A question for experts in inductors

Thank you very much Genome for your Round Wire Calculator. It gives amazing results! (BTW what is the meaning of the items Dia(L), Dia(S), Dia(H) and Dia(T) ?)

I finally found a test to demonstrate such kind of figures. I made an impulse test. I arranged the inductor in a LC parallel circuit (c=10nf) and I used as impulse a DC current suddenly decreasing from 50mA to 0. With the scope I catched the damped oscillation and I calculated the time constant of the exponential decay. From tc=2L/R, using the nominal value of L=12.6uH (measured @f=1KHz) I deduced R and also Q=omega(L/R). I tested also the circuit with different capacitor (C=0.47uf) so to investigate the behaviour at a lower frequency. Here are the results:

a) L=12.6 uH R=0.046 ohm Copper plait made of 8 wires diam. 0.315 mm wound on air.

C=10nf Oscillation frequency = 495KHz R= 6.8 ohm Q= 6

C=0.47uf Oscillation frequency = 69KHz R= 0.32 ohm Q= 17

b1) L=12.6 uH R=0.055 ohm Single copper wire diam. 0.64 mm wound on a toroidal core (Micrometals T68-2).

C=10nf Oscillation frequency = 470KHz R= 1.9 ohm Q= 21

C=0.47uf Oscillation frequency = 69KHz R= 0.19 ohm Q= 29

As you can see those figures are not very different from the ones elaborated by Genome's program. This definitely answers to my original question. My test oscillator requests an higher power supply current with inductor a) because it has a very poor quality factor (that is a more resistive losses). And the main reason of that is the multi-layer structure of the inductor.

A final observation. The inductor will be used together with a C=0.47uf to implement a low pass filter for a class D amplifier. The lower Q of this inductor will reduce the peak of the audio frequency response and it will produce a more flat bandwidth. At lower frequencies the equivalent R of the inductor approaches the DC value (0.046 ohm) ensuring in this way a fair damping factor. I think this kind of inductor could have unexpected benefits...

Thank you very much to all the guys of this thread for their precious hints.

Marco

Reply to
Guido
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Those are grades (thickness) of insulation. Light, Single, Heavy and Thick.

Glad to hear it produces 'meaningful' results. I'm just the monkey who plugged someone elses sums in and added the data.

Nice method for verifying your result. You are now the expert.

DNA

DNA

Reply to
Genome

Strange...... As I wrote that reply I was wondering, I wrote the program a while ago.

Part of the answer involves the spacing of wires between layers. It's probably taking things too far because they are not going to lie flat against each other but the insulation will separate the layers.

If you pick a copper diameter and then click on the L,S,H and T columns then you can see the AC resistance drop as the insulation thickness increases.

You should have roundedit.pas. You can open that in a text editor. Procedure RoundWireForm.Recalc is the one that does the sums and there is something in there called LayFact, the Layer Factor.

DNA

Reply to
Genome

its a good way of doing it, and can be done crudely with a current-limited DC power supply, by removing one wire....

pretty neat, huh?

as a general rule of thumb, always damp LC circuits. choosing just the right (ferrite) material can also do this, by making the losses equate to a suitable damping resistance.

Reply to
Terry Given

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