Filtering the facts

[John Y.]

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]

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]

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]

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]

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.]

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]

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]

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.]

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.]

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.]

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]

Love this Hartsfield !!
Where are the pictures we want to look ??

Eckhard

[Alex Lancaster]

JohnY: Look up Zobel in google, it will give You the answer, which is a little too long for here.

[John Y.]

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]

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