# Inductance of loop area

Discussion in 'Electronic Design' started by Pooh Bear, Jun 15, 2005.

1. ### Pooh BearGuest

Further to my post about the IR2110 gate driver, does anyone have a
nice simple equation for calculating the inductance of a given loop on a
pcb ?

Yes I know..... we looked at it when I was about 17 / 18. Never needed
it since then until now.

Cheers, Graham

2. ### John LarkinGuest

L = (a/100) * (7.353 * log10(16*a/d) - 6.386) uH

d = wire dia, inches

courtesy Reference Data for Radio Engineers

John

3. ### Terry GivenGuest

buy a copy of Terman's Radio Engineers Handbook, then use pp47-72 for to
solve for the geometry present. This was one of my first "old" books,
and I have used it for this purpose for a decade or so now. Grover is
pretty good too, but Terman suffices for most PCB-type problems.

Cheers
Terry

4. ### Pooh BearGuest

Lol @ Radio Engineer's Handbook. That was one area I never planned to get
involved in !

Cheers, Graham

5. ### Pooh BearGuest

Hmmmm..... so for wire diameter I presumably substitute pcb track cross
sectional area ? I hadn't imagined that to be a factor, just expected the
'cut area' to influence.

Graham

6. ### Terry GivenGuest

Laugh ye not:

rectangle of round wire, sides s1 & s2, diagonal g,
wire diameter d
L=0.02339[(s1+s2)*log(4*s1*s2/d)-s1*log(s1+g)-s2*log(s2+g)]
+ 0.01016*[u*delta*(s1+s2) + 2(g+d/2) - 2(s1+s2)]

rectangle of rectangular wire, thickness b, width c (into plane of loop)
L=0.02339[(s1+s2)*log(2*s1*s2/(b+c))-s1*log(s1+g)-s2*log(s2+g)]
+ 0.01016*[2*g - 0.5*(s1+s2) + 0.447*(b+c)]

all dimensions inches, L in uH. u is permeability (1.25e-6), delta is
skin depth

Cheers
Terry

7. ### Pooh BearGuest

Oh my God !

Kind of thinks " why did I ask " !

I think I prefer John's equation for simple rule of thumb !

All this 'high frequency' stuff is quite new to me. I can make analogue and
digital stuff work ok though !

For small variations in d where d is fairly small to begin with does it really
make much difference to the result ? E.g. where the cut area is a few square
inches, does a 40 thou ( mil ) track have significantly less inductance than a
20 thou one ?

Graham

8. ### Terry GivenGuest

it is little more than an LRC circuit. the trick is recognising L, R & C.
he gives a simple loop formula:

L = 0.00508*p*[2.303*log(4*p/d)-theta], uH

2.303*log() = ln() so

L = 0.00508*p*[ln(4*p/d)-theta], uH

p = perimeter of loop, d = diameter of wire in inches
(convert rectangle area into circular area)

theta = shape factor
= 2.451, circle
= 2.561, regular octagon
= 2.636, regular hexagon
= 2.712, regular pentagon
= 2.853, square
= 3.197, equilateral triangle
= 3.332, isoscoles right-anlge triangle

Terman claims 0.5% accuracy. clearly theta=2.65 is a pretty good compromise.

Cheers
Terry  