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Mariachi Mariachi is offline
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Default Replace positive wire on HU?

Does anybody think that a 16 gauge power wire is sufficient enough to
deliver 100 Watts (25 watts per speaker)? I was thinking of replacing
the factory 16 gauge power wire with a 14 gauge power wire. The only
problem is, I don't know my cars electrical system well enough in
order to run the 14 gauge where I'm supposed to. Some people say
there is a resistor between the head unit and the positive terminal of
the battery in order to limit the current... but isn't the head unit
supposed to do that by itself? I'm sure the head unit has it's own
switch to limit the current that is going into the HU. I don't want
to run a wire straight to the positive terminal of the battery because
I might ruin something that might be originally in parallel with the
head unit power wire. Right now, I'm trying to locate the 10 or 20 A
fuse that the power wire is connected up to. Any suggestions?

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D.Kreft D.Kreft is offline
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Default Replace positive wire on HU?

On Feb 24, 8:47 pm, "Mariachi" wrote:

Does anybody think that a 16 gauge power wire is sufficient enough to
deliver 100 Watts (25 watts per speaker)?


You'd be pretty lucky to get 25 Watts...like "superconductor" lucky--
you're likely getting closer to about eighteen or so.

At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)

I was thinking of replacing
the factory 16 gauge power wire with a 14 gauge power wire. The only
problem is, I don't know my cars electrical system well enough in
order to run the 14 gauge where I'm supposed to. Some people say
there is a resistor between the head unit and the positive terminal of
the battery in order to limit the current... but isn't the head unit
supposed to do that by itself?


Sounds like "ignernt" folk to me. The HU, like *any* other electrical
device, will only draw as much current as it needs.

I'm sure the head unit has it's own
switch to limit the current that is going into the HU.


Yeah, it's called a "fuse." :-)

And remember...current doesn't "go into" an electrical device--it is
*drawn* by the device. It's not like you can *force* current to go
through something (at least not without jackiung up the input
voltage :-).

I don't want
to run a wire straight to the positive terminal of the battery because
I might ruin something that might be originally in parallel with the
head unit power wire.


This doesn't really make sense.

Right now, I'm trying to locate the 10 or 20 A
fuse that the power wire is connected up to. Any suggestions?


The first question you need to ask yourself is "Why do I want to do
this?" Do you have an actual problem that you're trying to solve, or
are you creating problems for yourself to solve because you're bored?
It may sound like a wise-guy question, but it *is* what you need to be
asking yourself.

-dan

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Mariachi Mariachi is offline
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Default Replace positive wire on HU?

so is it alright to drag a 12 or 14 gauge wire to the positive power
source wire of the HU from the positive terminal of the battery just
to test it out? Not like I need to replace it or anything... I just
thought I might get like 2-4 watts more power for each speaker. Plus
I replaced all the other speaker wires with 12 or 14 gauge so why not
do the same thing with the power wires. Since I don't have money to
buy an amp at the moment, I just thought do all that I can with the
money and equipment I have. Hobby of mine...

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Captain Howdy Captain Howdy is offline
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Default Replace positive wire on HU?

In article om, "D.Kreft" wrote:


What does all of this mean? Doesn't the current drawn by the device not flow
into the device?LOL From what Edison would tell you, you can lower current
demand by jacking up input voltage. Europeans even bought into this idea.



And remember...current doesn't "go into" an electrical device--it is
*drawn* by the device. It's not like you can *force* current to go
through something (at least not without jackiung up the input
voltage :-).


-dan

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Mariachi Mariachi is offline
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Default Replace positive wire on HU?

On Feb 25, 3:16 am, (Captain Howdy) wrote:
In article om, "D.Kreft" wrote:

What does all of this mean? Doesn't the current drawn by the device not flow
into the device?LOL From what Edison would tell you, you can lower current
demand by jacking up input voltage. Europeans even bought into this idea.



And remember...current doesn't "go into" an electrical device--it is
*drawn* by the device. It's not like you can *force* current to go
through something (at least not without jackiung up the input
voltage :-).


-dan


i think what he means is that the head unit has it's own resistance
control to limit the current. If the radio is turned off by the
switch wire then the radio has a resistance infinity, therefore no
current is allowed through ("no current" meaning only the current that
keeps the clock running is allowed through, which is in the microamps,
or you can call this the "bleeder" current). When you turn the radio
on, the switch wire turns on the radio by closing the electrical
switch and therefore current is allowed. How much current? Well
depends on the potentiometer (the volume knob)... Technically there is
a minimum resistance in order to safe guard how much current goes
through in order not to blow up anything. If Voltage = Current *
Resistance. If you can't change resistance, then increase the voltage
across the load (the applied voltage) in order to increase the
current.



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Mariachi Mariachi is offline
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Default Replace positive wire on HU?

At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...

Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)

If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.

So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.

So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2

R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms

4.194 / 2.634 = 1.59

Resistance decreased by 1.59 times

R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6

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Matt Ion Matt Ion is offline
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Default Replace positive wire on HU?

Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.


Mariachi wrote:
At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)



yes I agree...

Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)

If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.

So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.

So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2

R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms

4.194 / 2.634 = 1.59

Resistance decreased by 1.59 times

R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6

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Mariachi Mariachi is offline
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Default Replace positive wire on HU?

On Feb 25, 4:01 pm, Matt Ion wrote:
Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.

