Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

Subject:
[Ohms Law] Watts and Impedance?
Date:
Sat, 07 Feb 2004 13:01:36 GMT
From:
Computer Prog
Organization:
Optimum Online
Newsgroups:
rec.audio.tech
References:
1 , 2 , 3 , 4




Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both
amps
draw the same current from the car alternator when they are producing the 200W
off
the 14V input voltage?

My reason for asking this question in the first place was because my brother
just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra
Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @
8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be
rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm
14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input
voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

Subject:
[Ohms Law] Watts and Impedance?
Date:
Sat, 07 Feb 2004 13:01:36 GMT
From:
Computer Prog
Organization:
Optimum Online
Newsgroups:
rec.audio.tech
References:
1 , 2 , 3 , 4




Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both
amps
draw the same current from the car alternator when they are producing the 200W
off
the 14V input voltage?

My reason for asking this question in the first place was because my brother
just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra
Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @
8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be
rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm
14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input
voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

Subject:
[Ohms Law] Watts and Impedance?
Date:
Sat, 07 Feb 2004 13:01:36 GMT
From:
Computer Prog
Organization:
Optimum Online
Newsgroups:
rec.audio.tech
References:
1 , 2 , 3 , 4




Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both
amps
draw the same current from the car alternator when they are producing the 200W
off
the 14V input voltage?

My reason for asking this question in the first place was because my brother
just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra
Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @
8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be
rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm
14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input
voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

Subject:
[Ohms Law] Watts and Impedance?
Date:
Sat, 07 Feb 2004 13:01:36 GMT
From:
Computer Prog
Organization:
Optimum Online
Newsgroups:
rec.audio.tech
References:
1 , 2 , 3 , 4




Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both
amps
draw the same current from the car alternator when they are producing the 200W
off
the 14V input voltage?

My reason for asking this question in the first place was because my brother
just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra
Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @
8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be
rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm
14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input
voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

Computer Prog,

To some extent you are asking "how heavy is a ton of bricks?" Are they
heavier if I stack them in a single brick column, in a one high row, or
some combination? The obvious answer is that they always weigh a ton,
but one stacking configuration may be more convenient to truck around
than another.

In your amplifier situation, unless you have some sort of exotic
process running inside (be sure to market it), an audio amplifier
cannot create or destroy energy. It can only transform the energy from
one form to another. And --- there is always a "handling charge" for
this transformation. In general, only 30-50% of the amplifier's input
energy from the battery is available for running the speakers.

---

With respect to your voltage issues. A common trick in automotive
amplifiers is to use two amplifiers per channel. One amplifier runs 0
to +14 and the other is turned on it's head to run 0 to -14 volts. The
speaker, connected between both amplifiers, doesn't know or care about
this trick and thinks the amplifier is actually a 28 volt unit.

Other possibilities include switching power supplies that will step the
battery voltage up before applying it to the amplifier.

---

At your level, assuming that the amplifier has reasonable design and
build qualities and was fairly rated by some known standard, the major
issues a does the unit have enough power for your application; can
your battery charging system come up with the required power; and can
the vehicle wiring support the required current? The rest is your
judgement about how the system looks, operates, and sounds.

Don't worry too much about what is inside the box. A common audio
marketing strategy is to convince the customer that a unit benefits
from some exclusive circuit innovation. Checking around, you'll find
that no other product has this innovation. Once you are convinced about
the innovation's benefit, you'll have to purchase that product.

I rememberer one product line that touted an exclusive design approach.
They were very decent products from an excellent manufacturer, but it
was mostly a (well executed) standard textbook design with a few
impressive sounding terms thrown into the consumer literature. From my
point of view, the only novelty was that they published some graphs in
the consumer literature. Very similar graphs had been published almost
a decade prior by another manufacturer and before that in my textbooks.
From a marketing standpoint the program was a success because it
generated some excitement for the sales staff who were more likely to
mention the product to customers.

-----------------------------------------------------------
spam:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------

Computer Prog,

To some extent you are asking "how heavy is a ton of bricks?" Are they
heavier if I stack them in a single brick column, in a one high row, or
some combination? The obvious answer is that they always weigh a ton,
but one stacking configuration may be more convenient to truck around
than another.

