Correct the pair of 8ohm as parallel set. The pair of 4 ohm as a parallel set. Then each set ran together in series on 1 channel. Am I correct that this yeilds a 6ohm load and that the 4ohm speakers will receive ⅔ of the available power?Sorry. it's not clear what you are trying to do or what your question is.
Are you trying to connect the speakers in series/parallel?
So is there any way to get the majority going to the 6.5in 4ohm ?You effectively have a 2 ohm load in series with a 4 ohm load. As they are in series the current through them will be the same but there will be twice as much voltage developed across the 4 ohm load. So 67% of the power will go to the 4 ohm load and 33% to the 2 ohm load. As the two 5.25" speakers are the 8 ohm ones (Forming the 4 ohm load.) they will get 67% of the power. This is the opposite way to what you have shown in post #1
Put them in series.So is there any way to get the majority going to the 6.5in 4ohm ?
Thank you very much.Parallel the two 8 ohm speakers and connect these two in parallel in series with both 4 ohm speakers in series.
So the two 8 ohms in parallel make 4 ohms and the two 4 ohms in series make 8 ohms. So the total resistance will be 12 ohms. The total output power will be less than the original configuration for the same voltage output from the amplifier.
Thank you so much sir! You have no idea how much that helps! I haven't looked at the link yet but is there anything you can send me on how you came to that solution?If you put the two 4 ohm speakers in series with the two 8 ohm speakers in parallel, 2/3rds of the power will go to the 4 ohm speakers.
But the total impedance will be 12 ohms, so the maximum amp power will be reduced.
View attachment 183772
If you put the two 4 ohm speakers in series, in parallel with the two 8 ohm speakers in series, then 2/3 of the power will go to the 4 ohm speakers with a total impedance of 5.3 ohms, allowing the amp to put out more power.
This configuration will also give better speaker damping, and likely better sound.
View attachment 183771
It's just using Ohm's law (if you don't know that you can look it up).I haven't looked at the link yet but is there anything you can send me on how you came to that solution?
I have looked everywhere! There isn't much information when it comes to series/parallel on the internet. What you can find doesn't help much with mixing impendence. Once again thank you very much for your time and assistance. I really appreciate your help.It's just using Ohm's law (if you don't know that you can look it up).
I looked at the various ways to wire the speakers and calculated the resulting impedance and relative speaker power.
Which is why if you want a 110Uf for a highpass on a speaker you actually need a 220Uf cap. I'm actually fairly intelligent but I'm new to this. Honestly I'm probably a little over my head with the electronic side for now, but I also want to build and have a passion for designing and building audio systems. So here I am asking stupid questions! At least I wasn't completely wrong with the solution I originally had. I got the power distribution incorrect which is the least important in a audio application. Now if I had gotten the resistance wrong and destroyed my amp that would be a big deal. Thanks for all of you guys help. You have managed to not make me feel like a complete idiot and went out of your way to help. This is a really good forum you guys have here!Also remember that equal items in parallel is 1/2 the item. 8||8 = 8/2 = 4
and two items in parallel results in a number less than the lowest item. 1||4 will be less than 1.
the formula is usually written as 1/Rt=1/R1+1/R2....+1/Rn
And resistors in series uses the same formula as capacitors in parallel.
capacitors in series uses the resistors in parallel form.
1/Ct=1/C1+1/C2....+1/Cn (Capacitors in series)
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by Aaron Carman