Bi-Amping Impedance Question

Ceruleance · February 2003

Alright,
so when you Biamp, you take the jumper out and run the tweeter with one amp and the woofer with another (let's keep it a two-way example for simplicity) Now, my question is, are the driver's in a speaker with a jumper hooked up in parallel or in series?

The way I see it, if you have a 4ohm woofer and a 4 ohm tweeter (these are nominal or average ratings i'm speaking of) if they are in series, you will have an 8 ohm nominal speaker when you use the jumper and a single amp, but when biamping, you will need 2 hi-current amps, because each of them will see a 4 ohm load once the drivers are seperated.

On the other hand, if they are hooked up in parallel, then by the rule 1/r1 + 1/r2 = 1/rTotal , in order to have an 8 ohm speaker, you will need a 16 ohm tweet and a 16 ohm woofer, meaning when you bi-amp each amp will see a 16 ohm load, which doesnt really make sense either.

So what am I missing? Are there resistances in the crossover that makes it work out?

So the 2 main questions:
Drivers - Series or parallel when using jumper?
Crossover - Does it play a big role in the resistance (it shouldn't, right?)

tjmmdennis · February 2003

I asked the same question a couple of months ago and never got a response.

Ceruleance · February 2003

Well I will keep bumping it till someone pays attention, with all these fights over correct terminology of bi-amping, bi-wiring, "Mantis-wiring" and all that nonsense, someone better know the answer

stevew14 · February 2003

Call Ken S at Polk and ask him. I did. See if you understand what he tells you and let us know. I'm still confused as well. I bi-amped any way and it sounds great so, I guess it's all good, but it would be nice to really grasp it

MxStYlEpOlKmAn · February 2003

Bi amping -- where u plug the tweeters in an amp, and the woofers in another....thats my take......or in others, i guess u could plug the tweeters in the receiver, and the woofers in a seperate amp or vice versa. I dunno - thats my take

Ceruleance · February 2003

You're definitely not answering the question, either that or you are answering some other irrelevant question. Thanks for your input though!...

polkatese · February 2003

Ceruleance,
I am going to take a stab answering your question. I think there is a difference between impedance and resistance, even though both were subject to ohm's law, although they might not exactly behaving the same way. If you are dealing with resistor, your points apply exactly as you stated. In term of impedance, it may not exactly work the same way. I have to believe that drivers has to be parallel when using jumpers, and crossovers should have nothing to do with resistance in that sense......it's time to dig up my old elec.eng. 101 textbook, perhaps

HBombToo · February 2003

Originally posted by tjmmdennis
I asked the same question a couple of months ago and never got a response.

I have never taken the time to call Polk and ask for the crossover circuit so I can't give much detail. If time permits tomorrow I'll call and if Polk is willing to share the crossover I'll be able to give a better answer than the following...

I pulled the crossover out of my RT55 which is biamp capable and saw (2) 5 watt resistors across the pair of posts. The resistors had slightly different resistances, **** I already forgot but I think they were 5 and 3 Ohms.

"Forgive me cause the baby just got out of the bathtub and ran into the livingroom wet and Nekkid so I scrambled to get the speaker back together. "

If I had to reverse engineer the circuit I believe the reason for the different resistors is because the tweets and woofers have slightly different impedances across the band so the 2 resistors add impedance in the network as the straps are removed. So all in all when the straps are attached the effective network is 8 Ohms and when removed it is 8 Ohms. I assume this is true because I'm sure Polk would put a note in the owners manual if a reduction of impedance would result because of potential damage to the speaker or amp and vice a versa.

This sounds reasonable to me but again I don't have the schematic and if I'm wrong I'll humbly apologize.

Resistance and impedance are really the same thing with 1 exception... we can't roll up inductive and capacitive reactance in a circuit with linear mathematics or only useing resistance. The non-real "a + bj" is the domain in which ELI the ICE man is quantified.

HBomb

HBombToo · February 2003

OK, I tied off with Ken this morning and have to Humbly apologize slightly. I was incorrect regarding the resistors... They are not across the binding posts but are in series in the high and low pass circuit. These resistors are ment to reduce the output of the tweeter so that it is not to bright or the drivers such that they are not to boomy. Makes for a nice smooth sound across the band.

What does this mean and what have I learned??? A 2 way crossover is really a combination of 2 filters high and low pass in parallel to the power source and has an 8 Ohm impedance overall but should be treated mathematically as 2 seperate circuits. I'll now state the obvious... Across the audio band the cross over splits the low and high frequency. The reactance of the high pass circuit looks like an open circuit to frequencies below the crossover point. Conversely the low pass filter is an open circuit to the high frequencies. So while removeing the straps on the the speaker we have seperated in whole the 2
high and low pass circuits and now have the ability to power each circuit seperately which are 8 Ohm circuits.

The crossover that I received from Polk for a 55i clearly showes the 6.5" woofers in series which says to me that they are 4 Ohms
drivers and the single tweeter is an 8 Ohm device.

I hope this helped and again I apologize for the confusion this rookie started.

If I have not explained this well enough please ask questions.

Regards
HBomb

Ceruleance · February 2003

Im still not quite grasping it. So what I am getting so far is:
(Let's stick with the 55i, since I have a pair too

)
-The woofers are in series with each other.
-The tweeter and woofers are two seperate circuits
-I deduced myself that there is no way to use jumpers and connect the two circuits in series, so they must be in parallel.