Mariachi wrote:
At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6


my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.

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Matt Ion Matt Ion is offline
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Default Replace positive wire on HU?

Mariachi wrote:
On Feb 25, 4:01 pm, Matt Ion wrote:

Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.

Mariachi wrote:

At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6



my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.


Yes, but "rated" with what test criteria? This is the problem with car audio,
there are no real solid "standards" for testing and specifying power output (I
understand there have been some ad-hoc standards introduced that are adhered to
*voluntarily* by SOME manufacturers, but even that is borderline meaningless).

Your ACTUAL output will vary more with your RPM and alternator output voltage
than it will with increased wire size to your HU, and even then, you're not
usually going to hear the difference between 12.5V and 14.5V, let alone with the
few fractions of a volt difference you'll see with the larger wire.
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Mariachi Mariachi is offline
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Posts: 174
Default Replace positive wire on HU?

On Feb 25, 4:22 pm, Matt Ion wrote:
Mariachi wrote:
On Feb 25, 4:01 pm, Matt Ion wrote:


Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.


Mariachi wrote:


At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6


my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.


Yes, but "rated" with what test criteria? This is the problem with car audio,
there are no real solid "standards" for testing and specifying power output (I
understand there have been some ad-hoc standards introduced that are adhered to
*voluntarily* by SOME manufacturers, but even that is borderline meaningless).

Your ACTUAL output will vary more with your RPM and alternator output voltage
than it will with increased wire size to your HU, and even then, you're not
usually going to hear the difference between 12.5V and 14.5V, let alone with the
few fractions of a volt difference you'll see with the larger wire


It doesn't really depend on the rpm's of the engine or the alternator
if your battery is properly charged... you're talking about subwoofer
systems that need 300 Watts or more that constantly requires the
battery to be recharged... i'm just talking about around 80 Watts
total. Unless you have a crappy battery maybe it would depend more on
your alternator than usual. But still, it's a head unit amp... sure
it can drain the battery but not as quick as a 800 Watt amplifier. It
takes about 4 hours of music playing to kill my battery and that's
with the stereo running at full blast. If you are running a higher
voltage battery, you are able to get more voltage to the head unit and
therefore more power. Of course the head unit amp has it's limits...
but with a higher voltage battery you can help the head unit amp reach
those limits. That's why they say 22 Watts RMS per 4 ohm speaker
"using a 14.4 V battery". I wouldn't put a 14.4 V battery in my car,
but I do like to provide good connections for all the audio equipment
in my car in order to get the most out of what I have.



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Matt Ion Matt Ion is offline
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Default Replace positive wire on HU?

Mariachi wrote:
On Feb 25, 4:22 pm, Matt Ion wrote:

Mariachi wrote:

On Feb 25, 4:01 pm, Matt Ion wrote:


Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.


Mariachi wrote:


At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6


my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.


Yes, but "rated" with what test criteria? This is the problem with car audio,
there are no real solid "standards" for testing and specifying power output (I
understand there have been some ad-hoc standards introduced that are adhered to
*voluntarily* by SOME manufacturers, but even that is borderline meaningless).

Your ACTUAL output will vary more with your RPM and alternator output voltage
than it will with increased wire size to your HU, and even then, you're not
usually going to hear the difference between 12.5V and 14.5V, let alone with the
few fractions of a volt difference you'll see with the larger wire



It doesn't really depend on the rpm's of the engine or the alternator
if your battery is properly charged... you're talking about subwoofer
systems that need 300 Watts or more that constantly requires the
battery to be recharged... i'm just talking about around 80 Watts
total. Unless you have a crappy battery maybe it would depend more on
your alternator than usual. But still, it's a head unit amp... sure
it can drain the battery but not as quick as a 800 Watt amplifier. It
takes about 4 hours of music playing to kill my battery and that's
with the stereo running at full blast. If you are running a higher
voltage battery, you are able to get more voltage to the head unit and
therefore more power. Of course the head unit amp has it's limits...
but with a higher voltage battery you can help the head unit amp reach
those limits. That's why they say 22 Watts RMS per 4 ohm speaker
"using a 14.4 V battery". I wouldn't put a 14.4 V battery in my car,
but I do like to provide good connections for all the audio equipment
in my car in order to get the most out of what I have.


Er... dude, you're WAY out to lunch here.

Your battery is ALWAYS 12V (give or take a bit). Your ALTERNATOR can output
anywhere between 12 to 15 volts depending on load and RPM. You have little to
no control over it (unless you use an ACCUVOLT like MOSFET).

In normal operation, with the engine running, NOTHING should be drawing on the
battery - the alternator should be providing ALL your car's power needs (yeah
yeah, the battery filters power, yada yada - semantics).

Unless your amp is running a regulated power supply (which your head unit likely
isn't), your max output power will vary with the power input voltage, and most
manufacturers will test and spec their amps at 14.5V because that gives higher
numbers - it's marketing, plain and simple.
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Mariachi Mariachi is offline
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Posts: 174
Default Replace positive wire on HU?

On Feb 26, 11:34 am, Matt Ion wrote:
Mariachi wrote:
On Feb 25, 4:22 pm, Matt Ion wrote:


Mariachi wrote:


On Feb 25, 4:01 pm, Matt Ion wrote:


Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.


Mariachi wrote:


At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6


my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.


