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I have two identical sets of ceiling speakers connected to a cheap 2 channel Chinese micro amp - one 8 ohms and one 4 ohms, wired in series per channel. The 8 ohm speaker is too loud when the volume is set ok for the 4 ohm speaker.

How can I reduce the volume of the 8 ohm speaker so they are more evenly matched?

Would a resistor make sense here? If so, what rating?

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    \$\begingroup\$ I suspect you should consider using a transformer designed for audio use. I haven't used such a device in years and never one that I think you need. But you are looking for 4 Ohm:8 Ohm. Hang the 8 Ohm speaker on its secondary and put the 4 Ohm primary in series with your other 4 Ohm speaker. Just found this article which discusses what I was trying to say. \$\endgroup\$ – jonk Oct 30 at 20:28
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When wired in series the same current flows through both speakers. Power = current2 x resistance, so the 8 Ohm speaker will get twice as much power as the 4 Ohm speaker. The human ear's response is logarithmic so this will sound like the 8 Ohm speaker is abut 40% louder.

To make them draw equal power you can put a resistor in parallel with the 8 Ohm speaker. This will reduce voltage and current in the 8 Ohm speaker, and increase voltage and current in the 4 Ohm speaker.

schematic

simulate this circuit – Schematic created using CircuitLab

The resistor value is tricky to calculate because it involves solving two simultaneous equations. I cheated and used LTspice with a voltage controlled resistance to simulate a resistance from 1 to 40 Ohms. The speakers got equal power at ~19.3 Ohms. A standard value of 18 or 22 Ohms should be close enough.

In the graph below Green is power in the 4 Ohm speaker, Red is power in the 8 Ohm speaker, and Blue is power in the resistor. The horizontal axis represents the variable resistance.

enter image description here

The power dissipated by the resistor depends on the amplifier's rated output power and load impedance, but is about half the power that the speakers have to handle. If the amplifier was rated for 3W into an 8Ω load then (at maximum output) it would put ~1.1W into each speaker and 0.5W into the resistor. For a different amplifier power rating (into 8Ω) scale accordingly. For safety you should rate the resistor for about double the expected power dissipation, ie. 1W in this example.

Note that real speakers are not prefect resistances and different speakers may have different impedance curves (depending on diameter, cone stiffness etc.) so they may not sound equally load at all frequencies. The speakers may also have different efficiencies which could make one sound louder than the other, so you might need to experiment with different resistor values to get the best balance.

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  • \$\begingroup\$ It isn't all that tricky -- the exact value is \$\frac{8}{\sqrt{2}-1} = 19.3137 \Omega\$. Also, if the amp is rated for 3W into 8Ω, then it's only going to deliver about 2.5W into 9.66Ω -- we can't know whether it hits its voltage limit or its current limit first, so we have to assume both. That means 1.03W is going to each speaker and 0.44W is going into the resistor. \$\endgroup\$ – Dave Tweed Oct 31 at 13:10
  • \$\begingroup\$ "That means 1.03W is going to each speaker and 0.44W is going into the resistor" - slightly lower than my numbers because I rounded from 3.125W (the actual power in my simulation) to 3W. Most amps are designed to drive an 8 Ohm or lower load so at 9.66 Ohms I bet it reaches the voltage limit first. But this is only an example anyway - we don't know anything about the OP's amp. \$\endgroup\$ – Bruce Abbott Oct 31 at 13:31
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Sure, add a resistor parallel with the 8 ohm so that it sees the same power as the 4 ohm. 12 ohms should be about right.

the resistor here will reduce the loudness of the 8 ohm and increase the 4 ohm so you may need to turn the amplification down a little, lower resistances will have a greater effect.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ First of all, you're assuming that the two types of speaker have the same sensitivity (SPL for a given power input). Second, where did you come up with the value for the resistor? With that value, for every watt that the 4-ohm speaker gets, the 8-ohm speaker only gets 0.72 watt. To get equal power, you'd need a resistor of 19.3 ohms. \$\endgroup\$ – Dave Tweed Oct 30 at 22:17
  • \$\begingroup\$ I'm not assuming an exact match, that I'm assuming they are fairly close, with the existing setup the 8 ohms gets four times the power. I guessed 12 ohms, assuming it would be fairly close to balancing the power (-1.5 dB is close to balanced) as you say the speaker efficinecy will probably be different and the deisired loudness may also be different, some experimentation is probably needed, I gave 12 ohms as a starting point. \$\endgroup\$ – Jasen Oct 31 at 3:16
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By now you realize that it was a dumb thing to do — you should have used identical speakers to begin with. And obviously, that would be one way to fix this with minimum fuss — just replace two of the speakers.

Another way to address this would be to put the two 4Ω speakers together on one channel of the amplifier and the two 8Ω speakers together on the other. Then you'd be able to adjust their volume independently.

But if you want a workaround that doesn't involve so much rewiring, a speaker volume control is called an "L-pad". Search using that term in order to find suitable products.

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