A Test. How much Voltage (power) do your speakers need? - Page 11 - diyAudio
 A Test. How much Voltage (power) do your speakers need?
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 Multi-Way Conventional loudspeakers with crossovers

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 View Poll Results: I measured the test tone at: 2 volts or less 220 38.26% Between 2-5 volts 185 32.17% Between 5-10 volts 77 13.39% Between 10-20 volts 40 6.96% Over 20 volts. 53 9.22% Voters: 575. You may not vote on this poll

 27th January 2012, 11:31 AM #101 diyAudio Member   Join Date: Jan 2009 It definitely works. For my test (details in the full range section), I recorded 0.75V with my 98dB @ 1W speakers. Squared gives 0.5625W. If you work it out long hand, 0.75V @ 8 ohms gives 0.0703W (V^2 / R) at -12dB. Each 3dB is double power, so 0.1406W at -9dB, 0.28125W at -6dB and 0.5625W at -3dB (biggest sine wave possible). So what happened there, well to work it out Pano's way, I squared the voltage. To do it long hand, I squared the voltage, divided by 8 (speaker impedance) to work out power, then multiplied by eight (3 doubles) to get from -12dB to -3dB. Cool, wish I had thought of that. Brian.
 27th January 2012, 12:21 PM #102 diyAudio Member     Join Date: Jul 2005 Location: leeuwarden I also did the test. "normal" listening-level gave me 0.5v, Loud (at the treshold of being annoyingly loud) gave me 2-3v. This was on ~88dB/2.83v/1m loudspeakers, 4 ohms impedance. (3*3)/4=2.25 times 8 gives me 18 watts.
 27th January 2012, 01:43 PM #103 diyAudio Member     Join Date: Jan 2011 Location: Ladysmith, BC Ok I get it now. Whew!
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Quote:
 Originally Posted by ra7 And 6.5" Tannoy dual concentrics in a Voight pipe gave me 4.0V. That would mean a peak of 16V is required. By your equation, that means a 250W amp for clean peaks. Hmmm... kinda high.
You went a step too far. All you needed to do is 4x4=16. That's the power rating of the amp you need. (see my post above) You made the mistake of squaring that again to get to your result. Good effort, but one step too far.

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Like Ra7, you are making it too complicated.
Quote:
 Originally Posted by back since i already know my speakers lowest impedance(have measured it) it would be. 7x7=49/2.7(ohm)=18.14AX49Volt=888.86watts.
If you measured 7V, then your maximum sine level would be 19.8V (9dB higher). You need that much voltage at the maximum sine wave level. Peak will be 3dB higher.

If you want to know how many watts that is into your 2.7 ohm load, do this:
19.8x19.8/2.7=145 You need an amp that can supply about 150 watts RMS into a 3 ohm load. See?

 27th January 2012, 03:09 PM #107 diyAudio Member     Join Date: Sep 2002 Location: Lakewood, Ohio Don't you need to use the same digital source to play the test tone as you used to play the high volume music? It's important that no volume adjustments are made between playing the two tracks. __________________ Kevin
 27th January 2012, 03:29 PM #108 diyAudio Member   Join Date: Mar 2009 2.5v rms into 4 ohm load. I had it cranked up listening to some rock. I may not listen to it at this level for long because it is hard on the ears. The speakers are quite efficient in the 93 db range. This DIY chipamp will do about 10v rms into 4 ohm resistors continuous, both channels driven before clipping. I will have to try my other listening system. Fun test!
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Quote:
 Originally Posted by Pano Like Ra7, you are making it too complicated. If you measured 7V, then your maximum sine level would be 19.8V (9dB higher). You need that much voltage at the maximum sine wave level. Peak will be 3dB higher. If you want to know how many watts that is into your 2.7 ohm load, do this: 19.8x19.8/2.7=145 You need an amp that can supply about 150 watts RMS into a 3 ohm load. See?

unless if an 150watt rms amp can give 890 watt peak power.

 27th January 2012, 03:56 PM #110 diyAudio Member   Join Date: Feb 2009 ok, I got it from Brian's explanation. I was put off by your first post that said peak volts will be 4X the measured value. Anyway, all is well. So, my 25W amp is just enough.

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