Z
ed does not recommend the use of power distribution blocks for the purpose of
distributing the +12volt voltage to several amplifiers. The reason is that the vehicle’s
battery is the lowest AC impedance point in the power grid of the vehicle. We want each
amplifier to draw it’s current from this low impedance point. Thus any modulation on any
+12v power cable (which is inevitable) is then shunted to ground by the massive
capacitance of the battery. This is the reason that “star” grounding is used in grounding
circuits/equipment so that ground current is drawn from a common point and thus no
ground loop can occur. Fortunately for us, the body of a vehicle made of steel is so large,
and is thus a very low impedance path for ground currents, that it is not necessary to
ground all equipment at one point. In fact we do not advocate it at all as this would then
necessitate the head unit’s ground running all the way to the battery location and the
amplifier’s ground(s) also running all the way to the battery.
If multiple amplifiers are being used we highly recommend the use of separate ground
points at the amplifiers’ location. This spreads out the amount of current being drawn
through one bolt connection.
tiffening capacitors - These are of NO use with our amplifiers due to the fact that the
S
amplifiers have fully regulated power supplies. The power supplies will compensate
for small volt drops which exist on the +12v power cable. The amount of current drawn by
a particular amplifier would drain a fully charged 1 Farad capacitor almost instantly.
Consider the theory. Energy(Joules) = Power(Watts) x Time(Seconds). The energy in a 1
Farad capacitor = 0.5CV.V
= 0.5x1x12x12= 72 Joules. Let us assume an amplifier such
as
Let us assume that we are playing it such that the channels(into 4 ohms
each) are just clipping on the loudest musical peaks. This means that we are delivering
500 watts on peaks. The amplifier’s average efficiency is about 60%. The peak to
average power ratio is about 10% so average power is 10% of 500 = 50 watts. The input
power is therefore 83 watts. If the 1 Farad capacitor was charged to 12 volt and we
remove the main source of power -- the battery, the amp would remain playing for 0.86
seconds! (Put the numbers in the formula (E=PxT above and solve for time T). Now
compare this to the battery. The amplifier will play for some hours (depends on actual
battery of course) as compared to 0.86 seconds! So what good is a 1 Farad capacitor?
KRONOS.
Page 6
