[Alan]

Let's say I have a 4-channel amp which has a rating of 200x4 @ 2-ohms.
Let's say I connect a pair of *2-ohm* component speakers to the front
channels of the amp. Additionally, let's say I also connect a pair of
*4-ohm* coaxials to the rear channels of the amp.

Now my question is: do the rear coaxials get 200 watts each since they
are only 4-ohm speakers, or will they only get 100 watts each?

[g]

In article , Alan wrote:
Let's say I have a 4-channel amp which has a rating of 200x4 @ 2-ohms.
Let's say I connect a pair of *2-ohm* component speakers to the front
channels of the amp. Additionally, let's say I also connect a pair of
*4-ohm* coaxials to the rear channels of the amp.

Now my question is: do the rear coaxials get 200 watts each since they
are only 4-ohm speakers, or will they only get 100 watts each?

100.
Most amps when lightly loaded tend to output more peak volts, so
it may really be 120 watts, just playing with figures.
It would depend on the efficiency of the speakers as to what set
plays louder, but a 3dB difference is not much. Will only shift
the sound slightly to the front or rear. It does not matter much anyway,
because you would have to start clipping the sound if you got anywhere
near max output.


greg

[Matt Ion]

G wrote:
In article , Alan wrote:
Let's say I have a 4-channel amp which has a rating of 200x4 @ 2-ohms.
Let's say I connect a pair of *2-ohm* component speakers to the front
channels of the amp. Additionally, let's say I also connect a pair of
*4-ohm* coaxials to the rear channels of the amp.

Now my question is: do the rear coaxials get 200 watts each since they
are only 4-ohm speakers, or will they only get 100 watts each?

100.
Most amps when lightly loaded tend to output more peak volts, so
it may really be 120 watts, just playing with figures.
It would depend on the efficiency of the speakers as to what set
plays louder, but a 3dB difference is not much. Will only shift
the sound slightly to the front or rear. It does not matter much anyway,
because you would have to start clipping the sound if you got anywhere
near max output.

And if you really want to be accurate, you can also point out that
power/load calculations like this are approximate only - IN THEORY an
amp rated 200W into 2 ohms will only drive 100W into 4 ohms, and IN
THEORY would drive 400W into a 1 ohm load... but theory alas must always
bow to reality, which mostly comes in the form of inefficiency and
losses to heat.

As noted, the difference will not usually be noticeable or even audible,
so we tend to speak "in theory" most of the time...

[bob wald]

i'm so tired coming here n seeing you trying to figure out car audio
stuff. its like watching monkeys try to figure out a simple task.lol
100w at 4ohm are equal to 200watts 2ohm.
which means niether is louder. theyre equal!
same with 1ohm. 400 1ohm is equal to 200 2ohm.

[Mariachi]

On Nov 13, 2:34 am, Matt Ion wrote:
G wrote:
In article , Alan wrote:
Let's say I have a 4-channel amp which has a rating of 200x4 @ 2-ohms.
Let's say I connect a pair of *2-ohm* component speakers to the front
channels of the amp. Additionally, let's say I also connect a pair of
*4-ohm* coaxials to the rear channels of the amp.

Now my question is: do the rear coaxials get 200 watts each since they
are only 4-ohm speakers, or will they only get 100 watts each?

100.
Most amps when lightly loaded tend to output more peak volts, so
it may really be 120 watts, just playing with figures.
It would depend on the efficiency of the speakers as to what set
plays louder, but a 3dB difference is not much. Will only shift
the sound slightly to the front or rear. It does not matter much anyway,
because you would have to start clipping the sound if you got anywhere
near max output.

And if you really want to be accurate, you can also point out that
power/load calculations like this are approximate only - IN THEORY an
amp rated 200W into 2 ohms will only drive 100W into 4 ohms, and IN
THEORY would drive 400W into a 1 ohm load... but theory alas must always
bow to reality, which mostly comes in the form of inefficiency and
losses to heat.

As noted, the difference will not usually be noticeable or even audible,
so we tend to speak "in theory" most of the time...

not necessarily true. Two ohms could be the peak output of the
amplifier, which means; if you decrease the load resistance further,
most of the amplifier power will be allocated to heat dissipation of
the amplifier, instead of 50/50. It all depends on impedance
matching. If it matches best with 2 ohms, then 1 ohm will actually be
worse and it might burn the components in your amplifier.

