radiogram Posted August 28, 2010 Author Share Posted August 28, 2010 Al: This still does not answer my quesion or rather I cannot comprehend and infer answer for my questions regarding: 1. if I decide to subtract Le from L2, then should I not be be designing the T Section for Re (not nominal impedence) and then subtract Le from the L2 value obatined based on Re? 2. Conversely if I choose to use some nominal impedence around crossover instead of Re,and compute the T Section paramters (L1,C,L2), I have then by defintion already taken into account the inductive AC resistance and hence I will just use compued L2 as is (no subtraction)? Thanks Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 28, 2010 Author Share Posted August 28, 2010 Gill, Lee: What is your take on this ? Please look at Posts Dated 08.28.10 9:12 PM thru 10:31PM Thanks Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted August 28, 2010 Share Posted August 28, 2010 Gram, It looks like we are not on the same page! Maybe it would be best if you gave me a call on the phone (410 546-5573). Text communication leaves a bit to be desired! Al K. Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 28, 2010 Author Share Posted August 28, 2010 Al: Thanks for offering me your time to speak with me on the phone. It is prettly late now (India time 12 AM) and I just about hitting the sack. I will try calling you Sunday your up time. Thanks Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 28, 2010 Author Share Posted August 28, 2010 Al: BTW, my plan is to build Butterworth (0dB ripple) and not Chebyshev. Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted August 28, 2010 Share Posted August 28, 2010 Gram, OOPS! I didn't realize you were in India! So much for telephone idea. If you plan to go with Butterworth you are definitely stuck with whatever output inductor the computer program gives you! There is really no advantage to Butterworth. It's just going to limit your flexibility. Could you try again to reword what is confusing you. Having nearly 35 years designing L-C filters makes it difficult for me to understand the thinking of a new-comer to the field. I take too may very basic things for granted. Al k. Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted August 28, 2010 Share Posted August 28, 2010 Gram, I simply did a sample design that might do what you need. It's a 2nd order Butterworth (12 dB / octave). The 1.1 mHy is the woofer Le. the Zobel of 6.8 Ohms and 27 uF NonPoler electrolytic The main elements are 27 uFd polypropylene and 2.4 mHy. The element values are the same in the highpass and lowpass which is normal for N=2 Butterworth. The crossover is roughly 620 Hz. The computer design is below, BUT: the plot shows 6 dB / octave steeper slope than you will get becasue the analysis is assuming the Le (1.1 mHy) is part of the filter. It's NOT! It's part of the woofer driver. Al K. Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 29, 2010 Author Share Posted August 29, 2010 Gill: We know the standard equation for impedance or X or Z of an inductor L. It is Z of inductor = 2*pi*F*L We have Z and we want to find the value of L My result was that L = 0.000955 Henrys The total impedence of the woofer is Zw = Re + jZl (where Zl being 2*pi*f*L). Is'nt then my measured impedence at the input of Re+Le circuit actually |Zw| = SQROOT(SQ(Re) + SQ(XL)) and NOT (Re + Zl) ? If I go by |Zw| and plug in my measured impedence @ 630Hz to be 7.7 Ohms with Re=3.7, I end up with Le=1.7mh. Please clarify. Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 29, 2010 Author Share Posted August 29, 2010 Al, The 1.1 mHy is the woofer Le. the Zobel of 6.8 Ohms and 27 uF NonPoler electrolytic Can you explain how arrived at R=6.8 and C=27uF for the Zobel? Myunderstanding was that C=Le / Re (DC Resistance) and if I assume Le as 1.1mhy and use Re=3.7, I end up with 80uF. So my zobel has R=3.7 and C=80; Further some people sugget that the R in zobel must be 1.25 * Re. Thanks Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted August 29, 2010 Share Posted August 29, 2010 Gram, I'll let Gil explain his numbers. I didn't follow his post carefully as I trust what he says. I'm sure he's right. The 27 uF + 6.8 Ohms was done by computer optimization. The 6.8 ohms is simply the closest standard value to the design impedance of 7 Ohms. The 27 uF happens to do the best of tuning out the Le over the widest frequency range. I have seen the formula for Zobel networks and it works pretty well, but this method nails it! Al K Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 29, 2010 Author Share Posted August 29, 2010 Al: After comparing your two Zobel plots and I got pretty excited