What do I get with Acoustic B100 combo + B115 cab IN SERIES?

[GatorDogDoc]

After spending 2 nights pouring over the amp threads (especially IvanMike/Kelly Lee/bassybill's super sticky) I think I've got this ohm/watts thing down... but I'd still feel better asking a question before I link up some equipment:

I've got an Acoustic B100 combo amp. The amp/head part pushes 100W @ 4 ohms. The speaker is 1x15, and actually is stamped 250W @ 8 Ohms. On the back is a speaker out jack which the company's website states is a SERIAL output.

The company website states you can add a B115 or B410 cab to the serial output, no problem.

I know the ohms isn't an issue, since they are serial... If I understand this right, each speaker would independently see the 4 ohms of the head.

What will I get (wattage) if I hook up an Acoustic B115 cab (1x15, 250W @ 8 ohms--identical speaker as the combo) to the serial output?


Would the combo put out 100W (amp's max) and the cab put out 100W (the amp's max) since the head can only put out 100W @ 4 ohms?


Or would the cab put out something less since both speakers are 8 ohms?

Whoa... got a little dizzy there!

[username1]

I know the ohms isn't an issue, since they are serial... If I understand this right, each speaker would independently see the 4 ohms of the head.

What will I get (wattage) if I hook up an Acoustic B115 cab (1x15, 250W @ 8 ohms--identical speaker as the combo) to the serial output?


The speaker wont see any ohms from the head, the speaker is the resistive load, not the head. In series with the internal 8 ohm speaker "if thats what is in there" your total load will be 16 ohms. Your head will likely be putting out about 60 to 70 watts with this load but it will still be louder and have better low end than with just the one speaker.

[GatorDogDoc]

Originally Posted by username1

The speaker wont see any ohms from the head, the speaker is the resistive load, not the head. In series with the internal 8 ohm speaker "if thats what is in there" your total load will be 16 ohms. Your head will likely be putting out about 60 to 70 watts with this load but it will still be louder and have better low end than with just the one speaker.

Thanks... that's what I was kind of thinking... I'll trade the minimal wattage cut for 2 speakers any day.

But... it turns out that I was incorrect:
the Combo has a 4 ohm speaker, and
the Cab has a 8 ohm speaker,
giving a total ohm load of 12 ohms.

--Will the 4 ohm speaker put out more watts than the 8 ohm speaker using the 100W 4 ohm head due to the unequal resistance?

--Or is the total resistance shared equally between the speakers?

[SurferJoe46]

First - I play a lot of Acoustic gear - but NOT the B-100 combo.

If the website says the internal driver is in series with the output jack and NOT parallel, then the Ohms are additive, which is NOT what we do to express the total Ohms of two drivers in parallel.

OK - series resistance, since it's additive - would be like this:::

4Ω + 8Ω = 12Ω

Parallel resistance goes like the following:::

4Ω & 8Ω (in parallel) = 2.6Ω ± a very tiny bit

Forget that the head 'see's Ohms since what's really happening is the ultimate load on the final side of the amp and it will 'see' what's happening no matter what the values.

If you exceed the minimum rating, then poof! You get to see smoke.

The only reason we talk about Ohms is to correlate the total load on the amp for a given Wattage - that's the true bottom line.

So - if your amp is capable of true 100 Watts @ 4Ω, then you can not go below that rated resistance since that approaches a dead short (0Ω) to the finals of that particular amplifier and they WILL go smoke!

Adding more drivers in series will numerically raise the Ohms in an additive way if indeed, you really are in series with the onboard driver.

Any time you numerically raise the Ohms, you will concurrently drop the total Wattage that is driving the speakers (the amp will 'relax' if you'll allow that phrase) - so you go down a corresponding amount that is simply algebraically expressed through Ohm's Law.

It's all a cosmic see-saw and when one goes up (and all other things stay equal) then the other goes down.