Yes, but "rated" with what test criteria? This is the problem with car audio,
there are no real solid "standards" for testing and specifying power output (I
understand there have been some ad-hoc standards introduced that are adhered to
*voluntarily* by SOME manufacturers, but even that is borderline meaningless).


Your ACTUAL output will vary more with your RPM and alternator output voltage
than it will with increased wire size to your HU, and even then, you're not
usually going to hear the difference between 12.5V and 14.5V, let alone with the
few fractions of a volt difference you'll see with the larger wire


It doesn't really depend on the rpm's of the engine or the alternator
if your battery is properly charged... you're talking about subwoofer
systems that need 300 Watts or more that constantly requires the
battery to be recharged... i'm just talking about around 80 Watts
total. Unless you have a crappy battery maybe it would depend more on
your alternator than usual. But still, it's a head unit amp... sure
it can drain the battery but not as quick as a 800 Watt amplifier. It
takes about 4 hours of music playing to kill my battery and that's
with the stereo running at full blast. If you are running a higher
voltage battery, you are able to get more voltage to the head unit and
therefore more power. Of course the head unit amp has it's limits...
but with a higher voltage battery you can help the head unit amp reach
those limits. That's why they say 22 Watts RMS per 4 ohm speaker
"using a 14.4 V battery". I wouldn't put a 14.4 V battery in my car,
but I do like to provide good connections for all the audio equipment
in my car in order to get the most out of what I have.


Er... dude, you're WAY out to lunch here.

Your battery is ALWAYS 12V (give or take a bit). Your ALTERNATOR can output
anywhere between 12 to 15 volts depending on load and RPM. You have little to
no control over it (unless you use an ACCUVOLT like MOSFET).

In normal operation, with the engine running, NOTHING should be drawing on the
battery - the alternator should be providing ALL your car's power needs (yeah
yeah, the battery filters power, yada yada - semantics).

Unless your amp is running a regulated power supply (which your head unit likely
isn't), your max output power will vary with the power input voltage, and most
manufacturers will test and spec their amps at 14.5V because that gives higher
numbers - it's marketing, plain and simple.- Hide quoted text -

- Show quoted text -



why would it use the alternator??? You might be talking about the
engine.. but the stereo uses the battery. It uses the alternator
indirectly because the battery get's energy from the alternator.
wow... this is why your amplifiers are connected to the battery, so
they get power from the battery, not the alternator. The alternator
is not a stable power source, the battery is. Otherwise you would be
getting alternating current into your amplifier and resulting in
noise. This is why audio equipment uses DC voltage, not AC voltage
(like your alternator). Don't you know this?

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Default Replace positive wire on HU?

No, the stereo uses the dominating power source which would be the alternator
when the engine is running. Yes amplifiers are connected to the battery, as is
the alternator. Your alternator does not use or make AC voltage, the only
place you'll find AC voltage in your car is in your speaker wires.

You're out of your box and you're way off track.............




why would it use the alternator??? You might be talking about the
engine.. but the stereo uses the battery. It uses the alternator
indirectly because the battery get's energy from the alternator.
wow... this is why your amplifiers are connected to the battery, so
they get power from the battery, not the alternator. The alternator
is not a stable power source, the battery is. Otherwise you would be
getting alternating current into your amplifier and resulting in
noise. This is why audio equipment uses DC voltage, not AC voltage
(like your alternator). Don't you know this?

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Default Replace positive wire on HU?

On Feb 26, 2:28 pm, (Captain Howdy) wrote:
No, the stereo uses the dominating power source which would be the alternator
when the engine is running. Yes amplifiers are connected to the battery, as is
the alternator. Your alternator does not use or make AC voltage, the only
place you'll find AC voltage in your car is in your speaker wires.

You're out of your box and you're way off track.............





why would it use the alternator??? You might be talking about the
engine.. but the stereo uses the battery. It uses the alternator
indirectly because the battery get's energy from the alternator.
wow... this is why your amplifiers are connected to the battery, so
they get power from the battery, not the alternator. The alternator
is not a stable power source, the battery is. Otherwise you would be
getting alternating current into your amplifier and resulting in
noise. This is why audio equipment uses DC voltage, not AC voltage
(like your alternator). Don't you know this?- Hide quoted text -


- Show quoted text -


All automotive alternators naturally produce AC current. But they
have a set of rectifiers, diodge bridges, to convert to DC.

Let me ask you this? If the alternator doesn't supply enough
amperage to power up your stereo system, doesn't it take excess energy
from the battery to compensate for the alternator... Does it solely
depend on the amperage that the alternator is putting out... or both
the battery and the alternator. So if my amplifier requires 30 amps
and my alternator only supplies 12 amps at the moment... doesn't the
battery supply the other 18 amps?

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Default Replace positive wire on HU?




In article . com, "Mariachi"
wrote:


All automotive alternators naturally produce AC current. But they
have a set of rectifiers, diodge bridges, to convert to DC.


Yes, this is true, but with that said you still won't see any AC voltage in
your car.


Let me ask you this? If the alternator doesn't supply enough
amperage to power up your stereo system, doesn't it take excess energy
from the battery to compensate for the alternator... Does it solely
depend on the amperage that the alternator is putting out... or both
the battery and the alternator. So if my amplifier requires 30 amps
and my alternator only supplies 12 amps at the moment... doesn't the
battery supply the other 18 amps?


Yes is does, but also keep in mind that if the alternator does not keep up
with the power demands of the car and battery included, you will end up with a
no start dead battery.

