Magnepan Ribbon Tweeter Question
Ok, hate to ask such a dumb question, but never really thought to ask until now...
On a typical, say Magnepan Tympani IVa Ribbon Tweeter, it appears from the schematic it is 2 ohms with a 1 ohm series resistor...no coil to bring up the impedance... There is only one pleated "vertical run" for the tweeter (no return passes). How the hell is it 2 ohms? For it to be 2 ohms it would have to be quite long for being only 1/2" wide...what gives? I have a pair of trashed Apogee Caliper Tweeters that I was going to run some ribbon experiments on after regluing al the magnets...I am preparing a ribbon made of 12 micron Aluminum metalitized mylar I used on previous ESL experiments. Also getting a pleater and (don't laugh) planning on using my wife's spaghetti machine to cut the 1/2" wide strips. Pretty smart huh? Just don't tell the wife. 
Is it really that wide? The MG3.6/R ribbons are only 5/32 inch wide, 0.00015 inch thick, and 55 inches in length.
Anyway, it depends on the crosssectional area, the length, and the conductivity of the material. Resistance of an aluminum conductor is R = pL/A, where R is in Ohms, the conductivity of aluminum, p, is 2.6 x 10^(8) Ohmmeters (where 2.6 x 10^(8) = 0.000000026), L is length in meters, and A is crosssectional area in square meters. From that, the Ohms per meter would be .026 / (area in sq mm). So, for example, one meter of .0254 mm thick x 2.547 mm wide foil would have a resistance of .026/.0647 = .4019 Ohms per meter. So, to get 3.9 Ohms of resistance it would require 9.705 meters of foil, or 31.84 feet. Let's see... Ohms/meter = .026 / (thickness x width). And (Ohms/meter) x length(in meters) = Ohms [so Ohms/meter = Ohms/length(in meters) and length = Ohms / Ohms/meter)]. So, with aluminum foil length in meters and foil thickness and width both in mm, we have: (1) width (mm) = (.026 x length_m) / (Ohms x thickness_mm) and (2) thickness (mm) = (.026 x length_m) / (Ohms x width) So, with those equations, we could calculate the needed width or thickness of the foil, if we specified the thickness or width and knew the length and resistance that we were replacing. (Also, remember that the length is in meters but the thickness and width are both in mm.) Also: (3) Ohms = (.026 x length_m) / (thickness_mm x width_mm) (4) length (m) = Ohms x (thickness_mm x width_mm) / .026 Handy conversion equations: mm = inches x 25.4 inches = mm / 25.4 m = inches / 39.37 inches = m x 39.37 Example: MG3.6/R ribbon is 55 inches long, 5/32 inch wide, and 0.00015 inch thick. What is its resistance? L = 55 / 39.37 = 1.397 m W = 5/32 x 25.4 = 3.969 mm T = 0.00015 x 25.4 = 0.00381 mm (3) Ohms = (.026 x length_m) / (thickness_mm x width_mm), from above. Ohms = (0.026 x 1.397) / (0.00381 x 3.969) = 2.402 Ohms 
By the way, my other main posts about foil resistance are at:
http://www.diyaudio.com/forums/plana...ml#post2857842 and RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ...  tom_gootee  Planar Speaker Asylum (where you might also want to click on "All" at the top and then Edit>Find all occurrences of "gootee"). 
Cool, that explains it. Too bad I do not know the thickness of that Aluminum Metalitized Mylar is. I guess I could work backwards (find resistance in 10' long, 1/2" wide strip section).
Another thing I have been thinking is puting sections of ribbon in series (mids  wide + tweeter  thinner) to get the required resistance. In other words, 3/4" wide section (Midrange) strip of 42" clamped to a 1/2" (Tweeter) 42" strip to get somewhere over 2 ohms , then add a single resistor to get near 3 ohms. This, to me, should work for a mid/tweeter...no? 
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Good point, Roger. And since the ribbon foil's thickness and width are known (giving the crosssectional area), the resistance could be measured and used to calculate how much length the pleating added, i.e. what length of unpleated ribbon was used to make the 55inch pleated ribbon.

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I wonder if that material will work well for a tweeter. The mylar is fairly massive. I think that Magnepan's ribbon tweeters are just foil alone, with nothing else attached (unlike their "quasiribbon" tweeters, which are on mylar, but are not a separate ribbon). But they get 22 kHz frequency response with foil on mylar (but 40 kHz with just the foil), so it appparently CAN work wellenough. However, their foilonmylar tweeters use at least a couple of inches of width, and multiple 0.1inchwide foil conductors, whereas the true ribbon (without mylar) only needs a single 5/32ndsinchwide foil conductor. 
The ribbons on my Apogee Duetta Signatures are foil on Kapton. I wanted to experiment a little with both foil and aluminum Metalitized Mylar.
I do not remember my TIVa tweeter ribbons being that narrow (5/32") 
So I took the wifes spaghetti machine and inserted a 5' wide standard aluminum foil, and it came out at a perfect 1/2" wide. I then used a pleaster the wife got for me from Michaels crafts. It puts nice clean pleats on the Aluminum strip. I made two 40" long in series, and clamped them into an Apogee Caliper tweeter magent assembly in series. The resistance laying down was 19 ohms, standing up straight was around 8 ohms. I did not tension teh ribbons, just snug. Put a ramshakle crossover I slapped together for my TIVas, (2nd order two caps, two inductors) with a xover point at around 150hz (I know, too much for the ribbons  just was too lazy to look for a higher value series High Pass cap) and it sounded impressive... The ribbons danced around a little too much on high volume (all the while being careful not to have the two ribbons touch, shorting out the amp).
For a first pass proof of concept, it was great. I will experiment with putting these two ribbons in parallel and if needed a series resistor to get at least 3 ohm, and put a higher xover HP cap for a xover of 500  600hz and see how she sounds. So in short, stock Reynolds kitchen foil (.00064" thk  16 micron) , a spaghetti machine and a $12 pleater. Pics later. I love this stuff. 
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