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Inductor unwinding calculations?
Inductor unwinding calculations?
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Old 3rd June 2017, 01:53 PM   #1
filthyone is offline filthyone
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Default Inductor unwinding calculations?

Hello people of diyaudio!
I was given by a friend a set of generic Visaton crossover networks (HW2/70NG). I would like to put them in use for a pair of 2 way speakers. I have already established the values for the 2nd order filters using XSim, and i need to lower the 1mH low pass inductor to around 0.56mH (exact value is not critical, what matters most is that i reduce both by the same amount).
The inductor has 16 verical layers (or stacks?) by 14 horizontal layers (hope the pictures make more sense). So there should be around 224 turns of 1mm copper wire. The coil internal diameter is 18mm, the outer diameter is 45mm, and the coil hight is 15mm.
Is there anyone cunning enough to tell me how many turns should i unwind in order to achieve this value (0.56mH)?
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Old 3rd June 2017, 03:14 PM   #2
jan.didden is online now jan.didden  Europe
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It is probably possible to calculate the required windings for that value using an average winding diameter. Assuming that you are able to measure, I would proceed to remove say 1/3rd of the layers, then measure, then again remove some windings, etc. Should be able to home in to the required value in a few iterations.

After all, this is diy ;-)

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Old 3rd June 2017, 03:22 PM   #3
Lojzek is offline Lojzek  Croatia
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Approximately calculating, there is dcr values on Visaton web for the SP1 air core coils. 0,68 mH= 0,4 ohm; 0,47 mH=0,3 ohm; since there is no 0,56 mH to choose from, the middle dcr value of the two would be about 0,35 ohm. 1 mH = 0,5 ohm, so you'd need to remove the excess of 0,15 ohm of wire to arrive closer to 0,56 mH. Find the characteristic resistance per meter for the copper wire of 1 mm diameter and you can solve for the length of wire by dividing the two values.

You could build this jig to measure directly with a 1 kHz mp3 pure sine wave or search for a coil calculator.
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Old 3rd June 2017, 05:17 PM   #4
filthyone is offline filthyone
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after a little more search, i found this really old post http://www.diyaudio.com/forums/multi...tml#post293905
which helped me calculate the number of turns to be removed.
If i did this correctly, i should remove around 56 turns.
Thanks all!

Last edited by filthyone; 3rd June 2017 at 05:25 PM.
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Old 5th June 2017, 08:52 PM   #5
augerpro is offline augerpro  United States
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Decent LCR meters are only $40-50
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Old 5th June 2017, 09:22 PM   #6
Andrew Eckhardt is offline Andrew Eckhardt  United States
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If i did this correctly, i should remove around 56 turns.

That sounds about right.
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Old 5th June 2017, 11:59 PM   #7
system7 is online now system7  United Kingdom
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56 turns, or about a 1/4 of the 224 sounds dead right.

Inductance is the square of the number of turns, disregarding varying size of turns. 3/4 squared is 1mH X 9/16, which should be about 0.56mH.

I'd always take off a bit less, then measure.

SP coils

The data for 1mm windings is saying the resistance should drop from 0.5R to about 0.37R.
Best Regards from Steve in Portsmouth, UK.
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Old 6th June 2017, 02:37 AM   #8
JMFahey is offline JMFahey  Argentina
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You have 16 layers of 14 turns each, so 224 turns total

To reduce from 1mH to 0.56mH you must reduce inductance to 56% of original value.
Since inductance varies with the square of turns, you need to reduce winding to square root of 56% so to 0.75 of turns , so to 168 turns.

You must unwind and discard 224-168=56 turns.
Design/make/service musical stuff in Buenos Aires, Argentina, since 1969.
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