Zed Audio Draconia Instruction & Installation Manual - page 7
Zed 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 a medium size
amplifier such as
Draconia. Let us assume that we are playing it such that all four
channels(into 4 ohms each) are just clipping on the loudest musical peaks. This means
that we are delivering 300 watts on peaks. The amplifier’s average efficiency is about
45%. The peak to average power ratio is about 20% so average power is 20% of 300 = 60
watts. The input power is therefore 133 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 0.54 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 0.54 seconds! So what good is a 1 Farad
capacitor? Assign the above example to a
Megalith playing into a 1 ohm load. Going
through the mathematics the 1 Farad capacitor would keep the amplifier playing for 0.14
seconds. So if the voltage at the amplifier’s power terminal dropped by say 1volt the
amplifier’s power supply will correct for this on a continuous basis. The stiffening
capacitor however has nowhere close to the energy reserve to compensate for a 1 volt
drop.The main internal power supply capacitors hold 100 Joules of energy and these are
more effective than any stiffening capacitor since they do not have to contend with the in-
efficiencies of the amplifier’s power supply.
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