Due to having unused old inductor of 2.25mH air coil diameter of 2.5CM
wire diameter 0.8mm
How long I need to cut the wire to drop to 1.5mH?
Such as I need to cut about 2 meters wire to drop from 2.25mH to 1.5mH
Thank you
wire diameter 0.8mm
How long I need to cut the wire to drop to 1.5mH?
Such as I need to cut about 2 meters wire to drop from 2.25mH to 1.5mH
Thank you
Unwind 2 turns at a time and measure using LCR meter. As you approach 1.8mH, reduce it to 1 turn at a time.
There's no shortcut around this. The inductance is a combination of core permeability, wire diameter and number of turns. The number of layers also has an influence. If you knew all of these in exact terms, you'd know how many turns you'd need for 2.25, and for 1.5, and then subtract the two. It's not a linear relationship so you can't extrapolate from one to the other. Both are needed.
There's no shortcut around this. The inductance is a combination of core permeability, wire diameter and number of turns. The number of layers also has an influence. If you knew all of these in exact terms, you'd know how many turns you'd need for 2.25, and for 1.5, and then subtract the two. It's not a linear relationship so you can't extrapolate from one to the other. Both are needed.
Dear supernut,
I assume you're asking this question because you do not have an LCR meter in the first place.
If you know how many turns make 2.25mH, you could use the following equation to arrive at 1.5mH, with L1 = 2.25, L2 = 1.5, N1 = current turns & N2 = required turns for 1.5mH.
L1 / L2 = (N1 / N2)^2
However, if you do not know how many turns you have in all (i.e. for 2.25mH), you may have to use the DCR and wire diameter resistivity (rho) to estimate it. I don't think it's an easy job though, since could be several layers, indicating non-constant diameter. Still, you should be able to get close to the required value, just by using your brain a little.
EDIT: If I were you, I would do the following:
Diameter difference (outer minus inner dia.) is proportional to the number of turns. Therefore, N1/N2 = (OD1-ID)/(OD2-ID), which brings us to:
L1 (OD1-ID)
--- = (----------)^2;
L2 = (OD2-ID)]
where L1 = 2.25mH, L2 = 1.5mH, OD1 = 2.5cm, OD2 = to be calculated, ID = inner dia (which you could easily measure)
Now, I think you should be able to get going.
I assume you're asking this question because you do not have an LCR meter in the first place.
If you know how many turns make 2.25mH, you could use the following equation to arrive at 1.5mH, with L1 = 2.25, L2 = 1.5, N1 = current turns & N2 = required turns for 1.5mH.
L1 / L2 = (N1 / N2)^2
However, if you do not know how many turns you have in all (i.e. for 2.25mH), you may have to use the DCR and wire diameter resistivity (rho) to estimate it. I don't think it's an easy job though, since could be several layers, indicating non-constant diameter. Still, you should be able to get close to the required value, just by using your brain a little.
EDIT: If I were you, I would do the following:
Diameter difference (outer minus inner dia.) is proportional to the number of turns. Therefore, N1/N2 = (OD1-ID)/(OD2-ID), which brings us to:
L1 (OD1-ID)
--- = (----------)^2;
L2 = (OD2-ID)]
where L1 = 2.25mH, L2 = 1.5mH, OD1 = 2.5cm, OD2 = to be calculated, ID = inner dia (which you could easily measure)
Now, I think you should be able to get going.
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2.25mh / 1.5mh = 1.5
Of 45mm - ID 25 = 20 mm
Square root of 20 / 8.888 = 1.5
Hence same ID of 25mm
OD = 25+8.888 = 33.888 to get 1.5mh
Thx
Of 45mm - ID 25 = 20 mm
Square root of 20 / 8.888 = 1.5
Hence same ID of 25mm
OD = 25+8.888 = 33.888 to get 1.5mh
Thx
An inexpensive way to measure inductance, and the calibration and modification to improve accuracy.
Digital LC100-A LCD High Precision Inductance Capacitance L/C Meter Tester | eBay
Digital LC100-A LCD High Precision Inductance Capacitance L/C Meter Tester | eBay
Oh dear,
If OD = 45mm for 2.25mH and ID = 25mm, then for 1.5mH it has to be like:
2.25 / 1.5 = [(45-25)/(x - 25)]^2 => 1.225 = 20 / (x - 25) => x = 41.32mm.
However, if it's the inner dia. that you're changing (i.e. unwinding from the inside) then it's:
L1 / L2 = [(OD - ID1) / (OD - ID2) ]^2
However, note that only the diameter difference (thickness of the copper turns) matters in either case.
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