Am I hurting my speakers?

It's probably in the FAQ

That's how it works - you have a 4ohm and a 2ohm load in series.
The 4ohm will get twice the power as the 2 ohm load.
If total wattage is 300W - the 4ohm will be getting 200w, and the other two 50w each (100w total)

You should plug the missing speaker hole. It's not a breather - it's a big leak.
Since the speakers and box aren't designed together it's a coin toss on if they will sound any good.

 

Yes - the single 4 ohm gets twice the wattage of the other 2 combined.

But the hole could cause a lot of damage.

 

Actually in parallel/series with 3 speakers, all 3 will see the same power. In your example, the 2 in parallel (2 ohm) will see DOUBLE the power of the 1 in series (4 ohm), but then they have to divide that in half between them, meaning that they will all see 100W. Seeing as the driver groups are in series, the total load will be 6 ohms.

If the drivers were in series then paralleled, you would get an 8 ohm load with a 4 ohm load giving you a total of 2.66 ohms (32/12).

I can't tell which way it is from the drawing... but be careful. If it is measuring like you said, then I would imagine that it is the first scenario.

I may be way off, but if I am then I am sure that I will soon be corrected!

Ox.

 

I think that this is only true if the speakers are in series then paralleled with another. Then, in your example, both sides would see 150W, but the paralleled speakers would split the wattage to put them each at 75W.

Ox.

EDIT - And for that to work, then the one paralleled would have to be an 8ohm speaker.

 

Absolutely. A cab that's not properly tuned will unload the drivers, with over excursion leading to driver destruction.

Ever hear about the guy who jumped off a 50 story building? Residents heard him say, as he passed their windows on the way down, "So far so good".

 

Without digging into ohms law, Feeding in 300W power, each driver "X" being 4ohms, I'm just pointing out it breaks out like this:

Code:

        50W
      +--X--+
      |     |  200W
------+     +---X---
      |     |
      +--X--+
        50W

 

This is where digging into Ohm's Law is a good idea. When all the loads are parallel it's easy to mentally calculate power distribution, but when you add series loading into the mix the better method is to calculate voltage drop across each component and calculate the power from that.

 

You STILL have to block the unused holes or you will damage those drivers installed.

Paul

 

Bingo! Read BillF post again he explains it better than I.

Personally, in your place, I would leave the three speakers in there, block the remaining hole - a piece of 1/2" ply drilled to mimic the mounting holes of a speaker and mounted in its place will do fine temporarily. I'd wire the speakers in series for 12Ω and run that until you get your fourth driver. That way all the drivers get the same power.

Paul

 

I should have said "Without explaining Ohms law"
My calculations were based on current, but same results as voltage.

 

...and your calculations were correct. I did mine in my head and they were wrong. I looked at it today and felt pretty silly. Anyways... sorry for the confusion.

Ox.

 

So with regards to the OP's drawing... is it 2x4 ohm in series, then in parallel with a 1x4 ohm? Or is it 2x4 ohm in parallel, then in series with a 1x4 ohm? And how do you determine that?
Thank You