Cut-Throat Posted March 9, 2008 Share Posted March 9, 2008 Is there a list of combinations of attenuation that you can get withthe T2A autoformer that also includes the correct resistor to use tomaintain impedance?For instance, if I decided that -8 db would be the right attenutation for my Midrange horn. Would I still use a 15 ohm resistor across the Midrange? Quote Link to comment Share on other sites More sharing options...
greg928gts Posted March 9, 2008 Share Posted March 9, 2008 This is a good question. I was once directed to put a resistor in line on the positive terminal to the mid horn to lower the output by 3db. I tried it and it seemed to work. I don't remember what size resistor. Greg Quote Link to comment Share on other sites More sharing options...
Guest " " Posted March 9, 2008 Share Posted March 9, 2008 read gil's response on this thread. http://forums.klipsch.com/forums/p/47861/452646.aspx#452646 I'm in the group that puts the resistor across the autoformer as indicated in the referenced thread. Quote Link to comment Share on other sites More sharing options...
greg928gts Posted March 10, 2008 Share Posted March 10, 2008 You wouldn't happen to have a picture of what that looks like? Schematic? What size resistor? Greg Quote Link to comment Share on other sites More sharing options...
Cut-Throat Posted March 12, 2008 Author Share Posted March 12, 2008 Maybe I should make this question a bit more specific. If someone was running a Klipsch Type "A" crossover. What we be the various combinations of attenuation on the autoformer? And then to keep the impedance to the amplifer similar the various combinations of resistors across the midrange/and or cap sizes? Need a crossover expert here. Quote Link to comment Share on other sites More sharing options...
BEC Posted March 12, 2008 Share Posted March 12, 2008 I can help you with part of that. The attenuation values available on the T2A are: Tap 4 - 3db, tap 3 - 6db, tap 2 - 9db and tap 1 - 12db. Now you need to find someone who believes in adding resistors to answer the rest of that for you. Bob Crites 1 Quote Link to comment Share on other sites More sharing options...
Guest " " Posted March 12, 2008 Share Posted March 12, 2008 Maybe I should make this question a bit more specific. If someone was running a Klipsch Type "A" crossover. What we be the various combinations of attenuation on the autoformer? And then to keep the impedance to the amplifer similar the various combinations of resistors across the midrange/and or cap sizes? Need a crossover expert here. I don't think the math will hold up if you added resistors to the driver end as you change the tap settings. in essence, as you went from -3db, -6db, -9db, etc, you would be doubling the impedance at each step, 2x,4x, 8x, etc. Putting resistors on the driver end to bring the impedance back down would be a massive signal drain. Think about it. If you went from -3db to -6db, thats moving from 2x to 4x. Your at 64 ohms. To bring it back down to 28 ohms, would probally take a 7 ohm resistor across the driver. Resulting in 66% of the signal going to the resistor and 33% going to the driver. So your not at -6db anymore. All numbers approximate. Most folks that use resistors do it across the main input connections of the autoformer. They use 1 resistor . Look up swamping resistor on this forum for reason why they do it. Quote Link to comment Share on other sites More sharing options...
Deang Posted March 12, 2008 Share Posted March 12, 2008 ! Quote Link to comment Share on other sites More sharing options...
Deang Posted March 12, 2008 Share Posted March 12, 2008 ~ Quote Link to comment Share on other sites More sharing options...
Deang Posted March 12, 2008 Share Posted March 12, 2008 With the Autoformer swamped, a floating ground, and the primary cap scaled for 8 ohms -- the T2A gives you -3.8, -4.6, -6.7, -7.4, and -9.5. More options with the 3619 -- even more with the 3636. Quote Link to comment Share on other sites More sharing options...
Cut-Throat Posted March 12, 2008 Author Share Posted March 12, 2008 Most folks that use resistors do it across the main input connections of the autoformer. They use 1 resistor . Look up swamping resistor on this forum for reason why they do it. Yup, I have read a lot of posts on this. I understood it pretty well at the time I read it. That is why I put a 15 ohm resistor on the midrange. - I dug it out of the archieves. Quote Link to comment Share on other sites More sharing options...
