Mixing power rating of speakers in a cabinet

Regarding to the total power handling of the cabinet. You can imagine that amp is a source of water, which is fed through just one valve into two (or more) containers, through pipes of the exactly same size, diameter, angle etc., so the water flow is the same (this is impedance). One of the containers has volume of 30 liters (or gallons ;) ), another one is 500 liters. You open the valve, containers are being filled at the same rate. You have to stop the valve when the smallest one is full, otherwise it will overflow, even the bigger one (ones) are not full yet. So the total capacity of such a container battery in my example is 60 liters, even the bigger one could take 500.

This example converted to speakers applies whether those speakers are connected in series, parallel, or series/parallel and all of them have the same impedance.
 
1-2-3-5 said:
Take this with a grain of salt, because I obviously need help understanding this stuff:

You said "surge", so maybe he was talking about a possible power surge from your electrical source. I think when loads are connected in series, the first one gets hit with the most voltage. So if it gets blown it acts as kind of a fuse and doesn't allow the circuit to pass current, thus protecting the remaining loads. Maybe, I don't know...
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Negative, sir. There is no first one, even if it seems so. Speakers in series (even unlimited number of speakers) are a closed circuit, which is ready to go immediately after the first sign of voltage appears on terminals. Speaker coils and wires are full of electrons, which are ready to move immediately (current flow), as soon as voltage appears. All speakers are equal independently of the position and the current flow is the same in all of them. This is of course the same for in parallel or in series/parallel connected speakers.

Example with water is closed water circuit (pipes, closed containers), which is already completely filled with water, so as soon the valve on the input is opened, water starts to flow immediately in all parts. If pipes are of the same diameter, water flow is the same in all of them.
 
Got it. 30 in one and 30 in the other (even though it's got space for 500), and that's 60. I overlooked a third item, though, so if you could please indulge me one more...

My amp is a combo with a 90W Black Shadow. Does that get factored into the total? A 30W and a 60W plugged into one port, and the on-board 90W plugged in to the other port. 30W x 3 speakers = 90W, right? From what I can tell, I went through all this hand-wringing unnecessarily.
 
mcstinger said:
Negative, sir. There is no first one, even if it seems so. Speakers in series (even unlimited number of speakers) are a closed circuit, which is ready to go immediately after the first sign of voltage appears on terminals. Speaker coils and wires are full of electrons, which are ready to move immediately (current flow), as soon as voltage appears. All speakers are equal independently of the position and the current flow is the same in all of them. This is of course the same for in parallel or in series/parallel connected speakers.

Example with water is closed water circuit (pipes, closed containers), which is already completely filled with water, so as soon the valve on the input is opened, water starts to flow immediately in all parts. If pipes are of the same diameter, water flow is the same in all of them.
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Bad ass facts. Hella cleared it up. Thank you, man. Love it!
 
1-2-3-5 said:
Got it. 30 in one and 30 in the other (even though it's got space for 500), and that's 60. I overlooked a third item, though, so if you could please indulge me one more...

My amp is a combo with a 90W Black Shadow. Does that get factored into the total? A 30W and a 60W plugged into one port, and the on-board 90W plugged in to the other port. 30W x 3 speakers = 90W, right? From what I can tell, I went through all this hand-wringing unnecessarily.
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What are impedances of individual speakers?
 
Speakers in external cab in parallel, that will give 8 Ohms total impedance of the cab. Should set the amp to 4 Ohms, because external cab in parallel with internal speaker will result in 4 Ohms.

There will be half power on the internal speaker and half power on the external cab. So each of the speakers in external cab will run on 25% of the total power of the amp. That's not optimal, but no solution for this, as impedances of individual speakers are different.
 
mcstinger said:
That's not optimal, but no solution for this,
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Well, I haven't assembled the cab yet. Nevertheless, I can't think of a plan for a solution with optimal results. My simple goal is to have an easily portable cab for rehearsal purposes. Something that I can load easily on my own. I have a 4x12 that's a perfect electrical match, but that thing is HEAVY. I'm going to continue scheming. The outputs from my amp are an 8-ohm (single) and two 4-ohm ones that are wired in parallel (intended to accomodate two 8 ohm loads). Maybe I can devise something with the principles that you've taught me. Thanks a lot!
 
I only meant that 50% of the power will be on the speaker in the combo and 25% on each speaker in the external cab. But it is more or less impossible to do it differently with an odd number of speakers.
 
