive of the power supply's typical
output impedence not in addition to. If it were in addition to, I would
think that max efficiency would be at the higher impedence settings.

My 2 cents is make sure the load impedence matches the amp impedence
setting. Put a 2 ohm load on an amp set at 4 ohms will waste power in
the amp, put an 8 ohm load on an amp set at 4 ohms will drop the current
to the speakers and loose efficiency, put a 4 ohm load on a 4 ohm amp
and you have optimized the system (for the most part).
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at the risk of going down in flames...

I beleive that Lizard's statements were based on the idea that speakers
are voltage dependent devices (ie double the voltage = double the sound
output of the speaker) which would validate his 400W@2ohm = 200W@4ohm
example.

I still don't see how you can explain 400 watts = 200 watts under any
circumstances. If you're increasing power, then it doesn't matter what the
voltage or current is doing in this context.


I beleive that Mark Zarella's statements were based on speakers being
current dependent (ie double the current = double the sound output of
the speaker).

So I guess the $10,000 question is "Is a typical car speaker's sound
output voltage or current dependent?" I am inclined to beleive that
sound output of a speaker is a function of current through the coil.

Ohm's law says it's non-separable.

With regards to impedence matching, I would think that the 2 ohm/4 ohm
setting on the amp would be inclusive of the power supply's typical
output impedence not in addition to. If it were in addition to, I would
think that max efficiency would be at the higher impedence settings.

My 2 cents is make sure the load impedence matches the amp impedence
setting. Put a 2 ohm load on an amp set at 4 ohms will waste power in
the amp, put an 8 ohm load on an amp set at 4 ohms will drop the current
to the speakers and loose efficiency, put a 4 ohm load on a 4 ohm amp
and you have optimized the system (for the most part).

Rarely will you find 2 ohm/ 4 ohm settings on amplifiers these days.

Not really. Aside from the fact that I've measured output Z for many many
amplifiers. However, if you look at test results for amplifiers, or if
you
believe manufacturer specs to be on target to some degree, you need to
only
look at the damping factor value to determine output impedance.

Another way to do it is to look at the power measurements from 4 ohm and 2
ohm testing (carsound.com for instance). Note the losses and then calculate
an effective output impedance from that.

at the risk of confusing people... here is my interpretation of
lizard's example.

the assumption - sound output is directly dependent on voltage...
400 watts @ 2 ohms P=VI; V=IR; P=(I^2)R; P=(V^2)/R
I = 14.14
V = 28.3

200 watts @ 4 ohms
I = 7.07
V = 28.3
therefore if sound is directly dependent of voltage, the sound would
have been equal in these two examples...

NOTE: I do not beleive that the sound output or voice coil movement is
solely dependent on voltage. I do beleive that voice coil movement is a
function of the current through the coil.

$10,000 question, you are correct voltage and current are
inseparable per ohms law. I guess were I was going with the $10k
question "Does a typical audio amplifier amplify voltage and current
increases as a byproduct or does it amplify current with the increase in
voltage as a byproduct?"
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In article , The Lizard wrote:

BINGO! That's because amperes law doesn't define the "strength" of a
magnetic field, it just predicts it's density.

Dude, your ass is sucking wind. Really.

Density is the whole deal. The denser the field the stronger the magnet
is, all other things being equal. Add coils or increase the amperage and
you get a stronger solenoid. If you want to try to argue that then find a
contrary reference from a reputable source.

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Mike Graham | Metalworker, rustic, part-time zealot.
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http://www.metalmangler.com| Caledon, Ontario, Canada

In article , The Lizard wrote:

You need to have a talk with your ISP then.

It must have been off-line. It worked now. The page was rather
lackluster, though. No info that really pertained to the discussion at
hand.

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Mike Graham | Metalworker, rustic, part-time zealot.
|
http://www.metalmangler.com| Caledon, Ontario, Canada