[Mariachi]

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?

[D.Kreft]

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

[Mariachi]

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...

[Captain Howdy]

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

[Mariachi]

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.

[Mariachi]

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

[Matt Ion]

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

[D.Kreft]

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

[Captain Howdy]

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

[D.Kreft]

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