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.

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.

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.

Responding to myself here, parts made no sense.

(Svante) wrote in message . com...
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

Should read:
Yes they would. In both cases a voltage supply of 14 volts to the
amplifier is not enough, 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.

Should read:
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. (Non-bridged design
assumed.)

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.

Sorry for the double-post.

Responding to myself here, parts made no sense.

(Svante) wrote in message . com...
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

Should read:
Yes they would. In both cases a voltage supply of 14 volts to the
amplifier is not enough, 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.

Should read:
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. (Non-bridged design
assumed.)

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.

Sorry for the double-post.

Responding to myself here, parts made no sense.

(Svante) wrote in message . com...
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

Should read:
Yes they would. In both cases a voltage supply of 14 volts to the
amplifier is not enough, 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.

Should read:
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. (Non-bridged design
assumed.)

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.

Sorry for the double-post.

Responding to myself here, parts made no sense.

(Svante) wrote in message . com...
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

Should read:
Yes they would. In both cases a voltage supply of 14 volts to the
amplifier is not enough, 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.

Should read:
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. (Non-bridged design
assumed.)

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.

Sorry for the double-post.

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

On Sat, 07 Feb 2004 13:02:27 GMT, Computer Prog
wrote:

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?


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Even allowing for the exaggeration applied to wattage of car systems,
a "200W" system probably draws a little more than that. Therefore
the power supply to the\ amp isn't the raw 14.4v delivered by the
battery. It's inverted and transformed up to a higher voltage.

On Sat, 07 Feb 2004 13:02:27 GMT, Computer Prog
wrote:

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?


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Even allowing for the exaggeration applied to wattage of car systems,
a "200W" system probably draws a little more than that. Therefore
the power supply to the\ amp isn't the raw 14.4v delivered by the
battery. It's inverted and transformed up to a higher voltage.

On Sat, 07 Feb 2004 13:02:27 GMT, Computer Prog
wrote:

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?


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Even allowing for the exaggeration applied to wattage of car systems,
a "200W" system probably draws a little more than that. Therefore
the power supply to the\ amp isn't the raw 14.4v delivered by the
battery. It's inverted and transformed up to a higher voltage.

On Sat, 07 Feb 2004 13:02:27 GMT, Computer Prog
wrote:

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?


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Even allowing for the exaggeration applied to wattage of car systems,
a "200W" system probably draws a little more than that. Therefore
the power supply to the\ amp isn't the raw 14.4v delivered by the
battery. It's inverted and transformed up to a higher voltage.

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

"Laurence Payne" wrote in message
...
On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only
consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what
the
current rating is and I always reply with a question asking what the
supply
voltage is. They get confused because they do not understand how power,
and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.

"Laurence Payne" wrote in message
...
On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only
consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what
the
current rating is and I always reply with a question asking what the
supply
voltage is. They get confused because they do not understand how power,
and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.

"Laurence Payne" wrote in message
...
On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only
consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what
the
current rating is and I always reply with a question asking what the
supply
voltage is. They get confused because they do not understand how power,
and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.

"Laurence Payne" wrote in message
...
On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:

My point was only in regards to power ratings. A person cannot only
consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what
the
current rating is and I always reply with a question asking what the
supply
voltage is. They get confused because they do not understand how power,
and


If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.

On Sun, 08 Feb 2004 05:08:04 GMT, "flint"
wrote:

If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.


Oh, right. These techs work globally, and forget to tell you what
country they're in at the time. Fair enough ;-)

On Sun, 08 Feb 2004 05:08:04 GMT, "flint"
wrote:

If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.


Oh, right. These techs work globally, and forget to tell you what
country they're in at the time. Fair enough ;-)

On Sun, 08 Feb 2004 05:08:04 GMT, "flint"
wrote:

If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.


Oh, right. These techs work globally, and forget to tell you what
country they're in at the time. Fair enough ;-)

On Sun, 08 Feb 2004 05:08:04 GMT, "flint"
wrote:

If a tech asks the question, I expect it has some practical
application. Techs are like that :-)

What do you suppose he wants to know? Other than what you're
prepared to tell him ;-)

The Tech wants to know what to tell the electrician to install. If I don't
know what the line is going to be, I cannot answer. I don't want to give him
five answers for the five most common AC voltages all over the world,
because the tech will get them mixed up most of the time. I need to know if
their location uses 120V, 100V, 220V or 240V. Without that, I cannot answer
accurately.


