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Crossover Capacitors and Crossovers In General


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2 hours ago, Curious_George said:

 

“Contradictions do not exist. Whenever you think that you are facing a contradiction, check your premises. You will find that one of them is wrong.” ~ Ayn Rand

A premise is "facts and/or data". Not all data is accurate and not all facts are facts.

 

I know we are talking about capacitor data in this case and it is hard to dispute manufacturer data such this. But a blanket statement like "facts & data are not biased" is not necessarily accurate. 

 

 

yes i agree that not all data is always accurate, too many variables, probably should have worded that better.  for this topic let's say any data provided is as accurate as it can be.   

 

by definition a fact is a fact until proven untrue, & for audio capacitors or test results of audio components, those are not likely to change anytime soon.  but i see your point. 

 

 

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8 hours ago, captainbeefheart said:

 

What he means is the output impedance from most SET amps is high due to them not using any negative feedback from the output to lower it. Since the output impedance is so high with any variation in load impedance you get large voltage differences. An amplitude vs frequency plot shows this well and isn't a straight line. The amp behaves more like a constant current source than the ideal voltage source.

 

With a voltage source the voltage stays relatively the same and current though the load changes. With a current source the current through the varying load stays relative the same but voltage changes.

 

He is saying his networks are flat for impedance vs frequency.

 

Yes, this part we know. High output impedance means the frequency response will track with the impedance of the loudspeaker, resulting in non-flat response.

 

The real question has always revolved around the use of the swamping resistor on the autotransformer. What is the cost? As someone here used to say, quite often actually -- there is no free lunch. 

 

The resistor gets hot. This should be adding to the insertion loss, and costing the speaker some sensitivity. I've never really understood what might be happening on the amplifier end. I did try the network with a Wright Labs 2A3, and it really did seem like it ran out gas real quick. 

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8 hours ago, Schu said:

Simply studying something fundamentally changes that thing - inserting a bias

 

7 hours ago, 001 said:

 

that may be true, but the facts & data are not biased...

 

i usually prefer a more straightforward approach to looking at facts & data, for the most part i am capable of setting aside any personal bias when presented with facts.  having an open mind allows one to learn new things.  

 

I've decided that some subjects and/or topics are so complex and nuanced that it's better to stay fluid in your thinking.

 

Ayn Rand is correct in saying “contradictions do not exist”.

Ayn Rand is not correct in saying “contradictions do not exist”.

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54 minutes ago, Crankysoldermeister said:

The real question has always revolved around the use of the swamping resistor on the autotransformer. What is the cost? As someone here used to say, quite often actually -- there is no free lunch. 

 

The resistor gets hot. This should be adding to the insertion loss, and costing the speaker some sensitivity. I've never really understood what might be happening on the amplifier end. I did try the network with a Wright Labs 2A3, and it really did seem like it ran out gas real quick. 

 

The resistor costs the speaker zero loss in regard to sound pressure out. It only costs the amp some power that wouldn't be going through the squawker anyway.

 

It's purpose is the auto-former is there to reduce the signal to the squawker, the issue is it squares the squawker impedance by the amount of turns; i.e. it increases the impedance the amp sees to 30 ohms. The resistor is only there so the amp sees a flat response at 8 ohms instead of 30 ohms. The same amount of current flows into the squawker with or without the resistor there so sound pressure remains the same. The amp needs to supply extra current which passes through the resistor. I.e. it doesn't take any power away from the drivers, it only pulls power from the amp.

 

 

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Basically look at it this way. Given the same exact output voltage the speakers will have the same exact output in sound pressure with or without the resistor. The difference is at certain frequencies (midrange) the amp will produce more current but the output pressure is the same from drivers.

 

So yes overall system efficiency is reduced, but it doesn't cost you any sound pressure out of the speakers with you having the volume control at the same exact spot. E.g. The speakers will have the same sound pressure output at the same spot on the volume control. You don't need to turn the amp up any louder like you would with a harder to drive set of speakers. The amp works a little harder is all at those frequencies.

 

 

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It's confusing but let's look at it in more depth with specific amplifiers.

