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
I encourage anyone wanting to follow along, to grab XSim crossover simulator, and examining the Frequency Response, Watts Dissipated and Impedance charts. 

http://www.diyaudio.com/forums/software-tools/259865-xsim-free-crossover-designer.html

The "blank" driver is an ideal 8 Ohm driver with an 80 dB efficiency. This will put to rest any questions about power, efficiency and sensitivity you may have. 

Best,

E
This jives with measurements I've made of acoustic output of drivers in series.
Sounds like you are running a solid state amp. Solid state amps, generally speaking, are usually built to behave as a voltage source. If that is the case, it will make double the power into half the impedance, and thus also half the power if the impedance is doubled. So such an amplifier will make 1/4 of its 4 ohm power into 16 ohms (drivers in series).

@atmasphere

I'm the only one in this thread actually citing equations based on Kirchoff's law while you hand wave generalizations about tube amps.
I'm the only one in this thread actually citing equations based on Kirchoff's law while you hand wave generalizations about tube amps.
Really? Hand-waving seems to be what you are doing with this quote...

Somehow you don't seem to make the connection that current has to flow if a circuit is complete. Its Circuit Basics 101 first day stuff.