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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
erik_squires
15,632 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
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almarg
9,670 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

erik_squires
15,632 posts
 
@cj1965

Where in my post did I mention power? I mentioned the relationship between output dB and input voltage.

Having said that, as @almarg has alluded, both formulas are true. Much like

P = V x I

and

P = (V x V) / R

They are mere re-writing of each other. Again, I encourage you to grab XSim to validate any formulas. It is pretty accurate. 

Best,

E
erik_squires
15,632 posts
 
Al - Yes indeed. These are all just rewrite.  Since power is proportional to the square of the voltage:

(0.5 x 0.5) = 0.25

Now, calculating power difference: 

10 log (0.25) = -6 dB

Calculating voltage difference:

20 log (0.5) = -6 dB

In either case, the SPL at a reference distance, measured in dB, changes in proportion to the power OR voltage when either is expressed as dB assuming there is no compression in the driver. 

Said another way, for a single driver:

Delta V dB = Delta W dB = Delta SPL dB

That's what's so cool about dBs! 

Best,

E
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