Using Two or More Cabs, Is Wattage Cumulative/Additive?

[2F/2F]

Basically speaking, the answer is no.

A cab is rated where it is rated, period. When adding cabs, it might seem like it is increasing what the entire speaker system can handle, but it is really just changing what the amp puts in to each cab.

If you happen to have an amp that perfectly doubles its power when you have two cabs hooked up to it, and you have two identically spec'd cabinets, then you end up with each cab getting what it would be getting if used alone.

Bottom line, real world: When mixing cabs with different power handling ratings, but the same impedances, you need to play to the limits of the lower rated cabinet.

 

[selowitch]

Bottom line, real world: When mixing cabs with different power handling ratings, but the same impedances, you need to play to the limits of the lower rated cabinet.

Right, I intend to use two cabinets of the same wattage and resistance or a combo and an extension speaker also both of the same wattage and resistance.

 

[Wasnex]

You need to understand Ohms law to really grasp what is happening.

The cabs respond to the voltage the amp produces. Conversely, the impedance of the cabs limits the amount of current the amp can deliver. Therefore, power is the product of voltage and current (P=VI)

Let's assume the amp makes 400W at 8 ohms and 800W at 4 ohms like a Mesa Subway.

First we will figure out the voltage.
V=sq rt(PZ)
V=sq rt(400X8)=~56.57V and
V=sq rt(800X4)=~56.57V

FYI used ~ to mean approximately because I am rounding to the nearest 100ths.

I'll add a comment here: In this case since the amp makes the same power at both 4 and 8 ohms, the voltage holds at ~56.57V regardless the impedance; this is a bit unusual. The power supply in most amps sags a bit when the impedance goes down, so the amp is not capable of doubling power when the impedance is cut in half.

What you should see here is the amount of power the speaker will dissipate is dependent on how much voltage the amp produces. So you have to be careful to keep the voltage below the whatever voltage will produce the power level equal to the rating of your weakest cab.

So if both cabs are the same impedance and one cab is rated for 500W and the other cab is rated for 200W. The total power rating of the system is double the weakest cabs rating, or 2x200=400W total.

If each cab has a different impedance rating it gets a bit more complicated. Lets assume the the 500W cab is 4 ohms and the 200W cab is 8 ohms.

First we need to voltage that will produce 200W at 8 ohms.
P=sq rt(200X8)=40V.

Next we calculate how much wattage the 4 ohm cab will dissipate at 40V. If this is below the cabs power rating were good to go.
P=(V^2)/Z
P=40^2/4=400W

Remember the 4 ohm cab has a 500W power rating, but if we feed it more voltage the 8 ohm cab will be pushed beyond it's 200W power rating.

So the system power limit in this example is 200+400=600W total.

In the event the voltage pushed the 4 ohm cab over it's power rating, you would have to solve the problem going the other direction. In other words, find the voltage where the 4 ohm cab hits its power limit and then calculate how much power the 8 ohm cab makes at the same voltage.

 

[Matty Koff]

Yes two 8 ohm cabinets rated for 300w should be able to handle an amp rated for 600w @ 4 Ohm. The amp is going to send an equal amount of power to each connected cabinet.

You've got some things backwards it sounds like though. Typically, when you get a 200w amp, or a 500w amp, or an 800w amp, that's the power it outputs at 4 Ohm. I.E. my old Momark 500W was capable of 500W RMS @ 4Ohms, and 300W RMS @ 8Ohm. My Mesa Subway WD-800 with the switch set to 4Ohm is 800W at 4 Ohm and 400W at 8Ohm.
When using two 8 Ohm cabinets, ideally you want to make sure that each cabinet can handle at least half of what the amp is capable of @ 4Ohms. The combined total power handling of the cabinets are irrelevant. Also bare in mind what I said about the Markbass amp, 500W @ 4Ohm, 300W @ 8Ohm. If you were using two 250W RMS capable 8Ohm cabinets with that, and decided you only needed one of them for a set, @300W you're capable of more power than the cab is rated for. It would probably be okay though as long as you were mindful of the input gain and volume controls. Still, it might be worth making sure how many watts RMS the amp is rated for with an 8 Ohm speaker load and ensuring both cabs meet or exceed that power handling.

For an amp rated for 400W @ 4Ohm two 8 Ohm cabinets rated for a maximum of 200W each should be fine. You could safely use cabinets rated for more power as well. Even so, always use your ears. There are many factors which could lead to a blown speaker or amp if you don't use common sense.