So each section is 8 ohm when you Bi-amp, fair enough, but then how do you get an 8-ohm load when they are connected with jumpers? You seem to be saying something about the filters making the low and high frequencies respectively not able to flow into the wrong driver. So what is the effect of this exactly?

HBombToo · February 2003

Originally posted by Ceruleance
Im still not quite grasping it. So what I am getting so far is:
(Let's stick with the 55i, since I have a pair too )
You seem to be saying something about the filters making the low and high frequencies respectively not able to flow into the wrong driver. So what is the effect of this exactly?

Your summary is right on!

Your question or confusion is regarding a very tricky concept and even for me its a tough one. To try and clarify this is very difficult but I'll take another stab at it.

In a purely resistive direct current network you are right on track... meaning 2 circuits in parallel @ 8 Ohms would be 4 Ohms but, and this is a big but, the crossover network itself is not just resistive. The high pass and low pass circuits can be thought of as either being 8 Ohms in their respective pass bands or infinite/open in impedance outside their respective pass bands.

So for instance at low frequencies the highpass circuit looks like an Open circuit and the overall circuit load remains 8 Ohms. To simplify this, imagine an 8 Ohm resistor in parallel with an open circuit... that open circuit does nothing in the way to change the impedance of the existing circuit.

Now back to the straps... With the straps connected the amplifier must provide the current to drive both independent circuits simultaneously. When we remove the straps and add an additional amp the 2 amps work independently and provide the current demand for both the high pass and low pass circuits. The biggest benefit of this type of configuration is the bass frequencies or lows do not consume the amp resources so that we exaust the amp and don't provide adaquite power for the highs in the combined settup.

If I have confused the issue even more please let me know because that is not my intent here....

HBomb

Ceruleance · February 2003

Ahhh I think I have almost got it.

I understand the concept that the filter has basically infinite resistance above or below a certain frequency (I'd be interested to know what they are made of and how this works etc., if you want to take this thread in that direction)

So I think the final assumption is that, (With the jumpers on) with two filters calibrated so that (ok bad example but just bear with me) 20hz-200hz passes through one at 8 ohms, and 200hz-20khz passes through the other at 8ohms, the combined resistance combines to 8 still?

It makes sense yet it doesnt, but it does.. ok sorry here's what I am saying: If you assume that its two seperate circuits connected in parallel by the jumpers then it doesnt make sense because 8 + 8 in parallel = 4 total.
However, If (and I am assuming that this is the way it works in reality, or else stuff wouldnt work the way it does) you think of the circuit as one entity, with 8 ohms resistance below 200hz(arbitrary) and 8 ohms above, it makes sense that over the entire audio frequency the resistance is 8 ohms. Just as if you were to hook up a single 8 ohm driver, the resistance over the entire frequency would be 8 ohms (ideally)

Ok So I think i've pretty much got the answer I wanted, and I definitely learned a lot, so thanks for sticking through it with me. Like I said I would be interested to know how the filters work, if anyone knows and cares to share!

Thanks again

HBombToo · February 2003

I'm not sure where the exact break is between the high and low pass circuits are, "Members like Russ and the Doc and Mr. Grand and BDT would be better answering this", but in general its probably around 1.5 Khz. The crossovers in the 55 are designed as second order but that is a little deaper than I would like to go.

In a general discussion I'll stay engaged though.

The high and low pass circuits are both series RLC circuit with the tweeter as the load across the inductor in the high pass circuit and the drivers in series across the capacitor/resistor in the low pass.

As far as how this works I'll try to keep it very simple.

Capacitors are electron storage devices and Inductors are magnetic field storage devices. What this means is that as voltage changes across the plates of the capacitor electrons are stored to try to maintain that voltage potential. In the inductor as current changes a magnetic field is induced such that it wants to maintain that level of current. The rate at which charge is transferred or magnetic fields are induced are the driving force between the high and low pass filters.

Long story short, capacitors look like an open to very slow changes in Potential but react to rapid changes in potential and hence a high pass filter. Inductors don't react to fast changes in current but react very well to slow changes and hence a low pass circuit.

There is a ton of math involved in the actual understanding of this and to be honest I forgot alot due to time away but in general what I have described is the physical criteria of why these things work.

Hope this helps and again I'm sorry if I caused more confusion.

Real tough topic...

HBomb

Ceruleance · February 2003

Im aware of basic capacitor and inductor theory but I didnt realize how they were used in crossover design. Pretty intelligent I would say.

-what's an RLC circuit?
-What determines the order of a crossover network?

HBombToo · February 2003

Resistor, Inductor and Capacitor Network.

How many bandpass circuits are in the crossover. Plus I think the way I ment is was what order the differential equation is when actually doing the math.

Yes the speaker designers are smart plus there is a lot of experience that is rolled up into a black art... speaker design is as probably moreso an art rather than just raw calculations. I have alot of respect for that talent because creativity makes things work not a calculator.

HBomb

I'm going to the BAR NOW... Talk to you guys when the evil twin has taken over and I have my Rooster on.

HBomb

Ceruleance · February 2003

3 way speaker equals third order crossover network? Seems like if the number of bandpass circuits and the order of the DE coincide then some of those 4 way monsters must be pretty crazy to design, 4th order DE's are a mess

Bi-Amping Impedance Question

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