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John Y.
10-11-2004, 05:27 AM
Let's say, for the sake of argument, that one were to take a 3115 crossover and separate the HF section from the LF section to yield passive filters.

Further suppose that one were to have two amplifiers, each fed a full range signal, with one amp connected to the HF (high pass) filter and HF driver, the other connected to the LF (low pass) filter and woofer.

Without getting into an active vs. passive filter discussion, is there anything wrong with doing as I described? I'll tell you why I am asking later, but I don't want the discussion to deviate from the fundamentals.

Thanks for your input.

John Y.

speakerdave
10-11-2004, 05:32 AM
I believe what you describe is called "biwire" and is built into some speakers, like my LSR 32's.

David

John Y.
10-11-2004, 05:45 AM
Originally posted by speakerdave
I believe what you describe is called "biwire" and is built into some speakers, like my LSR 32's.

David

Doesn't biwire utilize the same amplifier channel to drive both sections over different wires? I'm asking about using two completely separate amplifiers, each of which has an input fed with the same full range signal.

John

speakerdave
10-11-2004, 05:54 AM
My LSR 32's have two sets of connectors which are normally strapped. When the straps are removed an amp channel can be connected to each set. One would power the woofer the other would power the midrange/tweeter. Between the woofer and its connectors is a low pass network; between the mid/tweeter and its set of connectors is a high pass network which also divides frequencies for the midrange and tweeter. Both networks may also provide other response tailoring compensation. I believe this is called biwire to distinguish it from biamping which requires external crossovers.

David

speakerdave
10-11-2004, 06:31 AM
Some speakers use a combination of the methods. My 4333A's have two sets of terminals and a switch. For use with a single amp channel the switch is set in one position which connects the drivers through the full network, providing frequency division and compensation; only one set of terminals is used. In the second switch position the terminals normally used connect only to the woofer, and the network to the woofer is truncated; the low pass filter and response tailoring are removed, but there remains a zoebel across the woofer to smooth impedance variations. This setup eliminates the big inductor from the woofer circuit, thereby improving transient response in the low frequencies. An external low pass network would be used between the preamp and the power amp.

An amplifier for the mid and high frequencies would then be connected to the other set of terminals. The internal high pass network for those drivers remains in the circuit.

In the literature these speakers are called "biampable", but it is a mixture of biamping and what came to be called biwiring as I described in my earlier post.

The advantage both of this method of biamping and biwiring is to be able to choose different amps for high and low frequencies, as is often desireable.

David

John Y.
10-11-2004, 06:55 AM
Originally posted by speakerdave
My LSR 32's have two sets of connectors which are normally strapped. When the straps are removed an amp channel can be connected to each set. I believe this is called biwire to distinguish it from biamping which requires external crossovers.

David

David,

When you say "an amp channel can be connected," are you saying separate amplifier channels, or the same amp channel connected to each of the two terminals? The LSR32 manual (pdf from the JBL site) talks about "passive bi-amping", which I take to mean separate amplifier channels - one to each terminal.

Do you have access to the LSR32 schematic?

John

Earl K
10-11-2004, 07:01 AM
Hi John


is there anything wrong with doing as I described? I'll tell you why I am asking later,

I don't see anything wrong with this approach - after all - it's just taking advantage of the inherent "parralleled circuit topology for this sort of passive crossover .

You could/might run into some unprediticable phase issues if the two amps are of different makes ( or types ). I speculate that direct coupled amps vs transformer coupled may not be in relative phase to each other . On the input side ; transformer balanced vs capacitor isolated/active could easily be in a 90 phase quadrature. A few extra 90 phase rotations, unknowingly induced by different manufacturers' amps could break apart any "designed-in" cohesion ( in the crosssover/speaker system ). A person should check out ( think/follow through ) all these phase issues, if going this route .

