The value of the swamping resistor is entirely up to you. Feel free to choose a value about 1.05 to 1.1x higher than the Z-average of the driver that your highpass crossover is designed for.
Since the resistor is in parallel with the complex impedance appearing at the primary of the transformer, if you choose a value just slightly higher, it'll drop right in.
Small is beautiful when it comes to HF performance from transformers. Since the highpass filter thoughtfully removes all of the LF content, why not optimize for HF performance?
I think you'll like what you hear with the nickel-core autoformers. What I heard with the early prototypes was a synergy between the clarity and directness of the AH425 and the beautiful tone that autotransformer attenuation gives you. Please tell us what your impressions are when you get it working.
Since the resistor is in parallel with the complex impedance appearing at the primary of the transformer, if you choose a value just slightly higher, it'll drop right in.
Small is beautiful when it comes to HF performance from transformers. Since the highpass filter thoughtfully removes all of the LF content, why not optimize for HF performance?
I think you'll like what you hear with the nickel-core autoformers. What I heard with the early prototypes was a synergy between the clarity and directness of the AH425 and the beautiful tone that autotransformer attenuation gives you. Please tell us what your impressions are when you get it working.
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just take typical Tannoy autoformer ;
data is around ; knowing L and Rdc ( that one as pretty easy target to beat) , it's easy to calculate air choke for exact L of lowest tap , then just add rest of taps as percentage of starting winding , according to calc from voltage db increments
I intend to do exactly that for my fun with this
so , recipe (using Lalena online calc , in past confirmed with measurements) :
Inductance 2.56 mH
DC Resistance 0.28 Ohms
Wire Gauge 14 AWG
Wire Diameter 64.1 mils (1 mil = .001 in)
Coil Length 0.65 in
Coil Inner Diameter 1 in
Coil Outer Diameter 3.44 in
Average Turn Diameter 2.14 in
Wire Length 106.89 feet
Copper Weight 1.33 pounds=600gr
Turns 191
Levels 18.84
Turns/Level 10.14
that's black-green as in original Tann autoformer
add 38-40 turns for yellow tap
for more background , look at atachedf file
Tann autoformer displayed just as example , that's what I have data for (and ,incidentally, what I'm messing with lately
)
edit : I hope IW capacitance will not make a mess with additional 1.5db , which is my goal ....
data is around ; knowing L and Rdc ( that one as pretty easy target to beat) , it's easy to calculate air choke for exact L of lowest tap , then just add rest of taps as percentage of starting winding , according to calc from voltage db increments
I intend to do exactly that for my fun with this
so , recipe (using Lalena online calc , in past confirmed with measurements) :
Inductance 2.56 mH
DC Resistance 0.28 Ohms
Wire Gauge 14 AWG
Wire Diameter 64.1 mils (1 mil = .001 in)
Coil Length 0.65 in
Coil Inner Diameter 1 in
Coil Outer Diameter 3.44 in
Average Turn Diameter 2.14 in
Wire Length 106.89 feet
Copper Weight 1.33 pounds=600gr
Turns 191
Levels 18.84
Turns/Level 10.14
that's black-green as in original Tann autoformer
add 38-40 turns for yellow tap
for more background , look at atachedf file
Tann autoformer displayed just as example , that's what I have data for (and ,incidentally, what I'm messing with lately
)edit : I hope IW capacitance will not make a mess with additional 1.5db , which is my goal ....
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Passive crossover - a question
Hello,
I have one question regarding the crossover between 2. (horn mid) and 3. (woofer). You seem to use symmetrical 3rd order/3rd order slopes. So how do you deal with time alignment, if at all?
By looking at the picture you posted in a previous thread it looks as if in your speakers, like in virtually all Woofer + horn systems in fact, the acoustic centre of the woofer sits quite a bit closer to the listener than that of the compression driver.
It is my understanding that the common way to deal with this using passive crossovers is to increase the slope on the speaker unit (the woofer in this case) that sits closer. TAD and Rey Audio, for instance, use a 6th order low-pass on the Woofer coupled with a 2nd order on the Horn (the acoustic transfer function of the high-passed horn then transitions to a 6th order-plus below the horn's Fc, but that is another matter...).
Cheers,
Marco
Hello,
I have one question regarding the crossover between 2. (horn mid) and 3. (woofer). You seem to use symmetrical 3rd order/3rd order slopes. So how do you deal with time alignment, if at all?
By looking at the picture you posted in a previous thread it looks as if in your speakers, like in virtually all Woofer + horn systems in fact, the acoustic centre of the woofer sits quite a bit closer to the listener than that of the compression driver.
