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Inductance of loop area

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

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  1. Pooh Bear

    Pooh Bear Guest

    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 Larkin

    John Larkin Guest

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

    a = radius, inches

    d = wire dia, inches

    courtesy Reference Data for Radio Engineers

    John
     
  3. Terry Given

    Terry Given Guest

    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 Bear

    Pooh Bear Guest

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

    Cheers, Graham
     
  5. Pooh Bear

    Pooh Bear Guest

    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 Given

    Terry Given Guest

    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 Bear

    Pooh Bear Guest

    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 Given

    Terry Given Guest

    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
     
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