0
\$\begingroup\$

I can see that there are speakers of various sizes with various power ratings. However, they usually seem to be 8 ohm or 12 ohm. What does this refer to? And why are they created to be this way? I mean why do 7 ohm or 10 ohm not exist?

\$\endgroup\$
  • \$\begingroup\$ They're not, they're found to be 8 or 15-16 ohms, approximately, bearing in mind it's an impedance not a resistance, and that it varies with frequency. You will certainly find speakers that are 7ohms at some frequencies. \$\endgroup\$ – user207421 Apr 21 '15 at 23:00
  • \$\begingroup\$ so like 50-ohm PCB tracks, this also is just an arbitrary number? \$\endgroup\$ – quantum231 Apr 22 '15 at 18:37
1
\$\begingroup\$

They are usually that way, partly for standards, partly for physics. From a standards perspective, it helps to have a particular set of impedances for which you can design amplifiers for, knowing the load is important in designing one. From a physics standpoint, the impedance of an audio driver is largely determined by its mechanical resonance. Not only is the impedance frequency based, but in speakers the impedance tends to be it's lowest (excluding DC signals) around its resonant frequency, I believe just slightly after resonance. As a related aside, speakers are composed of drivers and those drivers are connected to the amplifier by a crossover network. This networks purpose is to separate the signal frequencies into the appropriate driver, and ideally, present a nicer load to the amplifier as well.

\$\endgroup\$
  • \$\begingroup\$ To add a comment, since I am not confident on this aspect. However, I would assume, during mechanical resonance, the driver has the least control, and thus the higher the (if not highest) impedance will be at that point. \$\endgroup\$ – Jarrod Christman Apr 22 '15 at 0:54
  • 1
    \$\begingroup\$ A multimeter will just be passing a DC voltage to it, so it wouldn't be useful. You'd need to do a frequency sweep from 1 to 20kHz and measure the impedance. The lowest value (greater than DC) impedance will be your nominal impedance. I believe you should be able to treat the speaker as an inductor, even though it has a crossover network the load should still largely be inductive. \$\endgroup\$ – Jarrod Christman Apr 22 '15 at 19:25
  • 2
    \$\begingroup\$ The crossover network can be composed inductors, resistors, capacitors. These components can cause phase shifts (leading or lagging) and a mixture of complex interactions that will make it seem like more than a lump of wire. \$\endgroup\$ – Jarrod Christman Apr 23 '15 at 0:23
  • 1
    \$\begingroup\$ en.m.wikipedia.org/wiki/Audio_crossover \$\endgroup\$ – Jarrod Christman Apr 23 '15 at 1:07
  • 1
    \$\begingroup\$ I assume by 'speaker' you're referring to an entire assembly and not just a single audio driver. A driver by itself would just be seen as an inductor. \$\endgroup\$ – Jarrod Christman Apr 23 '15 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.