In your amplifier situation, unless you have some sort of exotic
process running inside (be sure to market it), an audio amplifier
cannot create or destroy energy. It can only transform the energy from
one form to another. And --- there is always a "handling charge" for
this transformation. In general, only 30-50% of the amplifier's input
energy from the battery is available for running the speakers.

---

With respect to your voltage issues. A common trick in automotive
amplifiers is to use two amplifiers per channel. One amplifier runs 0
to +14 and the other is turned on it's head to run 0 to -14 volts. The
speaker, connected between both amplifiers, doesn't know or care about
this trick and thinks the amplifier is actually a 28 volt unit.

Other possibilities include switching power supplies that will step the
battery voltage up before applying it to the amplifier.

---

At your level, assuming that the amplifier has reasonable design and
build qualities and was fairly rated by some known standard, the major
issues a does the unit have enough power for your application; can
your battery charging system come up with the required power; and can
the vehicle wiring support the required current? The rest is your
judgement about how the system looks, operates, and sounds.

Don't worry too much about what is inside the box. A common audio
marketing strategy is to convince the customer that a unit benefits
from some exclusive circuit innovation. Checking around, you'll find
that no other product has this innovation. Once you are convinced about
the innovation's benefit, you'll have to purchase that product.

I rememberer one product line that touted an exclusive design approach.
They were very decent products from an excellent manufacturer, but it
was mostly a (well executed) standard textbook design with a few
impressive sounding terms thrown into the consumer literature. From my
point of view, the only novelty was that they published some graphs in
the consumer literature. Very similar graphs had been published almost
a decade prior by another manufacturer and before that in my textbooks.
From a marketing standpoint the program was a success because it
generated some excitement for the sales staff who were more likely to
mention the product to customers.

-----------------------------------------------------------
spam:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------

Computer Prog,

To some extent you are asking "how heavy is a ton of bricks?" Are they
heavier if I stack them in a single brick column, in a one high row, or
some combination? The obvious answer is that they always weigh a ton,
but one stacking configuration may be more convenient to truck around
than another.

In your amplifier situation, unless you have some sort of exotic
process running inside (be sure to market it), an audio amplifier
cannot create or destroy energy. It can only transform the energy from
one form to another. And --- there is always a "handling charge" for
this transformation. In general, only 30-50% of the amplifier's input
energy from the battery is available for running the speakers.

---

With respect to your voltage issues. A common trick in automotive
amplifiers is to use two amplifiers per channel. One amplifier runs 0
to +14 and the other is turned on it's head to run 0 to -14 volts. The
speaker, connected between both amplifiers, doesn't know or care about
this trick and thinks the amplifier is actually a 28 volt unit.

Other possibilities include switching power supplies that will step the
battery voltage up before applying it to the amplifier.

---

At your level, assuming that the amplifier has reasonable design and
build qualities and was fairly rated by some known standard, the major
issues a does the unit have enough power for your application; can
your battery charging system come up with the required power; and can
the vehicle wiring support the required current? The rest is your
judgement about how the system looks, operates, and sounds.

Don't worry too much about what is inside the box. A common audio
marketing strategy is to convince the customer that a unit benefits
from some exclusive circuit innovation. Checking around, you'll find
that no other product has this innovation. Once you are convinced about
the innovation's benefit, you'll have to purchase that product.

I rememberer one product line that touted an exclusive design approach.
They were very decent products from an excellent manufacturer, but it
was mostly a (well executed) standard textbook design with a few
impressive sounding terms thrown into the consumer literature. From my
point of view, the only novelty was that they published some graphs in
the consumer literature. Very similar graphs had been published almost
a decade prior by another manufacturer and before that in my textbooks.
From a marketing standpoint the program was a success because it
generated some excitement for the sales staff who were more likely to
mention the product to customers.

-----------------------------------------------------------
spam:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------

Computer Prog,

To some extent you are asking "how heavy is a ton of bricks?" Are they
heavier if I stack them in a single brick column, in a one high row, or
some combination? The obvious answer is that they always weigh a ton,
but one stacking configuration may be more convenient to truck around
than another.

In your amplifier situation, unless you have some sort of exotic
process running inside (be sure to market it), an audio amplifier
cannot create or destroy energy. It can only transform the energy from
one form to another. And --- there is always a "handling charge" for
this transformation. In general, only 30-50% of the amplifier's input
energy from the battery is available for running the speakers.

