Jump to content
The Klipsch Audio Community

mboxler

Regulars
  • Content Count

    224
  • Joined

  • Last visited

Community Reputation

43 Excellent

About mboxler

  • Rank
    Forum Veteran

Profile Information

  • Gender
    Male
  • Location
    Golden CO

Recent Profile Visitors

3687 profile views
  1. I wonder what the 24uf labeled cap actually measures. It could very well measure 26uf, which is within +- 10% tolerance.
  2. Two 70uf caps in parallel to get the 140uf value. Yep...they are pretty big.
  3. This link will take you to Klipsch Crossovers. The AK-2 schematic is three-quarters down on the first page. https://community.klipsch.com/index.php?/topic/113804-klipsch-crossovers/
  4. I think this is the document describing the issue. I don't remember anyone figuring out how to fix it. If you run differential inputs to the amp and bypass the input op amps, I believe you circumvent the issue. http://www.ti.com/lit/an/slaa719/slaa719.pdf Mike
  5. About three-quarters down this page...maybe? https://community.klipsch.com/index.php?/topic/113804-klipsch-crossovers/page/8/ Looks like it was pulled from here... https://audiokarma.org/forums/index.php?threads/crossover-upgrade-on-rf-83s.588494/
  6. Another way to look at it... The two rooms are the two capacitor plates. The wall between them is the dielectric. Ideally, the door between the rooms is so small that everyone (the current) will need to go through the hallway (the circuit) to get to the other room (the other plate). In reality, some people do get through the small door (leakage current). Sorry...only door analogy I could come up with.
  7. I use this for capacitors http://www.pronine.ca/capimp.htm and this for inductors http://www.pronine.ca/indimp.htm Just enter the "resistor" value in the impedance field and it will calculate the -3db frequency.
  8. 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
  9. 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
  10. 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
  11. 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.
  12. Did you end up soldering a 15 - 16 ohm resistor across the squawker and move the squawker to taps 0 - 3?
  13. 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
  14. 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
  15. 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
×
×
  • Create New...