
Home  Forums  Rules  Articles  The diyAudio Store  Gallery  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Planars & Exotics ESL's, planars, and alternative technologies 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
5th January 2012, 03:50 PM  #171  
diyAudio Member
Join Date: Jan 2007
Location: wigston leics england

Quote:
__________________
HENRY 

6th January 2012, 02:42 PM  #172 
diyAudio Member
Join Date: Jan 2007
Location: wigston leics england

Finished the latest diaphragm for my 50 x 12.7 x 6 mm magnet layout. Just got to cut MDF etc. Then wire up.
__________________
HENRY 
8th January 2012, 11:49 PM  #173 
diyAudio Member
Join Date: Jan 2007
Location: wigston leics england

Have fixed new diaphragm to base, although not quite finished .Top and bottom not in position yet. Couldn't wait to try it out for sensitivity compared to the ferrite one on right, which was louder than my previous design using neos on the left. Shown in last picture hanging up. WOW what a big increase in sensitivity. Completely buries the ferrite one , which means I will definitely have to rebuild all my 20+ designs Give me strength, still it will keep me out of mischief!!
__________________
HENRY Last edited by JAMESBOS; 8th January 2012 at 11:52 PM. 
9th January 2012, 07:31 PM  #174 
diyAudio Member
Join Date: Jan 2007
Location: wigston leics england

FINISHED THIS ONE NOW STARTING ON ITS PARTNER. Can't wait.
__________________
HENRY 
11th January 2012, 06:24 PM  #175  
diyAudio Member
Join Date: Mar 2004
Location: Den Haag

Quote:
if you etch the aluminum in smaller strips you will end up with several omh for a cm of coil. if you can find any laminated mylar that would be good and could work great for etching the coils. i searched high and low for this stuff , not able to find any yet 

12th January 2012, 12:37 AM  #176  
diyAudio Member
Join Date: Jan 2007
Location: wigston leics england

Quote:
__________________
HENRY Last edited by JAMESBOS; 12th January 2012 at 12:44 AM. 