---

With respect to your voltage issues. A common trick in automotive
amplifiers is to use two amplifiers per channel. One amplifier runs 0
to +14 and the other is turned on it's head to run 0 to -14 volts. The
speaker, connected between both amplifiers, doesn't know or care about
this trick and thinks the amplifier is actually a 28 volt unit.

Other possibilities include switching power supplies that will step the
battery voltage up before applying it to the amplifier.

---

At your level, assuming that the amplifier has reasonable design and
build qualities and was fairly rated by some known standard, the major
issues a does the unit have enough power for your application; can
your battery charging system come up with the required power; and can
the vehicle wiring support the required current? The rest is your
judgement about how the system looks, operates, and sounds.

Don't worry too much about what is inside the box. A common audio
marketing strategy is to convince the customer that a unit benefits
from some exclusive circuit innovation. Checking around, you'll find
that no other product has this innovation. Once you are convinced about
the innovation's benefit, you'll have to purchase that product.

I rememberer one product line that touted an exclusive design approach.
They were very decent products from an excellent manufacturer, but it
was mostly a (well executed) standard textbook design with a few
impressive sounding terms thrown into the consumer literature. From my
point of view, the only novelty was that they published some graphs in
the consumer literature. Very similar graphs had been published almost
a decade prior by another manufacturer and before that in my textbooks.
From a marketing standpoint the program was a success because it
generated some excitement for the sales staff who were more likely to
mention the product to customers.

-----------------------------------------------------------
spam:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------

On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote:

Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

---
If both amps are capable of delivering 200W into their respective loads
then, yes, for the same power output they will draw identical currents
from the same supply voltage.
---

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

---
Just looking at the link reveals that neither amp is capable of putting
out 200 watts, so, yeah, if that's the part the salesman's talking about
he's wrong (to put it politely) to start with.

But, you bring up a good point, which is:

if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply,
how much power can you pump into the speaker?

Assuming that the amp has an "H" bridge output stage, it can make the
louspeaker think it's got 28 Volts, peak-to-peak, across it, which is
9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its
impedance, we can say:

P = E²/R = 9.9VRMS²/8R = 12.25W

So the best you can do with a 14V supply is to put about 12W into an 8
ohm speaker. Since there are four speakers total, if we assume two per
channel and assume that they'll be connected in parallel, then we can
say:

P = E²/R = 98/4 = 24.5W

and since the amp is a stereo amp, if we assume 24.5 watts per channel
we come out with about 50 watts, somewhat higher than the basic amp is
rated for. To find the voltage we need to drive a 4 ohm load to 39
watts, we can say:

E = sqrt(PR) ~ 12.5V,

which is right in the ballpark for a 12V automotive-type application.

But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....

--
John Fields

On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote:

Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

---
If both amps are capable of delivering 200W into their respective loads
then, yes, for the same power output they will draw identical currents
from the same supply voltage.
---

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

---
Just looking at the link reveals that neither amp is capable of putting
out 200 watts, so, yeah, if that's the part the salesman's talking about
he's wrong (to put it politely) to start with.

But, you bring up a good point, which is:

if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply,
how much power can you pump into the speaker?

Assuming that the amp has an "H" bridge output stage, it can make the
louspeaker think it's got 28 Volts, peak-to-peak, across it, which is
9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its
impedance, we can say:

P = E²/R = 9.9VRMS²/8R = 12.25W

So the best you can do with a 14V supply is to put about 12W into an 8
ohm speaker. Since there are four speakers total, if we assume two per
channel and assume that they'll be connected in parallel, then we can
say:

P = E²/R = 98/4 = 24.5W

and since the amp is a stereo amp, if we assume 24.5 watts per channel
we come out with about 50 watts, somewhat higher than the basic amp is
rated for. To find the voltage we need to drive a 4 ohm load to 39
watts, we can say:

E = sqrt(PR) ~ 12.5V,

which is right in the ballpark for a 12V automotive-type application.

But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....

--
John Fields

On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote:

Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

---
If both amps are capable of delivering 200W into their respective loads
then, yes, for the same power output they will draw identical currents
from the same supply voltage.
---

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

---
Just looking at the link reveals that neither amp is capable of putting
out 200 watts, so, yeah, if that's the part the salesman's talking about
he's wrong (to put it politely) to start with.

