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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
timlub
2,045 posts
 
Wow, I am soooo glad that I didn't spot this thread before now..... 
giving way to all the theory and mathematics,  
@williewonka 
If all things are equal,  meaning a 4 ohm 8 ohm and 16 ohm driver would all sound equivalent in a given cabinet, Paralleled wired speakers sound better overall than series.  I would also recommend that you add a simple impedance correction circuit.   Many amplifiers react better to a consistent impedance.  Not all, but No amplifier sounds worse with it. (unless you are dropping impedance too low for a few tubes) still within tubes impedance operating range,  even tubes sound better with impedance correction circuits on speakers.  
NO PROOF,  this is my experience. 
Good Luck,  Tim 
almarg's avatar
almarg
9,670 posts
 
CJ1965, I believe that what your analysis about phase differences between voltage and current in reactive circuit elements may not be taking into account is that the motion of a dynamic driver results from the current it is conducting, rather than from the voltage that is applied to it. And the current in a series circuit is of course identical at all points in the circuit, aside from the consequences of the propagation delay that will exist along that path. Which in turn will be completely negligible for path lengths that would be present in a home environment, putting aside reflection effects for which timing may have relevance in digital applications.

Regards,
-- Al

almarg's avatar
almarg
9,670 posts
 
CJ1965 4-17-2018
"So if you have a 97 db 1 watt/1 meter 8 ohm driver, two in series will have the same efficiency (since each is absorbing 1/2 watt) while the sensitivity is 94 db. If you put the two drivers in parallel for a 4 ohm load, the efficiency is the same as 1/2 is absorbed by each driver if 1 watt is applied. However the sensitivity is now 100db. " - atmasphere

With that, you clearly blew it.
Ralph’s (Atmasphere’s) statement is entirely correct, if (as he is assuming, consistently with commonly seen usage) efficiency is defined on a 1 watt basis. And if sensitivity is defined on the basis of 2.83 volts.

Strictly speaking, I would define speaker efficiency as being the ratio of acoustic power out to electrical power in. But that definition would have little practical use, and the term is widely (and IMO very reasonably) used to refer to the SPL produced at 1 meter in response to an input of 1 watt.

Regards,
-- Al

kijanki's avatar
kijanki
4,477 posts
 
@cj1965 Current in the circuit is still the same. You are confusing voltage across the speaker with current. Imagine simple circuit consisting of voltage source and bunch of inductors and capacitors in series. Now you insert speaker into it. Do you think it will sound differently in different places of insertion? There is only one current in the circuit and two speakers in series have to respond at the same time (unless there is place where "faster current" can escape).
cj1965
124 posts
 
@almarg 

Do the math. Take two 8 ohm nominal woofers, connect them in series, and apply 2.83v.

2.83 divided by 16  = .176875 Amps (current "I")

power dissipated into each woofer in this circuit then becomes:

R(I X I) or resistance of each woofer times the square of the current running through it. Thus
8(.176875 X .176875) = .25 Watts or 1/4 watt - not 1/2 watt as atmasphere and you have suggested -

 " two in series will have the same efficiency (since each is absorbing 1/2 watt) while the sensitivity is 94 db "- atmasphere

The above can be expressed another way.
Power dissipated in two woofers in series = net voltage drop across both times the net current flowing through both.
2.83 Volts times .176875  = .5 watts or 1/2 watt total between both drivers in series. If together both woofers are dissipating 1/2 watt, then individually they must dissipate 1/4 watt - thus agreeing with the above calculation. At half the voltage drive level (voltage divided when connected in series), each woofer should produce half the acoustic output and consume 1/4 the power.When their acoustic outputs are summed, the net acoustic result is the same as if for one driver - however the power consumed is cut in half.

This jives with measurements I've made of acoustic output of drivers in series.


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