Inductor unwinding calculations?
2 Attachment(s)
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)? 
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 ;) Jan 
1 Attachment(s)
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. 
OK,
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! 
Decent LCR meters are only $4050

If i did this correctly, i should remove around 56 turns.
That sounds about right. 
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. 
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 224168=56 turns. 
All times are GMT. The time now is 01:19 AM. 
Search Engine Optimisation provided by
DragonByte SEO (Pro) 
vBulletin Mods & Addons Copyright © 2021 DragonByte Technologies Ltd.
Resources saved on this page: MySQL 17.65%
vBulletin Optimisation provided by
vB Optimise (Pro) 
vBulletin Mods & Addons Copyright © 2021 DragonByte Technologies Ltd.
Copyright ©19992021 diyAudio