Altec Lansing 9441A - Manual (Page 4)

Operating and Service Instructions for the Altec Lansing 9441A Power Amplifier
WHITE (A.a HECEPTAia^
X I ^
BIACK BUWHT BLUE
nV/HT BROWN
A. 120 V ac Primary Wiring
WHITE (A.C RECEPTACLE)
8.240 V ac Primary Wiring
BLACK BLIWirr BLUE
_______________ BRIWHT BROWN
Figure 2 Primaiy Wiring Configuration for 120 V ac and 220/240 V ac
SIGNAL CONNEC­
TIONS
3.1 Input Connection.s
Balanced input connec­
tions may be made to either the
V a
phono (TRS) or the female
XLR connectors. For single-ended
inputs, strap the low (—) input to
ground (pin 3 on XLR or
Ring on
V a " phone). Otherwise, the elect­
ronically-balanced input stage will
see 6 dB loss input signal level
than with a balanced input. Refer
to Figure 3 for typical input con­
nections.
3.2 Line Output Connec­
tions
TliD XLR and
Vi" phone
connectors are wired in parallel.
Pin 2 of the XLR is the Tip of the
V a " phono comiector, and pin 3 is
the Ring. Since the input imped­
ance of the electronically-balanced
input stage is high (15 kohms),
there is minimal loading on the
signal source, Wien the input
connections are made to one con­
nector, the other may be used as
an auxiliary line output to feed
other high input impedance equip­
ment. Refer to Figure 3 for po.ssi-
ble applications.
3.3 Output Connections
Output connections are
made to the four terminal barrier
strip connector located on the rear
of the unit. Refer to Figure 4 for
tj'pical output connections.
3.4 Output Cable Selection
Speaker wire size plays an
important part in quality sound
systems. Small wire gauges can
waste power and reduce the
damjiing factor at the speaker
terminals. This can add coloration
and muddincss to the sound. To
help offset this problem. Table I
has been assembled to enable you
to calculate the power losses in
the speaker cable.
3.4.1 Calculating Power
Losses with 8 ohm
Loads
To calculate the total
power loss in the spealcer cable,
multiply the power loss per foot
(or meter) of the 2-wire cable sel­
ected from Table 1 by the length
of the cable in feet (or meters).
For example, suppose an installer
uses 160 feet of 10 GA 2-wire
cable with an 8 H speaker system.
The total po%ver loss in .U. cable
is;
Total Power Loss in cable
= 0.0191 watts/foot
X
lOr feet
= 3.0 watts
Does this mean that whena er the
amplifier produces 75 walls of
output power, 72.0 watts (75 watts
minus 3.0 watts) will be delivered
to the 8 ohm load? N'^d The
actual load impedance is ohms
plus the resistance of Iht cable
(0.00204 ohms/foot times L'"' feet)
for a total load imped.nice of
8.3264 ohms. At the 8 lated
output power, the output, /oltage
is 24.4 V rms. Therefore, I t ■ amp­
lifier produces
71.5 wati - with
this load instead of 75 wa.‘ . This
was calculated by squaring the
voltage and dividing by I load
impedance (24.4* divid. i by
8.3264 ohms). As a result, ue act­
ual power delivered to thf !-ad is
68.5 watts (71.5 watts minus 3.0
watts).
Had 18 GA wire been used hi the
above example, the lass in the
cable would have been Ifl.r ivatts.
Tliis example lllustrale.s I im­
portance of using the pro; . wire
size.
3.4.2 Calculating <wer
Losses with 4 ohm
Loads
To calculate the- i sses
when using a 4 ohm speak .-; os­
tein, multiply the loss at hms
by 3. In the above exampk-. the 10
GA wire would consume 9,'' watts
of power while the 18 GA '.■.•ire
would waste 58.5 watts
more
than half of the amplifier’’^ ohm
power rating.
3.5 Damping Factor
The higher the da aping
factor rating of an amplill'-!' the
greater the ability of the am; lifier
A1.TEC lANSING' CORPORATION * a Mark T\' Company