6 ohm and 4 ohm cabs together

[Core Creek]

I think my math is right here but just want to make sure.

I have a Markbass 3x10 at 6 ohms and I’m thinking of getting the matching 2x10 at 4 ohms for smaller gigs.

If I ever decide to run them together the setting on my amp would need to be on 2 ohms, right?

6 ohms x 4 ohms 24
————————— —
6 ohms + 4 ohms 10 = 2.4 ohms required

Fortunately my Carvin BX700 has a switch between 2 and 4 ohms so it can be done. I doubt I’ll do it often but before I get the 2x10 I just wanted to confirm it can be done.

Thanks!

 

[Not Insane]

Close enough. Frankly, if I was worried about impedance, I'd wire them in series. I homebuilt a few "Bose stick" type speakers and I did all sorts of "parallel/series" wiring in the sticks to keep the impedance within reason.

But my final answer is still, "close enough".

 

[agedhorse]

Use the 2 ohm mode on your amp.

 

[Richard Lindsey]

Just be aware that in that scenario, more power will be going to the 2-10 cab than to the 3-10 cab, which may not be what you want.

 

[Redbrangus]

Exploring that further: it's very likely that the 310 box has three 16Ω drivers in parallel and the 210 box has two 8Ω drivers. Plugging the two boxes together is basically connecting all those drivers in parallel - the three 16Ω AND the two 8Ω. ones, so those each of those two 8Ω drivers will be dissipating more than twice as much power as each of the 16Ω ones. This may indeed not be what you want to do, even if your amp will drive the resultant 2.3Ω nominal load.

Is the 310 cab just not loud enough?

Edit: Just re-read the OP, and realize this is a tentative proposal. Yes, you can run those two cabs together, but it's not necessarily advantageous. OTOH, either of those cabs makes a dandy speaker stand.

 

[agedhorse]

I think the maximum power of his amp will be fine no matter what the combination, all of the drivers will be well within their rated power handling.

 

[FatOwlBass]

2.4 ohms. So yes, the 2 ohm setting.

Parallel Resistor Calculator - Engineering Calculators & Tools

Calculate the equivalent resistance of up to six resistors in parallel with ease while learning how to calculate resistance in parallel and the parallel resistance formula.

www.allaboutcircuits.com

 

[HolmeBass]

Someone check my math, but I think the 4 ohm cabinet will get 60% of the power, the 6 ohm will get 40%.

 

[Core Creek]

Yeah, it was just an idea. I put the 2x10 in my eBay watchlist and got an offer for $100 less, but I’m not entirely sold the 2x10 is enough for most of my gigs. I just kind of wanted the portability. But the 3x10 has been working perfect for all my gigs and weighs close to nothing. But it only fits on the front seat of my little Fiat!

 

[Rip Van Dan]

I think my math is right here but just want to make sure.

I have a Markbass 3x10 at 6 ohms and I’m thinking of getting the matching 2x10 at 4 ohms for smaller gigs.

If I ever decide to run them together the setting on my amp would need to be on 2 ohms, right?

6 ohms x 4 ohms 24
————————— —
6 ohms + 4 ohms 10 = 2.4 ohms required

Fortunately my Carvin BX700 has a switch between 2 and 4 ohms so it can be done. I doubt I’ll do it often but before I get the 2x10 I just wanted to confirm it can be done.

Thanks!

You got to the right answer but an easier way to do it is find the common denominator and convert each one to that. Then add up the total of the converted nominators. Then divide the common denominator by the total nominator and you have it.

• 12 is the lowest number that can be divided by both 6 and 4, so 12 will be the common denominator.
• 4 goes into 12 three times, so that would be 3/12. 6 goes into 12 twice so that would be 2/12
• 3+2=5 So now you have 5/12.
• Divide the common denominator "12" by the total nominators "5" and you get a 2.4 ohm load.
- 12 divided by 5 = 2.4

Like @agedhorse said, be sure you select 2Ω load on your amp. If you set it to 4Ω and run a 2.4Ω load on it, you can damage your amp if your protection circuits don't kick in soon enough to turn your amp off. Since it has a 2Ω/4Ω switch, it has different circuits for both loads and you want to use the safest minimum load circuit, which would be the 2Ω load.

