[Chewbacca]

Just a correction to my above statement:

Danny in his video made it very clear he feels this way, or he incorrectly explained this to be the case.

Meant to say: Danny made it clear he feels these electrons will go back into the driver against the current.

[Jaytor]

As Norman points out, the current is a closed loop - meaning that any current flowing from the amp is split between the parallel paths and then is combined to flow through the return path to the amp. So anything the perturbs the current flow (such as dielectric absorption) through one parallel path will have the opposite affect on the other paths since the sum of the currents must always equal the current coming from and returning to the amp.

This is a bit of a simplification since the amplifier is not a perfect voltage source, meaning that the current from/to the amp will be affected by speaker/crossover to some extent, but this doesn't change the fact that anything that affects the current flow through one parallel path directly affects the current flow through the other paths.

[Chewbacca]

Hi, Norman! Thanks for the response! This makes total sense to me, and absolutely... there's a ton to consider, and an infinite amount of variables!

I know a capacitor in parallel you will want the proper value (tolerance) to not affect the frequency response. Also the fastest cap possible, to release all the energy, and not cause a time delay/smearing of the signal. This I knew. But I'm still lost as how even the leaky DA will go against the flow of electrons/electromagnetic waves (Personally, I'm actually in the waves camp lol) and end up back at the driver as Danny explained.

Will it not instead load the other lead of the crossover with these frequencies/waveforms/electrons? Then when the flow switches directions, the cap will grab the same frequencies and load the other lead of the crossover again? - Thus the additional output of the driver? Does that make sense? In the most basic of terms.

[Chewbacca]

Hey, Jaytor - Thanks again for the response!

I don't want to keep saying "I get it" "I get it", because regardless this has been an awesome review, and I'm sure I've learned a few things along the way!

But, I am specifically referring to what Danny says in this video, beginning at 9 minutes.

https://www.youtube.com/watch?v=RrpIBQ7gW5Q

More or less, He says that after the current alternates, the flow will go up the shunt capacitor, then back to the driver... Again, this being a closed loop, going one way or another, how can this be possible with the current coming from the driver in the opposite direction when it switches flow? Or vise versa on the opposite side.

But absolutely, I can see how it could leak some back into the signal with the imperfect DA/time needed to dissipate. But he makes it seem it's actually going against the flow from the video.

[Tyson]

Finally visualize that solder joint where the parallel leg of the crossover circuit joins the minus conductor. In one direction we have 8 inches of wire to the tweeter’s terminal. In the other direction we have 8 feet (20 vs 244 cm in metric) of wire until the minus leg gets to the amplifier terminal. So in AC World where hifi lives there is ~1/10 the impedance between the DA caused noise/distortion voltage and the tweeter terminal vs the amp terminal. Ah rats, it’s another voltage divider hidden in plain sight.

To ameliorate this a bit, could we use very small wire from the crossover to the tweeter (say, 24 gauge) and very heavy wire from the amp to the crossover (say 12 gauge)?

[Norman Tracy]

Chewbacca commented and asked:
“I'm still lost as how even the leaky DA will go against the flow of electrons/electromagnetic waves (Personally, I'm actually in the waves camp lol) and end up back at the driver as Danny explained. Will it not instead load the other lead of the crossover with these frequencies/waveforms/electrons? Then when the flow switches directions, the cap will grab the same frequencies and load the other lead of the crossover again? - Thus the additional output of the driver? Does that make sense? In the most basic of terms.”

And then referring to a YouTube video with DR explaining how a cap in parallel with the tweeter adds area under the frequency response curve Chewbacca adds “More or less, He says that after the current alternates, the flow will go up the shunt capacitor, then back to the driver... Again, this being a closed loop, going one way or another, how can this be possible with the current coming from the driver in the opposite direction when it switches flow? Or vise versa on the opposite side. But absolutely, I can see how it could leak some back into the signal with the imperfect DA/time needed to dissipate. But he makes it seem it's actually going against the flow from the video.”

I think we are getting close here, you seem to be grasping it. In the video Danny R is referencing how capacitors and inductors are reactive components. In the electrostatic charge of a cap’s plates or the coil of the inductor they exhibit the property to store charge at one time and then release it later at another time. Reminder here time and frequency are related f (frequency) = 1 /T (period). In the video DR shows using this to bring up a frequency band so the tweeter does not roll off as gradually as a 1st order example. So that is an example of the quality of the parallel shunt cap being imprinted on that energy in the region being boosted. My example using DA leakage voltages was looking more at “after the current alternates” i.e. later in time when the waveform has cycled when effects like DA and ESR (equivalent series resistance) push the voltage at a given node at a given time away from the ideal value. Both effects bring the quality of the parallel connected parts into play. Both the desired action (DR using a C to tailor frequency response) and distorting action (my DA example) are injecting charge at a given time into the circuit where it acts against the impedance of the circuit at that node at that time to raise (or stated with more rigor ‘add or subtract’) a voltage; desired or otherwise.

Tyson asked “To ameliorate this (division of error voltage between short internal wiring and long speaker cables) a bit, could we use very small wire from the crossover to the tweeter (say, 24 gauge) and very heavy wire from the amp to the crossover (say 12 gauge)?”

I think not, let’s remember the crossover to the tweeter wire also and in fact primarily needs to carry all the desired signal voltage. One needs to attack the noise/distortion voltages at their source.

[Chewbacca]

Thanks, Norman!

I sincerely appreciate your responses! And that all makes perfect sense to me, and aligned right with what I'm thinking with greater clarity.

I'm just still hungup (and totally willing to drop it at this point lol) that just the way Danny illustrated the example, it made it seem like he was directly saying this:



That once the current alternates the current inside the capacitor will literally go against the other currents direction and back to the driver.