Why do you ask ?

regards <> Earl K

Robh3606
10-11-2004, 07:04 AM
Sure you can do that it's Passive Biamping like in the LSR manual. On the down side running both full range you won't get the increased headroom and lower distortion levels that an active crossover set-up would give you. Curious to know why you are asking.

Rob:)

still4given
10-11-2004, 07:26 AM
It would seem to me that driving low frequencies with the amp that is driving the HF driver is just wasting horsepower. I doubt that driving the LF amp at full range would matter all that much but still, why make the amp work to produce frequencies that aren't going to be heard?

Blessings, Terry

Alex Lancaster
10-11-2004, 07:50 AM
It looks like You could burn up both sections of the separated xover, where is the energy that is filtered going to go?

Mr. Widget
10-11-2004, 09:27 AM
John,

As long as you fully separate the two sections, no common point (neg), there will be no problem at all. Both amps need to be inverting or non inverting. If they are different you could always flip the polarity at one crossover input.

I could see this being useful if you were going to use a low power tube amp or a small class A solid state on the HF section and wanted a solid state brute on the woofer. You would want to remove the padding built into the crossover for the HF section.

In this sort of set up you would gain some of the benefits of biamping but not all. It seems like a lot of trouble and expense to achieve only some of the benefits. I would build a passive network in front of the amps. That would give you all of the benefits.

Widget

Mr. Widget
10-11-2004, 09:29 AM
Originally posted by Alex Lancaster
It looks like You could burn up both sections of the separated xover, where is the energy that is filtered going to go?

The same place it always goes. It is dissipated as heat by the inductors and capacitors of the respective network.

Widget

Alex Lancaster
10-11-2004, 09:51 AM
Widget:

Yes, but in a normal passive xover, the bass energy goes to the woofer, the HF to the tweeter, if You run the amps full range, You will overload the LRC components, it is not the same.

John Y.
10-11-2004, 09:51 AM
Originally posted by speakerdave
Some speakers use a combination of the methods. My 4333A's have two sets of terminals and a switch.
In the second switch position the terminals normally used connect only to the woofer, and the network to the woofer is truncated; the low pass filter and response tailoring are removed, . . . This setup eliminates the big inductor from the woofer circuit, thereby improving transient response in the low frequencies.

An external low pass network would be used between the preamp and the power amp.

An amplifier for the mid and high frequencies would then be connected to the other set of terminals.

The advantage . . . is to be able to choose different amps for high and low frequencies, as is often desireable.

David

David,

That answers my last post, which I wrote after you sent the above.

I understand that you have to apply a low pass active (desirable) or passive (less desirable) filter to keep the HF out of the woofer path, unless the woofer has a natural roll off at the intended crossover.

John

Zilch
10-11-2004, 10:27 AM
Originally posted by Alex Lancaster
It looks like You could burn up both sections of the separated xover, where is the energy that is filtered going to go? It doesn't "Go" anywhere, it's simply blocked. It is never drawn out of the amp.

Consider, for example, a capacitor in series with a tweeter. Is there any "dissipation" of low frequencies there? Nope, they don't pass through, is all.

Depending upon the design of the the crossover, there may be attenuation of the high frequency leg(s) to balance the inherent higher efficiency of those drivers. There seems to be little value in retaining that function if they are to be driven by a separate amp.

Similarly, is there any dissipation of high frequencies in an inductor in series with a woofer? Well, maybe, but minimal. The primary dissipation there is the IR drop across its impedance by the low frequency current that IS passed to the woofer.

So, as Terry suggests, what is "wasted" is the full-range capability of each of the several amps, assuming that is what is used. Presumably, however, one can still employ different kinds and sizes of amps for low vs. mid/high drivers, as desired.

"Biwire" to me implies separate amps or separate channels thereof, at least as much as any advantage (dubious) that might accrue from running different wires from the same source to low vs. mid/high drivers.

Bottom line: Sure, it'll work just fine....

John Y.
10-11-2004, 10:28 AM
Originally posted by Mr. Widget
The same place it always goes. It is dissipated as heat by the inductors and capacitors of the respective network.