The power is drawn from the dominating power source which in most cases in the
alternator, because most batteries lose voltage fairly fast when under load.
The alternator also has an easier time dealing with loads because it has next
to no recovery time and operates at a higher voltage then the battery. The
alternator also loses output under low rpm's and this is where people notice
headlight dimming the most. Stereo's aside, you might notice that most cars
get major headlight dim just with the heater and wipers on when the car comes
to a stop, again that's because the alternator output drops while at idle and
in gear and the battery starts to take the load on, bigger batteries tend to
remedy this problem to a point since they tend to have higher output vs.
smaller batteries.


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Default Replace positive wire on HU?

On Feb 25, 12:16 am, (Captain Howdy) wrote:
In article om, "D.Kreft" wrote:

What does all of this mean? Doesn't the current drawn by the device not flow
into the device?


The original post inferred that the current was not *drawn* but rather
*pushed* into the device.

From what Edison would tell you, you can lower current
demand by jacking up input voltage. Europeans even bought into this idea.


From what Georg Ohm would tell you, if you jack up the input voltage

you will necessarily increase the input current given a non-regulated
or purely-resistive load. You can continue to crank up the input
voltage and the input current will continue to increase until
something lets out its "magic smoke" and the device presents an open
circuit.

Jacking up the input voltage in order to reduce the *need* for current
only works when transforming or "inverting" (converting DC to AC)...it
does not hold true in general, and even for those cases where it does
work, it will generally only do so over a finite range of voltlages--
go too high and you let the magic smoke out.

And let's not forget that in the specific case of a head unit, the
device in question is not purely a transformer or inverter--there are
also components in the device that operate on the unregulated DC input
voltage, so I stand by what I've stated.

-dan


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Have you not ever hooked up a dual votage electic motor, if you ever come
across one see the amp specs at 110v and at 220v, Whoever told ya that this
only works when transforming or inverting was just trying to mess with your
mind.



Jacking up the input voltage in order to reduce the *need* for current
only works when transforming or "inverting" (converting DC to AC)...it
does not hold true in general, and even for those cases where it does
work, it will generally only do so over a finite range of voltlages--
go too high and you let the magic smoke out.

And let's not forget that in the specific case of a head unit, the
device in question is not purely a transformer or inverter--there are
also components in the device that operate on the unregulated DC input
voltage, so I stand by what I've stated.

-dan


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Default Replace positive wire on HU?

On Feb 26, 3:37 pm, (Captain Howdy) wrote:

Have you not ever hooked up a dual votage electic motor, if you ever come
across one see the amp specs at 110v and at 220v, Whoever told ya that this
only works when transforming or inverting was just trying to mess with your
mind.


I'm not an electric motor expert, but I do hold a degree in Electrical
Engineering (I suppose that one could argue that my EE profs were all
out to mess with my mind). Granted, I'm not an EE by profession, but I
do have enough edukashun and real-world experience with electronics to
have something of a clue as to how these things work.

From what I understand about dual-voltage electric motors, they

achieve their dual-voltage capability via alternate wirings in the
motor itself or via an internal transformer (such as would be found in
a computer power supply or home electronics box) which is used to
convert the to the desired level. Some devices, like laptop computer
power converters do this voltage sensing automagically so that you
don't have to flip as switch as you would on a typical desktop
computer power supply or home audio receiver.

So even here, you're still talking about using a transformer to handle
the voltage conversion. The reason that a doubling of input voltage
results in a halving of input current is simple--because assuming
negligible losses in in the transformer itself the law of the
conservation of energy (from the first law of thermodynamics) applies:

Pin = Pout

Substituting VI (voltage * current) for each side:

Vin * Iin = Vout * Iout

Then divide both sides by Vin:

Vout * Iout
Iin = -------------
Vin

So yeah, doubling Vin leads to a halving of Iin. There's nothing new
here.

Ultimately, though, we're still dealing with DC inputs so this
comparison to dual-voltage AC motors, while entertaining, isn't
entirely relevant and does more to support my point than detract from
it.

-dan

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Alright with that said and with your degree in Electrical
Engineering this shouldn't be hard. All you need is a regulated power supply,
any cheesy amplifier will do, although an unregulated amplifier will show
clearer results. Run a low test tone into the amplifier with a dummy load on
it at 14.5 volts and at 10.5 volts and watch the current draw. The tick here
is to keep the same output voltage on the amplifier at both input voltages.
See what happens to the current draw. And you're right about the dual voltage
motors, they achieve their dual voltage capability via alternate wiring in the
motor.



In article om, "D.Kreft"
wrote:
On Feb 26, 3:37 pm, (Captain Howdy) wrote:

Have you not ever hooked up a dual votage electic motor, if you ever come
across one see the amp specs at 110v and at 220v, Whoever told ya that this
only works when transforming or inverting was just trying to mess with your
mind.


I'm not an electric motor expert, but I do hold a degree in Electrical
Engineering (I suppose that one could argue that my EE profs were all
out to mess with my mind). Granted, I'm not an EE by profession, but I
do have enough edukashun and real-world experience with electronics to
have something of a clue as to how these things work.

From what I understand about dual-voltage electric motors, they

achieve their dual-voltage capability via alternate wirings in the
motor itself or via an internal transformer (such as would be found in
a computer power supply or home electronics box) which is used to
convert the to the desired level. Some devices, like laptop computer
power converters do this voltage sensing automagically so that you
don't have to flip as switch as you would on a typical desktop
computer power supply or home audio receiver.

