So I have a 3W amp that I want to use, and I've been told that that means each speaker channel outputs 1.5W. If that's correct, I have my mind set on two 1.5W speakers, but I don't know if I need 4 ohm or 8 ohm speakers. I've been told that the ohms shouldn't matter, but I've been told by another that it is what determines volume range. The datasheet of the amp itself says 4 ohms, so should I match the speaker ohm rating with that? Obviously I want to have the widest volume range of the volume knob possible.

  • \$\begingroup\$ You might want to check the assumption that a 3W amplifier is only 1.5W per channel. Read the data sheet. \$\endgroup\$ – Simon B May 24 '17 at 10:36

Almost all audio amplifiers today are approximately voltage sources -- they have relatively low output impedance. If you change the speaker impedance, to first order, you change the power delivered, which goes as $$P = V^2/R$$.

This applies both to the power delivered for a given input signal and volume setting, and also to the maximum possible power output.

You can't get infinite power out by lowering the speaker impedance. There is a minimum possible impedance that the amplifier can effectively drive. This will be specified in the data sheet, and will generally also allow the maximum power output.

Beyond that, there isn't much difference -- a similarly designed 8 ohm vs. 4 ohm speaker will operate mostly the same, as long as you drive it with the right type of amplifier.

There are often multiple ways to connect a speaker to an amplifier. For instance, many multi-channel amplifiers can be connected in "bridge tied load" mode, where two ampliifer channels drive opposite signals to the speaker + and - (rather than having - at ground). Other amplifiers always work this way, and still others allow multiple outputs to be connected in parallel to drive more output power. Each of these modes may have a different minimum impedance and power vs. impedance curve, so make sure to look at the data sheet for the exact topology you use.

  • \$\begingroup\$ From P=V²/R, it follows that if the amplifier is specified for 4 ohm speakers, you will only get half the power using 8 ohm ones. \$\endgroup\$ – Simon B May 24 '17 at 10:35

The speaker "load" is the average impedance of pumping AC air pressure doing real work, and not reactive or inductive impedance that would only result from an inductor.

As such the coil DC Ohms is typically near 1/2 of the rated impedance over the spectrum it is designed to operate. Speakers in enclosures tend to have a frequency where it is high impedance and low current at that bass frequency becuase it is mechanically resonant at that frequency and amplifies sound by this property and may sound "boomy" and thus gets louder with less effort or work being done meaning higher impedance. However that is the exception and lower impedance implies mroe current and more force moving the coil and thus acceleration of the coil.

But cone displacement,d is the 2nd integral of acceleration vs frequency so d drops inverse-squared with rising frequency.

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So what is the best impedance?

It depends!

THD rises with current and lower R but higher power high impedance demands very high voltages, so if the amp is high power and excellent linearity with high current then it will have low THD at 4 Ohms and this will give more power for same voltage than 8 Ohms.

But if THD rises too sharply with current, then it may be louder but horribly distorted. So 8 Ohms has often been a standard but for Mobiles high current "can be" inefficient, so internal speakers are much higher R.


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