But, you bring up a good point, which is:

if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply,
how much power can you pump into the speaker?

Assuming that the amp has an "H" bridge output stage, it can make the
louspeaker think it's got 28 Volts, peak-to-peak, across it, which is
9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its
impedance, we can say:

P = E²/R = 9.9VRMS²/8R = 12.25W

So the best you can do with a 14V supply is to put about 12W into an 8
ohm speaker. Since there are four speakers total, if we assume two per
channel and assume that they'll be connected in parallel, then we can
say:

P = E²/R = 98/4 = 24.5W

and since the amp is a stereo amp, if we assume 24.5 watts per channel
we come out with about 50 watts, somewhat higher than the basic amp is
rated for. To find the voltage we need to drive a 4 ohm load to 39
watts, we can say:

E = sqrt(PR) ~ 12.5V,

which is right in the ballpark for a 12V automotive-type application.

But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....

--
John Fields

On Sat, 07 Feb 2004 13:01:36 GMT, Computer Prog
wrote:

Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

---
If both amps are capable of delivering 200W into their respective loads
then, yes, for the same power output they will draw identical currents
from the same supply voltage.
---

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

---
Just looking at the link reveals that neither amp is capable of putting
out 200 watts, so, yeah, if that's the part the salesman's talking about
he's wrong (to put it politely) to start with.

But, you bring up a good point, which is:

if you've got an 8 ohm loudspeaker and you've only got a 14VDC supply,
how much power can you pump into the speaker?

Assuming that the amp has an "H" bridge output stage, it can make the
louspeaker think it's got 28 Volts, peak-to-peak, across it, which is
9.9Volts, RMS. Now, knowing the voltage into the loudspeaker and its
impedance, we can say:

P = E²/R = 9.9VRMS²/8R = 12.25W

So the best you can do with a 14V supply is to put about 12W into an 8
ohm speaker. Since there are four speakers total, if we assume two per
channel and assume that they'll be connected in parallel, then we can
say:

P = E²/R = 98/4 = 24.5W

and since the amp is a stereo amp, if we assume 24.5 watts per channel
we come out with about 50 watts, somewhat higher than the basic amp is
rated for. To find the voltage we need to drive a 4 ohm load to 39
watts, we can say:

E = sqrt(PR) ~ 12.5V,

which is right in the ballpark for a 12V automotive-type application.

But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....

--
John Fields

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

of course it will go much lower when you only use
the battery voltage of 12V, then you get 18/36W

for any higher wattage the amplifier must make use
of a voltage inverter i.e. 12V 40V for 200W/8 Ohms
or 28V for 200W/4 Ohms

The current drawn from the battery/generator will be
about the same whatever the speaker impedance is when
driven by the same power of 200W, around 60A at full
power stereo 2x200W RMS...

Peter

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

of course it will go much lower when you only use
the battery voltage of 12V, then you get 18/36W

for any higher wattage the amplifier must make use
of a voltage inverter i.e. 12V 40V for 200W/8 Ohms
or 28V for 200W/4 Ohms

The current drawn from the battery/generator will be
about the same whatever the speaker impedance is when
driven by the same power of 200W, around 60A at full
power stereo 2x200W RMS...

Peter

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

of course it will go much lower when you only use
the battery voltage of 12V, then you get 18/36W

for any higher wattage the amplifier must make use
of a voltage inverter i.e. 12V 40V for 200W/8 Ohms
or 28V for 200W/4 Ohms

The current drawn from the battery/generator will be
about the same whatever the speaker impedance is when
driven by the same power of 200W, around 60A at full
power stereo 2x200W RMS...

Peter

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

of course it will go much lower when you only use
the battery voltage of 12V, then you get 18/36W

for any higher wattage the amplifier must make use
of a voltage inverter i.e. 12V 40V for 200W/8 Ohms
or 28V for 200W/4 Ohms

The current drawn from the battery/generator will be
about the same whatever the speaker impedance is when
driven by the same power of 200W, around 60A at full
power stereo 2x200W RMS...

Peter

On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote:


But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....


Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's
more like it!

--
John Fields

On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote:


But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....


Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's
more like it!

--
John Fields

On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote:


But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....


Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's
more like it!

--
John Fields

On Sat, 07 Feb 2004 10:02:54 -0600, John Fields
wrote:


But how about that 70 watt amp? It's not likely they've got a DC to DC
converter in there to boost the supply voltage, and the best you can do
with a 14V supply and two 4 ohm loads is 50 watts, so they must be
playing with the speaker impedance. For a nomonal 12.5V supply, then,
let's look at what the speakers need to look like to get us 70 watts
out. Since:


P = E²/R

we can rearrange that to:

R = E²/P = 8.8VRMS²/70W ~ 1.1 ohms

Which, for a stereo rig means two, 2.2 ohm speakers in parallel per
channel. Hmmm....


Aaarrghhhh!!! ...two ~ 4 ohm speakers in parallel per channel. That's
more like it!

--
John Fields

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

--
John Fields

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

--
John Fields

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

--
John Fields

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

--
John Fields

On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote:

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

---
Another error, dammit!

with a 14.4V supply and no bridge, the maximum signal swing into the
load will be 14.4VPP, which is ~ 5VRMS

Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4
ohm load, 25/4 ~ 6.25 watts.

--
John Fields

On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote:

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

---
Another error, dammit!

with a 14.4V supply and no bridge, the maximum signal swing into the
load will be 14.4VPP, which is ~ 5VRMS

Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4
ohm load, 25/4 ~ 6.25 watts.

--
John Fields

On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote:

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

---
Another error, dammit!

with a 14.4V supply and no bridge, the maximum signal swing into the
load will be 14.4VPP, which is ~ 5VRMS

Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4
ohm load, 25/4 ~ 6.25 watts.

--
John Fields

On Sat, 07 Feb 2004 12:49:13 -0600, John Fields
wrote:

On Sat, 07 Feb 2004 17:24:54 +0100, Peter Völpel
wrote:

Computer Prog wrote:

Thanks for the reply, but I think you misinterpreted my question. In my example
I
wanted both amps to run off the same input voltage, like a car audio
application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms


the maximum you can get out of 14,4V is:

26W into 8 Ohms and 52W into 4 Ohms

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is

(14.4V/sqrt(2))²/8 ~ 13W

and in a 4 ohm load

(14.4V/sqrt(2))²/4 ~ 26W

Assuming, of course, that the loudspeaker is AC coupled to the amp and
that the voltage swing into the loudspeaker is +/- 7.2V.

---
Another error, dammit!

with a 14.4V supply and no bridge, the maximum signal swing into the
load will be 14.4VPP, which is ~ 5VRMS

Then, since P = E²/R, that'll be P = 5²/8 = 25/8 ~ 3.125W, and for a 4
ohm load, 25/4 ~ 6.25 watts.

--
John Fields

Computer Prog wrote in message ...
Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load.
Will both amps draw the same current from the car alternator when they are
producing the 200W off the 14V input voltage?

Yes. In both cases a voltage supply of 14 volts to the amplifier, even
if it is bridged. So in the first case the 14 volt has to be stepped
up to

U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case
to 40 volts, plus losses in the design.

Stepping up voltage is commonly done in high(er) power car audio power
amps. There are mainly three classes of amplifiers for car audio (or
more?).
-The first uses a single output stage directly connected to the
speaker, max output (assuming 14 volt supply and 4 ohm speakers) =
(14/2/sqrt(2))^2/4=6 watts.
-The second uses two output stages in a bridge and can deliver
(14/sqrt(2))^2/4=24 watts
-The third class uses step-up of the voltage supply and can in
principle deliver any power.

In rare cases loudspeaker with lower impedance is used, but not very
often.

Computer Prog wrote in message ...
Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load.
Will both amps draw the same current from the car alternator when they are
producing the 200W off the 14V input voltage?

Yes. In both cases a voltage supply of 14 volts to the amplifier, even
if it is bridged. So in the first case the 14 volt has to be stepped
up to

U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case
to 40 volts, plus losses in the design.

Stepping up voltage is commonly done in high(er) power car audio power
amps. There are mainly three classes of amplifiers for car audio (or
more?).
-The first uses a single output stage directly connected to the
speaker, max output (assuming 14 volt supply and 4 ohm speakers) =
(14/2/sqrt(2))^2/4=6 watts.
-The second uses two output stages in a bridge and can deliver
(14/sqrt(2))^2/4=24 watts
-The third class uses step-up of the voltage supply and can in
principle deliver any power.

