Is it too low, 2-Ohm drivers?

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I’m going to start a project of building a pair of large floor-standing speakers. I plan to use a pair of 12” woofers per cabinet.

I intended to go with 4 Ohms for the woofer section, so I could save money on inductors. I’m, thus, thinking of parallel connecting two 8-Ohm drivers together to bring the impedance/ or resistance to 4 Ohms.

I found 2 pairs of 8 Ohms drivers now. Regrettably, they are out of stock when I ask the seller and the 12” woofers + 8 Ohms are hard to find locally. Yet, I found 4-Ohm drivers instead.

So, I’m thinking to change the plan. I’d like to ask everyone that if I use 4 Ohms and parallel connect them to make it 2 Ohms totally, will it have any problems?

I’m concerning on amplifier that may be danger - because of the load on 2-Ohm woofers. But I’m not going to connect them serially since I don’t want to use too large coils - too much internal resistance and too expensive - for 8-Ohm configuration.

Actually, I saw the vintage speakers: Acoustic Research AR9, use the same configuration as my project — 2 x 12” woofers. When I search for their info, I found their woofers are parallel connected and, moreover, they are 4 Ohms. Thus, the impedance/ resistance may be 2 Ohms for woofer section. Can I conclude that it’s okay to use 2-Ohm configuration for woofer connection?
 
The two AR9 woofers are wired in parallel to the same filter circuit, however they use a clever filter circuit.

Although each driver is nominally 4 ohms, it is much higher at its resonance frequency. With two such drivers in parallel, you would expect the impedance to be well above 4 ohms at resonance but down to 2 ohms or less at other frequencies. But this is altered by the arrangement of inductors and capacitors which increase the series impedance above and below the system bass resonance of 28 hz so that the combined impedance never goes much below 4 ohms at any frequency. This protects the amplifier and flattens the frequency response of the woofers at the same time.
 

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Concerning the bass portion of the AR-9 crossover. I suggest removing the 2500 uF cap - but NOT the 470 uF cap shunt to ground. The 470 is a critical part of the 200 Hz low pass filter. The 2500 uF cap was placed there to raise the impedance at resonance - otherwise the two 12" woofers are effectively in parallel with an impedance of about 2.8 ohms. An amp killer - particularly in the world of 1979. So if you do remove the 2500 uF cap make sure you have a modern solid state amp - no tubes need apply - that can handle low impedance loads.

I have found two amps that are comfortable with that load; Odyssey Audio Khartagos and Pass XS150.5. I am relatively certain that ANY Pass amp in his XS or XA models would work well with that kind of load. There may well be other solid state amps that can handle such a load - but DO NOT try to drive the modified Nine with any "vintage" receiver - the result will be smoke, possibly flames and a dead receiver. Nor try to use an H/T receiver - those things are at best rather flimsy in terms of actual power capability.
By valkyrie (Posted November 7, 2018)
"AR 9 Crossover Rebuild - Page 2 - Acoustic Research - The Classic Speaker Pages Discussion Forums" AR 9 Crossover Rebuild - Page 2 - Acoustic Research - The Classic Speaker Pages Discussion Forums
 
Could anyone help to explain the principle and the calculation of this circuit, please?

Is this filter be a 3rd-order low-pass filter?

If I would go for 2nd-order with 85 Hz x/o point, could I simply calculate it by using a normal 2nd-order method and only paralleling those large capacitor and coil to the first inductor?
 

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You ae not easily put off, are you? :D

Using two 4 ohm woofers in parallel is a bad idea, unless you are an accomplished loudspeaker designer.

The AR-9 bass filter not only takes into account the electrical parameters of the woofers, but also their acoustical interaction with the enclosure.

It's complicated! :eek:
 
Hi all,

Let's look at, finally, what I've found. I found a calculation of that L-C circuit! :eek::D

It may be a kind of bandpass zobel circuit. The formula are already shown thoroughly. Here is a full link; Zobel network - Wikipedia

I'm not a big fan of math, so I'm needing a help on it. :p I'd like to try to reverse engineering that AR9's filter.

Please educate me, what is R-naught? Where can I find it? Thanks
 

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presscot, here is an Xsim comparison. The difference is the notch. It also affects the knee of the filter but that's an incidental effect which you just need to adjust for.
So, Allen, are you advocating that presscot wire two 4 ohm woofers in parallel? :confused:
I thought you were doing quite well yourself on that front but if it helps I don't disagree with you (maybe a little on the severity, but no I wouldn't go around suggesting members do this).
 

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So, I’m thinking to change the plan. I’d like to ask everyone that if I use 4 Ohms and parallel connect them to make it 2 Ohms totally, will it have any problems?

Assuming you're still planning to use cheap woofers and XO low, then much better to series wire them since I seriously doubt the coils can handle the extra current of 2 ohms and even if they don't burn up will still obviously distort even at modest power whereas in series its inductance will double, but with a low XO point it won't matter.
 
Hold up.

What's the amplifier that will be used?

Crown MA5002VZ? Fine. Heck, go down to 1ohm if you like. At any sensible level, those things don't care.

If you've got a fairly delicate solid-state amp with no over-current protection, forget it. Build for 8ohm.

Chris
 
What if I will parallel two 4 Ohms woofers to be 2 Ohms totally, then, series with a 2 Ohms resistor? I know it will act as an l-pad, but here we have two drivers, instead of a single. Thus, the SPL may drop very little, am I correct?

Another way is to series the two woofers to get 8 Ohms, then, parallel with a 8 Ohms resistor. However, I have been warned for a long time ago about not to do this action. Still, I’m looking for a confirmation again. :rolleyes:
 
No, not recommended. Compared to the way Cal has suggested, the sensitivity will be the same for all options, and you will use double the power and half of that will be wasted.

Does same sensitivity mean I will hear sound at the same decibel compared to a conventional 4 Ohms configuration?

And, does “I will use double the power and half of that will be wasted” mean I will have to turn the volume up for that equal decibel heard?
 
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