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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
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almarg
9,670 posts
 
CJ1965, 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. So when efficiency is referred to, and defined as the SPL produced at 1 meter in response to 1 watt, the results of your calculation need to be normalized to 1 watt of supplied power. Which in turn makes Ralph’s statement that you alleged to have "blown it" entirely correct.
You [Kijanki] could know something that no one else on the planet knows and we’re all in for a new, amazing discovery. But then again, you just might be another clueless individual trolling around on the internet in search of a pointless argument....
For someone who has only been participating in this forum for a couple of months you certainly are quick to direct insulting comments at some of the forum’s most knowledgeable and widely respected members. And that certainly includes Kijanki, as well as Atmasphere. I suggest that some modification to your manner would result in your contributions being better received, and discussions in which you participate being more constructive.

Regards,
-- Al

cj1965
124 posts
 
" 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. " - atmasphere

What a clueless statement. When a capacitance is present, current flow is dependent on the time rate of change of voltage across the capacitor. Since it is a series circuit, this time rate of change affects the flow of current with time through both inductors - Circuit Basics 102.

I= Cdv/dt

Your simplistic view assumes that current through all reactive elements in a circuit is steady state (constant). And that is pretty scary for someone who has experience building amps. But then again, it’s tube amps and this is the year 2018 - so that kinda makes sense.
cj1965
124 posts
 
" For someone who has only been participating in this forum for a couple of months you certainly are quick to direct insulting comments at some of the forum’s most knowledgeable and widely respected members. And that certainly includes Kijanki, as well as Atmasphere. I suggest that some modification to your manner would result in your contributions being better received, and discussions in which you participate being more constructive. " - almarg

Perhaps you should direct your comments to the folks with vast, impeccable credentials who say things like this:

" You're telling Almarg to do the math? LOL, Man, you got big mouth. " - kijanki

Btw, I know my math is correct, thank you. I should hope after four years of electrical engineering, I would know  a little about basic addition, subtraction, multiplication, and division....


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.

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kijanki's avatar
kijanki
4,413 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. 
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