Widget

This is one of the reasons that I started this thread. I have been troubled by the reasoning of what happens to the energy.

Let's take a solid state non-transformer coupled amp. You could feed an AC signal all day at large level and nothing would happen if there was no connection to a load. Right? You have an infinite impedance load. Energy would remain as potential energy in the power supply capacitor. Is my thinking going astray?

I specified non-transformer because the primary does load the circuit and the transformer would draw current based on the primary impedance.

John

Mr. Widget
10-11-2004, 10:30 AM
Originally posted by Alex Lancaster
Widget:

Yes, but in a normal passive xover, the bass energy goes to the woofer, the HF to the tweeter, if You run the amps full range, You will overload the LRC components, it is not the same.

I am no expert here but my understanding is that if you hook an amp running full frequency to a tweeter with a cap in series (a simple first order crossover) the capacitor acts like an open circuit or extremely high value resistor at low frequencies. Wouldn't the amp simply see this as no load?

Widget

Earl K
10-11-2004, 10:32 AM
but I don't want the discussion to deviate from the fundamentals.


Okay:

"Capacitance Reactance" & "Inductive Reactance" are measured in ohms. The impedances ( in ohms ) are frequency dependant .

ie ;
A capacitor "shows/presents" a frequency dependant resistance to the output of an amplifier . Below it's Fc the amp starts to see a "dam" ( actually more like "weir" ) that regulates and slows the flow of current in the lower frequencies by presenting greater and greater resistance ( height of dam ) to the current.
Conversely above Fc, the current is allowed to flow fairly unimpeded ( with a 90 phase shift ) since a capacitor is essentially an "energy storing" device .

An inductor also "shows" a frequency dependant resistance to the output of an amplifier . Above its frequency dependant Fc point, an opposition to current flow is induced that becomes greater and greater with increasing frequencies .

It's this "frequency dependant" current regulation that does the filtering/crossover functions. Some energy is stored and dissipated as heat in both devices - & that's why they have ratings in DC & AC voltage ( physical size matters here )

The amount that "one-dam" effects ( interacts with ) the other "dam" is debated. As long as both "sluice-gates" are balanced in size ( impedance ) interaction will be negligable. Unbalanced "sluicegate" effects can be rebalanced at the amplifiers themselves .

( since I'm self-taught, I'll entertain any & all corrections to the above - please include your "learned" sources) Mine are from most of the relevant SAMS books .
<> Earl K

Mr. Widget
10-11-2004, 10:43 AM
That's the way I remembered it. I actually studied this stuff in school many, many, many.... years ago.

BTW, John I don't think there would be a problem with the transformer coupled amps either.

Zilch
10-11-2004, 10:51 AM
Originally posted by John Y.
This is one of the reasons that I started this thread. I have been troubled by the reasoning of what happens to the energy.

For the most part, nothing. It's never drawn out of the amp. If you disconnect the woofer in a system, does the high-pass still operate to the tweeter? Of course.

Is there some danger of "smoking" the crossover because it's blocking the low frequencies and they are not being used by the non-existant woofer? Nope.

With the exception of any attenuation components and the IR drop across inductors, passive crossovers are not dissipative, by design. No heatsinks required.

John Y.
10-11-2004, 12:27 PM
Originally posted by Mr. Widget
John,

I could see this being useful if you were going to use a low power tube amp or a small class A solid state on the HF section and wanted a solid state brute on the woofer. You would want to remove the padding built into the crossover for the HF section.

Widget

To all who were so kind as to give of their time and knowledge - Thanks. Widget sort of hit the nail on the head, but not exactly.

I am preparing to incorporate a pair of 2206H woofers into a center channel. My mains are 4648A-8's with 2440 HF units and 3115 XO's. I want to duplicate the sound in the center, particularly as to the crossover and the 2440's. Obviously, I don't have room for a 15" woofer set on top of my RPTV, but I can use a pair of 2206H's.