 

[FatOwlBass]

Someone check my math, but I think the 4 ohm cabinet will get 60% of the power, the 6 ohm will get 40%.

I believe that's close enough. The speaker efficiency will also play into how loud you hear one cab over the other. Some drivers are audibly louder with less power than others.

 

[Wasnex]

I think my math is right here but just want to make sure.

I have a Markbass 3x10 at 6 ohms and I’m thinking of getting the matching 2x10 at 4 ohms for smaller gigs.

If I ever decide to run them together the setting on my amp would need to be on 2 ohms, right?

6 ohms x 4 ohms 24
————————— —
6 ohms + 4 ohms 10 = 2.4 ohms required

Fortunately my Carvin BX700 has a switch between 2 and 4 ohms so it can be done. I doubt I’ll do it often but before I get the 2x10 I just wanted to confirm it can be done.

Thanks!

If the amp is rated 2 ohms minimum, you need to make sure the load (2.4 ohms) is > 2 Ohms (It is ).

The other thing that would be useful is to figure out is how the two dissimilar cabs share power.

Arrange the impedance values into a fraction; assuming 4 ohm and 6 ohm cabs.
4/6

Simplify the fraction
2/3

Add the numerator and denominator
2 + 3 = 5. This is the number of shares the power is broken into.

Divide the available power by the number of shares. Let's assume the amp makes ~583W at 2.4 ohms.
583/5 = 116.6W per share
This estimated is based on the assumption the amp makes 700W at 2 ohms and produces the same voltages into a 2.4 ohm load. I'll post the Ohm's/Watt's Law formula wheel and estimate at the end.

The available power is distributed inversely to the impedances. So the 4 ohm cab gets 3 shares and the 6 ohm cab gets 2 shares.
4 ohm cab power = 3 x 116.6 = 349.8W out of 583W total
6 ohm cab power = 2 x 140 = 233.2W out of 583W total

Check results
Total Power = 349.8 + 233.2 = 583W

Make sure each cab can safely handle the expected power.

Formula Wheel

For the power estimate, I assumed the amp is rated 700W at 2 ohms and it produces the same voltage at 2.4 ohms

From the formula wheel:
V = sq rt of PZ
Let's rearrange this to
V^2 = PZ

Insert known values (700W and 2 ohms)
V^2 = 700 x 2 = 1400

The next formula we will use is
P = V^2/Z

We just calculated V^2 = 1400. Solve for P with Z = 2.4 ohms
P = 1400/2.4 = ~583.333333...(W)

 

[_One//Man//Mutiny_]

Your switch is minimum, so it’s less than 4ohms, that means 2ohm is the setting it should be on.

 

[pepj]

Unless you are ragging the amp and assuming 700w...then you should be ok matching them both but in practical terms the 310 should handle that amp on its own.

 

[Core Creek]

Unless you are ragging the amp and assuming 700w...then you should be ok matching them both but in practical terms the 310 should handle that amp on its own.

Yeah, it does fine on its own. It’s just a dab big for my car. It only fits in the front seat.

 

[Bassamatic]

Your math is correct. I like to be sure people understand that in parallel, the total impedance is always LESS than the LOWEST impedance cab.

 

[All The Way]

It’s likely that engaging the 2 ohm switch simply turns on the fan and engages a voltage limiter (to thereby limit output current)

 

[agedhorse]

It’s likely that engaging the 2 ohm switch simply turns on the fan and engages a voltage limiter (to thereby limit output current)

Depends on the design, there are different ways of approaching this.

 

[All The Way]

True

Depends on the design, there are different ways of approaching this.

True. Just pushing back on the idea of getting “more” by engaging the switch

 

[agedhorse]

True

True. Just pushing back on the idea of getting “more” by engaging the switch

Or less…