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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
16,922 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
16,922 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
ptss
1,964 posts
 
Cj1965 has made mistakes. Just shows he’s no different than anyone else when it comes to knowledgeable discussion.
Here’s hoping he’s at least bright enough-from an EQ perspective- to realize he wouldn’t enjoy belittling, condescending, or harshly critical comments directed as a personal attack on him. If he is “bright” enough he will reflect on his “petty” attempts at personal disparagement and get on with his sincere apology. Man up. If he doesn’t apologize he’ll be remembered (forgotten) as an inconsiderate, tiny minded( as he just couldn’t see the big picture) cretin, who could do a bit of math-but not always correctly :(
ptss
1,964 posts
 
I get particularly displeased when a person is graciously made aware of a shortcoming— but carries blithely on.
I was taught that’s a true true sign of real ignorance. 
cj1965
124 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    - erik_squires

Wrong again. You cannot simply substitute the electrical power ratio into the sound power formula and call it a day.

The ratio of measured sound power to reference sound power is 1/2 - not 1/4. And if you multiply the area factor in the formula for sound power in a medium, the ratio goes back to 1 - yielding 0 db. And with that, I’m done giving out free lessons in first year electrical engineering. Find someone else to insult.
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