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About mboxler

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  1. Class D

    Curious how you came up with those voltages. The maximum TPA3255 input voltage is 7 volts p-p; however, at a 21.5 db voltage gain (or 11.9 times), that would equate to 83.3 volts p-p. With a 53.5 volt power supply, you're clipping. 4.5 p-p, or 1.6 rms gets you to max, so I would think your max input voltage would be 1.6 rms unbalanced and 3.2 rms balanced. Am I missing something? Mike
  2. I had tested this before, thought I'd try again. I find this stuff fun, but I'm weird. I applied a 5.05 volt, 1160 hz signal to a dummy crossover (yes, I named it after myself). I picked this frequency because at 1160 hz, the impedance of the 6.66 uf capacitor I used = 20.6 ohms, and the impedance of the 10.3 ohm resistor I had across taps 0-4 should double. This would be the resonance frequency, the frequency at which the voltage across the load (taps 0-5) should be down 3 dB. 6.66 uf cap between input positive and tap 5 of autoformer. Wire between input negative and tap 0 of autoformer. Wire between tap 0 of autoformer and one end of 10.3 ohm resistor. Wire between tap 4 of autoformer and the other end of the 10.3 ohm resistor. Input voltage, 5.05 volts at 1160 hz Voltage across cap, 3.61 volts Voltage across taps 0 - 5, 3.66 volts Voltage across taps 0 - 4, 2.58 volts Now the fun part. The impedance of a 6.66 uf cap in series with a 20.6 ohm load is 29.135 ohms. The current of 5.05 volts thru a 29.135 ohm load is .17334 amps. .17334 amps thru a 20.6 ohm load (the capacitor) = 3.57 volts...pretty close to the 3.61 I measured. Voltage across taps 0-4 should be 3.61 * .707, or 2.55 volts...again, pretty close to the 2.58 I measured. This confirms to me that the autoformer (at least when using it to attenuate the K-55 by 3 dB), doesn't have much if any affect. Mike
  3. Measuring K-33 impedance in K-horn.

    A first order filter is doesn't do much within a couple hundred Hz either side of resonance. If everything was perfect (2.4 mH inductor, 6 ohm load, 400 hz), the voltage to the K-33 is down 3 dB. At 500 Hz, it's down 4 db. At 600, around 5 dB. If both the K-33 and K-55 were perfectly crossed (first order) at 400 hz, they would each be down 3 dB. The voltage across both drivers combine at 400 hz, resulting in a 0 dB summed voltage. It would be a 3 dB bump when summed, but the voltages across each driver is 90 degrees apart. As you can imagine, even if the K-33 were crossed at 400 hz and the K-55 at 500 hz, the combination would be down less than .5 dB around 450 hz (I didn't do the math). Too little to make any difference. Again, this is voltage only. Driver sensitivity, cabinet, room, and kids will change everything. That's my theory anyway! Mike
  4. Measuring K-33 impedance in K-horn.

    Thanks for doing that! Odd that the impedance is lower at 50 hz in the horn than in free air (which I assume is the black graph). Mike
  5. Measuring K-33 impedance in K-horn.

    I'm bummed. At 60 Hz, I get around 4 ohms. At 1000 hz I get around 7 ohms. Could my K-horn's not be sealed well enough against the wall, or did I just waste time with this technique.
  6. Measuring K-33 impedance in K-horn.

    You caught on to my motivation. I saw that plot. Perhaps I'll try this at 50 hz and see if I'm still in the ballpark.
  7. I would like to measure the impedance of the K-33 in my K-horns at 400 hz. I've assembled a "patch cable" to place between my amp and the bass bin. The cable contains a 101.5 ohm 10 watt resistor that will end up in series with the woofer. The resistor actually measures 101.5 ohms. I have a free product on my Windows PC (fg_lite from Marchand Electronics) that allows me to generate the 400 hz signal. The chain will be PC-->DAC-->amplifier-->speaker cable-->patch cable-->right K-horn bass bin binding posts. These binding posts are connected directly to the K-33. When the signal is generated, I will turn the amp up until the tone becomes somewhat loud. I will use my Fluke 115 Trus RMS meter to measure the voltage across the right amplifier binding posts. I'll call the Vin. I will then measure the voltage drop across the 101.5 ohm resistor. I'll call this Vr. I will then calculate the current through the resistor. This will be Ir. Ir = Vr / 101.5. I will then calculate the voltage drop across the K-33, which should be Vin - Vr. I'll call that Vk33. Since the current passing through the resistor must be equal to the current through the K-33, I can calculate the impedance (resistance) of the K-33, I'll call that Rk33. Rk33 = Vk33 / Ir. This is what I ended up with... Vin = 6.325 volts Vr = 5.995 volts vk33 = .33 volts (man that's loud!) ir = .059 amps Vk33 = 5.887 ohms Does this method correctly measure the impedance of the K-33 at 400 hz? If so, I think I'll plot a few other frequencies just for fun. Sorry if this has been tried before. Thanks, Mike
  8. Class D