You probably already know this though.

[Eddie Runner]

bob wald wrote:
i'm so tired coming here n seeing you trying to figure out car audio
stuff. its like watching monkeys try to figure out a simple task.lol
100w at 4ohm are equal to 200watts 2ohm.

Equal? Your saying its equal, but what is equal?
The watts aren't equal, the ohms aren't equal,
The current isnt equal!
That's not very equal ! ;-)

which means niether is louder. theyre equal!

Louder? How can you say that without knowing the
sensitivity of the speakers? We only know the speakers
are one set of 4 ohms coaxials and one set of 2 ohms
components. There are WAY too many unknown variables
for you to make the statement that the speakers will
be equal in loudness.


If ANYTHING were equal, it would be the amp output voltage,

ASSUMING were using the same sound and not touching the
volume control or gains, and assuming the amp can keep
the same stability of the output voltage at 2 ohms that it
normally does at 4 ohms.

(You know if 2 ohms is more of
a strain the voltage could drop.)


way too many variables for you to make those statements!

Eddie Runner

[Matt Ion]

Mariachi wrote:
On Nov 13, 2:34 am, Matt Ion wrote:
G wrote:
In article , Alan wrote:
Let's say I have a 4-channel amp which has a rating of 200x4 @ 2-ohms.
Let's say I connect a pair of *2-ohm* component speakers to the front
channels of the amp. Additionally, let's say I also connect a pair of
*4-ohm* coaxials to the rear channels of the amp.
Now my question is: do the rear coaxials get 200 watts each since they
are only 4-ohm speakers, or will they only get 100 watts each?
100.
Most amps when lightly loaded tend to output more peak volts, so
it may really be 120 watts, just playing with figures.
It would depend on the efficiency of the speakers as to what set
plays louder, but a 3dB difference is not much. Will only shift
the sound slightly to the front or rear. It does not matter much anyway,
because you would have to start clipping the sound if you got anywhere
near max output.
And if you really want to be accurate, you can also point out that
power/load calculations like this are approximate only - IN THEORY an
amp rated 200W into 2 ohms will only drive 100W into 4 ohms, and IN
THEORY would drive 400W into a 1 ohm load... but theory alas must always
bow to reality, which mostly comes in the form of inefficiency and
losses to heat.

As noted, the difference will not usually be noticeable or even audible,
so we tend to speak "in theory" most of the time...

not necessarily true. Two ohms could be the peak output of the
amplifier, which means; if you decrease the load resistance further,
most of the amplifier power will be allocated to heat dissipation of
the amplifier, instead of 50/50. It all depends on impedance
matching. If it matches best with 2 ohms, then 1 ohm will actually be
worse and it might burn the components in your amplifier.

You probably already know this though.

Can you not read?? This is exactly my point: all these numbers being
tossed around are THEORETICAL calculations; the ACTUAL output will
depend on a lot of other factors, such as those you've listed.

[Mariachi]

On Nov 14, 12:57 pm, Matt Ion wrote:
Mariachi wrote:
On Nov 13, 2:34 am, Matt Ion wrote:
G wrote:
In article , Alan wrote:
Let's say I have a 4-channel amp which has a rating of 200x4 @ 2-ohms.
Let's say I connect a pair of *2-ohm* component speakers to the front
channels of the amp. Additionally, let's say I also connect a pair of
*4-ohm* coaxials to the rear channels of the amp.
Now my question is: do the rear coaxials get 200 watts each since they
are only 4-ohm speakers, or will they only get 100 watts each?
100.
Most amps when lightly loaded tend to output more peak volts, so
it may really be 120 watts, just playing with figures.
It would depend on the efficiency of the speakers as to what set
plays louder, but a 3dB difference is not much. Will only shift
the sound slightly to the front or rear. It does not matter much anyway,
because you would have to start clipping the sound if you got anywhere
near max output.
And if you really want to be accurate, you can also point out that
power/load calculations like this are approximate only - IN THEORY an
amp rated 200W into 2 ohms will only drive 100W into 4 ohms, and IN
THEORY would drive 400W into a 1 ohm load... but theory alas must always
bow to reality, which mostly comes in the form of inefficiency and
losses to heat.