Wouldn't it just load the other side once the cap dissipates, and follow the current, never making it back up to the driver?



Obviously if he's taking away frequency/electrons/waveforms/energy, they have to go somewhere. Something can never turn into nothing. So it would make sense to me that it would load other frequencies (the gain he's seeing in the response), but IMO that does NOT mean these specific electrons that have flowed and have been temporarily stored at the capacitor will somehow make its way up to the driver - the affect of it absolutely will though! - I'm not debating this! Okay.. yes, maybe some DA leakage, but, not by design (in a perfect world).

So - After the capacitor releases it's electrons it would make its way back to the amp (loading this side of the crossover with additional electrons which will cause an imbalance, which then the driver will balance out with a different frequency), current then alternates, then electrons back to the capacitor, stored temporarily, and it makes its way back to the amp in the opposite direction - loading then this side of the crossover, again creating an imbalance with the driver, which I assume would actually produce the sound, or basically leveling out the circuit.

Am I understanding this in at least the most basic of ways?

Believe me, I'm not a naysayer, or a "Danny doesn't know his stuff" kinda guy... I'm quite the opposite, as I'm sure most of you are aware with some of my previous praise of Danny and his work. His way of explaining on this, and me being bad at explaining things over text really brought this tread to a place I had no intention of taking it lol.

If my above was correct (and even if not, maybe it'll shed some light on why Danny's explanation and others misinterpretation of what I was asking (again, my fault) led me to this reasoning)... This then brings me back full circle, with me saying that these individual electrons that have been temporarily stored in the capacitor will only see the amplifier and the capacitor - They will not go through the driver itself. Is that correct? THAT was my (and apparently just my lol) definition of something that's in the signal path. Electrons that will actually go through the driver to cause it to create a signal - I think this single definition and different interpretations of "in the signal path" started this whole digression of the thread lol.

[Chewbacca]

Okay, so with what Jaytor said about the signal going out and coming back into the amp being the same - the cap will just push back what it took away before the tweeter to the other lead (behind the tweeter now) balancing out the signal. Not necessarily the driver balancing out the signal, as I stated before. But could this possibly be a balancing act of the two, given nothing happens instantly (caps & drivers need to take time to dissipate and will do so at different speeds)?

I believe I'm really starting to grasp the full picture, and fill in the voids of my understanding.

I sure hope... Because I guarantee I'm annoying myself more than anyone else with this thread

Regardless, I sincerely appreciate everyone's time and detailed explanations of this to me and everyone else reading this thread.

[Chewbacca]

Not to keep bringing this thread back to light, but I just wanted to say, I reread this multiple times yesterday, and ALOT of the gaps of my understanding have been filled. Even after me saying "I get it", "that makes total sense", but I was probably only 90% of the way there. So I apologize if I came off as a know it all, especially being surrounded by industry pros...

It's kinda funny, I had a good understanding of what each component does on it's own, but I was missing more of the big picture with some of it, and exactly how each component interacts with each other. I was just staring at my speakers/crossovers/amps last night imagining them working, and I feel like I really get it now. THANK YOU! I'm an engineer by trade (obviously not electrical) so I feel I have a very intuitive way of deductive reasoning to get to a valid solution of reality - and now I feel I have (for the most-ish part), at minimum, a very basic understanding of how everything is working in tandum. I even explained it all (more or less) to my wife in very basic ways, and she seemed to have a clear grasp on it. Which, that passes my "do I understand something test".

I apologize that I was simply getting hung-up with how Danny explained in his video. Which I think he simplified to to shut up a lot of the "not in the signal path" people. Which I can understand the confusion people may have in either camp. But it's in CLEARLY most definitely causing a direct affect on the signal. So... ehh, it depends on how you look at it lol.. So for the educational purposes of his video, it's just fine of course - and to the less deep dive people as myself, it leads them right to the correct conclusion, which at the end of the day is all that matters! Thanks again Danny, for all your wonderful videos!

What I feel like I was truly missing in the equation of parallel components (which... I hate the term parallel. I get it, it's between the parallel leads, but to me, it's more logical to call it the perpendicular part of the xover... I digress) is what Jaytor explained quite simply explained, and Norman really dove into:

Jaytor quote: this distortion is being subtracted from the signal being delivered to the load, which means that the load will see the negative of distortion A which is essentially the same thing.

Basically the capacitor in shunt/parallel is creating a negative (like a photo) of the signal that's being taken away, and everything that's wrong in that negative is wrong in the end picture. NICE! Once you know all of this with the other tidbits I didn't know on this thread, and really think about it with deductive reasoning, this is, yes - Fairly simple... and totally enthralling! My engineering skills are no doubt lacking the most in electronics, yet it's probably one of the most exciting for me.

I want to thank everyone again, for truly expanding my knowledge on this subject and being patient with me, even if I wasn't being patient with myself

I hope I didn't loose too much credibility on this site, and hope people understand I'm here to gain knowledge in this awesome/crazy/amazing hobby we call music reproduction! I also hope that some here can gain the knowledge I've learned at the sake of me exposing some of my gaps in knowledge. I'm more than happy to take a hit if someone else can glean even the smallest tidbit from this!

Again, thanks everyone!... Oh yeah, and buy the best components you can in the crossover, regardless of purpose!

[Jaytor]

We're all here to learn and have fun 

[NoahH]

I was lurking on this thread as I too am an engineer of a different ilk and realized I did not have an intuition for this when you asked, so I am glad for the thread.

[TRADERXFAN]

Thanks to all the contributors on this thread- was a good read.

[g3rain1]

Dude, don't get hung up on electrons. In an AC circuit the electrons in ANY component will NEVER reach the driver. They move incredibly slow. Under DC it might take an hour or more.

We care what the electric field does. The electrons don't matter.