Bi Wiring Speakers from Luxman L-509X
@oldhvymec You might find this discussion useful. In it we go over the impedance of a woofer in a sealed cabinet: https://speakermakersjourney.blogspot.com/2016/12/crossover-basics-zobel_8.html |
Hey @oldhvymec A single driver in free space produces a 1 hump curve, where the peak is called the driver's resonant frequency. You’ll see this in any woofer spec sheet: https://www.madisoundspeakerstore.com/approx-8-woofers/seas-excel-w22nx001-graph-e0077-8-graphene-co... Put that woofer (all alone) in a sealed cabinet cabinet and the impedance curve will have 1 hump with a new resonance point. Add a tweeter with the appropriate crossover and you will have 2 impedance humps. The Magico S1 Mk II for instance: https://www.soundstagehifi.com/index.php/equipment-reviews/973-magico-s1-mk-ii-loudspeakers Click on the link near the start of the article and scroll down. That peak just below 50 Hz is the woofer/cabinet system alone. The peak above 1kHz is the combination of the low and high pass filter sections. Now, port that woofer in a cabinet and you’ll get 3 peaks total. The single woofer peak will turn into two. Best, E |
I also see super high ohms in certain regions, Wouldn’t the amp see just that and be down XX db without a notch to bring the impedance down and volume UP. @oldhvymec Speaker output is proportional to voltage, and we assume an amp acting as a voltage source with little output impedance, so long as the speakers impedance is high enough ( say > 3 Ohms), raising the speaker impedance to a very high amount like 100 Ohms wont alter the voltage. Here is one of hundreds of examples of a two way speaker with that xo hump: https://www.stereophile.com/content/polk-legend-l100-loudspeaker-measurements Notice the xo impedance hump is not present in the frequency response plot. There are speaker designers who take more care in order to appeal to tube amp users, like Fritz, and they take extra care to avoid that xo hump. In Fritz’s case, he does so by using serial xo. Others may add an impedance compensation circuit, but since they soak up watts they are big and expensive to make. But that simple XO in a real speaker isn’t gonna look like that is it? The first plot on the post i shared was a simulated XO, but the actual XO measures (via DATS v 2) nearly identical so it is absolutely not worth posting. It is the same curve. The trick to getting the camel humps right is to measure the woofer in the cabinet with the port in place. The last two curves in my post are for simple first order filters so you can see how rising impedance blocks unwanted frequencies. |
Then answer me this what happen if the two frequencies are hit at the same time, What does the amplifier see then? It seems the sim is showing the the two different frequencies but not at the same time. Lets assume an ideal 8 ohm tweeter and mid-woofer, with a perfect infinite slope crossover at 2 kHz. So perfect that at 1,999.99999999 Hz 100% goes to the mid-woofer and at exactly 2kHz it all goes to the tweeter. You would have an amplifier load that is exactly like a single, ideal 8 ohm driver without a crossover. We know that current = V/R, but how would we ever calculate the current for complex music? Imagine a test tones with two notes, 500 Hz and 4 kHz. 8Vrms each, with the peak voltage 2x the peak of either by itself. In this case we have 2 A, because you take each section and add them. 8 Vrms @4kHz / 8 Ohms = 1Arms tweeter 8 Vrms @500 Hz / 8 Ohms = 1Arms mid woofer Now imagine we use the same test tone and add a true woofer below 200 Hz, converting into a 3-way: 8V / 8 Ohms = 1A tweeter 8V / 8 Ohms = 1A mid 0V / 8 Ohms = 0A woofer We added a 3rd driver, but the current did not rise. Calculating a total R of of 8/3 will not work here. Let’s this time use a new test tone at 200 Hz but nothing else. 0V / 8 Ohms = 0A tweeter 0V / 8 Ohms = 0A mid 8V @200 Hz / 8 Ohms = 1A woofer Again, the current is entirely dependent on the section it went to. In all of these cases, the answer is that the current and power is identical to a single driver because of the magic brick wall filter sections used. Real filters are not ideal, but not too far off these examples we can't use them, and where the impedance curve drops to the impedance of the single driver you can see it is working very much like these examples. Of course, multi way speakers and music is complicated and ideally resistive drivers almost never exist, and never match other driver types. :D You can’t really do this with music without a digital sampling mechanism, but i hope you can see that no, the drivers don’t add up the same way in an AC circuit with filters. In the post I sent you I use a VERY typicical type of speaker build and crossover design, one you’ll find 100s of examples of in Stereophile. That peak between the mid and tweeter is the combination of the low and high pass filters. Split them apart, they go to infinity. Put them back together, either with jumpers at the speaker or bi wiring to the amp, and they meet in the middle at an amount a lot less than invinity, but a lot more than either driver. |
I understand what Erik is talking about, that is impedance curves at the driver, it has nothing to do with load to the amplifier. The article is talking about how to CORRECT with a crossover design. I’m afraid you’ve misunderstood then. Nothing in that article presents the impedance of any driver or measurement at the driver. The first chart is an impedance curve of an actual speaker as would be seen by an amplifier while the next two impedance charts are the impedance, as seen by the amplifier if they were separated. It was my goal that these 3 curves (all from an amplifier’s view) would help explain why the tweeter and woofer are not really seen in parallel by the amp. I will endeavor to make that more clear. In the last two impedance curves I use an ideal 8 Ohm resistive speaker load so the reader can see the part that is a driver vs. the part that would be contributed by the filter (high or low pass). |
The low and high pass filters means the multiple drivers are not not seen in parallel by the amplifier. They are seen as separate entities. For a better view to understand this, please see my post here: https://speakermakersjourney.blogspot.com/2016/12/crossover-basics-impedance.html My original statement, that using the A+B outputs to bi-wire a speaker is equivalent to using just A or B stands. The amplifier won’t know the difference and hopefully you’ll see the explanation in the post. Using simple, DC analysis for equivalent resistances is not appropriate in this case. It would work (with a lot of slop) with parallel drivers of the same type in the same section, like having 2 woofers for instance. |