So even here, you're still talking about using a transformer to handle
the voltage conversion. The reason that a doubling of input voltage
results in a halving of input current is simple--because assuming
negligible losses in in the transformer itself the law of the
conservation of energy (from the first law of thermodynamics) applies:

Pin = Pout

Substituting VI (voltage * current) for each side:

Vin * Iin = Vout * Iout

Then divide both sides by Vin:

Vout * Iout
Iin = -------------
Vin

So yeah, doubling Vin leads to a halving of Iin. There's nothing new
here.

Ultimately, though, we're still dealing with DC inputs so this
comparison to dual-voltage AC motors, while entertaining, isn't
entirely relevant and does more to support my point than detract from
it.

-dan

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Default Replace positive wire on HU?

Mariachi wrote:
On Feb 26, 11:34 am, Matt Ion wrote:

Mariachi wrote:

On Feb 25, 4:22 pm, Matt Ion wrote:


Mariachi wrote:


On Feb 25, 4:01 pm, Matt Ion wrote:


Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.


Mariachi wrote:


At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6


my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.


Yes, but "rated" with what test criteria? This is the problem with car audio,
there are no real solid "standards" for testing and specifying power output (I
understand there have been some ad-hoc standards introduced that are adhered to
*voluntarily* by SOME manufacturers, but even that is borderline meaningless).


Your ACTUAL output will vary more with your RPM and alternator output voltage
than it will with increased wire size to your HU, and even then, you're not
usually going to hear the difference between 12.5V and 14.5V, let alone with the
few fractions of a volt difference you'll see with the larger wire


It doesn't really depend on the rpm's of the engine or the alternator
if your battery is properly charged... you're talking about subwoofer
systems that need 300 Watts or more that constantly requires the
battery to be recharged... i'm just talking about around 80 Watts
total. Unless you have a crappy battery maybe it would depend more on
your alternator than usual. But still, it's a head unit amp... sure
it can drain the battery but not as quick as a 800 Watt amplifier. It
takes about 4 hours of music playing to kill my battery and that's
with the stereo running at full blast. If you are running a higher
voltage battery, you are able to get more voltage to the head unit and
therefore more power. Of course the head unit amp has it's limits...
but with a higher voltage battery you can help the head unit amp reach
those limits. That's why they say 22 Watts RMS per 4 ohm speaker
"using a 14.4 V battery". I wouldn't put a 14.4 V battery in my car,
but I do like to provide good connections for all the audio equipment
in my car in order to get the most out of what I have.


Er... dude, you're WAY out to lunch here.

Your battery is ALWAYS 12V (give or take a bit). Your ALTERNATOR can output
anywhere between 12 to 15 volts depending on load and RPM. You have little to
no control over it (unless you use an ACCUVOLT like MOSFET).

In normal operation, with the engine running, NOTHING should be drawing on the
battery - the alternator should be providing ALL your car's power needs (yeah
yeah, the battery filters power, yada yada - semantics).

Unless your amp is running a regulated power supply (which your head unit likely
isn't), your max output power will vary with the power input voltage, and most
manufacturers will test and spec their amps at 14.5V because that gives higher
numbers - it's marketing, plain and simple.- Hide quoted text -

- Show quoted text -




why would it use the alternator??? You might be talking about the
engine.. but the stereo uses the battery.


They're all connected together! It's a parallel circuit! Measure the voltage
at the alternator, the battery, and your amp - you'll find them all the same
(within a fraction of a volt).

wow... this is why your amplifiers are connected to the battery, so
they get power from the battery, not the alternator. The alternator
is not a stable power source, the battery is. Otherwise you would be
getting alternating current into your amplifier and resulting in
noise.


What the hell are you drinking?

The AC output from your alternator goes through rectifier diodes and a regulator
BEFORE it's connected to anything else. Whether the regulator is internal to
the alternator or mounted somewhere else in the engine bay, ALL the car's power
goes through it first; NOTHING in the car uses AC power.

The battery exists for one reason: to crank the starter. Once the car is
running, the alternator provides the WHOLE CAR's energy needs, including keeping
the battery charged.

(Yeah yeah, I know some wank is gonna step in to explain how the battery helps
filter the power, etc. etc.; this poor Mariachi guy has NO clue about electrical
theory, so let's not confuse the issue just now).



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Default Replace positive wire on HU?

On Feb 26, 10:49 pm, Matt Ion wrote:
Mariachi wrote:
On Feb 26, 11:34 am, Matt Ion wrote:


Mariachi wrote:


On Feb 25, 4:22 pm, Matt Ion wrote:


Mariachi wrote:


On Feb 25, 4:01 pm, Matt Ion wrote:


Mathematically, there may be a theoretical difference... REALISTICALLY you can
be 99.9% sure the difference would barely be measurable in your deck's output,
and certainly not audible or noticeable, especially when you factor in the
ambient noise and the fact that your HU's built-in amp isn't actually giving you
anywhere near 25W/ch.


Mariachi wrote:


At any rate, you'd have to know the length of the wire--there's
nothing magical about the gauge of the wire, it's all a matter of the
total resistance that wire has. So, would be fine with 1" of 18 AWG on
a 750 Watt amplifier, but you'd be in really bad shape if you had to
go 15 feet with the same wire on that same amp. :-)


yes I agree...