 

With a solid state amplifier, voltage source amp like 99% of them. The speaker is rated for 8 ohms anyway and so the amp will have no issues with the resistor in there, it's as if it were driving a more flat speaker system rated for 8 ohms that doesn't use an auto-former. I don't see why Al is so hung up on making the amp care to see 8 ohms at those frequencies with a SS amplifier. Current is reduced through the mid horn from the higher series impedance of the auto-former which reduced it's output. Goal is achieved so why care what the amp sees? I don't see it as any issue whatsoever.

 

The only issue I can see is when using a tube amplifier (pentode/tetrode operation) with an output transformer. The transformer reflects the load of the speaker back to the tube. We would like to keep the load of the tube as consistent as possible for linear performance across all frequencies. With the 30 ohm load at the secondary of the output transformer it's going to reflect back a MUCH higher load to the tube. For a triode output tube this isn't an issue. Possibly why PWK's favorite amplifier was a Brook tube amp with output triodes. BUT, with pentodes/tetrodes if the load is too high it brings the load line down below the knee of the curves which gives worse performance. The load is no longer optimal, higher distortion is the product.

 

So solid state amps why waste power through the output transistors with the swamping resistors? I see no advantage whatsoever. If I'm wrong and their is an advantage please correct me. With a tube amplifier it's worth having the swamping resistor in the network so the reflected load back to the tube is in it's correct most linear part of it's operation.

 

So there you have it. If you run a SS amp the networks are fine without the resistor, I don't see the amp caring about seeing a 30 ohm load and less heat is wasted through the output devices. With a pentode/tetrode output tube amplifier with output transformer then you may want to try the swamping resistor to flatten out the load impedance and keep the tubes in their linear region. This is what Al refers to on his website, he incorrectly actually states a SET amp, which he assumes a high output impedance which is silly. You can of course use gobs of feedback with a SET amplifier if you have enough open loop gain and the output impedance would be very low, certainly low enough to be considered a voltage source to the speaker load and not track the impedance. He should correct it and say for Pentodes/tetrodes. Triodes typically will have reduced distortion as load impedance increases. Too high an impedance and you don't get much current change at all and little power into the load. We find happy middle grounds with triodes to achieve power at a satisfactory level of distortion. It's all about the plate load characteristics of different types of tubes.

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12 hours ago, 001 said:

some comments on this thread, a couple phone calls & the many PMs ive received have mostly answered the capacitor questions i was asking recently. 

 

Will you please post a summary of the answers to the questions you raised?

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On 8/30/2022 at 8:53 PM, Crankysoldermeister said:

Al used to tell me they were perfect for SET amps because of their benign impedance. I’m like, “Al, they’ve only got 3 watts and now they only have 1.” I honestly still don’t know who was right. 


I certainly tried my best in the 2016 thread I linked in my previous post and I am absolutely shocked at your confusion of who was right after my simplified examples and excellent explanations….!!! 😄

 

miketn

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Yeah, great thread. Don't forget, I'm the "I wish I knew all of this math" guy, that's always being squished between two or more people. Conceptually I get it, but I don't do well in the weeds. That was also a time when there was a lot of off forum stuff going on with Al, who kept reeling me back in with his take on things. Not always an easy place to be. I never recommend the network for those running low powered tube gear - I do have personal experience to rely on, and my experience was the same as yours. 

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@captainbeefheart Thanks for the detailed explanations. I'm pretty familiar with the principle of operation of that network, I've been building it for many years. My questions have always been more about what the amp is doing. I didn't mention anything about sound pressure, but about sensitivity. I realize that with sufficient power, you can overcome what the network is taking away from you. If the loss is 3dB, than that's half power, which is quite a bit. The upside is the ability to adjust attenuation on the fly, which is very useful for those playing with different horns and drivers.

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33 minutes ago, Crankysoldermeister said:

My questions have always been more about what the amp is doing. I didn't mention anything about sound pressure, but about sensitivity.

 

Whether there is a loss or not depends on how you define sensitivity.  Is it at 1 watt or at 2.83 volts?  If you use 2.83 volts, then the resistor will not change sensitivity, it will just double the current draw from the amp.  I believe that's Al's argument on the subject.

