Speaker sensitivity 2.83V/m question - Page 1
Forum section: electronics / beginners

author: dr_p
author_rank: Frequent Contributor
timestamp: September 03, 2013, 12:24:05 pm
content: Hi there!I'm a little confused by somespeakerspecifications.They are 3 ohm speakers with a rated Sensitivity of 91dB/2.83V/m. I would like this converted to /1W/m for comparison purposes.As fas as I can tell, the sensitivity at 1W/m is 91+10*log(3/8)=91-4.26=86.7dBNow these are 3-way speakers, so does the same sensitivity apply to all 3? I am particularly interested in the main membrane(not the tweeters), becauseI'm trying to compare them to a 1-way speaker that has 85db/1W/m.

author: madires
author_rank: Super Contributor
timestamp: September 03, 2013, 01:43:10 pm
content: Quote from: dr_p on September 03, 2013, 12:24:05 pmI think the datasheet is wrong because 2.83V should be used just for speakers with 8 Ohms. Assuming it would be correct you can calcuate the power:I = V / R = 2.83V / 3 Ohms = 0.943AP = V * I = 2.83V * 0.943A = 2.67Wor the short way:P = V^2 / R = (2.83V)^2 / 3 Ohms = 2.67WFor 8 Ohms: P = (2.83V)^2 / 8 Ohms = 1WSimply take the 91dB from the datasheet as 91dB for 1W and 1m distance.QuoteIt's sensitivity of the whole chassis measured at 1kHz usualy. You may neglect the tweeter at that frequency anyway ;-) I'd would compare the sensitivity (SPL) of the complete systems and not just single speakers because the speakers, filters and chassis/box all contribute to the end result.BTW, the formula for adding dBs is: sq((dB-Value #1)^2 + (dB-Value #2)^2 + ... )

author: dr_p
author_rank: Frequent Contributor
timestamp: September 04, 2013, 08:40:33 am
content: Super! Thanks a lot for that

author: fcb
author_rank: Super Contributor
timestamp: September 04, 2013, 08:55:12 am
content: There is no 'standard' speaker impedance, and Blaupunkt are correctly stating things (with the exception of specifying Vrms/Vpp etc), you will have to convert the 2.83v @ 3ohms into a power and scale accordingly.Automotive speakers are almost always 4 ohms nominal and domestic units 8 ohms. It's easier to develop more power from a 12v battery without requiring SMPS. Also most (modern) automotive speakers are driven as BTL (bridge-tied load).

author: jaxbird
author_rank: Frequent Contributor
timestamp: September 04, 2013, 09:04:01 am
content: I wouldn't assume 91dB@1W, I think more likely it's 91dB@2.6W but it's hard to say without knowing the impedance and actual frequency response.IMO, too much marketing is involved in speaker driver specs, e.g. what does 40-25kHz mean?, surely it will not do that within 0dB and likely not within +/-10dB, most likely it will be as much as 30-40dB down at 40Hz in a free air measurement and it doesn't say what the free air resonance frequency is, so very hard to predict how it will behave in an enclosure.Of course the question is if you are comparing to an 8 ohm driver whether it's important that the 3 ohm driver needs more current at the same voltage level, really depends on your amplifier.

author: dr_p
author_rank: Frequent Contributor
timestamp: September 04, 2013, 01:49:13 pm
content: Well, I bought them and they're now on the car. They sound better than OEM 15W speakers and I can crank some more volume out of them without distortions. But they don't feel like 3 times the power. I guess my ear's logarithmic response is to blame. But for 20$ a pair I'm happy with them.Since the package said hundreds of Wand 40Hz specs, I wasn't putting much weight in the manufacturer's figures, but I was just curious what their numbers said.

author: madires
author_rank: Super Contributor
timestamp: September 04, 2013, 02:36:18 pm
content: Quote from: fcb on September 04, 2013, 08:55:12 amIt's even more complicated :-)The impedance is also dependant on the frequency. Industry standard is to measure the SPL at 1kHz at a distance of 1m. The SPL of 2.83V @ 3 Ohms would be a valid measurement result but it's nonsense. We don't know if there's a linear relation between SPL and power or if that relation is much more complex for the chassis. I'm not able to scale the SPL to 1W without knowning that relation.Therefore my guess is that someone at Blaupunkt took the 2.83V from the 2.83V @ 8 Ohms (=1W) by mistake without knowing about the physics. Of course they could done that also on purpose to make the specs look nicer. In either case they deserve a palmface :-)

author: jahonen
author_rank: Super Contributor
timestamp: September 04, 2013, 03:06:31 pm
content: This specification has caused me much headache back when I prepared the electro-acoustic simulation code for WinISD. Actually, the sensitivity can be calculated from other parameters, and there is a law which basically prevents one to make infinitely efficient ultra-small driver with frequency response down to very low frequencies. Efficiency (between 0 and 1) of a driver can be calculated as (borrowed from documentation of Driver Parameter Calculator by my friend Claus Futtrup):no := rho*power(Sd*Bxl,2)/(power(Mms,2)*2*pi*c*Re);So it basically is a function of effective cone area (Sd), motor force factor (Bxl), moving mass including air load (Mms) and voice coil resistance (Re). Other factor rho is density of the air and c is the speed of the sound. But I think that formula is valid only for "normal" drivers, so we don't actually get an inexhaustible free energy source if we make the voice coil out of superconductorVoltage sensitivity is unambiguous, but power sensitivity is more difficult. As speaker impedance is not constant and and complex, saying 1W is very difficult to interpret what it actually means. If we assume that a speaker (driver or tweeter) is linear, SPL level is directly proportional to the input voltage at speaker terminals. Power taken from the amplifier varies widely along the frequency.This sensitivity figure might also be to full-space (4-pi) or half-space (2-pi). We don't know.Also, when summing output of two or more drivers, the result depends whether or not they are coherent. If they reproduce the same signal and are in close proximity, then the output sums linearly. If outputs are not correlated, then the mentioned quadrature law applies.Regards,Janne