WMcD Posted March 12, 2008 Share Posted March 12, 2008 WARNING. I see there is an error in three cells of calcs. C12 - C14. (They need an extra zero) I will work on this over the weekend (or the next) and edit this post to upload a spread sheet with a name indicating "corrected". Also, worked on an explanation, The principles are better explained with a few diagrams and I'll include that . = = = = = Attached is a spreadsheet I just made up which may answer questions. It does, however, take a bit of knowledge to understand. BTW the forum will accept files in zipped format which are not the permitted types. (Let's see if this works.) Auto transformer taps of -3, -6, -9, etc. dB, essentially work because the taps present increasing doublings of the driver impedance. That is the easy part. We have to first start with an assumed driver Z. I believe the K-55 is about 13 ohms. Note that you can play with that value on the spreadsheet. I've locked other cells to avoid inadvertent alterations. Next we have to calculate an appropriate cap to serve as a simple first order crossover. That depends on the Z presented by the auto transformer and also the crossover freq. That freq is something you can play with too on the spreadsheet. Note that the starting value (here) of the cap and crossover freq pretty much corresponds to the cap in an A for a K-Horn. = = = A deeper problem is what value of resistor should we put across the input to the auto transformer if we want to maintain the load on the cap (which is to say, the design of this simple x-over). This depends on where you start. Going one step is pretty easy. If we go down 3 dB we are doubling the effective load, and therefore the resistor has to be the same as the steped down value, to bring it back to where we started. Two equal resistors in parallel give you half the values. Going down two steps is not quite as easy but can be calculated, as you can see. I've put in another column of a "proof". If you are familiar with "rules" of parallel resistors, the calculations in the cells should make sense. Of course I invite criticism and comments. Gil Resistance to Balance - Corrected.zip Quote Link to comment Share on other sites More sharing options...
Guest " " Posted March 12, 2008 Share Posted March 12, 2008 Attached is a spreadsheet I just made up which may answer questions. It does, however, take a bit of knowledge to understand. BTW the forum will accept files in zipped format which are not the permitted types. (Let's see if this works.) Auto transformer taps of -3, -6, -9, etc. dB, essentially work because the taps present increasing doublings of the driver impedance. That is the easy part. We have to first start with an assumed driver Z. I believe the K-55 is about 13 ohms. Note that you can play with that value on the spreadsheet. I've locked other cells to avoid inadvertent alterations. Next we have to calculate an appropriate cap to serve as a simple first order crossover. That depends on the Z presented by the auto transformer and also the crossover freq. That freq is something you can play with too on the spreadsheet. Note that the starting value (here) of the cap and crossover freq pretty much corresponds to the cap in an A for a K-Horn. = = = A deeper problem is what value of resistor should we put across the input to the auto transformer if we want to maintain the load on the cap (which is to say, the design of this simple x-over). This depends on where you start. Going one step is pretty easy. If we go down 3 dB we are doubling the effective load, and therefore the resistor has to be the same as the steped down value, to bring it back to where we started. Two equal resistors in parallel give you half the values. Going down two steps is not quite as easy but can be calculated, as you can see. I've put in another column of a "proof". If you are familiar with "rules" of parallel resistors, the calculations in the cells should make sense. Of course I invite criticism and comments. Gil If only I could open the file......how big is it...can you post unziped? Quote Link to comment Share on other sites More sharing options...
Guest " " Posted March 12, 2008 Share Posted March 12, 2008 Most folks that use resistors do it across the main input connections of the autoformer. They use 1 resistor . Look up swamping resistor on this forum for reason why they do it. Yup, I have read a lot of posts on this. I understood it pretty well at the time I read it. That is why I put a 15 ohm resistor on the midrange. - I dug it out of the archieves. On the mid range as in the connections to the drivers? Quote Link to comment Share on other sites More sharing options...
WMcD Posted March 12, 2008 Share Posted March 12, 2008 I will have to look at problems opening the file. It opens for me but I have Excel on this laptop and IE Explorer. Let's hear from others out there. Gil Quote Link to comment Share on other sites More sharing options...
jtkinney Posted March 13, 2008 Share Posted March 13, 2008 Gil, I saved it to the desktop using Firefox and opened without problems with Excel. I haven' t studied it to understand it, but did open it. Quote Link to comment Share on other sites More sharing options...
Cut-Throat Posted March 13, 2008 Author Share Posted March 13, 2008 I will have to look at problems opening the file. It opens for me but I have Excel on this laptop and IE Explorer. Let's hear from others out there. Gil I was able to open it with Excel also with no problems. Thanks for sending this! After I study it awhile, I can ask a more intelligent question. Quote Link to comment Share on other sites More sharing options...
Cut-Throat Posted March 13, 2008 Author Share Posted March 13, 2008 I admit to not understanding your spreadsheet. I am not sure what you mean by 'steps'. If this is a step from -3db to -6db, then the 'one step above' and 'two steps above' are confusing me. Also the resistor values, Are they in Ohms. Perhaps you could give us an example. For instance I am using a Klispch type 'A' with a 13uf Cap and a 2uf cap for the tweeter. Using these values what resistor value would you use and across what taps of the T2A autoformer would you put it? Also, Are you also changing the cap value of the 13uf Cap? - I had thought the idea of using a swamping resistor, was that so you did not have to change the Cap? Lots of confusion on my part.[*-)] Quote Link to comment Share on other sites More sharing options...
Guest " " Posted March 13, 2008 Share Posted March 13, 2008 the spreadsheet had 2 dot's in the name....my unzip file could figure what kind of file it was....one dot removed and it opned fine. Quote Link to comment Share on other sites More sharing options...
greg928gts Posted March 13, 2008 Share Posted March 13, 2008 Opened fine. I think I'm as confused as Cut-throat. Greg Quote Link to comment Share on other sites More sharing options...
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