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15 minutes ago, CECAA850 said:

The resistor had long leads so I actually mounted it to the crossover at the terminal strip.

That's the easiest way to do it.  This is an easy/inexpensive way to lower the voltage across the driver AND keep the current "through" the 13uf capacitor close to a stock AA crossover.

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10 minutes ago, mboxler said:

That's the easiest way to do it.  This is an easy/inexpensive way to lower the voltage across the driver AND keep the current "through" the 13uf capacitor close to a stock AA crossover.

It's easily reversible as well.  The tap wire was soldered originally.  I unsoldered it and crimped a female spade connector on it so now I can return it to a stock AA configuration without even removing it from the speaker.  I doubt I ever will though but someone in the future might like that option.

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6 hours ago, Marvel said:

There are multiple threads on here about how much power a constant impedance network uses. The 'swamping' resistor across the the autoformer will work fine.

 

I understand that, but not that the AA was ever a "constant impedance" design.  To be made into one surely requires a change in reactive component values to match the new impedance constant.  Right?

 

When I said what I said I was assuming the reactive components were going to remain unchanged.

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1 hour ago, glens said:

 

I understand that, but not that the AA was ever a "constant impedance" design.  To be made into one surely requires a change in reactive component values to match the new impedance constant.  Right?

 

When I said what I said I was assuming the reactive components were going to remain unchanged.

 

I agree. I don't think the AA would be considered a constant  impedance design. If that is something someone feels the need for, then ALK has some networks that will approximate that goal. IMO, a decent amplifier should be able to deal with changes in impedance (within reason). Although there are some tube amps (not most) that will track the load impedance. 

 

Just as a general comment, why are folks wanting to drop the midrange output on the Klipsch balancing network? A 3 dB change is actually a significant difference. I imagine the designer spent some time developing that network. Do you think they got it wrong? Is it possible that the "Klipsch sound" is not your cup of tea? I am not being mean, but there are a ton of speakers out there, you should be able to find one that has a design (components, drivers etc - all working together) that meets an individual's satisfaction. 

 

Good Luck,

-Tom

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Tom... For over 15 years I've read posts of people saying the didrange on LaScalas through the k400/401 was too painful with the gain up. Dropping the mids down 3db resolves that, to some degree, and turning it up gives you more bass. To me, the balance just seems a lot better. I should geta mic and REW to check.

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10 minutes ago, Marvel said:

Tom... For over 15 years I've read posts of people saying the didrange on LaScalas through the k400/401 was too painful with the gain up. Dropping the mids down 3db resolves that, to some degree, and turning it up gives you more bass. To me, the balance just seems a lot better. I should geta mic and REW to check.

You don’t need a mic and REW.

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Hey guys, it's okay to turn down the mid-range. My point is that the balancing network is a fundamental piece of the speaker system and a 3dB shift is substantial. If things are that far off, then maybe Klipsch is not the system for you.. There is no "wrong answer". 

 

The Klipsch system may have been designed for a different sort of listener. There may be designs that are better designed for your preferences. Perhaps Klipsch is not one of them. 

 

Personally, unless you are willing to go to DSP, I think they got it right.

 

Good Luck,

-Tom

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26 minutes ago, PrestonTom said:

Personally, unless you are willing to go to DSP, I think they got it right.

 

I believe Tom Brennan called the LS midrange a 'chainsaw in my forehead'.

 

To each his own. I like it tuned down a bit. I don't have an equalizer in my main system.

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On 4/1/2020 at 3:39 PM, CECAA850 said:
On 4/1/2020 at 3:27 PM, mboxler said:

That's the easiest way to do it.  This is an easy/inexpensive way to lower the voltage across the driver AND keep the current "through" the 13uf capacitor close to a stock AA crossover.

 

It's easily reversible as well.  The tap wire was soldered originally.  I unsoldered it and crimped a female spade connector on it so now I can return it to a stock AA configuration without even removing it from the speaker.  I doubt I ever will though but someone in the future might like that option.

 

So what value resistor did you use and were exactly is it connected?

 

I've been mulling this over and there are too many unknowns at the moment.  You said "across the squawker" earlier, so does that mean across the same (new) taps?  Dropping a tap will double the impedance, so to restore that would require halving the impedance at the new tap to maintain filter position and profile.  You've lost half the signal at the new tap, and it'll be another half across the new resistor which would need to be the same value as the driver, so one -3dB tap nets you -6dB out the driver.  That's really substantial. 

 

If you used a 32 ohm resistor across the "whole" transformer (the primary) I'm thinking that would be the same as 8 ohms along with the driver at the -6 tap.  Assuming it's an 8 ohm driver anyway.  Either way would be correct for maintaining the crossover point, and would be -6dB from the starting point.

