I'm building an active speaker and would like the audio in signal to be replicated to another speaker, in this way being able to create a chained system with several speakers. I'm afraid a direct "Y" cabling at the speaker may damage the audio source equipment since the the output impedance would decrease with the added speakers... What solution would you recommend? Is there a small circuit board I can use?
2 Answers
Amplifier source impedance is defined by the damping ratio (e.g. 50 to 1k) divided into the rated speaker ohms.
So power sharing may be done in series or parallel. But at max power, the THD may be better in one or the other depending on the voltage range or the current limit. This cannot be predicted until you test it or have specs for these properties.
So it is not about matching impedances per se, rather to choose the maximum power and THD tradoff configuration. Thermal losses also depend on the topology with max VI drop across the driver depends on output lelvels.
Bass tends to consume a higher current but is often common to both channels. Your best performance is with heavy high braid cables but also split from the source. If max power or THD for high fidelity is not so critical, then daisy-chained parallel operation is likely your 1st attempt.
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\$\begingroup\$ Thank you Tony, I will go to daisy-chaining \$\endgroup\$– FarukAug 13, 2021 at 11:27
Most line level audio sources will easily be able to drive 2 (line level) inputs, so if both your speakers are to be active, there should be no problem with a "Y" splitter. It is common practice - as is "daisy-chaining" various speaker cabinets together.
Just be aware that if your source is, say a headphone output on a (cheapish) portable, or mp3 player, the quality may not be that good to start with - and may degrade when further units are added. It may be a good idea to add a buffer if that is the case.
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\$\begingroup\$ Thank you Jim. It looks adding a buffer would be the solution. Now, is there a module ready to use or I have to use the soldering tools? \$\endgroup\$– FarukAug 13, 2021 at 11:25