Re-winding an inductor to decrease inductance

Question

I found a bunch of good size inductors in a bunch of old boards (16 of them, more to desolder). Luckily I had the service manual of said boards and I know the specs of the inductors: 22uH 9A 500V.

I need 15uH and 7uH inductors. How can I calculate how many turns to remove from the coil to get to the needed inductance? I don't have an L meter (that's a future project).

I counted the turns. There are about 50 turns of 1mm wire in a core that's about 21mm outer diameter, 9mm inner diameter and 8mm thick (I'm guessing the values from measuring and subtracting/adding the wire diameter).

The color of the core is a dark reddish-brown.

Answer 1

Inductance

Inductance is proportional to turns squared, all things being equal.

L2 / L1 = N2^2 / N1^2 ... 1 so
N2^2 = N1^2 x (L2/l1) ... 2 so
N2 = N1 x sqrt (L2/L1) .... 3

Apply 3. above

For 15 uH:
L1 = 22 uH
L2 = 15 uH
N1 = 50 so
N2 = 50 x sqrt(15/22) = 50 x 0.826 =~ 41.3 turns

So 41 turns in theory and maybe 40 - 42 range.

Similarly for 7 uH N = 50 x sqrt(7/22) =~ 28turns

Current rating:

Core magnetisation depends on "ampere-turns" so current handling capacity will increase with the inverse of the turns ratios.
I2 = I1 x N1/N2

So 15 uH & 41 turns will handle
9A x 50/41 =~~~= 11A
...
Thicker wire may be required depending on actual currents used, as copper losses increase with I^2.

Treat all such results with care and test in practice.

Insulation:

You do not describe the core type or former used.
You say 500V rating - I don't know what voltage you are using in practice but, be sure that insulation is (more than) adequate for the task. As you intend to use existing windings, take care not to nick insulation or have wire rub against core edge. If a bobbin former or insulated core is used this should be easy enough to manage. Odds are a 500-V rated part does not use a bare core and depend on wire insulation alone :-)