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Air Wound Coil Value

时间:04-04 整理:3721RD 点击:
Anybody please help me to find the value of the following Air wound inductor.

E2-0.30-1.7-5TL

An empirical equation to find the inductance/mm of an Air stripline without ground is:

L(nH/mm)=0.2[ln[L/(W+t)]+1.93+0.2235(W+t)/L]

where W is the width of the line(mm), t is the metal thickness(mm), and L is the length of the line(mm).

Hi Vfone,

Thanks for info. In my question, E2-0.30-1.7-5TL, what is W(mm),t(mm) and L(mm) in the sample value.

Thanks,

I don't think srikanthvajjala is speaking about the inductance of a track, but he is looking for the value of a commercial fixed inductor.

I only remember that 5TL means 5 turns. The other figures should be the diameter and the length of the coil, but I don't remember they order in the part number. However, if you can measure those dimensions by means of a caliper you can use the wheeler formula.

Calling N the number of turns (in you case N=5)
R the radius of the coils, in inches
len the total lenght of the coil, in inches

the estimated inductance L, in uH, will be:

L= (R*N)^2/(9*R + 10*len)

Let's suppose the numbers means 0.3 --> diameter=0.3 cm --> 0.12 inches
1.7 --> lenght=17 cm --> 0.67 inches

then R=0.06 inches

L≈0.012 uH = 12 nH

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