    Great catch! Where did you find that document? Even TI's website shows revision A document.
  9. Class D

    I posted that document here a month ago, wondering the same thing. Once I got the EVM, it was hard to tell if it was incorporated or not.
  10. Class D

  11. Class D

    The three pins in your picture are as follows... Left pin...hot...pin 2 on XLR jack Center pin...ground...pin 1 on XLR jack Right pin...cold...pin 3 on XLR jack The op amp will "flip" the cold and add it's voltage to the hot, canceling out any noise each picked up. This is the input to one channel, repeat on the other side for the second channel. I think that's right...
  12. Class D

    PS arrived this morning. I actually had a couple of M4 screws to secure it to the chassis. I secured the amp with 1 4-40 screw, just to keep it somewhat in place. I quickly hooked it up to an old pair of Kef speakers...SB touch--->Schiit Bifrost--->Schiit Lyr preamp outputs-->EVM. I noticed a hum with no sound. I disconnected the Lyr...dead quiet. I replaced the Lyr with my Dodd Audio buffer and powered everything back up. Dead quiet with no signal! Letting them get play with each other for a while before I connect the amp to the chassis components. MIke
  13. Class D

    I have an empty amplifier chassis, so I ordered and got the EVM, $74.50 shipped! . I must have lucked out, because I received it 6 days after the order. The chassis is from ClassDAudio. It's well built and has good components. http://www.classdaudio.com/cda-chassis-cases/cda-enclosure-with-components/ I should be getting my MeanWell PS on Friday, $30.09 shipped. I cut a piece of paper to the size of the PS, and crossing my fingers that it will fit okay. Instead of desoldering the RCA jacks, I butchered a couple of Switchcraft plugs (they are sitting on the PS template). I'll solder wires to these plugs and to the chassis jacks, then I can push them into the red and white RCA jacks on the EVM. All other wires will connect to the EVM thru the bare wire holes of the speaker/PS binding posts. Hope it all works out! I am currently driving my RF-83's with a ClassDAudio amp (IRS2092). I love that amp. This should be interesting. Mike
  14. Help Interpreting impedance plots

    Here's another great article. It made me realize that drivers with 2nd order passive crossovers are not really 180* out of phase with each other. http://www.bcae1.com/compleximpedance.htm Mike
  15. Help Interpreting impedance plots

    In reality, reactive components in series/parallel yield weird(?) results. Yes, the voltage and current differ by 90*, but when placed in a complex circuit, like a crossover, this get a little complicated. Let's take a simple first order crossover, 8 ohm woofer, 8 ohm tweeter, crossed at 1000 hz. The capacitor in series with the tweeter would be 19.88uF and the inductor to the woofer 1.27mH. At the resonance frequency (1000 hz), the impedance of the capacitor is 8 ohms @ -90*, the impedance of the driver is 8 ohms @ 0* However, the total impedance thru this circuit is 11.318 ohms @ -45 degrees. The woofer circuit is 11.318 ohms @ +45 degrees. If you pass a 10 volt 1000 hz signal thru the 11.318 ohm circuit, it will draw .8836 amps. Since the current thru the series circuit is the same as the current thru each component in the circuit, then .8836 amps thru an 8 ohm driver will be 7.07 volts, or -3 db. If the phase shift between the two drivers were identical, then a 1000hz signal to each driver, even though down -3db, would combine to give a +3db output bump. But the 90* phase difference between the two drivers eliminates the bump, because when the voltage to one driver is at it's peak, the voltage to the other driver is zero. I hope this makes sense. Here's a link to an interesting calculator. http://keisan.casio.com/exec/system/1258032632 Sorry if this doesn't answer your question. Second, third, and so on order filters get really interesting. Mike