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A Question on Speaker Driver Efficiency

I have been tweaking my guitar amps, by upgrading the speakers.

I installed a larger speaker (was 8" now 10") in my bass amp, but I made sure it was very efficient - net result
- not only is the bass much deeper sounding,
- but because the new driver was more efficiant I now play at a lower volume.

So I am now considering upgrading my other amp (i.e. used for my 6 string) and got to thinking about building a new cabinet that houses two speakers.

I know that connecting the speakers in ...
- series will double the impedance, i.e. 2 x 4 ohms would have an onverall impedance of 8 ohms
- parallel will halve the impedance, i.e. 2 x 16 ohms would have an onverall impedance of 8 ohms

But what I have not been able to get my head around is...
- what will each connection method (i.e. series or parallel) have on the "combined" sensitivity rating?
- e.g. if both speakers are rated at 96db sensitivity, will the overall sensitivity change due to the connection method or remain at 96db?

Since I can get 4 ohm or 16 ohm drivers - which connection method would be best? series or parallel?

in case it is a factor
- the amp is 15 watts into 8 ohm
- I am looking at employing two identical drivers each rated at 96db sensitivity
- 96 db (or higher) is the target for the combined sensitivity

Any help is appreciated - Many Thanks Steve
williewonka
cj1965
124 posts
 
@erik_squires 

No, we're not far off from the essence of the question posed by the OP. Application of the decibel expression of voltage gain/loss to sound power gain/loss is not appropriate..

Sound power level, denoted LW and measured in dB, is defined by

LW = 10 log 10 ( P/P 0 )   d B

not 20 X log 10(P/Po) db as you suggested above.




 
almarg's avatar
almarg
8,673 posts
 
... not 20 X log 10(P/Po) db as you [Erik] suggested above.
CJ1965, Erik did not say that or suggest that.

He said 20 x the logarithm of the ratio of applied voltage, not 20 x the logarithm of the ratio of power. If half the voltage is applied power will be reduced to 1/4 of the original amount, resulting in a 6 db reduction in SPL according to your own 10log(P/Po) formula, which is correct. And that is what Kijanki was implicitly pointing out when he quoted your erroneous original comment on the matter.

Also, and more significantly, +1 to the very well said comments above by Peter (Ptss), which together with the earlier comments by Timlub I suggest that you (CJ1965) would be well advised to take to heart.

Regards,
-- Al

erik_squires
5,268 posts
 
Thanks @almarg - I should have said "voltage" a few more times I think. 

Best,


E
cj1965
124 posts
 
@almarg 

Please look up the definition of sound power. You will find it is not based solely on the net force applied but the area as well. If you run the calculations for sound power and put the values in the sound power db formula, you will get -3db. The graph below the calculator on the website Erik linked to also indicates this (voltage gain/loss  is-6db while, sound power is -3db).

The formula or calculator for voltage gain/loss is not supposed to be used for sound power. One has force units the other power - apples and oranges.
.
almarg's avatar
almarg
8,673 posts
 
Assuming that a speaker is operating in a reasonably linear manner, meaning for example that it is not being over-driven to the point that thermal compression becomes significant, it seems to me that the relation between acoustic power out and electrical power in will remain constant to a close approximation. And electrical power in will be proportional to the square of the applied voltage.

Therefore it would seem to me (and I believe also to Erik, Atmasphere, and Kijanki) that since a 50% reduction in applied voltage will result in a 75% reduction in electrical power in, which corresponds to a 6 db reduction in electrical power in, the result will be a 6 db reduction in acoustic power out.

Regards,
-- Al

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