 

Any other value resistor (for whichever connection scheme) will most likely cause a change in the squawker response profile (as shown in the following quote) and somewhere between -3 and -6 dB output.

 

 

In order to merely decrease the level of the squawker by 3dB (a substantial change!) you really need to do one of three specific things:

 

- drop one tap and double up on the input cap [edit: that would be two in series, perhaps a can of worms for the tweeter as suggested next, too]

 

- leave the cap and tap alone and install a 16 ohm resistor in series with the input side of the cap (and shift the tweeter input cap ahead of the resistor, but that may be another small can of worms itself)

 

- leave everything else alone and install two resistors in an L-pad configuration between the crossover and driver.  I'd be happy to do the math to find the values needed (but likely not exactly available)

 

Personally I feel option two is the best, and that I'd start substantially smaller than 16 ohms because -3 dB is a lot!

 

I've not done current research on crossovers with a swamping resistor and readily-changeable taps, but have to think that the reactive components are sized appropriately with/for that consideration, and that the components you have are not correct for such usage.

 

 

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By across the squawker, we mean in parallel anywhere between taps 0 - 3 and the K-55.  Placing the resistor across the + and - terminal block screws going to the K-55 is a convenient spot.

 

A parallel resistor in and of itself will not change the voltage across the squawker.  What it will do is draw more current from the amplifier.  If the resistor is equal to the squawker impedance, the current drawn will double.  Since the extra current passes "through" the 13uf capacitor, it charges twice as fast.  This, in turn, doubles the corner frequency from 400 hz to 800 hz.  We haven't reduced the voltage to the squawker, it just takes an octave longer to reach maximum voltage.

 

By dropping the output taps to 0 - 3, we cut the current across taps 0 - 5 in half.  Now the current through the 13uf capacitor is equal to the stock setup, and it's corner frequency is back to 400 hz.  The maximum voltage across the squawker, however, is down 3 db at all frequencies.

 

I assume a 14 - 16 ohm resistor was used???

 

L-pads work well because they drop the voltage across the squawker without changing the current drawn from the amplifier.  The 13 uf cap charges at the same rate.

 

Now let's place the so-called swamping resistor (usually 10 ohms) across taps 0 - 5.  It has no effect on the voltage across taps 0 - 5, but it will draw a significant amount of current, and that current stays pretty much the same no matter what output taps are used.   A 13 uf capacitor will charge way too fast.  It needs to be replaced with, say, a 48 uf cap that will take longer to charge. 

 

I don't understand option two above.  It would drop the voltage throughout the circuit, but it would also lower the corner frequency of the 13 uf capacitor because less current is "flowing" through it.

 

At least I think that's right.  No matter what, if the change sounds better to you, it is better!

 

Mike

 

 

 

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7 hours ago, mboxler said:

A parallel resistor in and of itself will not change the voltage across the squawker.  What it will do is draw more current from the amplifier.  If the resistor is equal to the squawker impedance, the current drawn will double.  Since the extra current passes "through" the 13uf capacitor, it charges twice as fast.  This, in turn, doubles the corner frequency from 400 hz to 800 hz.  We haven't reduced the voltage to the squawker, it just takes an octave longer to reach maximum voltage.

 

By dropping the output taps to 0 - 3, we cut the current across taps 0 - 5 in half.  Now the current through the 13uf capacitor is equal to the stock setup, and it's corner frequency is back to 400 hz.  The maximum voltage across the squawker, however, is down 3 db at all frequencies.

 

Okay, so the driver has an impedance of 16 ohms, not 8, is what I'm gathering.  

 

I'm getting my noodle wrapped around this a little better.  (No whiskey yet today!)

 

Given:  the "corner" frequency is defined as that at which the capacitor (alone, in this case) and the driver (in conjunction with the transformer, in this case) have exactly the same impedance each.  By merely dropping to the next output tap, the impedance of the load becomes double, in this case.  Doing nothing more would result in halving both the output of the driver and the corner frequency.  By doubling up the load on the new output tap you're restoring the corner frequency.  Because you're using pure resistance to do so you're merely throwing away as heat half the power in the midrange circuit, which overall power level remains the same as before any changes.

 

This whole concept of transformers in crossovers is new and alien to me, and I'd somehow gotten the notion (and I know better!) that the resistor in parallel with the driver would further reduce driver output.  I apologize for that.

 

The option "2" above would achieve the same results (lowering driver output while maintaining the balance between transformer+driver / capacitor impedances), yet allow for much finer granularity in driver output changes.  3 dB is a lot of change per step!

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16 minutes ago, glens said:

 

Okay, so the driver has an impedance of 16 ohms, not 8, is what I'm gathering.  

 

Yes.  Actually,  I believe Bob Crites measured a K-55 driver in a K400 horn and came up with closer to 14.5 ohms, but you get the point.

