I've always been a bit mystified about how putting two speakers in parallel actually increases their efficiency. How could you actually change how efficient a linear motor is by changing the input.

Anyway, I thought about it the other day. Wayne or others, please correct me if I'm wrong.

Here's how I think it goes, hooking up two motors in parallel does NOT increase the efficiency of the system itself.

Example: One nominal 8 ohm driver being driven at 1 Watt.

Solving this simple system of equations(ohms law) you get:
(W=VI) 1 = 2.83 * .354
(V=IR) 2.83 = .354 * 8
1 Watt, 2.83V, .354A

If you connect two drivers in parallel you now have a 4 ohm device.

Solving this simple system of equations(ohms law) you get:
(W=VI) 1 = 2 * .5
(V=IR) 2 = .5 * 4
1 Watt, 2V, .5A

Now you have 1 Watt flowing through two parallel motors. Meaning they both experience the whole 2V, but each gets half the current(thus producing half the sound output each). Totaling to still... 1 Watt.

The Reason we always say that two drivers in parallel doubles the efficiency (+3db) is simple. At a constant preamplifier level, if you connect a second 8 ohm device in parallel(total of 4 ohms), roughly twice as much wattage will flow through the system(depending on amplifier design), giving you twice the output SEEMING like the system is twice as efficient.

Though each driver still has the same efficiency, your output is now over a greater area, moving away from a point source type configuration (1/R^2 falloff rate) and getting closer to a plane (no falloff) or a line (1/R falloff rate). When standing back a bit, this may increase apparent efficiency even more.

Am I making something more complex than need be? This seems right, but is it?

First off, I'm not sure where you get the numbers in your equations. Your first one uses 2.83 volts, the second one uses 2 volts. The two equations aren't identical on any levels, they should only change in the impedance category which means resistance is halved.

In any event, I'll find the link where I asked a similar question and it was answered. Also, remember that we gain 3 db effeciency because the cone area doubles.

Adam

---the system of equations
Watts = Current * Volts
Volts = Current * Resistance

---expanded solution
so Watts = Current^2 * Resistance

so Current^2 = Watts / Resistance

so Current = sqrt(Watts / Resistance)

since Volts = Current * Resistance
Volts = sqrt(Watts / Resistance) * Resistance

so Volts = sqrt(Watts * Resistance)

so 2.83V = sqrt(1W * 8) for 8 ohms.
2V = sqrt(1W * 4) for 4 ohms.

If you measure sensitivity at 2.82V/1M, then the Voltage will be the same and Wattage will be different.

Adam

If you are talking "efficiency" then that would be the acoustic power vs. the electrical power applied to the driver(s) and would probably best be expressed as a percentage - generally less (way less?) than 10% I think.

"Sensitivity" would be the Sound Pressure Level (SPL) generated by a particular input voltage. Most speakers are actually rated for sensitivity such as 98db @ 2.83v rms @ 1 meter. Of course the 2.83v level is used because it corresponds to an actual 1 watt when driving an 8 ohm load; it just makes a nice standard.

The amplifier puts out a _voltage_ signal identical to the input except multiplied by some factor - a constant voltage source like Adam said. Adding another driver (hopefully) doesn't change this. Since each driver is now seeing the 2.83v, each driver is drawing 1w and the amp is delivering 2w total.

We doubled the power being delivered and therefore added 3db to the sensitivity of the system.

So we arrived at basically the same conclusion up to this point.

We also doubled the radiating surface area. That is... our 2w is not moving one driver with x square inches of surface area but two drivers - 2x surface area. So we are moving twice as much air - right? That means that we are delivering twice as much acoustical power into the room (as compared to putting 2w into one driver). Twice the power being delivered into the room - add 3db again.

So adding one more driver adds 6db to the "sensitivity" of the system at the cost of dropping the impedance to half of what it was before. So nominal 8 ohm speakers with a 4 ohm minimum become 4 ohm speakers with a 2 ohm minimum.

(And my decrepit receiver then emits a wonderful aroma into the room... Doh!)

Did that help? I'm sure if I were a slow typist I could have fit this into fewer words. Ha.

Dan Ferguson

but if we want to calculate max SPL, amps are rated at a certain Wattage RMS at a certain Ohm resistance.

If we have a typical amp that puts out 100W at 4 ohms and 50 W at 8.
100W into 4 ohms is 5 amps and 20V.
50W into 8 ohms is 2.5 and 20V.

wow... well crap. the voltage is the same even at max output with different impedances, never thought about that. hmm... and at 20V the two parallel dirvers are each usign 50W... giving twice the efficiency..

don't I feel sheepish, I never really thought about how amps are limited.

I feel a little bit better that this doesn't defy any laws of physics, the speakers aren't anymore efficient in terms of pure power per output, but they make use of the amp circuitry more efficiently.

Thanks bud,
Danny

Adam

"Compact, powerful reinforcement systems, the 4662A and 4663A offer high efficiency, vivid, natural sound (even at very high levels), and a controlled dispersion pattern for indoor or outdoor reinforcement applications. These systems are capable of high acoustic output, can handle large amounts of power, and are designed for reliability and durability. Outdoors, with no support from room acoustics, either system will produce an impressive 117 dB at 3 m (10 ft) when driven at its rated power of 150 watts continuous pink noise. Where higher sound pressure levels are desired, systems may be paired to produce up to 6 dB more SPL than a single unit, with consequent improvement in peak power capability". The "close coupled" phrase is from a similar blurb on their site.......