Resistance of the wire = Resistivity * Length / Area
R(wire) = p * (L/A)


If you lower the wire by 3 gauges then you have a cross sectional area
increase by a factor of 2.
Therefore, the total resistance decreases by a factor of 2 given that
the resistivity and length are still the same.


So if you want to make a power wire twice as long, you need to
increase A by 2 times also. Therefore it would be the same ratio and
therefore it would be the same total resistance.


So if I have positive power wire with a length 2 meters with a 16
guage cross sectional area...
Resistivity of copper = 1.7 microohm/cm = 1.7 microohm/cm * (1 *
10^(-6) ohms / 1 microohm) * (1 m / 100 cm) = 0.000000017 ohm/m
L = 2 meters
A (16 gauge)= 0.8107 m^2


R(16 GWG wire) = (1.7e-8)*(2/.8107) ohms = 4.194e-8 ohms
R(14 GWG wire) = (1.7e-8)*(2/1.291) ohms = 2.634e-8 ohms


4.194 / 2.634 = 1.59


Resistance decreased by 1.59 times


R(16 GWG) / R(13 GWG) = 2
R(16 GWG) / R(14 GWG) = 1.6


my head unit is rated to get a continuous power at 24 Watts per
speaker, but with a 5% THD. It gets 17 Watts per each speaker with a .
01 % THD.


Yes, but "rated" with what test criteria? This is the problem with car audio,
there are no real solid "standards" for testing and specifying power output (I
understand there have been some ad-hoc standards introduced that are adhered to
*voluntarily* by SOME manufacturers, but even that is borderline meaningless).


Your ACTUAL output will vary more with your RPM and alternator output voltage
than it will with increased wire size to your HU, and even then, you're not
usually going to hear the difference between 12.5V and 14.5V, let alone with the
few fractions of a volt difference you'll see with the larger wire


It doesn't really depend on the rpm's of the engine or the alternator
if your battery is properly charged... you're talking about subwoofer
systems that need 300 Watts or more that constantly requires the
battery to be recharged... i'm just talking about around 80 Watts
total. Unless you have a crappy battery maybe it would depend more on
your alternator than usual. But still, it's a head unit amp... sure
it can drain the battery but not as quick as a 800 Watt amplifier. It
takes about 4 hours of music playing to kill my battery and that's
with the stereo running at full blast. If you are running a higher
voltage battery, you are able to get more voltage to the head unit and
therefore more power. Of course the head unit amp has it's limits...
but with a higher voltage battery you can help the head unit amp reach
those limits. That's why they say 22 Watts RMS per 4 ohm speaker
"using a 14.4 V battery". I wouldn't put a 14.4 V battery in my car,
but I do like to provide good connections for all the audio equipment
in my car in order to get the most out of what I have.


Er... dude, you're WAY out to lunch here.


Your battery is ALWAYS 12V (give or take a bit). Your ALTERNATOR can output
anywhere between 12 to 15 volts depending on load and RPM. You have little to
no control over it (unless you use an ACCUVOLT like MOSFET).


In normal operation, with the engine running, NOTHING should be drawing on the
battery - the alternator should be providing ALL your car's power needs (yeah
yeah, the battery filters power, yada yada - semantics).


Unless your amp is running a regulated power supply (which your head unit likely
isn't), your max output power will vary with the power input voltage, and most
manufacturers will test and spec their amps at 14.5V because that gives higher
numbers - it's marketing, plain and simple.- Hide quoted text -


- Show quoted text -


why would it use the alternator??? You might be talking about the
engine.. but the stereo uses the battery.


They're all connected together! It's a parallel circuit! Measure the voltage
at the alternator, the battery, and your amp - you'll find them all the same
(within a fraction of a volt).

wow... this is why your amplifiers are connected to the battery, so
they get power from the battery, not the alternator. The alternator
is not a stable power source, the battery is. Otherwise you would be
getting alternating current into your amplifier and resulting in
noise.


What the hell are you drinking?

The AC output from your alternator goes through rectifier diodes and a regulator
BEFORE it's connected to anything else. Whether the regulator is internal to
the alternator or mounted somewhere else in the engine bay, ALL the car's power
goes through it first; NOTHING in the car uses AC power.

The battery exists for one reason: to crank the starter. Once the car is
running, the alternator provides the WHOLE CAR's energy needs, including keeping
the battery charged.

(Yeah yeah, I know some wank is gonna step in to explain how the battery helps
filter the power, etc. etc.; this poor Mariachi guy has NO clue about electrical
theory, so let's not confuse the issue just now).


Read the posts above you before you say anything... as I already said,
I know that the AC voltage from the alternator goes through rectifier
diodes. Yes, you can call me an 'idiot' for all I care if it makes
you feel any better. I'm just saying the battery does output current
when the alternator can't supply it all.

And yes, I'm way out of whack blah blah blah... w/e you just said...
feel better?

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Default Replace positive wire on HU?

Mariachi wrote:

Read the posts above you before you say anything... as I already said,
I know that the AC voltage from the alternator goes through rectifier
diodes. Yes, you can call me an 'idiot' for all I care if it makes
you feel any better. I'm just saying the battery does output current
when the alternator can't supply it all.


Okay, one more time:

The alternator creates current, converting mechanical energy into electrical energy.

