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


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


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

  • \$\begingroup\$ Why you want to use an array of very low power speakers rather than, say, a single 100W speaker or two 50W speakers? \$\endgroup\$
    – Finbarr
    Jun 21 '19 at 11:09
  • \$\begingroup\$ @Finbarr The reason is that the target sound field can only be generated by multiple loudspeakers together, not by only 1 or 2 speakers, though they may have the same power in total. \$\endgroup\$
    – Zhang Ze
    Jun 21 '19 at 11:25
  • \$\begingroup\$ What type of signal would you like the play. Sine waves or music or something else. The "optimal" solution is very different for sine waves than for music \$\endgroup\$
    – Hilmar
    Jun 21 '19 at 15:46
  • \$\begingroup\$ @Hilmar I want to generate sin waves where loudspeakers will have the same phase. And if only 1 or 2 loudspeakers are used, the filed will be 'point-source' like. \$\endgroup\$
    – Zhang Ze
    Jun 24 '19 at 9:28

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\$ VRMS.

The amplifier produces up to 15 VRMS, 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 ARMS, so you can have up to 5 A/.47 A = 10 strings before you overload that particular amplifier.

  • \$\begingroup\$ Hi, thanks for the answer. If I understood correctly, you are suggesting to use 8 parallel lines with 4 speakers in series per line if I want to use 32 loudspeakers? I think the wiring of this is easier and the same power can be achieved as from @Bimpelrekkie 's answer. \$\endgroup\$
    – Zhang Ze
    Jun 21 '19 at 12:08
  • \$\begingroup\$ Yes, and this makes it easier to vary the number of speakers being used -- in increments of 4 at a time, of course. \$\endgroup\$
    – Dave Tweed
    Jun 21 '19 at 13:13
  • \$\begingroup\$ The optimum solution is 4S 10P =30 speakers for Impedance May aching and MPT using the most speakers and not exceed source or load power ratings. However THD is another matter vs Zeq. \$\endgroup\$ Jun 21 '19 at 15:41

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.

  • \$\begingroup\$ Hi, thanks for the clear answer! The third way (the hybrid connection) seems feasible. My only concern of this is can this connection ensure the each loudspeaker has the exactly or very close to same phase of signal? Since I want to use the array to generate signals of the same phase at one moment. \$\endgroup\$
    – Zhang Ze
    Jun 21 '19 at 10:06
  • \$\begingroup\$ Besides, I also found there is a parameter for the power amplifier called 'gain at 1kHz', which is 40 dB. Is this something that I can use to boost up the power of the amplifier for the serial connection. Just from a practical sense, the serial connection is easier and less error-prone. \$\endgroup\$
    – Zhang Ze
    Jun 21 '19 at 10:09
  • \$\begingroup\$ The current through a speaker actually determines the position of the speaker cone and since the speakers will all have the same current (due to how they're connected) the will all have the same phase. So no worries about that. \$\endgroup\$ Jun 21 '19 at 10:14
  • \$\begingroup\$ That gain at 1kHz is just a "general property" so that we know how much the gain the amplifier has. They just measured it at 1 kHz. If it is a decent amplifier the gain at 100 Hz or 10 kHz will also be 40 dB. If you want to use all speakers in series, this is not the right amplifier, then you need one which can deliver a much higher voltage. \$\endgroup\$ Jun 21 '19 at 10:16
  • \$\begingroup\$ Okay. Based on your recommendation, I did the calculation and found out that max output voltage will be reached for the amplifier (not the max current), and each loudspeaker will have a power of 1.76 w which is fair. Thanks! The only thing here is it's too much wiring (50 cables?). Do you have any other ways except for changing the power amplifier? \$\endgroup\$
    – Zhang Ze
    Jun 21 '19 at 12:00

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.

  • \$\begingroup\$ That isn't matching impedances -- the output impedance is (at most in the spec) 0.08 Ohms. The minimum supported load impedance is 3 ohms. \$\endgroup\$
    – esilk
    Jun 21 '19 at 15:47
  • \$\begingroup\$ I stated MPT for max power rated (THERMAL LIMITS) not the source impedance, which is due to current distortion heating and more important damping factor, so 3 Ohms is matching to MPT for the above stated reasons which gives a damping factor of 3.2/0.08= 400 which is excellent. See the difference? \$\endgroup\$ Jun 21 '19 at 15:52
  • \$\begingroup\$ Sorry that I do not have an electrical engineering background. What is MPT? \$\endgroup\$
    – Zhang Ze
    Jun 24 '19 at 9:30
  • \$\begingroup\$ Max power transfer (MPT) for rated power uses impedance matching methods \$\endgroup\$ Jun 24 '19 at 11:36

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