12th January 2012, 06:17 AM  #177  
diyAudio Member

Quote:
It would take 31.84 FEET of aluminum foil to get 3.9 Ohms, if the foil was 0.0254 mm thick x 2.547 mm wide (.001 inch thick x 0.1 inch wide). By the way, the aluminumcoated mylar film is definitely available. Try searching for "metallized film". I too would like find an efficient DIY way to etch or otherwise form the conductors, starting with metallized film.  The following should be useful for those of you who might want to be able to calculate foil thickness, foil width, foil length, and foil's electrical resistance. Below are some equations that I derived back when I posted in a thread about replacing the aluminum wire with foil, in some Magnepan speakers, at 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 . The idea, in that case, was to end up with a much lower mass on the membrane by replacing the original wires with foil, essentially to make it more agile, while retaining the same total impedance as seen by the audio amplifier. As it turned out, it could be done as well by using thinner wire, because by lowering the mass per inch of the conductor the resistance is increased and you end up having to use multiple conductor runs in parallel, to get back to the original impedance.  Resistance of an aluminum conductor is R = pL/A, where R is in Ohms, p is 2.6 x 10^(8), i.e. .000000026 Ohmmeters, 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 one meter of the .0254 mm x 2.547 mm 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. But since we'd be replacing existing wire, the length would already be decided. 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: (1) width = (.026 x length) / (Ohms x thickness) and (2) thickness = (.026 x length) / (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) / (thickness x width) (4) length = Ohms x (thickness x width) / .026 EXAMPLE: My MG12s have 32 bass/mid wire runs of about 42.5 inches each, which is about 34.544 meters. The driver I measured had a DC resistance of 3.85 Ohms. (Also note that there are only 26 vertical wire positions, but six of those have two wires.) If I wanted to replace the MG12 bass/mid wires with foil and get 3.9 Ohms total, and wanted to use .001inchthick foil (0.0254 mm), then the foil's width would need to be: width = (.026 x length) / (Ohms x thickness) = (.026 x 34.54) / (3.85 x 0.0254) = 9.18 mm That obviously wouldn't be a practical width but I wanted to see if it came out to about the same crosssectional area as the original 23 AWG wire. And 9.18mm x .0254mm = .2332 sq mm, which is close to the .2588 sq mm of the original wire. Note that my measurements of the speaker's wire length were not very accurate. Since length = Ohms x (thickness x width) / .026, the length of the original wire should have been 3.85 x .2588 / .026 = 38.3 meters  Now armed with some numbers and equations, and GOING BACK to the example in my earlier post (at the link posted above, in this post), about using 1/4th the crosssectional area and using two paralleled drivers of half the original panel size and about 8 Ohms each, let's say that the original MG12 panel has room for 26 vertical runs of about 43 inches each, or 28.4 meters. Half of that would be 14.2 meters. To get 8 ohms with 14.2 meters of aluminum, the crosssectional area would have to be .026 x 14.2 / 8 = .04615 sq mm. The original 23 AWG was .2588 / .04615 = 5.6X larger! Anyway, to get .04615 sq mm with .0254mmthick foil the width would have to be 1.83 mm (.072 inch). THEREFORE, if we replaced the wires of an MG12/QR's mid/bass panel with 0.001inchthick foil that was 1.83 mm wide, and wired it as two half panels in parallel, we would get a 4 Ohm driver with only 17.8% as much conductor mass as the original wires had.  Since the total available foil length should be known (based on the panel where the wires are being replaced or installed), I derived an equation that can be used to find, from a desired total length of foil, the crosssectional area needed to get two identical parallel foil runs that result in a 4 Ohm load: Area = thickness x width = length x .001625 where length is in meters, Area is in square millimeters, and thickness and width are in millimeters. If the desired width or thickness were known, then that could be rearranged to one of the following: thickness = length x .001625 / width or width = length x .001625 / thickness Or, if you had foil with a known thickness and width, you could calculate what total length would need to be divided in half to get 4 Ohms when the two halves were run in parallel, with: length = thickness x width / .001625 The following conversion formulas can be used: mm / 25.4 = inches inches x 25.4 = mm meters x 39.37 = inches inches / 39.37 = meters Example: If the optimal width that Wendell mentioned turns out to be 0.1 inch, i.e. 2.54 mm, we can calculate what total lengths would give 4 Ohms when divided into two parallel runs, for each of several different available thicknesses: .0005" (.0127 mm) x 0.1" foil: .0127x2.54/.001625 = 19.85 m (65.13 ft) .001" (.0254 mm) x 0.1" foil: .0254 x 2.54 / .001625 = 39.7 m (130.25 ft) I'll stop there, since .0015inch by 0.1inch foil would need 59.55 meters split into two parallel runs, to get 4 Ohms.  With only a tiny bit more work, I was able to derive the equations for n aluminum foil runs in parallel: (1) Rn = .026 x Ltot / nnTW where Rn is the resistance in Ohms of n parallel foil runs, each with length Ltot/n (in meters), and Ltot is the total foil length in meters, T is the foil thickness in millimeters, and W is the foil width in millimeters. Note that nn is n x n which is "n squared". The derivation for n parallel foil runs was easy, if we can assume that we want the same current through all runs, and therefore we have to assume that their resistances are equal. For equal resistances, the parallel resistances equation [Rpar = 1 / (1/R1 + 1/R2 + 1/R3 + ... 1/Rn)] boils down to Rpar = R/n. (We could also use shorter lengths of thicker foil, for example, to get the same current, which would give the same force. But then we'd be accelerating a different mass, so the acceleration would be different, even though the current was the same. We could calculate the difference in mass needed for a different current, to get the same acceleration from a different length, but then other aspects of the behavior of the membrane would probably still be changed. Maybe someone should figure that out, though, because it would open up an infinite number of options, if it could be done that way while still retaining satisfactory characteristics.) If we assume that we want to end up with 4 Ohms, that gives us the following different rearranged versions of the same equation: (2) Area = T x W = .026 L / 4nn (3) T = .026 L / 4nnW (4) W = .026 L / 4nnT (5) L = 4nnTW / .026 Examples: 1) What total lengths of .0005 x 0.1 inch foil would give 4 Ohms when divided into different numbers of parallel runs? TW = .0127 mm x 2.54 mm = .032258 sq mm n = 2: L = 4 x 2 x 2 x .032258 / .026 = 19.85 m (whew, same as before) n = 3: L = 4 x 3 x 3 x .032258 / .026 = 44.665 m 2) What thicknesses would work to get 4 Ohms from n parallel runs of 0.1inchwide foil that totaled 28.4 meters in length? T = .026 L / 4nnW n = 2: T = (.026 x 28.4) / (4 x 2 x 2 x 2.54) = .01817 mm (.0007153 inch) n = 3: T = (.026 x 28.4) / (4 x 3 x 3 x 2.54) = .008075 mm (.000318 inch) It doesn't look too good for the 28.4 meters needed for the MG12s, SO far. 3) What widths of foil would work to get 4 Ohms from n parallel runs totaling 28.4 meters, for a couple of different thicknesses? W = .026 L / 4nnT T = .001 inch (.0254 mm): n = 2: W = .026 x 28.4 / 4 x 2 x 2 x .0254 = 1.817 mm (.0715 inch) n = 3: W = .026 x 28.4 / 4 x 3 x 3 x .0254 = .8075 mm (.0318 inch) T = .0005 inch (.0127 mm): n = 2: W = .026 x 28.4 / 4 x 2 x 2 x .0127 = 3.634 mm (.143 inch) n = 3: W = .026 x 28.4 / 4 x 3 x 3 x .0127 = 1.615 mm (.0636 inch)  Cheers, Tom Last edited by gootee; 12th January 2012 at 06:29 AM. 

12th January 2012, 08:33 AM  #178  
diyAudio Member
Join Date: Jan 2007
Location: wigston leics england

Quote:
__________________
HENRY 

12th January 2012, 01:02 PM  #179 
diyAudio Member
Join Date: Dec 2004
Location: Staffanstorp

Sorry if this has been posted before!
How do you tension the PETfilm? Do you glue (tape) the coil to the film before stretching it? /R
__________________
The probability of someone watching you is proportional to the stupidity of your action.  A. Kindsvater 
12th January 2012, 02:04 PM  #180 
diyAudio Member
Join Date: Jan 2007
Location: wigston leics england

I do it by hand, pulling it whilst I PUT THE SCREWS IN the wood surround,another 45 meters bites the dust.
__________________
HENRY 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
my 14 inch full range speaker!  chinese  Full Range  235  21st September 2010 02:30 PM 
Where to find magnets???For Diy Planar  floydturbo  Planars & Exotics  8  27th July 2010 11:28 PM 
16 Ohm FullRange Speaker  thelordash  Chip Amps  13  23rd January 2009 06:13 PM 
full range speaker for HT  Jazzz  Full Range  13  8th September 2006 09:27 PM 
beginner full range speaker  tacomaboy  Full Range  5  12th January 2004 08:38 PM 
New To Site?  Need Help? 