I know almost nothing about electronics, but I'd like to intercept the power going to a small speaker and use a relay switch to close a speaker connection on a completely different, but more powerful device (a home outdoor sound system). I used a multi meter to discover that the original speaker works with a voltage of 1.4 VDC but I can't figure out how to get the amperage. Is there a generic relay switch that would work in a situation like this?


2 Answers 2


This isn't really an answer for the question, but the comment field is too small.

Speakers don't actually use DC currents. Instead they use AC currents, so it could be that your measurement is incorrect. Sometimes some multimeters will detect some AC voltages as DC, so it's best to be sure when measuring.

So a good way to test this would be to generate a signal which would be easier for multimeter to read. A nice program is Two channels frequency generator which can be obtained here.

Next check the output characteristics of your speaker. If it can make 50 Hz or 60 Hz sounds, open the generator, set left and right frequency to 50 Hz or 60 Hz and click start (or if you can't connect the system to a computer, record the sound and then play it on the speaker) and be sure to set the volume to maximum both on the device and in program and on computer output.

So why 50 Hz or 60 Hz? Well those frequencies are commonly used for mains power and most multimeters are made so that they can easy and correctly read voltages at those frequencies. A good quality multimeter should be able to read voltages at other frequencies too, but this way you'll be sure that you get the correct reading.

If the speaker can't produce those frequencies, then try with whole number multiples of them.

Next, why frequency generator? Well, normal music produces complicated waveforms which change very quickly and are quite complicated. A single tone which is made by the frequency generator will be much easier for multimeter to read.

In the end check available settings on your multimeter. Good meters will have AC+DC voltage measurement option which will give you the real voltage at the speaker. If yours doesn't have such setting, try measuring both AC and DC voltage. Somewhere I read that you should calculate the actual voltage using following formula: \$V_r=\sqrt{V_{DC}^2+V_{AC}^2}\$. That should give you the real effective voltage.


I'm not sure what your proposing to switch here, but I can address the speaker current question.

Suppose you have 8 ohm speakers, and want to drive 100W into them. Power is impedance times the square of the current, so the current will be the square root of power over impedance.

\$P = I^2 * R\$

\$I = \sqrt{\frac{P}{R}}\$

So in the example, current would be \$\sqrt{\frac{100}{8}}\$ = 3.5A . Similarly, power is voltage squared over impedance,

\$P = \frac{V^2}{R}\$

\$V = \sqrt{P * R}\$

so in this example, the voltage is \$\sqrt{100*8}\$ = 28.3V .

However, these would be RMS values; the peaks would be \$\sqrt{2}\$ times as great, so you'd want to size your relay accordingly. A relay that could handle 5A and 50V would just suffice for the example.

You would actually want a safety margin, so using these example numbers, 10A and 100V would be a better choice. Finally, you might want to size the relay in anticipation of future upgrades. If you thought you might replace the 100W source with say, 500W, you'd definitely want a heavier device.

  • \$\begingroup\$ @JustJeff, what do you think of my edits? I put mathjax in, if you do not like it roll it back, but I thought you might enjoy the improved reading quality. \$\endgroup\$
    – Kortuk
    May 9, 2011 at 2:43
  • \$\begingroup\$ @Kortuk - yeah, I tried to invoke the LaTex thing to no avail. Not to go all meta, but does the text in the main answer use different formatting rules than the text in comments? \$\endgroup\$
    – JustJeff
    May 9, 2011 at 10:33
  • \$\begingroup\$ @JustJeff, I think it does, but does \$2^2\$ show up as tex here in a comment? If you type something you need to click out of the edit box and watch the preview for just a little before it will show the result(can be 10 or 20 seconds). This does not work in all situations, but it works for me in FF. \$\endgroup\$
    – Kortuk
    May 10, 2011 at 0:14
  • \$\begingroup\$ @JustJeff, I spent forever trying to figure out how to make a nice Square root, and then @Andrejako pulls it off right below. Thanks for the help @andrejako. \$\endgroup\$
    – Kortuk
    May 10, 2011 at 0:19
  • \$\begingroup\$ @Kortuk - well, thanks for the free radical =P seriously though, the formatting behaves oddly. I figured out how to escape \$ signs, so as to put two of them in a comment. On the 1st refresh, I saw the dollar signs as anticipated. Then less than 1 minute later, and I don't know if there was a 2nd refresh, the comment apparently got evaluated a second time, and the dollars disappeared but everything between them went italic. \$\endgroup\$
    – JustJeff
    May 10, 2011 at 0:23

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