It is my understanding that the common way to deal with this using passive crossovers is to increase the slope on the speaker unit (the woofer in this case) that sits closer. TAD and Rey Audio, for instance, use a 6th order low-pass on the Woofer coupled with a 2nd order on the Horn (the acoustic transfer function of the high-passed horn then transitions to a 6th order-plus below the horn's Fc, but that is another matter...).
Cheers,
Marco
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Amp recommendations and output impedance - a question
I've read this kind of recommendation several times now, but I would still appreciate a clarification:
All the amplifiers you mention as "recommended" will have a non-negligible (e.g. 1 ~ 2 Ohm) output impedance.
This INVARIABLY results in the frequency response being affected by the impedance variations of the loudspeaker (low frequency peak(s) due to the Woofer, and typically one mid-frequency hump due to the crossover).
[for those who may doubt this, see for instance:
- Amplifier Output Impedance
- Loudspeaker Enclosure
- Avalon Eclipse loudspeaker Measurements | Stereophile.com]
So, does this mean that the "Beyond the Ariel" loudspeaker system is "voiced" so as to somewhat compensate this behaviour by NOT being linear in its frequency response when purely voltage-driven (as would be the case when using a ~zero outupt-Z transistor amp)?
If this is the case, does this not leave a system whose performance, at least frequency response-wise, is a bit of an unknown, given that no two high-output-Z valve amps will have exactly the same output Z?
Or do you just like the other attributes of your preferred amps (harmonic structure, dynamic range, 'soft' clipping, etc.) so much that you choose to base your recommendation on those, while just accepting a somewhat unstable frequency response (and consequent 'voicing') as an inevitable trade-off?
This is an honest question, I am NOT being polemic!
Thanks,
Marco
I've read this kind of recommendation several times now, but I would still appreciate a clarification:
All the amplifiers you mention as "recommended" will have a non-negligible (e.g. 1 ~ 2 Ohm) output impedance.
This INVARIABLY results in the frequency response being affected by the impedance variations of the loudspeaker (low frequency peak(s) due to the Woofer, and typically one mid-frequency hump due to the crossover).
[for those who may doubt this, see for instance:
- Amplifier Output Impedance
- Loudspeaker Enclosure
- Avalon Eclipse loudspeaker Measurements | Stereophile.com]
So, does this mean that the "Beyond the Ariel" loudspeaker system is "voiced" so as to somewhat compensate this behaviour by NOT being linear in its frequency response when purely voltage-driven (as would be the case when using a ~zero outupt-Z transistor amp)?
If this is the case, does this not leave a system whose performance, at least frequency response-wise, is a bit of an unknown, given that no two high-output-Z valve amps will have exactly the same output Z?
Or do you just like the other attributes of your preferred amps (harmonic structure, dynamic range, 'soft' clipping, etc.) so much that you choose to base your recommendation on those, while just accepting a somewhat unstable frequency response (and consequent 'voicing') as an inevitable trade-off?
This is an honest question, I am NOT being polemic!
Thanks,
Marco
Hi Gary and Lynn,
Thank you.
What is/are the practical implication(s) of choosing between 8 or 16 Ohm drivers?
The fellow I talked to over the phone told me about 3 setups. His setup consists of single horn and single woofer in a wooden, recatngular enclose per side. His woofers have similar sensitivity to the horns, from memory, over 110dB. Another setup has additional 'super-tweeters' for frequencies well above 20KHz. In another setup, there is a single woofer which is a compression driver in a huge horn (probably not applicable to my small living room). It looks like I'm going to have very interesting listening sessions, possibly followed by changes to my own setup.
Have you posted details of the crossovers you use?
If not, will you please post it?
Thank you.
What is/are the practical implication(s) of choosing between 8 or 16 Ohm drivers?
The fellow I talked to over the phone told me about 3 setups. His setup consists of single horn and single woofer in a wooden, recatngular enclose per side. His woofers have similar sensitivity to the horns, from memory, over 110dB. Another setup has additional 'super-tweeters' for frequencies well above 20KHz. In another setup, there is a single woofer which is a compression driver in a huge horn (probably not applicable to my small living room). It looks like I'm going to have very interesting listening sessions, possibly followed by changes to my own setup.
Have you posted details of the crossovers you use?
If not, will you please post it?
some woofers in some horns have that type of voltage sensitivity. Direct radiators pretty much top out around 98 db @ 60 cycles
I was also told once that unicorns exist and pigs fly.
Sorry, I couldn't resist.
A 110dB/W woofer is a physical impossibility.
