0
\$\begingroup\$

I have a 0.3W (RMS), .8ohm speaker coil + cone that I want to test using various music. However, the 8ohm impedance is lower than typical earbuds (24ohm), and also I can't find headphone jack power specs for my computer (Macbook Air 2010) or phone (Nexus S), so I'm worried I'll blow out my speaker sample.

I found this amplifier that's for 8ohm speakers, but it outputs up to 20W! I could turn it to its lowest setting but I'm worried that in tweaking the volume I'd accidentally exceed the speaker's rating just by the error of my hand's motor control abilities. Could I use that amp plus a current limiter, or will its digital electronics become under-currented?

Alternatively, if I can find a "powered / active speaker" product such as for desktop computers which happens to have a 0.3W, 8ohm coil internally, I could just swap out the coil. Know any?

Last resort would be to breadboard an amplifier circuit myself based on an 8ohm-load, 0.3W amplifier IC like the LM4864MM, but that's extra work just for a throwaway test circuit.

What is the laziest way to safely accomplish this test?

\$\endgroup\$
  • 1
    \$\begingroup\$ This is an alternative. So I couldn't post this as an answer. Try to find out impedance matching transformer. Or wind a one using a troid. \$\endgroup\$ – Standard Sandun Oct 9 '12 at 5:38
  • \$\begingroup\$ @sandundhammika Thanks, good point! So you're saying with a 1:3 transformer and no other components I could safely run it off my phone without overpowering it? Are you sure a phone won't be putting out more than 0.3W?? \$\endgroup\$ – AlcubierreDrive Oct 9 '12 at 5:59
  • 1
    \$\begingroup\$ I'd worry more about overloading the MacBook than overpowering the speaker. Don't know if MacBooks are short protected at the audio jack. \$\endgroup\$ – jippie Oct 9 '12 at 7:23
  • 3
    \$\begingroup\$ If you don't need to test full volume just put a 15 Ohm resistor in series. \$\endgroup\$ – Wouter van Ooijen Oct 9 '12 at 8:06
3
\$\begingroup\$

Don't be so shy.

0.3W is pretty loud! And don't forget that audio equipment potentiometers are logarithmical ones, so when you turn it halfway, it is roughly one tenth of the output power. So a 20W amp, driven with its rated RMS input signal would output 2W RMS when turned up halfway. A quarter setting would result in 0.2W RMS accordingly. And you said you'd want to test with music - rated music power is around 2x the rated RMS power for most of the speakers.

Don't fret, speakers are quite difficult to destroy in so low power situations. I did try with a PC speaker and a ~20W amp when I was young and feeble, to the point when it got quite warm on the outside - but no pops (it was horrible, by the way)... Ok, PC speakers are not too flimsy. I did build a tube amp from scrap parts, and on that I regularly used a way too -electrically- underrated speaker, and to make the situation even worse, I used it as a guitar amplifier - driven into the maximal power tube saturation I could achieve...

If however you'd want a professional approach, use an LM386 power op-amp to play with, a few external components and you are ready to go. I think that is far easier to obtain than an impedance matcher transformator - not to mention, it degrades and colours the sound a lot less.

LM386 datasheet

\$\endgroup\$
  • \$\begingroup\$ Thanks, I didn't realize that audio pots are logarithmic! How do they wind that? \$\endgroup\$ – AlcubierreDrive Oct 10 '12 at 4:31
  • 1
    \$\begingroup\$ I'm not an expert regarding potentiometer manufacturing, but these small pots are not wound at all - they use a resistive layer, on which the middle wiper slides, shaped to give the necessary resistance between the terminals, and the desired characteristic. From those that I've taken apart to service, I definitely think that the logarithmic characteristic is achieved by having a non-constant thickness of the resistive layer. Though it would be possible to have constant thichkness, and varying width... \$\endgroup\$ – ppeterka Oct 10 '12 at 7:44
  • \$\begingroup\$ I think the pots are linear. Human ears just hear logarithmically: en.wikipedia.org/wiki/Psychoacoustics \$\endgroup\$ – jbord39 May 23 '16 at 3:52
  • \$\begingroup\$ @jbord39 So you imply that logarithmically tapered pots are just linears labeled as log ones? I disagree: for proper volume setting, you need logarithmic pots. \$\endgroup\$ – ppeterka May 23 '16 at 19:11
  • \$\begingroup\$ @pperterka Lol, no. If they specifically say logarithmic then they are logarithmic. If it doesn't say otherwise I would assume it is linear. \$\endgroup\$ – jbord39 May 23 '16 at 23:10

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.