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One relic of the tube days, though, is the relatively common misconception tha
output transformer on any amplifier that drives a CV line. In fact, given the sta
t you must use an
te of modem power
a
tage distribution systems.
ier’s
, let’s
lop 70 volts directly.
s,
V
iers of this power class are today quite common and relatively inexpensive. It
r is specifically
the
velop sufficient voltage.
f amplifier we would need to develop 25 volts RMS across a CV line:
CV
amplifier technology, a transformer is not necessarily required. In this article, we will explore
number of alternative methods for driving constant-vol
Amplifier Bridging
One of the primary reasons that output transformers are often employed is to boost an amplif
output voltage to the CV line’s standard operating level. To see why this would be necessary
consider what kind of amplifier we would need to deve
A derivative of Ohm’s Law states that power equals voltage times current: P = E*l
Substituting E/R for current (again in accordance with Ohm’s Law), we get: P = E
2
/R
To get the 8-ohm rating of a power amplifier capable of 70 volts RMS output:
P = (70 volts)
2
/8 ohms=about 600 watts
This is more power than most CV systems would require. Furthermore, an amp that is rated at
600 watts per channel would be expensive for a smaller distributed system installation.
The common solution has been, of course, to use dual-winding step-up transformers at the
outputs of an amplifier with a more modest rating. Such transformers carry their own limitation
however. Especially with the advent of foreground music systems, the power requirements for C
systems have risen somewhat — but output transformers capable of significant power handling
are large, expensive and heavy. They may also exhibit limited low-frequency response and might
impose substantial insertion losses.
How, then, does one deliver 70 volts to the CV line? One way is to use bridging. Because a
bridged power amplifier drives the load push-pull, the voltage across the load is effectively double
that of each channel’s individual output. If we need to develop a total of 70 volts across the line,
then, we can use an amplifier, which delivers half of that, or 35 volts, for each channel.
Using the previous equation:
P = E
2
/R = (35 volts)
2
/8 ohms = about 150 watts
Professional amplif
is important to note that amplifier bridging should only be done when the amplifie
designed for this purpose.
The manufacturer’s instructions should be followed when using this technique. You must take
amplifier’s bridged, 8-ohm power rating as the reference maximum power figure for loading
calculations. Remember that the minimum load impedance for a bridged power amplifier is twice
that of each individual channel. The bridging power figure will be specified accordingly, and may
also be de-rated to reflect performance limitations of the amp’s power supply circuitry. For this
reason, a 200 watt per channel amplifier (or thereabouts) may actually be required, even though
the 150-watt amplifier could de
Single-Channel Direct Drive
As we have seen, the 70-volt line standard discourages direct drive from a single amplifier chan-
nel, if for no other reason than economy. But the same is not necessarily true of the 25-volt
standard. As above, we can use the form of Ohm’s Law that solves for power to ascertain what
kind o
P =E
2
/R=(25 volts)
2
/8 ohms=about 75 watts
Certainly, a 75 watt-per-channel professional power amplifier is well within the range of many
system budgets. Given the state of contemporary amplifier technology, there is no reason why an
amp of this power class could not drive two 25-volt branches directly.
ElectroVoice/Dynacord BGM Guide Page 38