What to do with the crossover? The center is driven by a 75 watt channel on my Rotel. I could, as many have suggested, forego use of the center channel amplifier and use the center preamp out into an active crossover. Need a LF amp that would handle 4 ohms.

My first thought is to use the center amp on the Rotel (with upper part of the 3115 to drive the 2440, and utilize the pre center out to feed a 75 watt stereo amp with lower part of a 3115 and a 2206H on each channel .

A fall back position would be to use a JBL impedance matching transformer (which I have) to make the paralleled 2206H's 8 ohms and just drive the center with the Rotel center amp through the 3115 crossover. Probably a big downside there.

Anyway, I have learned a lot and thank you all for your help.

John Y.

Earl K
10-11-2004, 12:57 PM
Hi John


The center is driven by a 75 watt channel on my Rotel.

Will the Rotel drive a 4 ohm load ?

If so, I believe your least expensive option is to just rebuild one 3115 crossover - or - make the woofer leg from scratch ( the easiest & least instrusive).

Remember; when your speaker load impedance is halved so is the value of the necessary inductor . Conversely, when halving the woofer load impedance, one doubles the uf value of the paralleled shunt capacitor .

Now, This is all to keep the same @ crossover point as the 3115. I didn't look at the circuit to see if it has a Zoebel included and/or a load resistor - if so those parts would also need changes to their values .

<. EarlK

tomp787
10-11-2004, 02:11 PM
Hello Mr Yoder,

I'm a bit late in this thread but I'll answer your original question anyway. Audio power amps are generally considered as voltage sources - that is they will output a certain voltage into any load resistance (impedance). So if you increase the impedance the maximum power goes down. Power is generally defined as voltage squared divided by resistance (impedance). As explained earlier the crossover presents a high impedance to out of band signals. The constant voltage divided by the hign impedance becomes a low power signal, so low power (heat) is dissipated.

I think your rotel center channel will work great. Doesn't the center channel have resticted low frequencies anyway? The matching transformer should work as well but I don't think you will need it.

Did you ever build a companion for your Hartsfield? I'm still interested if you ever do!

Best Regards,
Tom

John Y.
10-20-2004, 09:28 AM
Originally posted by Earl K
Hi John


Will the Rotel drive a 4 ohm load ?

If so, I believe your least expensive option is to just rebuild one 3115 crossover - or - make the woofer leg from scratch ( the easiest & least instrusive).

<. EarlK

EarlK,

I took too long to answer this so I hope you see it.

The Rotel that I use will not drive 4 ohms. I may, however, use it to drive the HF through the 3115 high leg and take the center pre out to another 4 ohm capable amp loaded with a paralleled 3115 low leg and the pair of 2206Hs.

John

John Y.
10-20-2004, 09:38 AM
Originally posted by Earl K
Hi John


Remember; when your speaker load impedance is halved so is the value of the necessary inductor . Conversely, when halving the woofer load impedance, one doubles the uf value of the paralleled shunt capacitor .

Now, This is all to keep the same @ crossover point as the 3115. I didn't look at the circuit to see if it has a Zoebel included and/or a load resistor - if so those parts would also need changes to their values .

<. EarlK

EarlK,

So, constructing an identical low leg of the 3115 and paralleling it with the original does, in fact , halve the inductor and double the capacitor, setting it up for a 4 ohm speaker load.

There is a resister and series capacitor across the low output of the 3115. Should this be duplicated and paralleled in the dual 3115? What is a Zoebel?

John

John Y.
10-20-2004, 09:43 AM
Originally posted by tomp787
Hello Mr Yoder,

Did you ever build a companion for your Hartsfield? I'm still interested if you ever do!

Best Regards,
Tom

Tom,

Nice to hear from you. I still have the Hartsfield construction on hold. As you may know, the JBL tent sale provided me with 4648-8s which I am using with the 2440's/horns/lenses that were destined for the Hartsfield project. This is my current home theater setup, along with L100s for the surrounds.