about your 2nd plot with R=6.8 & C=27uF. I wanted to try it on the speaker and checked what closest parts values I had. The closed R I had was 6.3 (strangely it measures 6.3 Ohms but label says 5 Ohm). As far the Cap the only I coould get close to it was using 4 100uF Caps in Series to get 25uF. I connected this zobel across the Woofer and did a measurement test by using a known R=7.4Ohms in series between the Amp and the woofer with the paralled zobel and as usual reverse calculated impedence using voltage division. It seems pretty good except that it differs from your simulation in having a mild hump between 400Hz-900Hz. Please see attached Comparitive measurements. I would still like to investigate flattening aout the 400-900Hz region. Thanks so much. Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted August 29, 2010 Share Posted August 29, 2010 Gram, If you have the equipment to measure the Zo you can optimize the values for the best Zo on your speaker for the area of the crossover. You need to investigate you setup to verify that you are including the inductive component. The simple voltage plot through a series resistor doens't give you the complete picture. Even though I shut the "return loss" display off when I made the plots, I actually used return loss to do the optimization. Return loss includes the complete complex impedance. It's directly related to VSWR. The simple Zo can look worse even when the return loss is better. It looks like your already have it a lot better than most designers will wind up with it though! Al K. UPDATE: I just looked at the return loss to optimize it right at 620 Hz. It looks like 22 uF is actually better then 27 uF. The difference is small though. Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 29, 2010 Author Share Posted August 29, 2010 Al: I do not have any specific instrument to measure Z. I am using the old fashioned crude method as follows: Amp Out------Known R=7.4Ohms----------------- | | R Woofer C | | | --------------------------------------------------------------- With a known input Voltage of say 1Volt, I measure the voltage across the known resistor, then divide that voltage By 7.4 to get the Current. I compute Z of the load as = (Input Volts - Voltage across R=7.4) / Current. How much I will be off by this type of calculation? Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 29, 2010 Author Share Posted August 29, 2010 Al: What does your simulation say for R=6.3 & C=50uF. When I expereminetd that I seem to get a better impedance plot as below. Granted my measurement technique is crude, but the goal of zobel being to present a fairly contant load, ultimately if the measured voltage is fairly constant either across the know Resistor or Load, is not that a good indication or good enough? Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 29, 2010 Author Share Posted August 29, 2010 Comparison Plot Quote Link to comment Share on other sites More sharing options...
radiogram Posted August 29, 2010 Author Share Posted August 29, 2010 Al: When I tested the Zobels I had connected the R first and then C to the ground. In your plots you seem to connect C first then R to ground. Does it matter since they are in series? Thanks Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted August 29, 2010 Share Posted August 29, 2010 Gram, I doesn't matter, R - C or C - R. Like you say, they are simply in sereis. I ran 6.3 Ohms and 50 Uf on the computer. It looked quite bad. The best value will depend on exactly what your woofer driver actually is. To get any closer you need to actually measure the complex Zo both of the woofer itself and of the Zobel together with the driver. Your woofer may be closer to 6 Ohms than 7. I think they vary a bit. Al K. Quote Link to comment Share on other sites More sharing options...
WMcD Posted August 30, 2010 Share Posted August 30, 2010 The clarification is: I made a mistake and failed to show all necessary calculations. If you look back at my post I've made up a spread sheet and attached it showing the correct calculations. Sorry about that. Wm McD Quote Link to comment Share on other sites More sharing options...
Blvdre Posted August 30, 2010 Share Posted August 30, 2010 Gil You'd make a terrible engineer (tongue in cheek). You not only admitted to making a mistake, but you went back and corrected it [] Quote Link to comment Share on other sites More sharing options...
Al Klappenberger Posted August 30, 2010 Share Posted August 30, 2010 Gil, MAN, That's like cheating on you wife. You have got to DENY, Deny, deny! Al K Quote Link to comment Share on other sites More sharing options...
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