If I can believe my eyeballs and actually see the values on my ol' Ohmite slide rule, the true Wattage you might get with 12Ω, could end up at 60 Watts with that additional cab.

But then my eyes are tired, the numbers are small and I need a beer - so take that as a good guess until someone like BFM comes here to your rescue.

[GatorDogDoc]

Adding more drivers in series will numerically raise the Ohms in an additive way if indeed, you really are in series with the onboard driver.

Yes, both the Acoustic B100 and B200 combo cabs have the external speaker jack wired SERIAL... I confirmed this with the literature and the wiring diagrams supplied in the manuals.

Originally Posted by SurferJoe46

If I can believe my eyeballs and actually see the values on my ol' Ohmite slide rule, the true Wattage you might get with 12Ω, could end up at 60 Watts with that additional cab.

But then my eyes are tired, the numbers are small and I need a beer - so take that as a good guess until someone like BFM comes here to your rescue.

So I guess the relationship isn't linear?

I was thinking that placing this 100 watts @ 4 ohm amp under a 12 ohm load would produce 33.33 watts, via:

Amplifier Output = Amplifier Watts x (Amplifier Rated at Ohms / Speaker Chain Ohms)
Amplifier Output = 100 watts x (4 ohms / 12 ohms)
Amplifier Output = 100 watts x 1/3
Amplifier Output = 33.33 watts

SurferJoe46, if your beer-deprived eyes are correct (~60 watts), then I must be missing something, and the relationship isn't linear, but maybe logarithmic?

The only reason I wondering, is only because if it is a drop from 100 watts to 60 watts, but I get 2x15 speakers out of it, I'll take it. If it is a drop from 100 watts to 33.33 watts, then I don't want to do it!

[SurferJoe46]

Beer of not - my Ohmite (US Patent-Pending: 1952 ~ PRICE: 25¢) slide rule shouldn't be lying - I don't think anyway! lol

☺

When I enter the total Ohms, and the computation on one side of the slide rule, it reads the presumed (if all variables are within some logical range) Wattage output from the available total Watts derived under optimal conditions.

It shows that you may assume you'll get about 61.125 Watts, which I rounded DOWN for some arcane reason. Beer does that - makes the arcane - er, more arcane I think.

There's PEP and RMS and all sorts of nasties to compute - but I will concede the floor to Bill Fitz Maurice, if he doesn't have a hyphen in there somewhere.

But who am I to question a slide rule anyway!

Superman's hole-y underwear - maybe!

But a slide rule?

Never! ♫

[GatorDogDoc]

Originally Posted by SurferJoe46

Beer of not - my Ohmite (US Patent-Pending: 1952 ~ PRICE: 25¢) slide rule shouldn't be lying - I don't think anyway! lol

I'll trust your slide rule--besides, I like its number better than mine!

I've gotta get some shut eye... been up almost 39 hours now for work... but I'll check back on Sunday and see what else has been postulated between now and then!

Thanks for all your help!

[SurferJoe46]

I forgot about the child that lived a few doors down from my old place in San Jacinto has a B-100 and feeds a B-115 with it for some nice volume.

So-o-o-o - I'd say that the B-100 and the B-115 together should be a great combination.

[GatorDogDoc]

Originally Posted by SurferJoe46

I forgot about the child that lived a few doors down from my old place in San Jacinto has a B-100 and feeds a B-115 with it for some nice volume.

So-o-o-o - I'd say that the B-100 and the B-115 together should be a great combination.

[GatorDogDoc]

Well, it turns out I got a B200 for my birthday!

So, I'm not getting the B115 cab and am selling my B100.

Ending my subscription to this thread...ya'll keep talking if you want!

Thanks for all the advice guys!

[Bassdirty]

Originally Posted by GatorDogDoc

Well, it turns out I got a B200 for my birthday!

So, I'm not getting the B115 cab and am selling my B100.

I have the B200..its pretty Nice.



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