As noted, the difference will not usually be noticeable or even audible,
so we tend to speak "in theory" most of the time...

not necessarily true. Two ohms could be the peak output of the
amplifier, which means; if you decrease the load resistance further,
most of the amplifier power will be allocated to heat dissipation of
the amplifier, instead of 50/50. It all depends on impedance
matching. If it matches best with 2 ohms, then 1 ohm will actually be
worse and it might burn the components in your amplifier.

You probably already know this though.

Can you not read?? This is exactly my point: all these numbers being
tossed around are THEORETICAL calculations; the ACTUAL output will
depend on a lot of other factors, such as those you've listed.

But what I stated was electrical theory.

[bob wald]

eddie i guess i didnt write it simple enough for you.
if everything else is the same except the ohms n the watts = one another
but they are just being run on different ohms..even tho the numbers may
be different theyre EQUAL!!!!!!
as far as loudness....but not every tiny spec....

[bob wald]

its been great fun watching over the yrs. pros talk about they changed
thier 4ohm to 2 ohm with the same equipment n it sounds much
louder...hAHAhaHahaHAHaha....
i think you all need to look up the meaning equal.

[Richard[_3_]]

Reality Check:

The "rated impediance" and the actual impediance may be quite different.

The actual impediance varies with audio frequency. What is typically
reported is the lowest impediance measured over the audio frequencies,
typically a low audio frequency.

The first priority in matching an amp with a speaker with respect to
impediance is to avoid too low an impediance load on an amp because that can
lead to overheating and damage to the amp.

Additional potential power output is merely a minor secondary consideration.
Remember, one needs to double output to get even a just noticable increase
in volume (about 3dB).

Most of the specs you have come to rely upon are written by advertising guys
not the tech nurds at the company.

You would be shocked at how little power is needed to provide good clean
sound. The trick is to find an amp that puts out good clean sound. My home
system can deliver about 120 watts into 8 ohms and at typical sound levels
is called upon to deliver about 1 to 5 watts. When cranked up to levels my
wife would not accept, my sub can suck up all of the 350 watts into 4 ohms
my sub amp can deliver.

[Mariachi]

On Nov 16, 10:39 am, "Richard" wrote:
Reality Check:

The "rated impediance" and the actual impediance may be quite different.

The actual impediance varies with audio frequency. What is typically
reported is the lowest impediance measured over the audio frequencies,
typically a low audio frequency.

The first priority in matching an amp with a speaker with respect to
impediance is to avoid too low an impediance load on an amp because that can
lead to overheating and damage to the amp.

Additional potential power output is merely a minor secondary consideration.
Remember, one needs to double output to get even a just noticable increase
in volume (about 3dB).

Most of the specs you have come to rely upon are written by advertising guys
not the tech nurds at the company.

You would be shocked at how little power is needed to provide good clean
sound. The trick is to find an amp that puts out good clean sound. My home
system can deliver about 120 watts into 8 ohms and at typical sound levels
is called upon to deliver about 1 to 5 watts. When cranked up to levels my
wife would not accept, my sub can suck up all of the 350 watts into 4 ohms
my sub amp can deliver.

the only way to prove how much power you are using is to use an
oscilloscope. What do you consider typical sound levels? 90 Decibels
at your ears? If this is true, then you are probably using 2 Watts of
power if the speaker sensitivity is 90 dB for 1 W @ 1 m, because your
ear is probably much a little further away than 1 meter. But what if
you listen to your music at 96 dB to your ears. Then, I'm estimating
that the speaker would have an electrical power of around 10 Watts.
Additionally, different frequencies have different sensitivities, but
usually an audiophile speaker has the same sensitivity at each
frequency that its supposed to cover. However, the perceived loudness
of one frequency to another at a given SPL differentiates because of
our ear sensitivity. For example, humans' ears are less sensitivity
to 100 Hz and below and are more sensitive in the midrange. Humans'
ears are also less sensitive in the high range 10 kHz and above. This
is where subwoofers come in handy. Subwoofers intentionally boost the
SPL level of the low range to purposely make up for the ears low
sensitivity to bass, in order for the listener to perceive that all
frequencies are at the same perceived loudness.

[Eddie Runner]

Mariachi wrote:

the only way to prove how much power you are using is to use an
oscilloscope.

How do you do it with an oscope??
Just Curious.
Eddie

[Christopher \Torroid\ Ott]

"Eddie Runner" wrote in message
...
Mariachi wrote:

the only way to prove how much power you are using is to use an
oscilloscope.

