Increasing speaker impedance on the cheap?

[Serow]

Hi all, I recently received a pair of HK Soundstick speakers for free, ATPM 10.04 - Review: Harman Kardon Soundsticks II

They are mini line arrays, with 4 8-ohm speakers wired in parallel for a nominal 2-ohm impedance.

I'm trying to track down a cheap commercial/DIY amplifier that would give me 10-15watts into their 2-ohm load, but I think that might be tough...

The other alternative that comes to mind is increasing their impedance.

What would I lose if I rewired them in a 2-series/2-parallel manner, to get a nominal 8-ohm load? If I remember correctly the efficiency will go down (by 3 or 6db?), but I'm thinking it would be much easier to find a cheap amplifier that would give 50 watts into 8 ohms than 10 watts into 2 ohms

Another alternative would be to leave them as is but add a series resistor, I believe. Normally I know this isn't good because it would raise the effective Q, but since I'm planning on crossing these speakers over fairly high (150Hz?) maybe it would be OK in this case?

Lastly, I've heard about using autoformers/transformers. The former seem $$$ from a brief search. Is there a cheap way I could get these speakers up to 4 or 8 ohms using a toroid transformer, given that they're only going to be used for mid/highs?

Speaking of toroids, would it be possible to kill two birds with one stone and use an crossover inductor to both perform the high-pass and increase the impedance that the amplifier will see, so that I could use an amplifier designed for 4/8ohm loads? I know very little about passive crossover design

Thanks very much for your help,
Eric

[speaker dave]

You are absolutely right that a series parallel arrangement will give you an 8 ohm impedance. Your sensitivity will drop by 6dB but, as you say, a 50 watt amp can be found and will have more power than the units need.

Autoformers can be used. If you step down the voltage you will step up the effective impedance. You either need a large inductance autoformer to run full range or a smaller one where the inductance creates a crossover effect. You could look for cheap low power tube output transformers. Place your units across the 4 ohm secondary and the 16 ohm tap (and ground) would give you an 8 ohm effective impedance. Don't connect the primary and insulate it well! You might get by with something small (say a 10 watt rated transformer) since you could keep the lows out of it and prevent core saturation at low frequencies.

Really, the series parallel connection is easiest.

David S.

[MarcelvdG]

If the four drivers are identical, the series-parallel connection sounds like the best solution. The frequency response won't be affected and the sensitivity in dB at 1 m, 1 W remains the same, although you will need twice the voltage (and half the current) to get a certain power level.

[sbrads]

The series parallel combo is the best (cheap) practical solution but you may get a looser bass as there'll be little damping from the amp compared with the parallel 2 ohm arrangement.

Another way may be to have 2 separate 8R//8R=4R circuits, with 2 sets of cables to 2 separate 4 ohm amps, you could even use the 2 channels of a stereo amp.

[speaker dave]

Originally Posted by sbrads

The series parallel combo is the best (cheap) practical solution but you may get a looser bass as there'll be little damping from the amp compared with the parallel 2 ohm arrangement.

No, with series parallel you get exactly the same LF. Since the impedance curve of one unit equals the impedance curve of the next you can put two in series and they divide the input volts exactly in half. No response curve change equals no damping change.

David S.

[MarcelvdG]

I agree with David. Suppose you connect two identical loudspeakers in series. When you consider one loudspeaker, it will indeed experience less electromagnetic damping than when it is directly tied to the amplifier. This would normally lead to a reponse peak (loose bass), but in this case it doesn't because the peak is exactly compensated for by the impedance peak of the other loudspeaker, as explained by David.

[sbrads]

Originally Posted by MarcelvdG

I agree with David. Suppose you connect two identical loudspeakers in series. When you consider one loudspeaker, it will indeed experience less electromagnetic damping than when it is directly tied to the amplifier. This would normally lead to a response peak (loose bass), but in this case it doesn't because the peak is exactly compensated for by the impedance peak of the other loudspeaker, as explained by David.

Not my experience at all. Electrically what you are saying is correct as far as frequency response is concerned, but I think you underestimate the braking effect of the amplifier on the speaker cone. There'll be extra overhang after a bass note ends. How much depends on the speaker and it may not be noticeable but I bet it will be if the speaker quality is high.

[speaker dave]

Originally Posted by sbrads

Not my experience at all. Electrically what you are saying is correct as far as frequency response is concerned, but I think you underestimate the braking effect of the amplifier on the speaker cone. There'll be extra overhang after a bass note ends. How much depends on the speaker and it may not be noticeable but I bet it will be if the speaker quality is high.

People frequently get confused when they talk in terms of the amplifier's braking motion of the woofer. In the end damping is just about the amplifier maintaining flat voltage in the face of the woofer's rising impedance at resonance. If the voltage at the woofer terminals is flat then the amplifier has exerted "maximum damping". If it isn't, then full Qe isn't developed and Qt slides towards the Qm value. With flat terminal volts total system Q will be lowest and transient response will be fastest damped.

The second way to look at it is that "overhang of the bass note" is always tied to woofer frequency response. If two woofers in series have the same frequency response (as one) then they absolutely have to have the same transient response. (Its a Fourier transform thing.)

David S.

[sbrads]

We'll have to agree to disagree about this one.

I think more in terms of back emf current being shorted out for maximum damping through its electromotive braking effect from it's reaction with the speaker magnet. I mean, you only have to finger tap on a woofer cone with and without the terminals shorted to hear the difference - it's massive with most speakers. Put two woofer cones in series and short the outer two terminals and now tap either cone and there will be a bit of resonance where there wasn't as much before because back emf current flow is now halved. If the 2 cones were physically 'locked' together then you would get full damping as they would effectively become one big speaker with one coil, but they're not locked together in the real world, there's springy resonant air coupling them, so although they're driven by the same signal and 'should' be moving exactly the same, that's not necessarily the case with practical speaker cabinets with their internal resonances and nodes, not to mention the room. So I see your theoretical answer as a 1st order approximation, not wrong as such.

[sreten]

Originally Posted by sbrads

We'll have to agree to disagree about this one.

Hi,

Well the actual facts of the matter is the series/parallel arrangement
will have slightly more damping than the all parallel arrangement given
a finite amplifier output resistance.

Whilst what is above might be true it is of no practical consequence.

However the differences will be trivial, series / parallel will not affect
bass damping, not that these speakers have any bass in the first place.

rgds, sreten.