The battery stores electrical energy.

The battery and the alternator are in a parallel circuit. They do not operate
independently of each other.

When the car is running, the alternator provides ALL the car's electrical needs,
including for the sound system. The battery DOES NOT.

If you still refuse to understand this, then nobody here can help you, as
everyone else, including probably even bob wald, get this concept.
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Default Replace positive wire on HU?

On Feb 27, 8:54 am, Matt Ion wrote:

When the car is running, the alternator provides ALL the car's electrical needs,
including for the sound system. The battery DOES NOT.


Correct...right up until the point where the current demand from the
vehicle (presumably from a big honkin' stereo system) exceeds the
output capability of the alternator, at which point the battery starts
converting that yummy chemical energy into electrical engery...which
only makes matters worse because then the battery becomes an even
*larger* load on the alternator as it tries to not only run your
electronics, but also recharge the battery.

As Matt has spelled-out (several times now), it really is this simple.

-dan

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The alternator creates current, converting mechanical energy into electrical energy
duh

The battery stores electrical energy.

once again, duh, but more specifically it stores chemical battery acid
cells that a connected in series in order to make 12 V by seperating a
difference in charge.

The battery and the alternator are in a parallel circuit. They do not operate
independently of each other.

really? I didn't know (sarcastic)

When the car is running, the alternator provides ALL the car's electrical needs,
including for the sound system. The battery DOES NOT.


if the voltage of the alternator goes below the voltage of the
battery, then the battery does supply current for whatever the
alternator cannot handle... unless you have a 75 amp alternator that
never needs to use the battery. But you are saying that you can have
50 amps going into my amplifier and my little 40 amp alternator will
supply all the current, which just doesn't make sense to me. If the
battery and alternator are in parallel, then I can see the alternator
taking over all the electrical systems when the current demand does
not exceed the limit of the alternator. But if the current demand
exceeds the limit of the alternator, you are saying the battery does
not take over at all, even though the battery is connected in parallel
with the alternator. I'm not saying you are wrong, i'm just asking if
the current demand exceeds the current supply limit of the alternator,
what happens then? Please explain...


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On Feb 27, 12:43 pm, "D.Kreft" wrote:
On Feb 27, 8:54 am, Matt Ion wrote:

When the car is running, the alternator provides ALL the car's electrical needs,
including for the sound system. The battery DOES NOT.


Correct...right up until the point where the current demand from the
vehicle (presumably from a big honkin' stereo system) exceeds the
output capability of the alternator, at which point the battery starts
converting that yummy chemical energy into electrical engery...which
only makes matters worse because then the battery becomes an even
*larger* load on the alternator as it tries to not only run your
electronics, but also recharge the battery.

As Matt has spelled-out (several times now), it really is this simple.

-dan


finally, yes, this is what I was asking... thanks Kreft.



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On Feb 27, 9:56 am, "Mariachi" wrote:

if the voltage of the alternator goes below the voltage of the
battery, then the battery does supply current for whatever the
alternator cannot handle...


Careful, sir. You've made a misstep three words into your post.
Because the battery and the alternator are wired in parallel, it is
physically impossible for the voltage at the alternator to differ from
that of the voltage at the battery (assuming negligible internal
losses in the connecting wires).

Remember, that at all points in a parallel circuit, the voltage will
be identical everywhere, but current draw may differ at various points
in the circuit. In *series* circuits, the current is the same
everywhere, but the voltage drop across each device is what may
differ. Put into formulas:

Parallel:
Vs = V1 = V2 = V3 = V4
Is = I1 + I2 + I3 + I4

Series:
Vs = V1 + V2 + V3 + V4
Is = I1 = I2 = I3 = I4

Where 'Vs' and 'Is' are the voltage and current at the source,
respectively; and where 'V#' and 'I#' are the voltage and current
measured at each device in the circuit.

-dan


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Mariachi Mariachi is offline
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Default Replace positive wire on HU?

On Feb 27, 1:28 pm, "D.Kreft" wrote:
On Feb 27, 9:56 am, "Mariachi" wrote:

if the voltage of the alternator goes below the voltage of the
battery, then the battery does supply current for whatever the
alternator cannot handle...


Careful, sir. You've made a misstep three words into your post.
Because the battery and the alternator are wired in parallel, it is
physically impossible for the voltage at the alternator to differ from
that of the voltage at the battery (assuming negligible internal
losses in the connecting wires).

Remember, that at all points in a parallel circuit, the voltage will
be identical everywhere, but current draw may differ at various points
in the circuit. In *series* circuits, the current is the same
everywhere, but the voltage drop across each device is what may
differ. Put into formulas:

Parallel:
Vs = V1 = V2 = V3 = V4
Is = I1 + I2 + I3 + I4

Series:
Vs = V1 + V2 + V3 + V4
Is = I1 = I2 = I3 = I4

Where 'Vs' and 'Is' are the voltage and current at the source,
respectively; and where 'V#' and 'I#' are the voltage and current
measured at each device in the circuit.

-dan


Okay, I know all that... but if you put two car batteries in parallel,
you would get the same voltage but wouldn't the total internal
resistance of the batteries be less than the internal resistance of
one battery. Since you have one internal resistance in parallel with
another internal resistance, the total internal resistance has to be
less than the least internal resistance of one battery. So you would
actually have less voltage drop over the interal resistance of both
parallel batteries, therefore leaving more voltage left for whatever
series loads there are. Therefore you would actually measure a minute
increase in voltage across the two parallel batteries because the
total internal resistance of both batteries in parallel would be less
than one internal resistance of one battery.

