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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
almarg's avatar
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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
4,442 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.


cj1965
124 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)." - kijanki

Please read and study the equation I posted above for two inductors (speaker coils) in series with a resistance and capacitance. The equation doesn't lie. Current and voltage are constantly varying and as the equation shows  the voltage representations of each woofer are NOT equivalent to the applied votlage input [V(t)] ,minus the other woofer's voltage. The voltage represented by the capacitance must also be accounted for and it is time dependent. If you disagree with the equation and what it is saying - address your comments/concerns with that equation and its applicability to the subject at hand. I didn't invent Kirchoff's law - I'm merely reciting it in the context of a series connected loudspeaker pair.
kijanki's avatar
kijanki
4,442 posts
 
You're telling Almarg to do the math?  LOL, Man, you got big mouth.
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