LONG SPEAKER CABLE RUNS
WASTE POWER AND REDUCE PERFORMANCE
By Marty McCann
Recently I had a customer who had an application where he had
to run a loudspeaker line 1500Ft. I suggested that he run a
70-volt distributed line instead. Since he didn't understand
70-volt lines he opted to do it his way, so I did a little
math regarding speaker line losses, here are the facts:
Power in Watts equals the voltage squared, divided by the resistance
of the load (V x V / R):
40 x 40 = 1600 1600 / 4 ohms = 400 Watts 1600 / 8 ohms = 200
Watts
Power in Watts also equals voltage times the current (W = V
x I). It is more difficult to measure the current flowing through
a device; however voltage and resistance can be determined
or measured with a simple inexpensive multimeter. So, if you
want to know the current that is present in the circuit, you
can simply measure the voltage and resistance and calculate
the power from the first formula, and then use the second formula
to calculate the current present in the circuit by dividing
the power in watts by the voltage to find the current (I =
W / V). 400 Watts divided by 40 volts equals 10 Amperes of
current flow (400 / 40 = 10).
The following charts for the wire gauges are based on an amplifier
producing 40 volts, which would normally result in 400 Watts
of power at 10 amperes of current into a 4 ohm load, and into
an 8 ohm load the same 40 volts would produce 200 Watts of
power at 5 amperes of current flow. Using A.W.G. # 18 gauge
wire with 153.6 feet per ohm of resistance as an example, if
you had 1500 feet of the wire, the resistance value would be
1500 divided by 153.6, which equals 9.766 ohms of resistance
(1500 / 153.5 = 9.766). But that is only the resistance of
one of the two conductors in the speaker cable, in other words
there is 9.766 ohms of resistance in each wire running to the
speaker, or a total of 19.53 ohms of resistance in both conductors
(9.766 x 2 = 19.53).
The two conductors are essentially in series with the resistance
of the loudspeaker. So in the case of a 4 ohm speaker and 1500
feet of # 18 gauge wire, there is 23.53 ohms of resistance
in the circuit that the power amplifier would be driving (19.53
+ 4 + 23.53). Now if that amplifier if putting out 40 volts,
the power produced in the circuit is 67.99 Watts (40 x 40 =1600
1600 / 23.53 = 67.99). If we round that off to 68 Watts, the
current in the circuit would be determined by dividing the
power in watts by the voltage, which would equal 1.7 amperes
of current flow (68 / 40 = 1.7).
Using another version of ohms' law, if you know the resistance
and the current, you can calculate the power dropped across
a resistance by multiplying the voltage times the current squared
(R x I x I). The power that actually reaches the loudspeaker
would then be 1.7 times 1.7, times 4, or 11.56 Watts (1.7 x
1.7 x 4 = 11.56). Rounding off, we would have about 12 watts
actually getting to the loudspeaker. The amount of power lost,
or consumed by the cable would then be the total power produced
by the amplifier, minus the power that reached the loudspeaker,
or 54 Watts (68 – 12 = 54). I hope I haven't hurt anyone's
brain here with the 4th grade math, but I needed to show how
the calculations were derived. The rule of thumb for professional
sound reinforcement is to use the shortest possible loudspeaker
run and a heavy gauge wire. Check out the wire chart that follows,
then read on to learn more about the consequences of using
too small of a wire gauge.
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