Alexander Posted November 18, 2017 Share Posted November 18, 2017 Would like to learn some of the basics in reading the data of a impedance plot like this one to consider improving on a xover. This one is from a T5000 with updated parts still using the stock schematic values. Quote Link to comment Share on other sites More sharing options...
Alexander Posted November 18, 2017 Author Share Posted November 18, 2017 Here is one from a forte II same deal as above, new parts using stock schematic and values. 1 Quote Link to comment Share on other sites More sharing options...
mboxler Posted November 19, 2017 Share Posted November 19, 2017 Maybe this old thread from ALK will help??? https://community.klipsch.com/index.php?/topic/106102-developing-a-network-for-the-forte/&tab=comments#comment-1168913 Quote Link to comment Share on other sites More sharing options...
Alexander Posted November 19, 2017 Author Share Posted November 19, 2017 3 minutes ago, mboxler said: Maybe this old thread from ALK will help??? https://community.klipsch.com/index.php?/topic/106102-developing-a-network-for-the-forte/&tab=comments#comment-1168913 Thanks mboxler, I have read that thread and it has helped, I did build the forte ii version of that x-over and will be updating some of the components that were used in the first build next Quote Link to comment Share on other sites More sharing options...
Alexander Posted November 19, 2017 Author Share Posted November 19, 2017 Here are a few things I understand, but very rusty: In pure reactance (X) a cap will have current leading voltage by 90* and inductance has voltage leading current by the opposite 90* so this is why we have the phase shifts in a RC,RL or LCR x-over. Efficiency is reduced with a phase shift (70.7%?). Quote Link to comment Share on other sites More sharing options...
mboxler Posted November 19, 2017 Share Posted November 19, 2017 38 minutes ago, Alexander said: Here are a few things I understand, but very rusty: In pure reactance (X) a cap will have current leading voltage by 90* and inductance has voltage leading current by the opposite 90* so this is why we have the phase shifts in a RC,RL or LCR x-over. Efficiency is reduced with a phase shift (70.7%?). 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 Quote Link to comment Share on other sites More sharing options...
mboxler Posted November 19, 2017 Share Posted November 19, 2017 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 Quote Link to comment Share on other sites More sharing options...
WMcD Posted November 19, 2017 Share Posted November 19, 2017 I think there is not any info on the graphs which, alone, will allow you to improve a crossover circuit. The graph can be used to detect a failed driver, connection to a driver, or a filed crossover component (if you know what to look for). The most telling would be to have a graph of a good speaker to compare to a suspected one. Regarding what you can see on the graph: There are two sharp peaks in the bass region. This is actually the combination of two resonances. 1) The mechanical resonance of the woofer (it is moving more) would cause a more gentle peak in this electrical impedance. (If you look at the electrical impedance of the Heresy in a sealed box, you see the gentle peak or hump.) 2) The acoustic resonance of the port or passive radiator. This freq is placed very near the resonance of the woofer. In this way the passive radiator or port put a load on the back of the woofer diaphragm and reduce the movement of the woofer while the bass output comes frm the passive or port. The result, in the electrical impedance is a mountain with a valley though it. As we go further up in freq the electrical impedance falls to a minimum magnitude. This is where all reactances (induction and capacitance) balance out. We see the equivalent of the dc voice coil resistance of the woofer (plus resistance of any crossover coil). We must pause to recognize that we're seeing the combined impedances of the drivers and crossovers. If we looked at just a dynamic driver (these all are) there is a rising impedance at freqs well above their resonance. This is sort of like a low pass crossover which would be a coil in series with the input. The coil is the actual voice coil of the driver. Going above 700 Hz we see a big mountain up to above 50 ohms or so (small lettering doesn't help). That is the crossover capacitor allowing power into the mid-driver through the auto transformer. The auto transformer acts as step down transformer to reduce the level delivered the mid. Its input impedance is relatively high. Then going up again we see the impedance dropping. This is because the capacitor in series with the tweeter is allowing power though. WMcD 1 Quote Link to comment Share on other sites More sharing options...
Alexander Posted December 20, 2017 Author Share Posted December 20, 2017 Thanks gents. It is a bit clearer now than it was. Quote Link to comment Share on other sites More sharing options...
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