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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
cj1965
124 posts
 
" re your last post that was directed to me, your math is of course correct. However, once again, if efficiency is defined as the SPL produced at 1 meter in response to an input of 1 watt, if that 1 watt is provided to two speakers connected either in series or in parallel each speaker will absorb half of a watt. " - almarg

Please, I cannot take you seriously if you (or anyone else for that matter) continues saying things like - "if that one watt is provided" to loudspeakers. You can't apply or "provide a watt". All you can do is apply a voltage and the load draws whatever current it draws based on its resistance. The accuracy of my math above stands and it stands in stark contrast to the erroneous statement atmasphere made and you supported which suggested two 8 ohm speakers in series would each dissipate 1/2 watt with a total of 2.83V applied as input. Wrong is wrong. Math is either correct or incorrect. 1/2 watt DOES NOT equal 1/4 watt. Period.

.
kijanki's avatar
kijanki
4,479 posts
 
I'm voting to remove cj1965 from our forum.  Administrator should already observe that he calls respected members "clueless".  I wonder who would join me and what is the procedure to remove such obstacle. 
almarg's avatar
almarg
9,670 posts
 
CJ1965 4-18-2018
... the erroneous statement atmasphere made and you [Almarg] supported which suggested two 8 ohm speakers in series would each dissipate 1/2 watt with a total of 2.83V applied as input.
Neither I nor Ralph (Atmasphere) said or even implied that.
You can't apply or "provide a watt". All you can do is apply a voltage and the load draws whatever current it draws based on its resistance.
What we were referring to is applying whatever voltage is necessary to result in a total of 1 watt being consumed by the two speakers, whether they are connected in series or in parallel. Which is what is relevant if what is being referred to is the overall efficiency of the speaker combination, and if efficiency is defined as the SPL produced at 1 meter in response to 1 watt.

I'm not sure how that can not be clear, after all that has been said.

Regards,
-- Al


cj1965
124 posts
 
" 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 " - atmashpere

Each is not absorbing 1/2 watt when placed in series (with the required 2.83 V applied) and the net acoustical sum of output is approximately 97db - not 94db. Two obvious errors in one sentence. The math doesn't lie and frankly, I'm getting a little tired of the verbal gymnastics at this point. The math is simple and anyone who wants to read the details above can readily see where the error was established. Yeah, you can double the input drive voltage to 5.7v and then the series arrangement will have each woofer dissipating 1/2 watt. You can also multiply the input voltage by 10,000 - it won't tell us anything about the effect placing two identical drivers in series has. Sorry, wrong is wrong. And with that, I have to move on. I have better things to do with my time.
almarg's avatar
almarg
9,670 posts
 
As quoted by CJ1965:
" 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 " - atmashpere [sic]
Given that efficiency is defined as the SPL produced at 1 meter in response to 1 watt, when he said "since each is absorbing 1/2 watt" he clearly meant "since each is absorbing 1/2 watt if a total of 1 watt is being provided." Also, the paragraph in which that statement appeared certainly made no reference to 2.83 volts.

If that was not clear then it certainly should have become clear during the course of the subsequent posts, especially mine.
... you can double the input drive voltage to 5.7v and then the series arrangement will have each woofer dissipating 1/2 watt. You can also multiply the input voltage by 10,000 - it won’t tell us anything about the effect placing two identical drivers in series has.
Specifying the efficiency of a series combination of speakers, defining efficiency as the SPL produced at 1 meter in response to 1 watt, will tell us whatever can be told by an efficiency spec. As Ralph alluded to earlier, efficiency specs tend to be especially relevant in the case of tube amps, since for example in the case of an amp providing 4 and 8 ohm output taps maximum power ratings (in watts) will typically be the same or similar when an 8 ohm load is connected to the 8 ohm tap as when a 4 ohm load is connected to the 4 ohm tap.

Regards,
-- Al

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