With all due respect to those who think I've introduced a "fairy tale" or "internet fake speaker lore", I'm going to guess most of you haven't had 80 plus vintage cabs in your lifetime, like I have. So I'm not going to take any offense from your theoretical arguments, since you don't have the depth of experience to offer up your valid/verified statements.

Specifically what I said was that in my experience with over 80 vintage cabs, when I bought them, almost all of them had blown the first speaker that is closest to the input jack. Why? Because even though they "theoretically" should all get the same power, the failures show that the first speaker (with the shortest wire throw, closest to the jack) in a series/parallel wiring setup is always the first one to get hit with the full power before ms later the rest get the signal. This does not apply to a parallel wiring setup.

I've seen this failure, and discussed it with other amp builders for the last 20 years, as they've seen it, too. There is no other logical reason, God knows I've tried to find one.

There just isn't one that makes sense. Why do you think you've seen quads of vintage Greenbacks for sale, with three original cones and one that's been reconed?
 
Scumback Speakers said:
With all due respect to those who think I've introduced a "fairy tale" or "internet fake speaker lore", I'm going to guess most of you haven't had 80 plus vintage cabs in your lifetime, like I have. So I'm not going to take any offense from your theoretical arguments, since you don't have the depth of experience to offer up your valid/verified statements.

Specifically what I said was that in my experience with over 80 vintage cabs, when I bought them, almost all of them had blown the first speaker that is closest to the input jack. Why? Because even though they "theoretically" should all get the same power, the failures show that the first speaker (with the shortest wire throw, closest to the jack) in a series/parallel wiring setup is always the first one to get hit with the full power before ms later the rest get the signal. This does not apply to a parallel wiring setup.

I've seen this failure, and discussed it with other amp builders for the last 20 years, as they've seen it, too. There is no other logical reason, God knows I've tried to find one.

There just isn't one that makes sense. Why do you think you've seen quads of vintage Greenbacks for sale, with three original cones and one that's been reconed?
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Apologies if I butchered what you told me.

That BM-75 is working great, BTW.

Thanks Jim ! :rawk:
 
@Scumback Speakers
1. First of all, I apologize for those words about fairy tales, it was unnecessary.
2. The problem is that if two speakers are connected in series, then none of them is the first, because both are connected to the terminals. And the "first" has no signal on it, if the signal does not go to its terminals through the "second" one. Not to mention, we are talking about AC. The loudspeaker is a complex load, not a pure resistor, and of course there are current and voltage shifts, but they are not in ms times. Speakers would be out of phase.

Despite my 30+ years of experience of course I respect your experience. But then, it would be interesting to know, when the "first" speaker will burn. And why is even the one with more power destroyed before the others. Whether it is when turning on the amplifier (with signal), pulling out the cable while playing, when using stereo setup, etc. Because everything is explainable, there is no magic in electronics.
 
mcstinger said:
@Scumback Speakers
1. First of all, I apologize for those words about fairy tales, it was unnecessary.
2. The problem is that if two speakers are connected in series, then none of them is the first, because both are connected to the terminals. And the "first" has no signal on it, if the signal does not go to its terminals through the "second" one. Not to mention, we are talking about AC. The loudspeaker is a complex load, not a pure resistor, and of course there are current and voltage shifts, but they are not in ms times. Speakers would be out of phase.

Despite my 30+ years of experience of course I respect your experience. But then, it would be interesting to know, when the "first" speaker will burn. And why is even the one with more power destroyed before the others. Whether it is when turning on the amplifier (with signal), pulling out the cable while playing, when using stereo setup, etc. Because everything is explainable, there is no magic in electronics.
Click to expand...
No worries. As I understand it, electricity takes the shortest path (or least resistance to complete the circuit), that's where the "closest to the jack" reasoning comes in. In a series wiring, the path goes to the first speaker, through it out to the next speaker, so the closest speaker gets the initial signal, even if it's only ms, it does get it first. So actually, the one closest to the jack gets the signal first, and this is borne out by all the failures I've seen in old Marshall cabs.

This is probably a good reason they went to that hated switching plate thing they use. Not only does it suck tone, but they fail quite often. I can't tell you when the first speaker will burn in a timeframe, only that when presented with the full signal load, it carries that for a few ms before the signal flows through to the other speaker in line after it. There is potentially some merit to the idea that you turn the amp on, at low volume to establish the signal, then turn it up to your playing volume after 30-60 seconds.

Of course, I know that they're all supposed to get the full signal right away, but after seeing so many old cabs with the first speaker blown, this is the only realistic conclusion I (and other amp builders I know) can come up with.

If you can come up with some logical reason to explain the dozens of failures like this I've seen, I'm all ears.
 
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