 

16 minutes ago, glens said:

I'm getting my noodle wrapped around this a little better.  (No whiskey yet today!)

 

Given:  the "corner" frequency is defined as that at which the capacitor (alone, in this case) and the driver (in conjunction with the transformer, in this case) have exactly the same impedance each.

 

If you mean the frequency at which the impedance of the capacitor is equal to the impedance of taps 0 - 5,  yes.  Using 14.5 ohms across taps 0 -4 as an example, the load across taps 0 - 5 would be double that, or 29 ohms.  A 13 uf capacitor reaches 29 ohms around 422 hz.  

 

16 minutes ago, glens said:

  By merely dropping to the next output tap, the impedance of the load becomes double, in this case.  Doing nothing more would result in halving both the output of the driver and the corner frequency.  By doubling up the load on the new output tap you're restoring the corner frequency.  Because you're using pure resistance to do so you're merely throwing away as heat half the power in the midrange circuit, which overall power level remains the same as before any changes.

 

A 14.5 ohm resistor in parallel with the 14.5 ohm driver gets you a 7.25 ohm load.  I realize the driver is not a pure resistive load, but...  By moving these to taps 0 - 3, the reflected load across taps 0 - 5 is 4 times that load, or 29 ohms, right back to stock.

 

 

16 minutes ago, glens said:

 

This whole concept of transformers in crossovers is new and alien to me, and I'd somehow gotten the notion (and I know better!) that the resistor in parallel with the driver would further reduce driver output.  I apologize for that.

 

No apology needed.  You had me questioning myself.

 

16 minutes ago, glens said:

 

The option "2" above would achieve the same results (lowering driver output while maintaining the balance between transformer+driver / capacitor impedances), yet allow for much finer granularity in driver output changes.  3 dB is a lot of change per step!

 

I'm not saying that the series resistor won't work, it just works differently.  In my example, adding a 14.5 ohm resistor in series with 29 ohm impedance of taps 0 - 5 results in a 43.5 ohm load on the 13 uf cap.  As mentioned before, this load is in parallel with the k-77 circuit, so that will change some as well.

 

Again, if it sounds good, it is good.

 

Mike

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15 ohm mills resistor installed on the crossover terminal strip from squawker + to squawker -.

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On 4/1/2020 at 7:07 PM, PrestonTom said:

 

Just as a general comment, why are folks wanting to drop the midrange output on the Klipsch balancing network? A 3 dB change is actually a significant difference. I imagine the designer spent some time developing that network. Do you think they got it wrong? Is it possible that the "Klipsch sound" is not your cup of tea? 

The mids just seemed to overwhelm the highs and lows to me.  It's possible that it's  a function of my aging ears inability to hear upper frequencies as well as I used to but others seem to like the change also.  They seem more balanced now.

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The mids just seemed to overwhelm the highs and lows to me. 


Same here. For quite a few years I've been EQing in the digital domain and it works for me but some day I'd like to mod the xover to see if it's any better.

It seems I come across so many that despise the Klipsch sound. I just had this discussion with my tech last week. However, I've seen a few now that have had a total change of heart when they've heard Klipsch done right by whatever means neccessary.

Klipsch stock, flat is not my cup of tea but I gotta say that I heard the new Cornwall IV's recently (no EQ) and I was impressed. I'm wondering what they changed......



Sent from my SM-G950U using Tapatalk

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The mids just seemed to overwhelm the highs and lows to me.  It's possible that it's  a function of my aging ears inability to hear upper frequencies as well as I used to but others seem to like the change also.  They seem more balanced now.

Well, also the same frequencies as a displeased female voice, so there”s that.

Did I just say that?
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On 4/1/2020 at 7:07 PM, PrestonTom said:

 

Just as a general comment, why are folks wanting to drop the midrange output on the Klipsch balancing network? A 3 dB change is actually a significant difference. I imagine the designer spent some time developing that network. Do you think they got it wrong? 

 

Good Luck,

-Tom

 

29 minutes ago, CECAA850 said:

The mids just seemed to overwhelm the highs and lows to me.  It's possible that it's  a function of my aging ears inability to hear upper frequencies as well as I used to but others seem to like the change also.  They seem more balanced now.

Klipsch designs for the masses (and do a great job at it). Many speakers and horn loaded ones especially, react differently in different environments and positions within that environment.To tweak your speakers once in place to your situation (and in Carl's case, age) is not claiming anyone got it wrong. There just may be a better setting for your set of parameters. Plus, it is a hobby, have fun with it. I for one like the midrange where it is and my ears are older than Carl's

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6 minutes ago, codewritinfool said:


Well, also the same frequencies as a displeased female voice, so there”s that.

Did I just say that?

I am unaware of that sound. I always please.

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4 minutes ago, MookieStl said:

I am unaware of that sound. I always please.

You need to send me a cheat sheet.

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