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In some parts of the U.S., safety regulations regarding conduit
use became stricter, forcing distributed systems to adopt a 25 volt
rms standard. This saves conduit, but adds considerable copper cost
(lower voltage = higher current = bigger wire), so its use is restricted
to small installations.
Calculating Losses — Chasing Your Tail
As previously stated, modern constant-voltage amplifiers either
integrate the step-up transformer into the same chassis, or employ
a high voltage design to direct-drive the line. Similarly, constant-voltage
loudspeakers have the step-down transformers built-in as diagrammed
in Figures 2 and 3. The constant-voltage concept specifies that
amplifiers and loudspeakers need only be rated in watts.
For example, an amplifier is rated for so many watts output at 70.7
volts, and a loudspeaker is rated for so many watts input (producing
a certain SPL). Designing a system becomes a relatively simple matter
of selecting speakers that will achieve the target SPL (quieter
zones use lower wattage speakers, or ones with taps, etc.), and
then adding up the total to obtain the required amplifier power.
For example, say you need (10) 25 watt, (5) 50 watt and (15) 10
watt loudspeakers to create the coverage and loudness required.
Adding this up says you need 650 watts of amplifier power –
simple enough – but alas, life in audioland is never easy.
Because of real-world losses, you will need about 1000 watts!
Figure 4. Transformer & Line Insertion Losses Figure 4 shows
the losses associated with each transformer in the system (another
vote for direct-drive), plus the very real problem of line-losses.
Insertion loss is the term used to describe the power dissipated
or lost due to heat and voltage-drops across the internal transformer
wiring.
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This lost power often is referred to as I2R losses, since power
(in watts) is current-squared (abbreviated I2) times the wire resistance,
R. This same mechanism describes line-losses, since long lines add
substantial total resistance and can be a significant source of
power loss due to I2R effects. These losses occur physically as
heat along the length of the wire.
You can go to a lot of trouble to calculate and/or measure each
of these losses to determine exactly how much power is required3,
however there is a Catch-22 involved: Direct calculation turns out
to be extremely difficult and unreliable due to the lack of published
insertion loss information, thus measurement is the only truly reliable
source of data. The Catch-22 is that in order to measure it, you
must wait until you have built it, but in order to build it, you
must have your amplifiers, which you cannot order until you measure
it, after you have built it!
The alternative is to apply a very seasoned rule of thumb: Use
1.5 times the value found by summing all of the loudspeaker powers.
Thus for our example, 1.5 times 650 watts tells us we need 975 watts.
Wire Size – How Big Is Big Enough?
Since the whole point of using constant-voltage distribution techniques
is to optimize installation costs, proper wire sizing becomes a
major factor. Due to wire resistance (usually expressed as ohms
per foot, or meter) there can be a great deal of engineering involved
to calculate the correct wire size.
The major factors considered are the maximum current flowing through
the wire, the distance covered by the wire, and the resistance of
the wire. The type of wire also must be selected. Generally, constant-voltage
wiring consists of a twisted pair of solid or stranded conductors
with or without a jacket.
For those who like to keep it simple, the job is relatively easy.
For example, say the installation requires delivering 1000 watts
to 100 loudspeakers. Calculating that 1000 watts at 70.7 volts is
14.14 amps, you then select a wire gauge that will carry 14.14 amps
(plus some headroom for I2R wire losses) and wire up all 100 loudspeakers.
This works, but it may be unnecessarily expensive and wasteful.
Really meticulous calculators make the job of selecting wire size
a lot more interesting. For the above example, looked at another
way, the task is not to deliver 1000 watts to 100 loudspeakers,
but rather to distribute 10 watts each to 100 loudspeakers.
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