What you really are asking is when to I need to use heavier Power/Ground
wires?

It all comes down to Volt/Amp requirements of a particular system setup.
There is also one other caveat. You do not want to **starve** your audio
system for its power. There is an equation that you might want to keep in
mind when sizing your wiring:

Power out (Watts) = Power in (Watts) - Losses (Watts) ; Losses being the
one that always trips everybody.

No amplifier is ideal wherein all power into it comes out of it and no wire
is ideal with zero resistance/ft. So, most class AB amplifiers are only
about 50% efficient and all wire exhibits some resistance/ft. When should
a Power or Ground wire size be increased? Simple, when the calculated or
measured voltage drop between the amplifiers terminal and the battery's
terminal is greater than .25 Vdc. The .25 Vdc figure is the generally
agreed upon industry standard.

When you calculate the voltage drop for a wire, length of wire now becomes
an issue. Remember wire has resistance/ft. So now another equation is
needed, V= IR. Where V is the voltage difference for one end of the wire
to the other. I is the **maximum** current through the wire. R is the
total resistance in the wire. To find I, we must now calculate the maximum
input power requirements for an amplifier. If a stereo amplifiers maximum
output is 500 Watts/channel, then the total output for this amplifier is
#channels x 500W, which equals 2 x 500W or 1000W. This is the amplifiers
total power output, not its input power (which comes from the power source).
The input power is equal to the amplifiers total output power times the
reciprocal of its efficiency. So if this amplifier is 50% efficient, then
its power input requirements will be 1000W x 1/.5, or 2000W! We can now
calculate the amplifiers current (I) requirements with the equation P=VI,
where P is the amplifiers power requirement, V is the source voltage(
usually a 12V battery) and I is the current draw from the voltage source
(battery). We rearrange the last equation to read I=P/V. Now, I = 2000W/12
V. This gives an current draw of 167 Amperes. We can now solve for the
maximum **total resistance** for the Power/Ground wire run. We know the
voltage drop, .25V. We know the maximum current requirement, 167A. So,
..25V=167A x R which reduces to R=.00149 Ohms or 1.49E-3 Ohms. BTW you
cannot measure this resistance with a standard multimeter. It is far too
low a value.

We can now determine what wire size is appropriate for our application. If
we know the length of our wire run in feet and the total resistance of this
run, we can now consult a table or chart of "resistance per foot for
standard wire gauges". Look around, they exist. Or, call your wire
supplier. For this application we would be looking for a Ohms/ft of 1.49E-3
Ohms / length of run or **lower**. I stress lower because your caluculated
value might not be exactly that of a standard wire gauge.

If you use your chassis as the return ground, you will not be able to
empirically determine the total resistance for this run.

Also, you must calculate this value for every amplifier in your system. And
if you intend to use only a single wire run from the battery to a terminal
distribution block, then this wire run will have to have a resistance/ft of
the summation of all individual wire run resistance in parallel. This is
given by the formula of 1/Rt = (1/R1+1/R2+1/R3+......1/Rn), where Rt is the
total effective resistance and R1 thru Rn are the individual total
resistance of the individual wire runs.

Hope this helps!

Taffer Garrett


"TWEVOLTMAN" wrote in message
...
with all the hype about large guage power wire i have a few questions.

tell me why i would need heavy guage power wire

tell me why i should switch from 8 guage to 4 guage

if i'm not blowing 40a fuses then in reality the 4 would be a waste, right

how many have really checked the amps while running your system

40 amps is alot of power, my home theatre is only on a 15 amp breaker.


just some good answers would help