Figured I'd chime in using layman's terms, since they're all I really know.

 

[g-dude]

You need to understand Ohms law to really grasp what is happening.
View attachment 4223374

The cabs respond to the voltage the amp produces. Conversely, the impedance of the cabs limits the amount of current the amp can deliver. Therefore, power is the product of voltage and current (P=VI)

Let's assume the amp makes 400W at 8 ohms and 800W at 4 ohms like a Mesa Subway.

First we will figure out the voltage.
V=sq rt(PZ)
V=sq rt(400X8)=~56.57V and
V=sq rt(800X4)=~56.57V

FYI used ~ to mean approximately because I am rounding to the nearest 100ths.

I'll add a comment here: In this case since the amp makes the same power at both 4 and 8 ohms, the voltage holds at ~56.57V regardless the impedance; this is a bit unusual. The power supply in most amps sags a bit when the impedance goes down, so the amp is not capable of doubling power when the impedance is cut in half.

What you should see here is the amount of power the speaker will dissipate is dependent on how much voltage the amp produces. So you have to be careful to keep the voltage below the whatever voltage will produce the power level equal to the rating of your weakest cab.

So if both cabs are the same impedance and one cab is rated for 500W and the other cab is rated for 200W. The total power rating of the system is double the weakest cabs rating, or 2x200=400W total.

If each cab has a different impedance rating it gets a bit more complicated. Lets assume the the 500W cab is 4 ohms and the 200W cab is 8 ohms.

First we need to voltage that will produce 200W at 8 ohms.
P=sq rt(200X8)=40V.

Next we calculate how much wattage the 4 ohm cab will dissipate at 40V. If this is below the cabs power rating were good to go.
P=(V^2)/Z
P=40^2/4=400W

Remember the 4 ohm cab has a 500W power rating, but if we feed it more voltage the 8 ohm cab will be pushed beyond it's 200W power rating.

So the system power limit in this example is 200+400=600W total.

In the event the voltage pushed the 4 ohm cab over it's power rating, you would have to solve the problem going the other direction. In other words, find the voltage where the 4 ohm cab hits its power limit and then calculate how much power the 8 ohm cab makes at the same voltage.

There’s another way that I approach the mixed ohm problem.

Basically, because the 500W cab has a resistance of 4ohm, and the 200W one has a resistance of 8 ohm, when the amp pumps out power, the 4ohm cab will get 2W for every 1W that the 8ohm cab gets.

That would then suggest that if you supplied 500W to the 4ohm cab, you’d be pushing 250W to the 8ohm cab - no good!

Going the other way, if you supplied the full 200W to the 8ohm cab, you would then been pushing 400W to the 4ohm cab. Neither cab is getting power over it’s rated limit, so this ends up being the way to go. 400W + 200W then equals...600W, which is exactly what you arrived at.

 

[JimmyM]

I’m trying to plan ahead to minimize the chance of blowing the amp or the speakers.

Your power handling rate is only as good as the lowest power handling cab. But as long as you always err on the side of caution, you'll be fine. Don;t trust your ratings...trust your ears.

 

[Samatza]

Right, I intend to use two cabinets of the same wattage and resistance or a combo and an extension speaker also both of the same wattage and resistance.

OK, the cabs are matched and you won’t have an issue. The power may not double but it will be noticeable.

Generally you want the total power handling of the cabs to match or exceed the amp power at that impedance.

My GK 1001RB is rated at 700W @4 Ohms, the cabs are rated at 300W and are 8 Ohms so together the present a 4 Ohm load.

The cabs are slightly under the amp power rating @ 600W vs. 700W but since I don’t run the amp flat out it’s never been a problem. If I did run the amp hard I would probably get two 400W cabs.

EQ also plays a big part and boosting bass requires more power so it’s possible to reach the mechanical limits of the speaker before the power max.

With a combo if you match the external cab you won’t have any issues. So start with your volume requirements and match amp and cabs to suit that so you have headroom, if you are maxing out a 200W amp then maybe a 500W amp is better as you won’t be maxing it out and you’ll have some headroom for those transients.

With class D these days and much lighter cabs than were previously available it’s possible to have a powerful rig that is all to easy to transport. For decades we wanted lighter bass gear, it has arrived!

 

[themickster]

That’s a theoretical amp I don’t own yet. I was thinking it would provide 200w at 4ohms and 350w at 8ohms. My Markbass amps work that way, don’t they?

It's the other way about. More ohms mean less power. Markbass Little Mark 500 is 500w at 4 ohms and 300 at 8. The 801 Micromark is 60 at 4 but 50 at 8.