There are only three practical ways to increase efficiency:
- increase the radiating surface (Sd) - but there are practical limits
- decrease the moving mass (Mms) - but here too there are practical limits before the cone begins to flex uncontrollably
- increase the magnetic force factor (BL^2/Re) - but if increased too much, then the speaker loses all capability of reproducing low frequencies, i.e. it ceases to be a Woofer
If you actually do the calculations, the practical limit for, say, a 15" woofer over the intended frequency range (i.e. NOT in a narrow spike around 1kHz
) is ~100dB/W(m).Marco
110dB woofers! That is amazing! https://www.youtube.com/watch?v=-sbFhOeqTzY
I have been reading up on the subject in this article. http://www.alkeng.com/dload/swamper.pdf
In the example of the Klipsch AA Network he moves the pin from -3dB to -6dB, in attempt to lower the squawker's (K55) volume. This causes the slope to go from 450Hz to 250Hz. He doesn't want this to happen, so he introduces a 10 ohm swamping resistor, which makes the amp see a 9.8 ohm load instead of a 32 ohm load (with the -3dB setting).
Coincidently, 250Hz is where I want to cross my JA66981B 16 ohm driver later.
What the article does not say is how autransformers creates 1st order, 2nd order, or 3rd order slopes. Is the article just talking about an L-pad or is the autotransformer doing something for the filter?
I am guessing that it adds a 1st order, which together with the "autotrafo protection cap" before creates a 2nd order filter.
Interestingly the article uses a 13 uf cap, but in other examples I have seen 40 uf.
g3dahl, you mentioned a 3rd order filter: "Radian 745NEO/Be (16 ohm) driver, Azurahorn AH425, 3rd order high pass at ~700 Hz, connected to -17.5 dB tap on Slagle autoformer with 35-ohm swamping resistor."
Do you use another autotransformer to add a third order?
Also, why a 35-ohm swamping resistor? I presume you have a 16 ohm midrange compression driver like in the example (Klipsch K55 16 ohm). Why are you not using a 10-ohm swamping resistor like in the example?
From the article:
"Recall that resistors in parallel total up as 1 / (1/R1 + 1/ R2). What value of resistor could we connect in parallel with the 32 Ohms looking into the autoformer to bring the total down to 8 Ohms? The value turns out to be about 10 Ohms. 1 / (1/32 + 1/10) = 7.6 Ohms. We can connect it right across the 0 and 5 taps of the transformer. Now, if
we move the transformer tap all the way down to the 15 dB setting, where the turns ratio becomes 5.62, the impedance ratio becomes 31.6, or the square of the turns ratio. The 16 Ohm load now looks like 506 Ohms (31.6 X 16) looking into the transformer. Using the parallel resistance formula again the filter sees 1 / (1/10 + 1/ 506) = 9.8 Ohms. Any 8 Ohm system would be happy with that load! The impedance seen by the filter can’t go higher than 10 Ohms even with no load at all!"
I have been reading up on the subject in this article. http://www.alkeng.com/dload/swamper.pdf
In the example of the Klipsch AA Network he moves the pin from -3dB to -6dB, in attempt to lower the squawker's (K55) volume. This causes the slope to go from 450Hz to 250Hz. He doesn't want this to happen, so he introduces a 10 ohm swamping resistor, which makes the amp see a 9.8 ohm load instead of a 32 ohm load (with the -3dB setting).
Coincidently, 250Hz is where I want to cross my JA66981B 16 ohm driver later.
What the article does not say is how autransformers creates 1st order, 2nd order, or 3rd order slopes. Is the article just talking about an L-pad or is the autotransformer doing something for the filter?
I am guessing that it adds a 1st order, which together with the "autotrafo protection cap" before creates a 2nd order filter.
Interestingly the article uses a 13 uf cap, but in other examples I have seen 40 uf.
g3dahl, you mentioned a 3rd order filter: "Radian 745NEO/Be (16 ohm) driver, Azurahorn AH425, 3rd order high pass at ~700 Hz, connected to -17.5 dB tap on Slagle autoformer with 35-ohm swamping resistor."
Do you use another autotransformer to add a third order?
Also, why a 35-ohm swamping resistor? I presume you have a 16 ohm midrange compression driver like in the example (Klipsch K55 16 ohm). Why are you not using a 10-ohm swamping resistor like in the example?
From the article:
"Recall that resistors in parallel total up as 1 / (1/R1 + 1/ R2). What value of resistor could we connect in parallel with the 32 Ohms looking into the autoformer to bring the total down to 8 Ohms? The value turns out to be about 10 Ohms. 1 / (1/32 + 1/10) = 7.6 Ohms. We can connect it right across the 0 and 5 taps of the transformer. Now, if
we move the transformer tap all the way down to the 15 dB setting, where the turns ratio becomes 5.62, the impedance ratio becomes 31.6, or the square of the turns ratio. The 16 Ohm load now looks like 506 Ohms (31.6 X 16) looking into the transformer. Using the parallel resistance formula again the filter sees 1 / (1/10 + 1/ 506) = 9.8 Ohms. Any 8 Ohm system would be happy with that load! The impedance seen by the filter can’t go higher than 10 Ohms even with no load at all!"
