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4th March 2004, 06:29 AM  #1 
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Coverting dB/2.83w to dB/1w
What is the conversion to change an efficiency spec that is in dB/2.83w to one that is in dB/1w so that one can compare apples to apples.
Since the power would be 2.83 times less it would seem that the efficiency rating would be reduced by a little over 3 dB. But what is the actual correction? 
4th March 2004, 06:56 AM  #2 
Wizard of Kelts
diyAudio Moderator

8 ohm speaker: 2.83 volts = 1 watt
4 ohms speaker: 2.83 volts = 2 watts. At equal efficiency, the 4 ohmer will have a plus 3 dB higher rating at 2.83 Volts than an 8 ohms speaker. 2 ohms speaker: 2.83 volts = 4 watts. The two ohm speaker at equal efficiency will have a plus 6 dB higher efficiency at 2.83 volts than it will at one watt.
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4th March 2004, 10:01 AM  #3  
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Re: Coverting dB/2.83w to dB/1w
Quote:
It depends on the load handling capability of an amplifier. Take an amplifier 50W on the point of clipping into 8 ohms. With 16 ohms output will be typically ~ 28W into 16 ohms. With 4 ohms amplifiers vary a lot, typically producing between 55W and 90W into 4 ohms, the latter value is more load tolerant. Speakers of equal efficiency will give max outputs directly related to the power output of the amplifier into that load. But note that a 8 ohm speaker 3dB less efficient than a 16 ohm speaker will have nearly the same maximum output with the above amplifier. For 4 ohm tolerant amplifiers the the same reasoning can be used, a 4 ohm speaker 6dB less efficient than a 16 ohm speaker will have nearly the same maximum output with the above amplifier. If you also note that bass extension is generally inversely proportional to efficiency, less efficient more bass extension, it becomes clear amplifier matching is critical to optimise maximum performance. 2.83V is 1 Watt into 8 ohm. (dB/2.83V) For impedance Z, you change the dB/2.83V by 10 x log (Z/8) to derive the dB/W figure. sreten. 

4th March 2004, 02:59 PM  #4  
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I wrote:
Quote:


4th March 2004, 03:24 PM  #5 
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dB~=10*log10(Re/8)
If you have a 4ohm speaker with a 3 ohm voice coil, rated at 93dB/2.83V, its 1W sensitivity would be dB=10*log10(3/8) = 4.26dB or 88.74dB  The least ambiguous spec is the 2.83V spec. Manufacturers play around with the ambiguity of the 1W spec so I wouldn't trust it.
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4th March 2004, 03:46 PM  #6 
Wizard of Kelts
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Ron:
You gave us the following equation for comparing SPL at 2.83 volts to SPL at 1 watt: dB~=10*log10(Re/8) The only thing is: you assume that an 4 ohm rated speaker with a 3 ohm voice coil should be rated at 3 ohms. But that an 8 ohm rated speaker should be rated with a DC resistance, (Re), of 8 ohms. If you are going to use the DC resistance of the 4 ohm speaker, then you should use the DC resistance of the 8 ohm speaker. As you know, a speaker rated at 8 ohms is very likely to have a DC resistance, (Re) of between 5.5 and 6.5 ohms. Call it 6 ohms as an estimate. So your equation becomes: dB~=10*log10(Re/6) = 10*log10(3/6) =3.01 dB. So the 4 ohm speaker rated at 93 dB @ 2.83volts will be 89.9999dB at 1 watt. Or 90 dB for short. Am I correct on this?
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4th March 2004, 03:54 PM  #7  
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Quote:
Re values can be very misleading and the above doesn't take into account different impedance characteristics. But the above is correct, if you use Re values, the Re of 8 ohms nominal is nominally 8/root2 = 5.6R. Nominal Re for 4ohms is 2.8R. The correct ratio's are Z/8 or Re/5.6. sreten. 

5th March 2004, 02:20 AM  #8  
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2.828 Volts is 1Watt into 8 ohms
2.828 volts into 3 ohms is 2.828^2/3 = 2.666Watts into 3 ohms 10*log10(2.666)= 4.26dB I made no errors, Kelticwizard. The question was how to convert 2.83V ratings to a 1W rating. It does not matter if the 4ohm voice coil is actually 2 ohms or 3 ohms or 4 ohms. The conversion from volts to watts with an assumed resistive load is not difficult............... sreten wrote: Quote:
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5th March 2004, 04:54 AM  #9  
Wizard of Kelts
diyAudio Moderator

Quote:
So why are we using Re in that formula instead of nominal impedance? It is never going to play at 3 ohms. It seems to me that by the same token, if we have a speaker with a 6 ohm voice coil that is rated 90 dB @ 2.83 volts/1M, then at 1 watt we have: dB~=10*log10(Re/8) dB= 10 log (6/8) = 1.25 A speaker with an Re of 6.0 is certain be rated at 8 ohms. The eight ohm speaker is 88.25 db @ 1W/1M but 90 dB at 2.83V/1M Is that true? That just about every speaker rated 8 ohms is really less efficient than what the 2.83V specs would indicate?
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5th March 2004, 08:03 AM  #10  
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Quote:
I never assumed resistive loads, I assumed nominal loudspeaker loads. Speakers are tested with a constant voltage for dB/2.83V. A 4 ohm speaker with with the same voltage sensitivity as an 8 ohm speaker is basically 3 dB less efficient, not 4.26. If you want to compare relative efficiencies of speakers then Z/8 or Re/5.6 can be used makes no difference. The fact that 2.83V into an 8R resistor is 1 Watt is a nicety. If you insist using this then the relative efficiency according to your method of an 8ohm speaker is 10log 6/8, i.e. 1.26, and the relative efficiency of 4 and 8 ohm speakers with the same voltage sensitivity is still  3dB. People do not seem to understand the meaning of nominal speaker impedance : The average impedance of the speaker must be higher or = nominal. The lowest impedance must not be less than nominal/root2. One of the two will decide the nominal impedance, unless the manufacturers cheat in their specifications, which is not exactly unknown. It should be obvious that dB/2.83V is easy to measure. Real efficiency is must harder as it depends on the impedance curve and is an essentially useless figure. However using Z nominal of the speaker will give you an idea. Using Re of a driver will give an unrealistically low figure. sreten. 

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