Interview to two driver in parallel,
Peter Strassacker asked Joe d'Appolito (7/2004).
If I have a woofer with 90 dB SPL (2,83V/1m) and I add a second in parallel
(distance less than 1 wavelength), I measure 96 dB. So the input power doubles
but SPL goes up 6 dB what means an increase of factor 4 in power. Here an example
with two speakers of 10 inch; the diagramm shows the increase in dB.
pressure gain of
two 10 '' drivers
with a distance
of about 15 ''
compaired to one.
The simple answer is that due to mutual coupling between closely spaced
diaphrams, the effective radiation impedance is also doubled. Another way to
say this is that the efficiency of the combined drivers has doubled and the
total radiated power goes up by 4 or 6dB. This only occurs when the driver
spacing much shorter than the wavelength of the sound they are producing and
the drivers are essentially omnidirectional.
This increase in efficiency occurs only at low frequencies. At higher
frequencies the situation becomes more complicated.
If we consider the two
drivers oriented vertically, then pressure will double at every point in the
horizontal plane passing through the midpoint between the drivers. This is
clear from the priciple of superposition. If each driver produces a
pressure, P, then at any point on this plane the pressure from each driver
will be in phase and will add to 2P or 6dB. However, above or below this
plane the situation is quite different. Looking at a point above the plane,
for example, the distance to that point from the upper driver will be
shorter than the distance to that point from the lower driver. So there will
be a phase difference between the two signals arriving at that point and the
pressures will now add up to something less than 2P. In fact points will
exist where the two arriving signals are completely out of phase and the
pressure will be zero. The actual radiated power is now a complex function
of frequency, driver spacing and driver polar response.
Thank you Joe.