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Good question. I'd like to hear an answer to this.
LeCleach designed a crossover alignment for best impulse response with time alignment a factor, but that design was apparently not used here.
g3dahl: Okay, I am slowly getting this.
With 8-ohm or 16-ohm using Slagle's $350 autotrafo a 35-ohm swamping resistor is used.
With Critesspeakers $80 a 10-ohm swamping resistor is preferable.
Not sure what the German uses, but probably something similar. He also has a knob that in some way controls slopes.
But how do you alter the slopes in a network like this? The -3dB, -6dB, etc, pinouts seem to decide volume, and thanks to the swamping resistor, the filter is NOT changed. But I would like to change it to whatever frequency and order I need, which is 3rd or 2nd order high pass at 250Hz for a 16 ohm driver.
Edit:
Aha. Can it be that autotrafos has nothing to do with slopes?
"The autotransformer used in crossovers is essentially an impedance multiplier. In other words, if you have a 16 ohm driver and want to attenuate it's output by 3db, the autotransformer makes the driver look (to the amp) like a 32 ohm driver. And, if you attenuate the driver by 6db, it looks like a 64 ohm driver to the amp.
This provides attenuation without loss in the crossover. Most other methods of attenuation involve using resistors which "waste" some of the power so it never gets to the driver. With the autotransformer, that power never leaves the amp since the driver just appears to the amp as a higher impedance.
Bob Crites"
With 8-ohm or 16-ohm using Slagle's $350 autotrafo a 35-ohm swamping resistor is used.
With Critesspeakers $80 a 10-ohm swamping resistor is preferable.
Not sure what the German uses, but probably something similar. He also has a knob that in some way controls slopes.
But how do you alter the slopes in a network like this? The -3dB, -6dB, etc, pinouts seem to decide volume, and thanks to the swamping resistor, the filter is NOT changed. But I would like to change it to whatever frequency and order I need, which is 3rd or 2nd order high pass at 250Hz for a 16 ohm driver.
Edit:
Aha. Can it be that autotrafos has nothing to do with slopes?
"The autotransformer used in crossovers is essentially an impedance multiplier. In other words, if you have a 16 ohm driver and want to attenuate it's output by 3db, the autotransformer makes the driver look (to the amp) like a 32 ohm driver. And, if you attenuate the driver by 6db, it looks like a 64 ohm driver to the amp.
This provides attenuation without loss in the crossover. Most other methods of attenuation involve using resistors which "waste" some of the power so it never gets to the driver. With the autotransformer, that power never leaves the amp since the driver just appears to the amp as a higher impedance.
Bob Crites"
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I found this quote. Does this suggest that in some designs a autotransformer can be used in creating the crossover slopes and not just act as an L-pad?
"In the classic JBL crossovers, the autoformer inductance WAS the crossover inductance. This requires a precise and calibrated airgap in the core, so that the autoformer inductance can be specified with adequate precision."
Also, this perhaps explains the L-pad knob in the German's crossover:
"Use a fixed resistor after determining the 'build-out' resistor value with the pot or L-pad."
So you dial in what swapping resistor you want then them measure, and then get that fixed value. Nice.
"In the classic JBL crossovers, the autoformer inductance WAS the crossover inductance. This requires a precise and calibrated airgap in the core, so that the autoformer inductance can be specified with adequate precision."
Also, this perhaps explains the L-pad knob in the German's crossover:
"Use a fixed resistor after determining the 'build-out' resistor value with the pot or L-pad."
So you dial in what swapping resistor you want then them measure, and then get that fixed value. Nice.
Yes. The autotransformer can be specified to provide inductance for the filter, or it can be designed to be used only as an attenuator. I chose the latter approach, which allows me to make changes to the crossover without having to modify or replace the autotransformer.
I won't have time to respond to the other recent questions until later in the week, sorry!
Gary Dahl
I won't have time to respond to the other recent questions until later in the week, sorry!
Gary Dahl
Martin - Thank you for sharing.
If you feel like experimenting… Try lifting your horns a few inches above the bass cabinet so that there is clearance below them. Listen and share.
The difference I hear when doing this is not subtle!
Pierre
Hello,
If you can use the autoformer as a crossover inductor u can save 1 component.
Very important to save money in commercial systems.