Oh, right. These techs work globally, and forget to tell you what
country they're in at the time. Fair enough ;-)

On Sun, 08 Feb 2004 01:14:54 +0000, Laurence Payne
wrote:

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

---
Still bad.

The maximum power you can dissipate in a 4 ohm resistive load from a
14.4V _DC_ source is ~52W.

From a 14.4VPP _AC_ source, (which is the absolute maximum you'll be
able to feed into a loudspeaker from a non-bridged amp fed from a
14.4VDC supply, duhhh...) first you've got to find the RMS voltage
which, for a sine wave, is the peak-to-peak voltage divided by twice
the square root of two, then that quotient (which will be the RMS value
of the 14VPP source) needs to be squared and divided by the load
impedance, i.e.:

P = (Vpp/2sqrt2)²/R

for a 14.4V sine wave feeding a 4 ohm load, we'll have, then:

P = (14.4/2.828)²/4 ~ 6.48 watts

However, considering an "H" bridged final with each half-bridge capable
of swinging from rail to rail in both directions, that 14.4VPP signal
could be brought up to 28.8VPP, with a resultant maximum power output
capability of

P = (28.8/2.828)²/4 ~ 26 watts

a four-fold increase in output power with _no_ increase in supply
voltage!

For a stereo amplifier, that would translate to a 50 watt output
capablitity.
---

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

---
Like you know anything about it...

--
John Fields

On Sun, 08 Feb 2004 01:14:54 +0000, Laurence Payne
wrote:

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

---
Still bad.

The maximum power you can dissipate in a 4 ohm resistive load from a
14.4V _DC_ source is ~52W.

From a 14.4VPP _AC_ source, (which is the absolute maximum you'll be
able to feed into a loudspeaker from a non-bridged amp fed from a
14.4VDC supply, duhhh...) first you've got to find the RMS voltage
which, for a sine wave, is the peak-to-peak voltage divided by twice
the square root of two, then that quotient (which will be the RMS value
of the 14VPP source) needs to be squared and divided by the load
impedance, i.e.:

P = (Vpp/2sqrt2)²/R

for a 14.4V sine wave feeding a 4 ohm load, we'll have, then:

P = (14.4/2.828)²/4 ~ 6.48 watts

However, considering an "H" bridged final with each half-bridge capable
of swinging from rail to rail in both directions, that 14.4VPP signal
could be brought up to 28.8VPP, with a resultant maximum power output
capability of

P = (28.8/2.828)²/4 ~ 26 watts

a four-fold increase in output power with _no_ increase in supply
voltage!

For a stereo amplifier, that would translate to a 50 watt output
capablitity.
---

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

---
Like you know anything about it...

--
John Fields

On Sun, 08 Feb 2004 01:14:54 +0000, Laurence Payne
wrote:

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

---
Still bad.

The maximum power you can dissipate in a 4 ohm resistive load from a
14.4V _DC_ source is ~52W.

From a 14.4VPP _AC_ source, (which is the absolute maximum you'll be
able to feed into a loudspeaker from a non-bridged amp fed from a
14.4VDC supply, duhhh...) first you've got to find the RMS voltage
which, for a sine wave, is the peak-to-peak voltage divided by twice
the square root of two, then that quotient (which will be the RMS value
of the 14VPP source) needs to be squared and divided by the load
impedance, i.e.:

P = (Vpp/2sqrt2)²/R

for a 14.4V sine wave feeding a 4 ohm load, we'll have, then:

P = (14.4/2.828)²/4 ~ 6.48 watts

However, considering an "H" bridged final with each half-bridge capable
of swinging from rail to rail in both directions, that 14.4VPP signal
could be brought up to 28.8VPP, with a resultant maximum power output
capability of

P = (28.8/2.828)²/4 ~ 26 watts

a four-fold increase in output power with _no_ increase in supply
voltage!

For a stereo amplifier, that would translate to a 50 watt output
capablitity.
---

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

---
Like you know anything about it...

--
John Fields

On Sun, 08 Feb 2004 01:14:54 +0000, Laurence Payne
wrote:

On Sun, 08 Feb 2004 01:10:23 +0000, Laurence Payne
wrote:


You can't draw 200W into 4 ohms from a 14.4v supply. The theoretical
maximum is 36 watts.

Sorry, my bad. 51.84 watts.

---
Still bad.