How do you do it with an oscope??
Just Curious.
Eddie

If you want to measure power, you'll need a current sense resistor (non
inductive) and a two-channel digital o'scope with the ability to multiply
the two channels. I seem to recall that either Tek or Fluke had a scope
which was optimized for this, but that's been some time ago. Most of the PC
based scopes can do this as well. Pretty straightforward from there. Set the
volts/division as needed to correctly scale the CSR and hook up a signal
generator to the amp.

If you have HP Vee or Labview or something similar, it would be pretty easy
to setup a jig to automatically measure power for virtually any
configuration of amp and speaker.

As a side note, I spent 2 years at HP working on a tool to measure the
energy delivered to inkjet resistors (on the silicon printhead) which did
almost the same thing as I described above, only with a bunch more detail
and about $4million invested. Inkjet is big money baby!

Chris

[Eddie Runner]

Normally a bench tech would measure power with a dummy load and a volt
meter. There really isnt a need for a scope, but many use it just to
watch the waveform for anything strange during testing. I dont think
I have seen a tech try to measure power on a scope itself, although
you could take the time to calibrate the screen marks to voltage but
anyone with a scope probably have a wheelbarrow load of meters and that
way easier....

BTW, a little trivia...
Since you worked for HP, can you tell me what the first thing they
build commercially? Hint, it was test gear!


Eddie

Christopher "Torroid" Ott wrote:
"Eddie Runner" wrote in message
...
Mariachi wrote:

the only way to prove how much power you are using is to use an
oscilloscope.
How do you do it with an oscope??
Just Curious.
Eddie

If you want to measure power, you'll need a current sense resistor (non
inductive) and a two-channel digital o'scope with the ability to multiply
the two channels. I seem to recall that either Tek or Fluke had a scope
which was optimized for this, but that's been some time ago. Most of the PC
based scopes can do this as well. Pretty straightforward from there. Set the
volts/division as needed to correctly scale the CSR and hook up a signal
generator to the amp.

If you have HP Vee or Labview or something similar, it would be pretty easy
to setup a jig to automatically measure power for virtually any
configuration of amp and speaker.

As a side note, I spent 2 years at HP working on a tool to measure the
energy delivered to inkjet resistors (on the silicon printhead) which did
almost the same thing as I described above, only with a bunch more detail
and about $4million invested. Inkjet is big money baby!

Chris

[Christopher \Torroid\ Ott]

That would be an audio oscillator. It was the first product to come from
"the garage". I've heard it was in common use well into the 60's.

I've got the schematic around here somewhere. It was really simplistic by
today's standards.

Chris


"Eddie Runner" wrote in message
...
Normally a bench tech would measure power with a dummy load and a volt
meter. There really isnt a need for a scope, but many use it just to
watch the waveform for anything strange during testing. I dont think
I have seen a tech try to measure power on a scope itself, although
you could take the time to calibrate the screen marks to voltage but
anyone with a scope probably have a wheelbarrow load of meters and that
way easier....

BTW, a little trivia...
Since you worked for HP, can you tell me what the first thing they
build commercially? Hint, it was test gear!


Eddie

Christopher "Torroid" Ott wrote:
"Eddie Runner" wrote in message
...
Mariachi wrote:

the only way to prove how much power you are using is to use an
oscilloscope.
How do you do it with an oscope??
Just Curious.
Eddie

If you want to measure power, you'll need a current sense resistor (non
inductive) and a two-channel digital o'scope with the ability to multiply
the two channels. I seem to recall that either Tek or Fluke had a scope
which was optimized for this, but that's been some time ago. Most of the
PC based scopes can do this as well. Pretty straightforward from there.
Set the volts/division as needed to correctly scale the CSR and hook up a
signal generator to the amp.

If you have HP Vee or Labview or something similar, it would be pretty
easy to setup a jig to automatically measure power for virtually any
configuration of amp and speaker.

As a side note, I spent 2 years at HP working on a tool to measure the
energy delivered to inkjet resistors (on the silicon printhead) which did
almost the same thing as I described above, only with a bunch more detail
and about $4million invested. Inkjet is big money baby!

