[SI-LIST] Re: FW: Inductance of wire section
- From: "Ken Cantrell" <Ken.Cantrell@xxxxxxxxxxxxxxxx>
- To: "McCoy, Bart O." <McCoy.Bart@xxxxxxxx>,<Ken.Cantrell@xxxxxxxxxxxxxxxx>, <si-list@xxxxxxxxxxxxx>
- Date: Fri, 5 Sep 2003 09:34:30 -0600
Your right, of course, given that you know the answer(0.86)is in nH.
Grover's original formula statement, pg. 35, eq 7, was for an answer in uH,
thus the .002 multiplier.
u = 4*pi*10^-7 Wb/A-m = 4*pi*10^-9 Wb/A-cm.
u/(2*pi)= 2*10^-9 Wb/A-cm(answer in nH/cm).
or if you are non-metric:
u_in/(2*pi)= 5.08*10^-9 Wb/A-in
-----Original Message-----
From: McCoy, Bart O. [mailto:McCoy.Bart@xxxxxxxx]
Sent: Thursday, September 04, 2003 7:19 PM
To: Ken.Cantrell@xxxxxxxxxxxxxxxx; si-list@xxxxxxxxxxxxx
Subject: RE: [SI-LIST] Re: FW: Inductance of wire section
I'm fairly sure K = 2 when units are in cm.
Take, for example, a typical 1 mil diameter wirebond
1 mm in length. It's well known that they are usually
near (I'd say slightly less than) 1 nH per mm.
Diameter = 1 mil = 2.54E-3 cm
Length = 1 mm = 0.1 cm
K = 2
L = K * length * [ ln(4*length/Diam) - 0.75 ]
L = 2 * 0.1 * [ ln(4*0.1/2.54E-3) - 0.75 ]
L = 0.2 * [ ln(157.5) - 0.75]
L = 0.2 * [ 4.31 ]
L = 0.86 nH
That's very much in the ballpark-- I'd say extremely
close based upon electromagnetic sims I've done which
came out a bit less than 1 nH for a 1 mm wirebond.
If you use 0.002, that would make it 3 orders of magnitude
less, which I know for sure isn't correct. Even
1 order of magnitude smaller would be WAY off base.
- Bart
-----Original Message-----
From: Ken Cantrell [mailto:Ken.Cantrell@xxxxxxxxxxxxxxxx]
Sent: Thursday, September 04, 2003 5:04 PM
To: McCoy, Bart O.; si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: FW: Inductance of wire section
K should be = .002, shouldn't it?
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of McCoy, Bart O.
Sent: Thursday, September 04, 2003 3:11 PM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] FW: Inductance of wire section
Well, you'll probably get a million replies with the same
equation in 50 different forms, but here's one that I use
for a straight-wire of circular cross-section inductance
calculation that will give you first order effects:
L = K * length * [ ln(4*length/Diam) - 0.75 ]
L = Inductance in nanohenries
D = Diameter of wire in cm
length = Length of wire in cm
K = 2
22 Gauge wire.... I'll let you do the calculation given:
20 Gauge AWG = 0.0813 cm diameter
26 Gauge AWG = 0.0406 cm diameter
Hope that helps.
- Bart
-----Original Message-----
From: sdb@xxxxxxxxxx [mailto:sdb@xxxxxxxxxx]
Sent: Thursday, September 04, 2003 2:53 PM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Inductance of wire section
Hi gang --
Please pardon me if this is a FAQ, but does anybody have a quick
rule-of-thumb for the inductance to expect from a 1 cm length of 22
gauge wire-wrap wire in free space? Yes, I've searched the web &
looked at a few pages which are far too detailed to be really useful
-- all I want is an order-of-magnitude number, and the websites are
all about calculating the number by integrating Maxwell's equations.
So far, I've been using a figure of around 360 nH/m, which is the
inductance of a common 50ohm transmission line. If the real number is
larger or smaller by a factor of 2 to 5 I don't care a lot, but I would
like to make sure that the "rule-of-thumb" number is *not* orders of
magnitude larger or smaller than this.
What rule-of-thumb do you use?
Stuart
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