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I don't know power is handled through speakers. Can I replace this 1 W speaker with a 0.5 W speaker in a computer?

From what I understand, it's the load which determines how much power something will draw. So does that mean that this speaker will only draw 0.5 W or is the speaker receiving 1 W, possibly damaging the poor little thing?

Here is a photo of the two speakers. 1 W on the left, 0.5 W on the right.

enter image description here

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  • \$\begingroup\$ The main difference is twice the coil temperature rise if you put 1W into a 0.5W load , whip is like putting 10 lbs into a 5 lb. bag \$\endgroup\$ Commented Oct 9, 2018 at 1:54

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There's honestly no real way of knowing without testing/experimenting. There could be heaps of situations. The two general cases, but, are:

  1. The speaker amplifier is producing more than 2V amplitude output. This means the same Ohm speaker will, indeed, produce the same power (\$ P = \frac{V^2}{R} \$), and at a voltage greater than 2V it will be greater than 0.5W power. The rating on the speaker is the maximum power the speaker can take before damage will occur, and so your new speaker will not work in this case (may work for a while, but the damage may be gradual and over time).

  2. The speaker amplifier is producing less than 2V amplitude. In this case, your new speaker is safe to use, as the total power the speaker will dissipate is less than 0.5W.

Speakers and amplifier systems are usually done close to the limit, as to maximise the sound output possible (i.e. the amplifier may be designed for 0.8W output, and then they chose a 1W speaker for some headroom). As a result, chances aren't in your favour as your new speaker is 50% the power of the original.

Best way to check is with a multimeter set to AC mode, probing the amplifier output when it is playing a constant sound and plugged into the new speaker (yes it may get damaged). It's important to actually load the amplifier with something, as the open-circuit gain \$ A_{vo} \$ may be a lot more than when it is not loaded, and so the output voltage may be higher than what it actually is compared to when the speaker is plugged in.

Ideally an oscilloscope would be better, but I imagine that's out of the question.

If the output is greater than 2V, you can - as you have suggested - add a resistor in series to limit the voltage received by the speaker. The amplifier will not have trouble driving that, as this will actually decrease the total power being driven (adding a resistor in parallel, however, will potentially damage the amplifier as this adds a whole extra load).

As for what value resistor, this depends on the voltage you measure. Remember - we want no more than 2V ampltiude on the new speaker. If, for example, you measure on your multimeter that the amplifier is outputting 2.5V, then you should add a \$ 2 \Omega \$ resistor in series, rated for at least \$ 125mW \$. That's just a specific example. If you need more help, feel free to comeback and I can go into more detail on those calculations.

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    \$\begingroup\$ Thank you. I'll ping you back if I have any further questions. \$\endgroup\$
    – Zhro
    Commented Oct 8, 2018 at 0:48
  • \$\begingroup\$ If I performed this test, I'd slowly rise the amplifier output while measuring. If I reach said 2V way before reaching max output setting, I'd know that it will fry the speaker before continuing. Would that work? \$\endgroup\$
    – Jahaziel
    Commented Dec 17, 2020 at 22:55
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It will mostly depend on your speaker driver, but likely not.

Since both speaker has the same 8ohm impedance, the current driven by the driver will be the same on both speaker, and thus will power the same wattage.

Since the original speaker is 1W, if you replace with the 0.5W, the driver will still drive it at 1W, potentially destroying it.

Having said that, you still can replace it, as long as you keep the volume not too high it will work, but bear in mind if you crank up the volume you might destroy your speaker.

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  • \$\begingroup\$ I'm very new to electronics. Can I add another resistor in series to limit the driver to 0.5W? \$\endgroup\$
    – Zhro
    Commented Oct 8, 2018 at 0:30
  • \$\begingroup\$ You can try, it will increase the impedance to 16ohm if you add another 8ohm, so your amp might have trouble driving that. \$\endgroup\$
    – Damien
    Commented Oct 8, 2018 at 0:33
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Those tiny cheap speakers with no proper enclosure will sound awful producing no low frequencies. The rated amplifier output power into its rated speaker impedance determines the speaker power. A lower speaker impedance draws more amplifier power. A speaker impedance that is too low will destroy the overloaded amplifier.

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