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mboxler

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About mboxler

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  1. If you are like me, you understand what capacitors are doing in a crossover, but may not know how they are doing it. I attached two videos that helped me. The first describes the capacitor itself. It helped me understand how an open circuit (the capacitor) can allow current to flow. The second places a capacitor in an AC circuit. It describes inductor as well, but that's another topic. If you replace the resistor with a speaker driver, or taps 0 - 5 in a heritage crossover, you will see why the current flows through them 90 degrees before the voltage across the capacitor changes. You will also see why more current flows at higher frequencies. I realize this is 20 minutes of your life that you will never get back, but I hope you find them educational. Mike https://www.youtube.com/watch?v=f_MZNsEqyQw https://www.youtube.com/watch?v=zO7RZZW0wSQ&list=PLkyBCj4JhHt-4PnnwpbG-ZKV_EjX03DX8&index=15
  2. Yes. Actually, I believe Bob Crites measured a K-55 driver in a K400 horn and came up with closer to 14.5 ohms, but you get the point. If you mean the frequency at which the impedance of the capacitor is equal to the impedance of taps 0 - 5, yes. Using 14.5 ohms across taps 0 -4 as an example, the load across taps 0 - 5 would be double that, or 29 ohms. A 13 uf capacitor reaches 29 ohms around 422 hz. A 14.5 ohm resistor in parallel with the 14.5 ohm driver gets you a 7.25 ohm load. I realize the driver is not a pure resistive load, but... By moving these to taps 0 - 3, the reflected load across taps 0 - 5 is 4 times that load, or 29 ohms, right back to stock. No apology needed. You had me questioning myself. I'm not saying that the series resistor won't work, it just works differently. In my example, adding a 14.5 ohm resistor in series with 29 ohm impedance of taps 0 - 5 results in a 43.5 ohm load on the 13 uf cap. As mentioned before, this load is in parallel with the k-77 circuit, so that will change some as well. Again, if it sounds good, it is good. Mike
  3. By across the squawker, we mean in parallel anywhere between taps 0 - 3 and the K-55. Placing the resistor across the + and - terminal block screws going to the K-55 is a convenient spot. A parallel resistor in and of itself will not change the voltage across the squawker. What it will do is draw more current from the amplifier. If the resistor is equal to the squawker impedance, the current drawn will double. Since the extra current passes "through" the 13uf capacitor, it charges twice as fast. This, in turn, doubles the corner frequency from 400 hz to 800 hz. We haven't reduced the voltage to the squawker, it just takes an octave longer to reach maximum voltage. By dropping the output taps to 0 - 3, we cut the current across taps 0 - 5 in half. Now the current through the 13uf capacitor is equal to the stock setup, and it's corner frequency is back to 400 hz. The maximum voltage across the squawker, however, is down 3 db at all frequencies. I assume a 14 - 16 ohm resistor was used??? L-pads work well because they drop the voltage across the squawker without changing the current drawn from the amplifier. The 13 uf cap charges at the same rate. Now let's place the so-called swamping resistor (usually 10 ohms) across taps 0 - 5. It has no effect on the voltage across taps 0 - 5, but it will draw a significant amount of current, and that current stays pretty much the same no matter what output taps are used. A 13 uf capacitor will charge way too fast. It needs to be replaced with, say, a 48 uf cap that will take longer to charge. I don't understand option two above. It would drop the voltage throughout the circuit, but it would also lower the corner frequency of the 13 uf capacitor because less current is "flowing" through it. At least I think that's right. No matter what, if the change sounds better to you, it is better! Mike
  4. That's the easiest way to do it. This is an easy/inexpensive way to lower the voltage across the driver AND keep the current "through" the 13uf capacitor close to a stock AA crossover.
  5. Did you end up soldering a 15 - 16 ohm resistor across the squawker and move the squawker to taps 0 - 3?
  6. Although the video describes a transformer instead of an autotransformer, it will still work. If your primary inductance is 105.3 mH, then set your secondary inductance to 10.53 mH. The 10 db turns ratio is 3.165. 3.165 squared is 10. Since this a step down transformer, you divide the primary inductance (105.3) by 10 to get the secondary inductance (10.53). I haven't tried this, but I think I got the math right. If you want to model an autoformer, the inductor values would be 49.2 mH and 10.53 mH. Mike
  7. I would be happy to send you my version of a 12 tap auto transformer. It seems to work pretty well. It's currently part of a Type A simulated circuit. I assume I can attach an .asc file in a private message. Mike
  8. Looks good to me! You have now created two separate filter circuits. They will become one parallel circuit when the binding posts are strapped. They can also be paralleled when "strapped" at the amp...bi-wiring. They can also be run as separate circuits, each connected to its own amp...bi-amping. Mike
  9. I like the colors! I believe you should move the common wire that connects to tap 0 from input L - to input H -. The autoformer is part of the high pass filter. Mike
  10. This won't be too difficult... Relabel the input terminals as input low. Use the two empty terminal barrier strips as input high + and -. Move the wire going to the 13uf cap from the current input + to the new input high +. Cut the common wire between the output woofer - and the output mid -. If you cut it close to the woofer out - screw you may have enough to shift the remaining wire to the left. In the end, one common wire from input low - to woofer out -, one common wire from input high - to output tweeter - and output mid -. Now, connect your top binding posts to input high, and the bottom binding posts to input low. For single wire from your amp, you will need to strap your + binding posts together and your - binding posts together. For biamp/biwire, remove the straps. Actually, you made your speaker a little more versatile! I hope I visualized that right. Mike
  11. Crap! I've seen that thread but never scrolled down far enough to see Mr. Warren's schematic of a T2A. Since the inductances of the various taps are listed on the T2A datasheet, he made it easier by replacing one four tap autoformer with four one tap autoformers. I, on the other hand, have been struggling with the one four tap math. The inductance of the T2A taps 0 and 5 must be 45.6 mH...4 times the 11.4 mH inductance between taps 0 and 3.
  12. Coupling series inductors using the K command is the only way. While having a few beers with my brothers, I got an idea. I disconnected my 3619-ET from ALK crossover and measured the inductance from 0 - 5, 76 mH. Again I tried to measure the other taps, but the reading would always change each time I tried. I decided to calculate the inductances based on the turns ratios involved. I used 76 mH as my baseline inductance. I think it works. Maybe you can also try this and see if it's close. It's sure a lot cleaner. The response curve is similar, but the phase curve changed. The DCR values for inductors L1 thru L5 are .108 .076 .054 .038 and .093 ohms respectfully. Around .37 ohms total. Mike
  13. Here's the response curve of the 20 X 10 mH inductor (mutually coupled) T2A, using tap 4 (14 inductors). Green is voltage across taps 0 and 5, red across 0 and 4. Not sure how close it is the the voltage across a real K-77.
  14. This may be hard to see. I'd be happy to send you the .asc file if you want. Mike
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