In rare cases loudspeaker with lower impedance is used, but not very
often.

Computer Prog wrote in message ...
Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load.
Will both amps draw the same current from the car alternator when they are
producing the 200W off the 14V input voltage?

Yes. In both cases a voltage supply of 14 volts to the amplifier, even
if it is bridged. So in the first case the 14 volt has to be stepped
up to

U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case
to 40 volts, plus losses in the design.

Stepping up voltage is commonly done in high(er) power car audio power
amps. There are mainly three classes of amplifiers for car audio (or
more?).
-The first uses a single output stage directly connected to the
speaker, max output (assuming 14 volt supply and 4 ohm speakers) =
(14/2/sqrt(2))^2/4=6 watts.
-The second uses two output stages in a bridge and can deliver
(14/sqrt(2))^2/4=24 watts
-The third class uses step-up of the voltage supply and can in
principle deliver any power.

In rare cases loudspeaker with lower impedance is used, but not very
often.

Computer Prog wrote in message ...
Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load.
Will both amps draw the same current from the car alternator when they are
producing the 200W off the 14V input voltage?

Yes. In both cases a voltage supply of 14 volts to the amplifier, even
if it is bridged. So in the first case the 14 volt has to be stepped
up to

U=sqrt(P*R)*sqrt(2)=sqrt(200*8)*sqrt(2)=57 volts, in the second case
to 40 volts, plus losses in the design.

Stepping up voltage is commonly done in high(er) power car audio power
amps. There are mainly three classes of amplifiers for car audio (or
more?).
-The first uses a single output stage directly connected to the
speaker, max output (assuming 14 volt supply and 4 ohm speakers) =
(14/2/sqrt(2))^2/4=6 watts.
-The second uses two output stages in a bridge and can deliver
(14/sqrt(2))^2/4=24 watts
-The third class uses step-up of the voltage supply and can in
principle deliver any power.

In rare cases loudspeaker with lower impedance is used, but not very
often.

On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote:

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

You're muddying the water by making them car amps :-)

The basic power supply is a nominal 12 volts. Basic physics tells us
that there isn't 200 watts to be had into a 4ohm load. So the amp
must include some sort of voltage convertor between the battery and
the power amp's supply rail.

All you can do with an amplifier is present a voltage to the load.
The load will determine how much current is drawn. Until it tries to
draw more than the amp can provide. Which is by no means a
theoretical limit.

On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote:

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

You're muddying the water by making them car amps :-)

The basic power supply is a nominal 12 volts. Basic physics tells us
that there isn't 200 watts to be had into a 4ohm load. So the amp
must include some sort of voltage convertor between the battery and
the power amp's supply rail.

All you can do with an amplifier is present a voltage to the load.
The load will determine how much current is drawn. Until it tries to
draw more than the amp can provide. Which is by no means a
theoretical limit.

On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote:

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

You're muddying the water by making them car amps :-)

The basic power supply is a nominal 12 volts. Basic physics tells us
that there isn't 200 watts to be had into a 4ohm load. So the amp
must include some sort of voltage convertor between the battery and
the power amp's supply rail.

All you can do with an amplifier is present a voltage to the load.
The load will determine how much current is drawn. Until it tries to
draw more than the amp can provide. Which is by no means a
theoretical limit.

On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote:

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

You're muddying the water by making them car amps :-)

The basic power supply is a nominal 12 volts. Basic physics tells us
that there isn't 200 watts to be had into a 4ohm load. So the amp
must include some sort of voltage convertor between the battery and
the power amp's supply rail.

All you can do with an amplifier is present a voltage to the load.
The load will determine how much current is drawn. Until it tries to
draw more than the amp can provide. Which is by no means a
theoretical limit.

John Fields wrote:

---
Without using a bridging configuration, the maximum power you can
dissipate in an 8 ohm load from a 14.4VDC source is=20
=20
(14.4V/sqrt(2))=B2/8 ~ 13W


Actually, the maximum power you can dissipate in an 8-ohm load from a=20
14.4 VDC source is simply 14.4**2/8 =3D 26 W !

This is at DC though, of course.