You'd be safest going with two matched speakers. Markbass make speakers to match their combos. The same goes for Trace Elliott Elves.

One 4 ohm speaker will be cheaper than two 8s. 2 1x12s will take less lugging about than a 2x12 of the same magnet kind. My neodymium Markbass 1x12 combo is half the weight of my Trace 1x15. Believe me that the 15 is harder to pick up than the Markbass and its matching outboard speaker together, and it's only 250w.

 

[M0ses]

You need to understand Ohms law to really grasp what is happening.
View attachment 4223374

How have I never seen this image before??!?!? That's so brilliantly useful. I finally have a reason to change my desktop from the circle of fifths! SAVED!

There’s another way that I approach the mixed ohm problem.

Basically, because the 500W cab has a resistance of 4ohm, and the 200W one has a resistance of 8 ohm, when the amp pumps out power, the 4ohm cab will get 2W for every 1W that the 8ohm cab gets.

That would then suggest that if you supplied 500W to the 4ohm cab, you’d be pushing 250W to the 8ohm cab - no good!

Going the other way, if you supplied the full 200W to the 8ohm cab, you would then been pushing 400W to the 4ohm cab. Neither cab is getting power over it’s rated limit, so this ends up being the way to go. 400W + 200W then equals...600W, which is exactly what you arrived at.

This is also GOOD STUFF but it's technically putting the cart before the horse and wasnex's theoretical explanation is gonna be more helpful for versatile problem solving, like when the math doesn't work out so nice and even like your example, or if you needed to do three cabs...etc

 

[Boundary Effect]

I'll emphasize something said two posts ago. Get two cabinets of the same make and model, or as you stated, the matching extension cabinet for a combo amp. This thread has focused on matching basic electrical characteristics. Two identical cabinets will mostly correlate and be evenly additive across the spectrum. Two different cabinets, even though similar in their nominal impedance, will not necessarily correlate acoustically across the spectrum.

 

[selowitch]

Two different cabinets, even though similar in their nominal impedance, will not necessarily correlate acoustically across the spectrum.

Can you elaborate on this latter point?

 

[HolmeBass]

When cabinets have different nominal impedances, the power from the amp is split in the inverse of the proportion of the impedances.

So with a 4 ohm and 8 ohm cabinet paired (net 2.67 ohm load) the 4 ohm cabinet has half the impedance of the 8 ohm one; the inverse of 1/2 is 2, therefore the 4 ohm cabinet will get twice the power that the 8 ohm one gets.

This means the 4 ohm cab will get 2/3 of the amp’s output power, while the 8 ohm cab will get 1/3. Quite sensibly 2/3 is twice 1/3.

For cabs with the same impedance, the ratio of impedances is 1; the inverse of 1 is 1, therefore both cabinets receive the same amount of power; they split the amp’s output 50:50.

8 ohms and 2 ohms: ratio is 4, therefore the 8 ohm cab gets 1/4 the power that the 2 ohm cabinet gets. The 8 ohm cabinet gets 1/5 of the total output power, the 2 ohm cabinet gets 4/5ths of the total output power. 4/5ths is 4 * 1/5.

 

[g-dude]

How have I never seen this image before??!?!? That's so brilliantly useful. I finally have a reason to change my desktop from the circle of fifths! SAVED!



This is also GOOD STUFF but it's technically putting the cart before the horse and wasnex's theoretical explanation is gonna be more helpful for versatile problem solving, like when the math doesn't work out so nice and even like your example, or if you needed to do three cabs...etc

I’d say that my approach works for people that didn’t learn it in school, and who don’t have the chart or a calculator handy.

Certainly my way is a bit more difficult in some situations - like if I were to add an 8ohm cab to my 12ohm cab (500W). Combined they would be just under 5ohm, but we’ll go with the 4ohm setting on my TT-800 which is 800W. The 8ohm cab is going to get 1.5 times as much power as the 12ohm one. Again, we’re going to do a bit of rounding and basically say that for every 8 watts, my 12ohm will get 3 and the 8ohm would get 5.

Essentially, if I get a 500W 8ohm cab, I’ll be set.

Now, let’s go back to the true math:

12ohm + 8ohm in parallel = 4.8ohm
Amp output is 800W at 4ohm means that we get 56.57 volts.

My 12ohm 500W cab is capable of 41.67 volts, but since things are going in parallel, I need to solve for half of 56.57 volts. If I do that math of (56.57/2) * 8ohms, I get 452 watts. To check, I also see about what my 12ohm cab would be getting 339 watts. Totaling those, is 791.42 because I’m doing some rounding through truncating the decimals.