Chris

[GregS[_3_]]

In article , Eddie Runner wrote:
Normally a bench tech would measure power with a dummy load and a volt
meter. There really isnt a need for a scope, but many use it just to
watch the waveform for anything strange during testing. I dont think
I have seen a tech try to measure power on a scope itself, although
you could take the time to calibrate the screen marks to voltage but
anyone with a scope probably have a wheelbarrow load of meters and that
way easier....

I don't think I would measure power without a scope. How else
would you watch for the clipping point. To measure max power
I usually find the clip then back off. You would need a THD meter
to find out how much to back off and make the distortion
spec look good.

greg



BTW, a little trivia...
Since you worked for HP, can you tell me what the first thing they
build commercially? Hint, it was test gear!


Eddie

Christopher "Torroid" Ott wrote:
"Eddie Runner" wrote in message
...
Mariachi wrote:

the only way to prove how much power you are using is to use an
oscilloscope.
How do you do it with an oscope??
Just Curious.
Eddie

If you want to measure power, you'll need a current sense resistor (non
inductive) and a two-channel digital o'scope with the ability to multiply
the two channels. I seem to recall that either Tek or Fluke had a scope
which was optimized for this, but that's been some time ago. Most of the PC
based scopes can do this as well. Pretty straightforward from there. Set the
volts/division as needed to correctly scale the CSR and hook up a signal
generator to the amp.

If you have HP Vee or Labview or something similar, it would be pretty easy
to setup a jig to automatically measure power for virtually any
configuration of amp and speaker.

As a side note, I spent 2 years at HP working on a tool to measure the
energy delivered to inkjet resistors (on the silicon printhead) which did
almost the same thing as I described above, only with a bunch more detail
and about $4million invested. Inkjet is big money baby!

Chris

[Eddie Runner]

Yes, not too many folks know it was an audio oscillator.
audio test equipment seems strange to folks that think of
HP today. In the past I have had MANY pieces of HP
audio test gear.

Eddie

Christopher "Torroid" Ott wrote:
That would be an audio oscillator. It was the first product to come from
"the garage". I've heard it was in common use well into the 60's.

I've got the schematic around here somewhere. It was really simplistic by
today's standards.

Chris


"Eddie Runner" wrote in message
...
Normally a bench tech would measure power with a dummy load and a volt
meter. There really isnt a need for a scope, but many use it just to
watch the waveform for anything strange during testing. I dont think
I have seen a tech try to measure power on a scope itself, although
you could take the time to calibrate the screen marks to voltage but
anyone with a scope probably have a wheelbarrow load of meters and that
way easier....

BTW, a little trivia...
Since you worked for HP, can you tell me what the first thing they
build commercially? Hint, it was test gear!


Eddie

Christopher "Torroid" Ott wrote:
"Eddie Runner" wrote in message
...
Mariachi wrote:

the only way to prove how much power you are using is to use an
oscilloscope.
How do you do it with an oscope??
Just Curious.
Eddie
If you want to measure power, you'll need a current sense resistor (non
inductive) and a two-channel digital o'scope with the ability to multiply
the two channels. I seem to recall that either Tek or Fluke had a scope
which was optimized for this, but that's been some time ago. Most of the
PC based scopes can do this as well. Pretty straightforward from there.
Set the volts/division as needed to correctly scale the CSR and hook up a
signal generator to the amp.

If you have HP Vee or Labview or something similar, it would be pretty
easy to setup a jig to automatically measure power for virtually any
configuration of amp and speaker.

As a side note, I spent 2 years at HP working on a tool to measure the
energy delivered to inkjet resistors (on the silicon printhead) which did
almost the same thing as I described above, only with a bunch more detail
and about $4million invested. Inkjet is big money baby!

Chris

[Eddie Runner]

GregS wrote:
I don't think I would measure power without a scope. How else
would you watch for the clipping point. To measure max power
I usually find the clip then back off. You would need a THD meter
to find out how much to back off and make the distortion
spec look good.

greg

To measure power you would use a distortion analyzer, you dont
need a scope. I have worked in many service centers and have
done it 100s of times.

There is usually a scope there, because it is a part of all audio
test benches, but you run the oscillator into the amp, measure the
output voltage with a volt meter, and the THD on the distortion
analyzer. You dont read the power from the scope, you read it from
the volt meter. then voltage squared divided by the test resistors
(dummy load).

Since most amps are rated at very low THD, the clipping point
would not be an issue, you wont get close.

Eddie Runner