And I'm not arguing that the alternator (within its limit) supplies
all the current when the car is turned on, but why is that so? Just
curious...

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D.Kreft D.Kreft is offline
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Default Replace positive wire on HU?

On Feb 27, 10:42 am, "Mariachi" wrote:

So you would actually have less voltage drop over the interal resistance of both
parallel batteries, therefore leaving more voltage left for whatever
series loads there are.


But this is only because your current demand on each battery is but a
fraction of what you would otherwise demand of a single battery. For
example, let's assume an "internal resistance" (in quotes because it's
not really "resistance" per se--it is simply modeled that way to aid
in understanding) of 0.5 Ohms per battery. Given a battery whose
resting voltage is 12.0 and a load drawing 2A, you would get an
internal voltage drop of:

Vi = I * R
= 2 * 0.5
= 1 Volt

So, drawing 2 Amperes from a 12 Volt battery would yield 11 Volts at
the battery terminals (12 V - 1 V).

Now put two identical batteries in parallel with the same load. The
current draw from each battery would be halved, so the voltage drop
across each battery would read 11.5 Volts (12 V - 0.5 V):

Vi = 2/2 * 0.5
= 1 * 0.5
= 0.5

See also:
http://en.wikipedia.org/wiki/Interna...ance#Batteries

As a side note, this internal resistance is the reason why a "dead
battery" can measure a full 12.8 Volts--a voltmeter presents an
"infinite" load on the battery and therefore draws no current.
However, when you put a load on the battery, the internal resitance
gobbles-up all the power and leaves you with next-to-nothing to power
your electrical devices.

Therefore you would actually measure a minute
increase in voltage across the two parallel batteries because the
total internal resistance of both batteries in parallel would be less
than one internal resistance of one battery.


Correct.

And I'm not arguing that the alternator (within its limit) supplies
all the current when the car is turned on, but why is that so? Just
curious...


I believe the answer there is simply "laziness." An alternator
delivers current much easier than a battery.

-dan


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Matt Ion Matt Ion is offline
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Default Replace positive wire on HU?

Mariachi wrote:
The alternator creates current, converting mechanical energy into electrical energy


duh


The battery stores electrical energy.


once again, duh, but more specifically it stores chemical battery acid
cells that a connected in series in order to make 12 V by seperating a
difference in charge.


The battery and the alternator are in a parallel circuit. They do not operate
independently of each other.


really? I didn't know (sarcastic)


Then why have you been repeated claiming thus far that the battery is always
powering the amps because they're connected to it directly while the alternator
is powering everything else?

Nevermind... I'm through with you. You're too thick.
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Mariachi Mariachi is offline
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Default Replace positive wire on HU?

On Feb 28, 12:19 am, Matt Ion wrote:
Mariachi wrote:
The alternator creates current, converting mechanical energy into electrical energy


duh


The battery stores electrical energy.


once again, duh, but more specifically it stores chemical battery acid
cells that a connected in series in order to make 12 V by seperating a
difference in charge.


The battery and the alternator are in a parallel circuit. They do not operate
independently of each other.


really? I didn't know (sarcastic)


Then why have you been repeated claiming thus far that the battery is always
powering the amps because they're connected to it directly while the alternator
is powering everything else?

Nevermind... I'm through with you. You're too thick.


Yes, I was wrong, but I did ask you a question in which you didn't
answer at all, instead you completely avoided the question altogether
and somehow got the impression that I didn't believe you the first
time after you corrected me... I asked

If the alternator doesn't supply enough
amperage to power up your stereo system, doesn't it take excess energy
from the battery to compensate for the alternator... Does it solely
depend on the amperage that the alternator is putting out... or both
the battery and the alternator. So if my amplifier requires 30 amps
and my alternator only supplies 12 amps at the moment... doesn't the
battery supply the other 18 amps?


instead you decided to criticize me again (acting like I'm thick-
headed, and in some cases i am) and again decided to reinforce the
fact that the alternator provides all the current without no help from
the battery even when the current demand goes over the limit of the
alternator (never answering my question). Anyways, sorry for the
confusion... i don't like to argue, i just like to learn... arguing
turns me off.



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Matt Ion Matt Ion is offline
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Default Replace positive wire on HU?

D.Kreft wrote:
On Feb 27, 8:54 am, Matt Ion wrote:


When the car is running, the alternator provides ALL the car's electrical needs,
including for the sound system. The battery DOES NOT.



Correct...right up until the point where the current demand from the
vehicle (presumably from a big honkin' stereo system) exceeds the
output capability of the alternator, at which point the battery starts
converting that yummy chemical energy into electrical engery...


Ah, yes, granted... but as you say...

only makes matters worse because then the battery becomes an even
*larger* load on the alternator as it tries to not only run your
electronics, but also recharge the battery.


....ultimately the alternator is the only thing that is really CREATING
electricity in the system - the battery is merely storing it.

Anyway, we could go around forever coming up with rare or momentary or
intermittant scenarios where different things happen like this... but that's
just confusing the issue with theoreticals and rhetoricals and gets away from
the basic fact that...

As Matt has spelled-out (several times now), it really is this simple.


Yup.
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