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

You got it, add them together.

But keep in mind the amp will send the power equally to both if they both have the same impedence (8ohm and 8ohm for instance).

Power ratings for a combination are only as strong as the weakest link.

So using a 500w amp with one 400w cab and one 100w cab would not mean it can handle 500w from a power handling/ratings standpoint.

There are lots of complicating variables, not the least of which are the ratings themselves, but suffice to say the 100w cab might not be happy in that arrangement.
If 400w + 100W doesnt work with a 500W amp then you're not adding them together. You're contraficting yourself. Either you add them together and 400 + 100 = 500 or you're not. Can't have both.
 
If 400w + 100W doesnt work with a 500W amp then you're not adding them together. You're contraficting yourself. Either you add them together and 400 + 100 = 500 or you're not. Can't have both.


Here is the statement you are responding to.
So using a 500w amp with one 400w cab and one 100w cab would not mean it can handle 500w from a power handling/ratings standpoint.
This is awkwardly written but correct.

If you run a 500W cab in parallel with a 100W cab, the total power handling is based on the 100W cab and how the two cabs share available power. Assuming both are the same impedance then power handling is 200W. The idea is when the 100W hits the limit, the other cab is also getting 100W. 100W + 100W = 200W

Results are different if the cabs do not have the same impedance.

If the 500W cab is 4 ohms and the 100W cab is 8 ohms, then the 500W cab will draw twice as much power. So when the 100W cab is at its limit, the other cab gets 200W. 200W + 100W = 300W

If the 100W cab is 4 ohms and the 500W cab is 8 ohms, the 4 ohm cab will draw twice as much power. So when the 100W cab it at its limit, the other cab only gets 50W. 100W + 50W = 150W

In order for these cabs to safely share 500W, the impedance of the 100W cab must be 4x higher than the 8 ohm cab. For example the 100W cab could be 16 ohms and the 500W cab 4 ohms. The combined impedance is 3.2 ohms and the voltage required for 500W is 40V. At 40V the 4 ohm cab will get 400W and the 16 ohm cab will get 100W. 400W + 100W = 500W
 
If 400w + 100W doesnt work with a 500W amp then you're not adding them together. You're contraficting yourself. Either you add them together and 400 + 100 = 500 or you're not. Can't have both.
You are completely misunderstanding the statement you posted about.

The 500 watts are shared equally, which results in a serious overpowering of the cabinet rated at 100 watts.
 
My apologies for the double post, but it seems this thread is still being found via search and referenced. So I'd like to more clearly state what is probably a better rule of thumb on this front for posterity. Who knows when somebody will revive this thread in 2029 ;)

Assuming all cabs are of the same impedance, so that power is being shared equally between them, the best rule of thumb is probably to take the lowest power rating of the cabs at hand and multiply it by however many cabs you're running.

Examples:

1. Two 300w 8ohm cabs (per the OP): Minimum power rating is 300, so 300x2= 600w

2. One 400w 8ohm cab and one 100w 8ohm cab: Minimum power rating is 100, so 100x2= 200w

3. One 800w 8ohm cab, one 400w 8ohm cab, and one 100w 8ohm cab (assuming an amp that can handle it): 100 is still the minimum, so 100x3= 300w

As I said originally: "There are lots of complicating variables, not the least of which are the ratings themselves,", but this is probably the most sensible rule of thumb in order to protect the weakest link in your cab mix.
 
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For me it's very simple as Kro indicated. There is no difference between overpowering a single cab (whatever its impedance is) and combining multiple cabs that will share power equally and having more amplifier capability available to each cab than they are rated for.

When you combine cabs of different impedance ratings the equation for how much power each cab will get is a little more complicated. However, if the math says one or more of the cabs will get more power than they are rated for, you have the exact some problem (too much power for the rating of the cab or cabs).
 
My apologies for the double post, but it seems this thread is still being found via search and referenced. So I'd like to more clearly state what is probably a better rule of thumb on this front for posterity. Who knows when somebody will revive this thread in 2029 ;)

Assuming all cabs are of the same impedance, so that power is being shared equally between them, the best rule of thumb is probably to take the lowest power rating of the cabs at hand and multiply it by however many cabs you're running.

Examples:

1. Two 300w 8ohm cabs (per the OP): Minimum power rating is 300, so 300x2= 600w

2. One 400w 8ohm cab and one 100w 8ohm cab: Minimum power rating is 100, so 100x2= 200w

3. One 800w 8ohm cab, one 400w 8ohm cab, and one 100w 8ohm cab (assuming an amp that can handle it): 100 is still the minimum, so 100x3= 300w

As I said originally: "There are lots of complicating variables, not the least of which are the ratings themselves,", but this is probably the most sensible rule of thumb in order to protect the weakest link in your cab mix.
Agreed. This is good, accurate information.