Paul Burridge wrote...

I need to wind 180nH inductor for a parallel tuned circuit I'm
building, since the 180nH factory-made chokes I have don't look
up to the job power-handling wise. This coil needs to handle about
90mA p-p / 500mW maximum dissipation sine current and I've allowed
3 ohms for series resistance. Can anyone give me some steer on
dimensions, number of turns, core type and so on? Thanks,

As John has said, that's a very low inductance that should not
present any problems at such a low power level. But perhaps for
a more detailed answer you can tell us the frequencies your coil
will experience. At high frequencies skin and proximity effects
dominate, and these can be evaluated with an Rac/Rdc ratio. If
a ferrite is used its high-frequency core loss can also be modeled
as an inductor resistance.

Do you have any special size constraints? Unless you really need
a miniature size, an air core may be best for 180nH. You can use
the Wheeler equation to experiment with different coil designs.

I'll add some new grist for the mill, with a copy of a portion of
a posting I made 28 Dec 1997, about air-coil inductance equations.

-------------------------------------------------------------------

Throughout the discussion we'll use the same dimensional system, based
on the drawing below. Here and in the 14 formulas below, N = turns,
a = mean radius, b = length, and c = winding thickness, and all are
in inches, unless otherwise stated.

length
|------ b --------|
--- ,-----------------,
c | cross section | ------------ a = winding mean radius
--- '-----------------' a |
__________________________ |
axis D = 2a
,-----------------, |
| cross section | --------
'------------\----'
\ solenoid coil layout
N turns
--------
[ snip five formulas and discussion ]

To simplify our lives, Wheeler empirically derived his popular
single-layer solenoid equation, using Nagaoka's equation and
tables. Wheeler's equation is shown below in two different ways.

a N^2 a^2 N^2 / 10 b
(6) L = ---------- = -------------- uH / inch
9 + 10 b/a 1 + 0.9 a/b

Wheeler says this equation is accurate to about 1% for long coils, or
any coils with (b/a 0.8). [Confirmed with extensive measurements I
made and posted on s.e.d.] It's easy to solve this equation for N.

A simple re-arrangement adds the concept of winding pitch. This can be
very useful, in part because a low-winding-height multilayer coil can
be treated as a single-layer coil with a higher winding pitch.

a^2 p N 1
(7) L = -------- * ---------------- uH / inch
10 1 + 0.9 a p / N

Here p is my turn-density pitch parameter, in turns/inch. Incidentally,
this makes clear that for long coils, once you pick a coil-winding pitch,
the inductance scales by N, rather than by N^2. Of course, the length
scales as well. Now solving for N isn't as easy. I get,

10 b
(8) N =~ ----- ( 1 + 9 a^3 p^2 / 100 L ) turns
p a^2

Alan Fowler pointed out a version of Wheeler's equation, claimed more
accurate, in F. Langford-Smith's "The Radiotron Designer's Handbook,"
1942. In the 3rd edition only, the work of Esnault-Pelterie is detailed,
a Frenchman who followed the "des savants japonais" (i.e. Nagaoaka) for
his derivation of a simple Wheeler-like formula with a claimed accuracy
of 0.1% for values of diameter/length between 0.2 and 1.5. Rearranging,

a^2 N^2 / 9.972 b
(9) L = -------------------- uH / inch
0.9949 + 0.9144 (a/b)

[ snip more formulas and stuff ]
-------------------------------------------------------------------

OK, there you have a small panaply of equation forms to select from.

(7) is easier to use than it appears at first glance. Let's design
a coil for you. We'll pick wire size #22, which has a diameter of
0.020 inches, prompting us to pick a winding-spacing of 0.04 inches,
or a 25 turns/in pitch. Inspired by a small art brush in my pencil
cup, we'll pick a coil diameter of 0.2", so equation (7) reduces to

.. 0.01 25 N 1
.. L = --------- * ------------------ uH / inch
.. 10 1 + 0.9 0.1 25 / N
..
.. 1
.. = 0.025" N * ------------ uH / in
.. 1 + 2.25/N

This formula is more simple than it appears, because the second term
approaches unity for coils of more than 10 - 20 turns.

The first term says a 180nH coil requires about 180/25 = 7 turns, so
we'll try N = 9, and get L = 225nH * 0.8 = 180nH, right on the money.

That's a 9-turn coil 0.2" in diameter and 0.36" long. It uses less
than six inches of wire, has a DC resistance of about 0.008 ohms,
and can handle very high DC currents.

Plugging our coil into equation (6) as a test, we have a = 0.1" and
b = 0.36" and N = 9, so we get L = 8.1 / (9 + 36) = 0.180 uH, bingo.

Thanks,
- Win

whill_at_picovolt-dot-com