Mute circuit (short to ground) is parallel, but it definitely does affect the signal when engaged! LOL
Hi richidoo, thanks for the time with your post.
If you read above you'll see that I never debated the fact that the items in parallel have a direct impact on the signal... I think no one debates that this is obvious. My first question was how much does it have an affect on the signal, and how much should you spend on this part of the crossover. It is clear Danny feels this is not where you should spend your money on the crossover, otherwise he would offer upgrades for these items as well. So I'm trying to gain knowledge on this fact - which I feel has been fairly adequately answered.
Second question that arose in this thread, and THIS is what I have been asking recently, and it's comically obvious I have not made myself clear, or I am screwing up some terminology. When I say
in the signal path, I don't believe this is understood for how
I understand it and was talking about it above. I am not asking if an item in parallel has an impact on the signal. Again, this is obvious.
In the signal path to me, when used while talking about a crossover, meant the electrons
in the signal path will literally travel through the driver and excite it to vibrations and create sound.
So. As simply as I can possibly ask:
Please explain how an electron traveling through a capacitor in parallel/shunt/short to ground will end up at the driver.Because, to the best of my knowledge, this is basically what Danny was saying in his video. Or referring to the last picture I posted above. How can the signal go both directions
at the same time? This is not how electricity works in my little brain. I have always held the theory of the path of least resistance. There cannot be two directions of one path. Of course the path can be split, and balanced. Thus a crossover.
I'm trying to gain understanding on how this works, and how it's possible these electrons that are flowing in parallel endup at the driver? If it's just "bleeds some current to ground", I don't think theoretically these items are
in the signal path of electrons that will pass through the driver. To me bleeding current is not part of a design. Unless this is the design of parallel.