John

paragon
10-20-2004, 09:59 AM
Love this Hartsfield !!
Where are the pictures we want to look ??

:p Eckhard

Alex Lancaster
10-21-2004, 07:51 AM
JohnY: Look up Zobel in google, it will give You the answer, which is a little too long for here.

John Y.
10-21-2004, 10:08 AM
Originally posted by Alex Lancaster
JohnY: Look up Zobel in google, it will give You the answer, which is a little too long for here.

Alex,
Thanks. It appears that the Zobel is used to filter the impedance rise at the higher frequencies. According to the calculator in www.the12volt.com (which relates to car audio, but is basic) JBL used a Zobel (13.5mf, 10 ohms) in the 3115 to compensate for a doubling of an 8 ohm impedance at 1500 Hz. Since paralleled 2206H's are 4 ohms and the impedance doubling should occur at 1500 Hz, I need (26.5 mf, 5 ohms) according to the calculator. This would be what I should get just paralleling the low legs of two 3115's, and I will be exactly on target.
Thanks to all who have helped to enlighten me. :)
John

Earl K
10-21-2004, 10:17 AM
Hi JohnY

Run the numbers for calculating the Inductor and Capacitor sizes for a ( 7 to 8 ohm ) load, crossing over at @ 500 hz.

You'll find that the original 3115 won't give you a 500 hz crossover point . Those 3115 inductor & capacitor values, will only "work" with the older, higher impedance woofers that predate JBLs' move to the somewhat standardized 8 ohm woofers.

The 3115a has component values that are more applicable for an 8 ohm woofer .

<> Earl K

Earl K
10-21-2004, 11:29 AM
JohnY,


- This thread ( up until quite recently ) has been about whether or not one can "splitup" the LP & HP legs of the typical parallel-type crossover ( as opposed to the series topology ) . The answer was "yes" the legs can be split as long as they are truly split. Splitting is especially important when 2 different amplifiers are invovled.



Will the Rotel drive a 4 ohm load ?

If so, I believe your least expensive option is to just rebuild one 3115 crossover - or - make the woofer leg from scratch ( the easiest & least instrusive).


The above quote ( of my words ) is not a confirmation that a single ( or twin ) 8ohm woofer(s) would achieve a 500 hz crossover point from the Low Pass leg of an "unaltered" 3115 network.

FAR FROM IT: It won't since the Inductor & Capacitor values are set to operate with a higher impedance woofer.


You've made incorrect assumptions and drawn the wrong conclusions from my words.

Earl K

John Y.
11-14-2004, 01:16 PM
Earl K.,

From your comments:

Quote:
Will the Rotel drive a 4 ohm load ?

If so, I believe your least expensive option is to just rebuild one 3115 crossover - or - make the woofer leg from scratch ( the easiest & least instrusive).
</B> The above quote ( of my words ) is not a confirmation that a single ( or twin ) 8ohm woofer(s) would achieve a 500 hz crossover point from the Low Pass leg of an "unaltered" 3115 network.

FAR FROM IT: It won't since the Inductor & Capacitor values are set to operate with a higher impedance woofer.

"You've made incorrect assumptions and drawn the wrong conclusions from my words."

Let me dig myself out of this one. You stated that the woofer leg of a passive crossover can be independently used as a filter between an amp output and a speaker. I used the example of the 3115 without the implication of use with an 8 ohm woofer. JBL literature states that it may be used with 8 - 16 ohm woofers, which always bothered me. I think you are correct in your warning that with a 4 ohm load, the crossover point would not be 500 Hz. The woofer knee would shift. That said, I have a 3115A, which should be correct for my application.

In any event, two woofer legs that are paralleled (if they are separated from the tweeter legs) should result in halving the inductor values and doubling the capacitor values, which is exactly what we want in order to have paralleled woofers at the same crossover point as the single woofer with only one crossover. Right?