 

If you consider the K55 driver 16 ohms, then same outcome can be reached by placing a 16 ohm resistor across the driver, and placing an 8 ohm Lpad before the resistor/driver combo.  

 

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10 hours ago, Dave MacKay said:

 

Will you please post a summary of the answers to the questions you raised?

 

id like to,  but with the recent warnings of mentioning anything other than what klipsch approves, its probably best i dont.  some of the answers are embedded in the detailed explanations on this thread or the other recent ones where they were discussed, unfortunately some got deleted before too many could read them...

 

 

 

  

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6 hours ago, Crankysoldermeister said:

@captainbeefheart Thanks for the detailed explanations. I'm pretty familiar with the principle of operation of that network, I've been building it for many years. My questions have always been more about what the amp is doing. I didn't mention anything about sound pressure, but about sensitivity. I realize that with sufficient power, you can overcome what the network is taking away from you. If the loss is 3dB, than that's half power, which is quite a bit. The upside is the ability to adjust attenuation on the fly, which is very useful for those playing with different horns and drivers.

 

Yes I didn't want to imply you don't know what the resistor is there for. I'll try and explain from a different approach.

 

Let's just go by simple concepts that a normal person will notice using said equipment.

 

An amplifier has specific loop gain, I'll use an amplifier that has a sensitivity of 20db, which is a gain of 10x.

 

Let's say we have the volume control set to where there is 1v input, the output will then be 10v. Let's leave the volume control at this exact same place for both sets of speaker with the two different networks, one with a swamping resistor and the other without it. This is a complimentary emitter follower output stage like most amplifiers out there. We won't use a tube amplifier for this explanation because it's a completely different result and most people will be using modern SS amplifier. Voltage will be peak voltage btw.

 

With the network without a swamping resistor we will be about -6db out of the auto-former going into the mid-horn. That gives us 5v across the driver and if we assume 8 ohms then that's a peak current of 625mA. 

 

With the network with the 10 ohm swamping resistor added (don't forget the 13uF is now much larger) we have -4db at the output of the auto-former which is 6.3v and a current of 787mA through the driver.

 

So for the same exact amplifier, with the same exact level of volume control setting you actually will have slightly more output from the midhorn. Overall speaker to amplifier relation for a given volume control setting will yield similar sound pressure levels, actually slightly higher with Al's network so his midhorn will be a little hotter. Yes I did add the other components into my analysis like the .3mH series inductor and 2.2uF shunt cap, these are just to filter high frequencies from going to the mid horn anyway.

 

Yes the resistor gets hot, and yes the amplifier needs to supply that current to get it hot but given the same amplifier, with the same volume control you lose nothing in way of sound pressure out of the speakers. That extra energy going into the resistor would be comparable to a normal loudspeaker system without an auto-former so you do not need to increase amplifier power to get the same sound pressure.

 

Hope this clarifies things better.

 

EDIT: The analysis was for 1kHz signal

 

 

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22 minutes ago, captainbeefheart said:

 

Yes I didn't want to imply you don't know what the resistor is there for. I'll try and explain from a different approach.

 

Let's just go by simple concepts that a normal person will notice using said equipment.

 

An amplifier has specific loop gain, I'll use an amplifier that has a sensitivity of 20db, which is a gain of 10x.

 

Let's say we have the volume control set to where there is 1v input, the output will then be 10v. Let's leave the volume control at this exact same place for both sets of speaker with the two different networks, one with a swamping resistor and the other without it. This is a complimentary emitter follower output stage like most amplifiers out there. We won't use a tube amplifier for this explanation because it's a completely different result and most people will be using modern SS amplifier. Voltage will be peak voltage btw.

 

With the network without a swamping resistor we will be about -6db out of the auto-former going into the mid-horn. That gives us 5v across the driver and if we assume 8 ohms then that's a peak current of 625mA. 

 

With the network with the 10 ohm swamping resistor added (don't forget the 13uF is now much larger) we have -4db at the output of the auto-former which is 6.3v and a current of 787mA through the driver.

 

So for the same exact amplifier, with the same exact level of volume control setting you actually will have slightly more output from the midhorn. Overall speaker to amplifier relation for a given volume control setting will yield similar sound pressure levels, actually slightly higher with Al's network so his midhorn will be a little hotter. Yes I did add the other components into my analysis like the .3mH series inductor and 2.2uF shunt cap, these are just to filter high frequencies from going to the mid horn anyway.