The maximum power you can dissipate in a 4 ohm resistive load from a
14.4V _DC_ source is ~52W.

From a 14.4VPP _AC_ source, (which is the absolute maximum you'll be
able to feed into a loudspeaker from a non-bridged amp fed from a
14.4VDC supply, duhhh...) first you've got to find the RMS voltage
which, for a sine wave, is the peak-to-peak voltage divided by twice
the square root of two, then that quotient (which will be the RMS value
of the 14VPP source) needs to be squared and divided by the load
impedance, i.e.:

P = (Vpp/2sqrt2)²/R

for a 14.4V sine wave feeding a 4 ohm load, we'll have, then:

P = (14.4/2.828)²/4 ~ 6.48 watts

However, considering an "H" bridged final with each half-bridge capable
of swinging from rail to rail in both directions, that 14.4VPP signal
could be brought up to 28.8VPP, with a resultant maximum power output
capability of

P = (28.8/2.828)²/4 ~ 26 watts

a four-fold increase in output power with _no_ increase in supply
voltage!

For a stereo amplifier, that would translate to a 50 watt output
capablitity.
---

Actually, a very respectable power output for a tiny space like a car,
assuming reasonably efficient speakers.

Once you've converted advertiser's "Music Power" or whatever into real
watts, that might be near enough what you're getting :-)

---
Like you know anything about it...

--
John Fields

On Sun, 08 Feb 2004 01:00:48 +0000, Laurence Payne
wrote:

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

---
The mains voltage is immaterial except for what the example was supposed
to show, (and it evidently went right over your head) and that was that
for equal load dissipations, current from a source will vary with the
source voltage.
---

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

---
No, _you_ do the math, you snotty little ****.

But, first, read my post about the differences between putting AC and DC
into a load for a clue.
---

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

---
Yes, but I doubt that you'd know what they are or, if you knew, how to
implement them.

--
John Fields

On Sun, 08 Feb 2004 01:00:48 +0000, Laurence Payne
wrote:

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

---
The mains voltage is immaterial except for what the example was supposed
to show, (and it evidently went right over your head) and that was that
for equal load dissipations, current from a source will vary with the
source voltage.
---

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

---
No, _you_ do the math, you snotty little ****.

But, first, read my post about the differences between putting AC and DC
into a load for a clue.
---

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

---
Yes, but I doubt that you'd know what they are or, if you knew, how to
implement them.

--
John Fields

On Sun, 08 Feb 2004 01:00:48 +0000, Laurence Payne
wrote:

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

---
The mains voltage is immaterial except for what the example was supposed
to show, (and it evidently went right over your head) and that was that
for equal load dissipations, current from a source will vary with the
source voltage.
---

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

---
No, _you_ do the math, you snotty little ****.

But, first, read my post about the differences between putting AC and DC
into a load for a clue.
---

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

---
Yes, but I doubt that you'd know what they are or, if you knew, how to
implement them.

--
John Fields

On Sun, 08 Feb 2004 01:00:48 +0000, Laurence Payne
wrote:

On Fri, 06 Feb 2004 18:22:16 -0600, John Fields
wrote:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

Who cares what the mains supply voltage is? I don't think you'll find
many amplifiers that use 120v ac as the supply rails for the output
stage :-) The first thing the mains supply hits when it enters an
amp. is a big transformer, outputting whatever voltages the design
requires.

---
The mains voltage is immaterial except for what the example was supposed
to show, (and it evidently went right over your head) and that was that
for equal load dissipations, current from a source will vary with the
source voltage.
---

Consider a car system. 12v into 4 ohms will pass a maximum of 3 amps.
Watts = volts x amps. So the maximum theoretical power is....well,
you do the math :-)

---
No, _you_ do the math, you snotty little ****.

But, first, read my post about the differences between putting AC and DC
into a load for a clue.
---

If you want more, you need to start with more volts. A transformer
won't do it directly, as you're starting with a dc supply. But there
are ways of doing it :-)

---
Yes, but I doubt that you'd know what they are or, if you knew, how to
implement them.

--
John Fields

John Fields wrote:


It's not likely they've got a DC to DC
converter in there to boost the supply voltage, ...

For almost any aftermarket amp, it's unlikely the amp _doesn't_ have a DC-DC
converter these days. The no-converter amps seemed to have disappeared from the
marketplace ("new" that is). Too bad for the AM radio since some DC-DC
convertors are so crappy they jam the AM.