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mboxler

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About mboxler

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  1. My apologies @John Warren . I've never used LTspice, and didn't understand the schematic. I've been using iCircuit and XSim, but can't simulate transformers/autotransformers. This link is interesting. Inductors are used along the a K-statement to mutually couple the inductors. Both transformers and auto transformers are discussed. I'm playing with LTspice now. Slow learner. https://www.analog.com/en/technical-articles/using-transformers-in-ltspice-switcher-cadiii.html
  2. Looks like that autoformer simulation creates an electrical band pass across the 13 ohm resistor. Was that your intent? An accurate simulation would result in the voltage across the K-55 rising to 0db along with the K-77. I did find this, but I'm not smart enough to determine if it works. https://www.quora.com/What-is-the-equivalent-circuit-for-auto-transformers-impedances
  3. Unless I'm doing something wrong, this comes very close to what I'm seeing if you assume the K-55 is a 16 ohm load. 13uf cap impedance + ((double the K-55 impedance // (2uf cap impedance + K-77 impedance))
  4. Well, here's my results using a 2.83 volt 400 hz signal. My setup includes a 26.38uf mid cap, a 11.76uf tweeter cap, a 14.2 ohm resistor connected to tap 4 (tap 0 connected to ground), and an 8 ohm resistor. First, the two caps in series (or 8.136 ohms) results in .46 volts across the 8 ohm resistor. Nothing else connected during this test. Next, the classic Type A configuration with the tweeter cap connected to tap 5. .424 volts across the 8 ohm resistor 1.296 volts across the 14.6 ohm resistor Last, the new Type A configuration, with the tweeter cap connected to the amplifier input. .651 volts across the 8 ohm resistor 1.785 volts across the 14.6 ohm resistor The math required to reach the Classic Type A results ended up being for a parallel connection. It bothers me the the voltages the 8 ohm resistor are so close between the first and second tests, so I plan on running another one tomorrow at 800 hz. If anyone is interested in the math involved, I'd be happy to post it tomorrow as well. It's kinda nerdy. Right now my brain is fried. Mike .
  5. Like I said in another thread, I'm not sure how the autoformer changes the circuit, but I would also think that the current flowing across taps 0 and 5 would have an affect. Before I send @Robbie010 the autoformers, I think I'll test this out. I don't have the 13uf/2uf caps to test with, but any two caps should do. I'll rig up the circuit with a 26uf cap to the autoformer and a 12uf cap to the tweeter. I have 14 ohm and 8 ohm resistors to use as loads. I'll pass, say, a 400 hz 2.83 volt signal and measure the voltages across each resistor. I'll then move the input to the 12uf cap and retest. If the caps are truly in series, the voltages across the resistors should not change, right? I've been wanting to test this for a while. Maybe it's about time. Mike
  6. Likewise, the amplifier needs to be designed for this. 2.83 volts into an 11.38 ohm load = .7 watts. However, the only "work" being done is the 2 volt signal across the 8 ohm tweeter, or .5 watts. A capacitor is an open circuit, it does no work. The output transistors of the amplifier need to be able to dissipate the excess .2 watts as heat. In class d amps with single ended outputs, this excess work is "pumped" back into the power supply as increased voltage. At least I think that's right. Sorry, getting way off topic.
  7. Although the K-77 isn't a true resistor, for the sake of this explanation let's say it is... The impedance of a 2uf capacitor at 9947 Hz is 8 ohms at -90 degrees. That impedance added to the driver's 8 ohm 0 degree impedance = 11.38 ohm at -45 degrees. 2.83 volts through a 11.38 ohm load = 0.248682 amps. .248682 amps through 8 ohms = 1.99 volts, or -3db. There is a 2 volt drop across the capacitor and a 2 volt drop across the driver. Result. the current through the driver is 45 degrees ahead of the voltage at the crossover frequency. If there is a comparable low pass filter to, say, a woofer, the current through it will be 45 degrees behind the voltage at the crossover frequency. This is what creates the 90 degree difference between drivers in a first order filter. Mike
  8. I agree, except for the "half voltage each". The impedance of the 2uf cap will be 8 ohms at around 10 khz, the crossover point. Isn't it true that at that frequency, given a 2.83 volt signal, the voltage drop across the capacitor and the driver will be 2 volts each? Not half of 2.83, or 1.414 volts each. I always knew that the order of series components didn't matter, but you are right! The parallel placement doesn't really matter either. Looks like the impedance of the circuit gets wacky, but the ending voltage across the driver stays the same. Thanks! I learn something new every day.
  9. Capacitors in series don't add. They act like resistors in parallel. A 13uf cap in series with a 2uf cap looks like a 1.73uf cap in series with the tweeter. I've never been able to calculate, though, how the autoformer changes that calculation, if it does at all. I'd stick with the 2uf cap. It will be fine. Mike
  10. Yep! Think of it this way...the Lpad is now part of the driver. Let's say you want to reduce the voltage to the K-77 by 3db. That would require a 19.4 ohm resistor in parallel with the 8 ohm K-77. Those two resistances in parallel equal 5.7 ohms. Now put a 2.3 ohm resistance in series with that, and you are right back to 8 ohms. The capacitor sees an 8 ohm load. Mike
  11. Tweeter Lpad must be next to the K-77, just as the squawker Lpad is next to the K-55.
  12. Here's a link. The second post shows the k-55's impedance. Varies all over the place. Around 14 ohm @ 400 hz. https://community.klipsch.com/index.php?/topic/65935-impedance-plots-for-k-55-v-and-k-77-m/ This load dictates the current running through the squawker circuit. The lower the impedance, the more the current, the faster the capacitor will charge. Using the -3db tap on an autoformer doubles the impedance of the squawker circuit. A 13 uf cap is used. Using a -3db lpad designed for the driver will not change the impedance. Current will double. The cap size needs to be doubled so it will charge/discharge at the same rate. A 26uf cap is needed. Hope that helps. Yes, you will need a separate lpad for each driver. The 16 ohm lpad should be fine for the squawker. Mike k55mimp.pdf
  13. That's an 8 ohm Lpad. It will work for tweeter, but not the squawker, which is 14 ohms (?). If you make your own 14 ohm Lpad, then a 26uf cap will be needed. With a 16 ohm Lpad, it will be a little off, but now by much. Mike
  14. If the lpad you are using on the K-55 is designed to keep it's impedance as is (and just drop the voltage), then you will need a 26uf cap. The 2.2uf tweeter cap should not be an issue. It may measure 2.0 anyway, depending on the tolerance. Mike
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