Seems logical that this might be done just paralleling two crossover lower legs at the inputs and connecting their outputs individually each to an 8 ohm woofer. Should give the same result as paralleling each component (inductor, capacitor, woofer) internal to the crossover. Agree?

What do I do about the Zobel? Use the same values, or parallel them?

Thanks, John Y.

Earl K
11-14-2004, 02:18 PM
Hi JohnY

I'm buried in work for about a week or so. I'd like to address your questions but in fact I can't even really focus my eyes on the computer screen properly. So, to make sure all the "nuances" of the printed words are addressed properly, I'm going to have to put this off till next Sunday when the skies clear ( metaphorically ).

- Maybe someone else in the meantime will address this . ( I do think you have the logic correct, but I need to revisit your comments about the Zobels ).

regards < Earl K

John Y.
02-13-2005, 07:00 PM
Hi JohnY

I'm buried in work for about a week or so. I'd like to address your questions but in fact I can't even really focus my eyes on the computer screen properly. So, to make sure all the "nuances" of the printed words are addressed properly, I'm going to have to put this off till next Sunday when the skies clear ( metaphorically ).

- Maybe someone else in the meantime will address this . ( I do think you have the logic correct, but I need to revisit your comments about the Zobels ).

regards < Earl K

Earl,

I have had no further input to the discussion so I intend to proceed with a simplistic approach of figuring the inductor/capacitor values for my situation and replacing (adding) these to a 3115A crossover then testing to assure I have everything right for a 500 Hz crossover of 8 Ohm woofers in parallel.
Thanks for all your help.

John Y.

duaneage
02-13-2005, 09:41 PM
Earl K.,

From your comments:

Quote:
Will the Rotel drive a 4 ohm load ?

If so, I believe your least expensive option is to just rebuild one 3115 crossover - or - make the woofer leg from scratch ( the easiest & least instrusive).
</B> The above quote ( of my words ) is not a confirmation that a single ( or twin ) 8ohm woofer(s) would achieve a 500 hz crossover point from the Low Pass leg of an "unaltered" 3115 network.

FAR FROM IT: It won't since the Inductor & Capacitor values are set to operate with a higher impedance woofer.

"You've made incorrect assumptions and drawn the wrong conclusions from my words."

Let me dig myself out of this one. You stated that the woofer leg of a passive crossover can be independently used as a filter between an amp output and a speaker. I used the example of the 3115 without the implication of use with an 8 ohm woofer. JBL literature states that it may be used with 8 - 16 ohm woofers, which always bothered me. I think you are correct in your warning that with a 4 ohm load, the crossover point would not be 500 Hz. The woofer knee would shift. That said, I have a 3115A, which should be correct for my application.

In any event, two woofer legs that are paralleled (if they are separated from the tweeter legs) should result in halving the inductor values and doubling the capacitor values, which is exactly what we want in order to have paralleled woofers at the same crossover point as the single woofer with only one crossover. Right?

Seems logical that this might be done just paralleling two crossover lower legs at the inputs and connecting their outputs individually each to an 8 ohm woofer. Should give the same result as paralleling each component (inductor, capacitor, woofer) internal to the crossover. Agree?

What do I do about the Zobel? Use the same values, or parallel them?

Thanks, John Y.

I have a simple answer for you, maybe it will work or not.

Wire the woofers, tweeters, the crossover, and anything else you have. Connect a signal generator to this with a voltmeter and sweep through the desired range. Record what the impedances are and what the voltages are across all the drivers. After this is mapped out you can see what effect changes will have. I always do this no matter how I "model" crossover designs. Then you will know what works.

You are looking at quite a bit of interaction and phase relationships and we have time delay caused by driver distances affecting the polar response at the crossover frequency, with driver polarity reversed or not, depending on what slope and cutoff combined with filter Q and parasitic resistance caused by inductors.

Whew that was a long sentence, eh?