I forgot to mention, I'm hoping to build with what I have on hand. I have the 8-ohm version from Parts express.
Are you sure your Tweeters are 16 ohm? How do you know?
The L-Pad will maintain a constant (more or less) load, but that load depends on the attached driver being roughly 8 ohms. What it will be with a 16 ohm load, can probably be worked out, but it is anyone's guess.
Can you do this? Yes, as I said, you can do anything you want.
If you use the calculator I provided for a Fixed L-Pad we can get some sense of the difference -
Crossover Design Chart and Inductance vs. Frequency Calculator(Low-pass)
With an 8 ohm tweeter and a -6dB of attenuation, the values are -
4 ohms in Series
8 ohms in Parallel
In this case you will have a 8 ohm resistor in parallel with the 8 ohm driver, which is 4 ohms, then you have an additional 4 ohms in series to being the perceived impedance back up to 8 ohms.
The same attenuation with a 16 ohm tweeter are -
8 ohm in Series
16 ohms in parallel
Though this is only vaguely true, you will have 8 ohms in parallel with 16 ohms, which yields a combined impedance of 5.3333 ohms. If by chance the Series resistance was 2.7 ohms, then the impedance seen by the crossover would be 8 ohms again.
Bearing in mind, that unless you design your crossover for that impedance the crossover frequency is going to be off.
And the above would be fine if you used fixed resistors, but you can't control the value of the 8 ohm L-Pad, they a balanced assuming an 8 ohm load.
What you propose will work as an attenuation, I just can't predict how well or predict how well it will work with your existing crossover.
You can use the L-Pad Calculator like to determine the resistor values for several different degrees of attenuation, then work out the new impedance with a 16 ohm load instead of an 8 ohm load.
In the example I gave for the 8 ohm and 16 ohm above for -6dB attenuation.
Instead of 8 ohms in parallel with an 8 ohm driver, we would have 16 ohms in parallel with the resistor.
So, in a standard configuration, we have an 8 ohm resistances in parallel with an 8 ohm driver, for a combined impedance of 8 ohms.
With a 16 ohm tweeter we have 8 ohms in parallel with 16 ohms and an additional 4 ohm resistor in series. So, 8 in parallel with 16 is 5.333 ohms with 4 ohms in series for a combined impedance of 9.333 ohms. That will cause a slight shift in the crossover location.
Again using the crossover calculator I provide, and assuming a 6dB/Octave Crossover, the crossover frequency will shift down by 430hz. Assuming we are starting with starting with a 3000hz crossover point.
So, many things come into play, if you are designing your own crossover, then you can adjust the values of the crossover components to compensate. But if you have on off-the-shelf crossover, the crossover points is going to shift from the stated design values.
Steve/bluewizard
Yes, I have both an 8-ohm diaphragm and a 16 ohm one. The 16 ohm one measures roughly double the resistance over the 8 ohm one. I know, you're asking, "Well why dont you just use the 8-ohm one?" It's because I already have the remnants of a heavy-duty crossover from before which I'd like to use to make this new one. It has the correct caps and inductors for a 4th order at 1.2K, using a 16 Ohm tweeter. Those inductors are hugely expensive, and I'd rather not have them go to waste or order more.
It's sounding like I should just bite the bullet and order the 16-ohm version., or not use a pad at all. If I do it with resistors, they'd need to be able to stand up to at least a hundred watts. The other solution is to use the active crossover in my power amp with two channels. I was just hoping to be able to run this box with one input, and not have to use both sides of a power amp.
If you have a 16 ohm crossover, that is better but it still complicates things. The Tweeter section, if you use an 8 ohm L-Pad will not cross at the correct frequency. Though you can probably predict the frequency that it will cossover at.
Make life simple, buy some 16 ohm L-Pad and be done with it.
As to the power of a fixed or variable L-Pad ...no... they don't need to be 100w, but a 100w L-Pad is not that expensive. The Power Distribution across the frequency spectrum is very low for high frequencies. For example, and from memory, if we have 10w at 350hz, then we will have about 1/2w above 4khz.
Here is a chart of Power Distribution across the spectrum for full orchestral music -
Though I would certainly not make my Tweeters or L-Pad components 0.5w. 16 ohm Variable L-Pads are not that expensive, and typically come in 50w and 100w ratings.
50w, 16 ohms are about $11 each.
Speaker L-Pad Attenuator 50W Mono 3/8" Shaft 16 Ohm
100w, 16 ohms are about $13 each.
Parts Express Speaker L-Pad Attenuator 100W Mono 3/8" Shaft 16 Ohm
It would be worth it not to have to kluge your system with the wrong parts.
Steve/bluewizard
This is my determination as well. $13 shipped. Big deal, I have to wait a few more days.
I have been using the box with a Crown XTi 4K. Using that active X-over at LR24 1.3K, it's sounding better than the passive one ever did. I also am able to run the woofers at 4-ohms, to help overcome the efficiency differences with the tweeter. Additionally, I'm able to use EQ settings via my analyzer software to flatten out the peaks quite a bit. I had wanted to have the flexibility of using the other channel for lows on a compact rig, but this is sounding too good to ignore.
Still, I think I'll buy the $13 L-Pad at 16 ohms. Then it's done and I can just swap it in/out if I need that flexibility.
Maybe even make a switch on the back of the cabinet.
So then, I had a thought. What if I put TWO Speakon inputs on the back of my box instead of one. One a two-pole input, and another a 4-pole input.
Then somehow make a switch that would make the two-pole which would go to the crossover. Throwing the switch and connecting to the other would break the 2-pole, and make the 4-pole which would disconnect the crossover and switch in wiring straight to the drivers - channel one low and channel two high.
Best of both worlds.
Not sure they make a switch with that many poles on it though. The X-Over would have to be completely disconnected, and speakers re-wired to be parallel, not series...
Then somehow make a switch that would make the two-pole which would go to the crossover. Throwing the switch and connecting to the other would break the 2-pole, and make the 4-pole which would disconnect the crossover and switch in wiring straight to the drivers - channel one low and channel two high.
Best of both worlds.
Not sure they make a switch with that many poles on it though. The X-Over would have to be completely disconnected, and speakers re-wired to be parallel, not series...
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