Looking back at that, I retract what I said about my way being more difficult.

So for my real world situation, as I am indeed contemplating what I would need in a cab that could be added to my existing Barefaced Two10S, my rough estimate got me to 500 watts and the precise calculations got me to 452 watts.

The difference is that when I did my rough estimates, I didn’t need to look at the chart, but I did end up with a bit more headroom than I might need. If I had my heart seat on a 450 watt cab, I’d probably do the numbers precisely in a spreadsheet to make sure that I was spot on. If I was somewhere and needed MOAR POWAH, and someone offered up a spare cab, I’d go with my back of the envelope method and rock out.

 

[g-dude]

When cabinets have different nominal impedances, the power from the amp is split in the inverse of the proportion of the impedances.

So with a 4 ohm and 8 ohm cabinet paired (net 2.67 ohm load) the 4 ohm cabinet has half the impedance of the 8 ohm one; the inverse of 1/2 is 2, therefore the 4 ohm cabinet will get twice the power that the 8 ohm one gets.

This means the 4 ohm cab will get 2/3 of the amp’s output power, while the 8 ohm cab will get 1/3. Quite sensibly 2/3 is twice 1/3.

For cabs with the same impedance, the ratio of impedances is 1; the inverse of 1 is 1, therefore both cabinets receive the same amount of power; they split the amp’s output 50:50.

8 ohms and 2 ohms: ratio is 4, therefore the 8 ohm cab gets 1/4 the power that the 2 ohm cabinet gets. The 8 ohm cabinet gets 1/5 of the total output power, the 2 ohm cabinet gets 4/5ths of the total output power. 4/5ths is 4 * 1/5.

Yup.

Frankly, I find these sorts of questions are often best considered when you have different ohm ratings because the increase in difficulty helps you think through the problem.

 

[g-dude]

Can you elaborate on this latter point?

If you go and listen to speaker cabs with identical output and ohm ratings, they can often sound...different. I believe that’s the point he’s making.

 

[M0ses]

Can you elaborate on this latter point?

It seems like people continued mightily elaborating on your original question instead of this one XD I'll try but not gonna go into great detail.

"Nominal" impedance is just that, it's not exactly an "estimate", but it's far from "exact." The most important cause being that impedance is frequency-dependent. Any given cab will not respond linearly to the same voltage across the frequency spectrum. That can't be reduced to a single integer and has to be graphed.
So just because two cabinets are labeled with equal ohms doesn't mean they have actually have the same response from 20-15khz

For a real world example, let's take the same mixed cabinet numbers we were using; 500w @ 4 ohms, 200w @ 8ohms
Except let us suppose that the 200w cab is ported and the 500w cabinet is sealed. In this case, when the signal frequency approaches the ported cabinet's tuning frequency, a lot more power will start flowing into that cabinet because it's impedance goes down at that point. In the sealed box, when the driver moves backwards into the cabinet, it can't push air out of the box so it has to compress the air; that puts backpressure on the coil and bingo that's impedance. The ported box doesn't have to compress the air, the coil is unimpeded and freely moving, so impedance bottoms out.

 

[Boundary Effect]

Check out this Talkbass thread:
Mixing Cabinets and Phase Response.

 

[bobyoung53]

Kind of a dumb question, but if I am using two bass cabinets simultaneously with one head, do I add the wattage of each together to arrive at the total wattage the pair can handle? For example, if each cabinet is rated for 300 watts, can the two together handle a total of 600 watts from the head?

Yes, each cab takes half the output so each cab will still only be getting 300 watts as long as the two cabs are the same impedance, when I use two cabs I always use identical cabs, the difference between one and two cabs can sometimes be dramatic even it's a tube head which would put out the same amount of watts with two cabs as one if the impedance is correct. Having twice the cone area gives your sound a lot more presence and bottom all other things being equal.

 

[scanpire]

I am about to run an Ampeg PF 500 head out to a) Ampeg VB 210 - rated at 300W, and. b) Carvin 1x15 speaker rated at 400w) .
Theyre each 8 ohms so thats not an issue. AFter I switch out the female XLR speaker in on the Carvin to a Speakon, I’ll be ready to roll with Speakon connections for both.

Any danger in running a pair of speakers with 100W differential?

 

[agedhorse]

No because they will share power roughly equally and each cabinet will receive a maximum of ~250 watts.