How to match a power amplifier and a loudspeaker array?

Question

I am trying to use a power amplifier that is available to me to actuate an array of loudspeakers. The array has a total number of 30 loudspeakers (ways of connection can vary).

And I would like to know how I should connect the loudspeaker array to the amplifier to make sure there's enough power for the array and also not overload the array. Below are some parameters of the amplifier and the loudspeaker. Do you think serial connection in this case will be fine?

If possible, would you please show the chain of thought to reach the conclusion? So will help me when I want to change the no. of loudspeakers in the future.

Power amplifier:

• Power output capacity: 75 VA into a 3 ohm resistive load
• Input impedance: 15 kohm
• Output impedance: 0.04 ohm: [10Hz-5kHz], 0.08 ohm: [5kHz-20kHz]
• Max output current: 5A or 1.8A rms according to selected value
• Max output voltage: 15 V RMS

Loudspeaker

• Input power: 2W (rated), 3W (max)
• impedance: 8 ohm

Array

The array will be composed of the loudspeaker above with a total number of 30.

Answer 1

Your speakers are rated at 2W and 8Ω, which means that they can take a maximum of \$\sqrt{P\cdot R} = \sqrt{2 \cdot 8} = 4\$ V_RMS.

The amplifier produces up to 15 V_RMS, so you need to connect the speakers in strings of 4 in series across the amplifier terminals in order to make sure that the speakers are not overdriven. You can put any number of such strings in parallel.

Each string consumes 15 V/32 Ω = 0.47 A_RMS, so you can have up to 5 A/.47 A = 10 strings before you overload that particular amplifier.

Answer 2

Power output capacity: 75 VA into a 3 ohm resistive load

So your power amplifier can deliver the highest power into a 3 ohms load.

If you would connect all speakers in series you would get 30 x 8 ohms = 240 ohms, then the maximum power to the speakers is limited by the voltage that the amplifier can make. This is about 15 V, into 240 ohms that means P = U^2 / R = 15^2 / 240 = 0.94 Watt which is not much.

If you would connect all speakers in parallel you would get 8 ohms / 30 = 0.27 ohms, then the maximum power to the speakers is limited by the current that the amplifier can deliver. This is about 5 A, into 0.27 ohms that means P = I^2 * R = 5^2 * 0.27 = 6.8 Watt which is more than the 0.94 Watt but still not much. Also, this quite low impedance of 0.27 ohms is bound to cause problems, the amplifier simply might not like it.

Your best bet is to combine series and parallel to get close to 3 ohms. My guess is that 4 Ohms would be good enough as well and somewhat easier as two 8 ohm speakers in parallel equals 4 ohms. Then use series and parallel again to add more speakers but keep a total of 4 ohms, like so:

^{simulate this circuit – Schematic created using CircuitLab}

There are 8 speakers, between the top and bottom there's 4 ohms.

You can combine 4 of these in series and parallel and then you'd need 32 speakers.

If you must have 30 speakers maybe you can use 5 x 6 = 30 meaning make sets of 5 speakers in series, connect 6 of those sets in parallel, that would result in a total of 8 ohm * 5 / 6 = 6.7 ohms.

Answer 3

Another approach is to match impedances for maximum power transfer out without overdriving all your speakers.

If MPT impedance is 3 Ohm load ( due to thermal IC limits) and you have 30 x Zo=8 Ohm loads, that would wish to arrange as a 3 Ohm load, then you have several options.

For an array of xSyP where x is the qty in a series string and y is the qty of parallel strings;

\$Z_{equiv.}=3<= \dfrac{x}{y} \cdot Zo\$ where x*y=30 and Zo=8

Solve this :
4S10P=3.2 Ohms which is > 3 Ohms and gives you closest to MPT with a 75VA source gives you 75VA/30 = 2.5 VA per speaker below its MAX limit of 3W peak and rated limit of 2Wrms.