 

Yes the resistor gets hot, and yes the amplifier needs to supply that current to get it hot but given the same amplifier, with the same volume control you lose nothing in way of sound pressure out of the speakers. That extra energy going into the resistor would be comparable to a normal loudspeaker system without an auto-former so you do not need to increase amplifier power to get the same sound pressure.

 

Hope this clarifies things better.

 

 

 

I don't know who actually came up with the idea of using the "swaping resistor", but I like how it flattens the impedance curve. Some speaker designers (not many) actually aim for a flat impedance curve as part of their crossover network design, many do not. 

 

Since I am a "tinkerer", I love being able to switch auto-former taps and/or drivers and not worry about changing a capacitor. 

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3 minutes ago, Curious_George said:

I don't know who actually came up with the idea of using the "swaping resistor", but I like how it flattens the impedance curve. Some speaker designers (not many) actually aim for a flat impedance curve as part of their crossover network design, many do not. 

 

For SS amplifiers I don't really see any benefit to the swamping resistor but since you and I and many others use tube amplifiers with output transformers I think it's certainly an advantage to aim for flat impedance curve to keep the loading of the output tube consistent over the frequency range. Pentodes/Tetrodes will suffer the worst from the large 30 ohm impedance right in the midrange, nobody wants gross distortion in the most important part of the audible frequency spectrum. Triode output stages if output impedance is sufficiently low enough (<1 ohm) shouldn't be affected so badly and may even get a slight reduction in distortion from the higher load impedance.

 

Here is a trick some of you may know. With a Pentode/Tetrode output stage this rise in impedance can be mitigated by using a Zobel network at the output. The values may need to be tweaked from typical network values that expect a rise in impedance at higher frequencies. Whenever I build Pentode/Tetrode amplifiers I always have a Zobel network to keep loading flat. Some well designed Pentode/Tetrode amplifiers have these networks but MANY do not.

 

I truly believe that's why PWK liked the Brook amplifier so much with it's Triode output stage and negative feedback. It had low enough output impedance to handle the varying impedance and the Triodes handle it better with no Zobel compensation added.

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21 minutes ago, Curious_George said:

I don't know who actually came up with the idea of using the "swamping resistor",

 

Swamping with an autotransformer - Greg Timbers of JBL, but Al did come up with the idea independently. 

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51 minutes ago, captainbeefheart said:

So for the same exact amplifier, with the same exact level of volume control setting you actually will have slightly more output from the midhorn.

 

Acoustic output? Uh, no.

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16 minutes ago, captainbeefheart said:

 

For SS amplifiers I don't really see any benefit to the swamping resistor but since you and I and many others use tube amplifiers with output transformers I think it's certainly an advantage to aim for flat impedance curve to keep the loading of the output tube consistent over the frequency range. Pentodes/Tetrodes will suffer the worst from the large 30 ohm impedance right in the midrange, nobody wants gross distortion in the most important part of the audible frequency spectrum. Triode output stages if output impedance is sufficiently low enough (<1 ohm) shouldn't be affected so badly and may even get a slight reduction in distortion from the higher load impedance.

 

Here is a trick some of you may know. With a Pentode/Tetrode output stage this rise in impedance can be mitigated by using a Zobel network at the output. The values may need to be tweaked from typical network values that expect a rise in impedance at higher frequencies. Whenever I build Pentode/Tetrode amplifiers I always have a Zobel network to keep loading flat. Some well designed Pentode/Tetrode amplifiers have these networks but MANY do not.

 

I truly believe that's why PWK liked the Brook amplifier so much with it's Triode output stage and negative feedback. It had low enough output impedance to handle the varying impedance and the Triodes handle it better with no Zobel compensation added.

 

On my loudspeakers that are not Klipsch, I made Zobel Networks to ensure the impedance is flat for my tube amps. Dayton VATS software is great for this and affordable. 

 

https://www.parts-express.com/Dayton-Audio-DATS-V3-Computer-Based-Audio-Component-Test-System-390-807

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