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Skin Effect Calculations
Michael Mirmak of Intel Corp writes in with
this question about skin effect calculations in High-Speed
Digital Design: A Handbook of Black Magic.
| On page 155 of the Johnson book, a
formula is given for the "AC resistance"
(skin depth impedance) of a transmission line:
[4.45] Rac(f) = (2.16 x 10^-7)(f*pr)^0.5
---------------------------
2*(w+d)
where
w = square trace width, inches
d = square trace height, inches
f = frequency, Hz
pr = relative resistivity, compared to copper = 1.00
Rac = AC resistance, ohms/inch
Quad tools use a version of this quantity/formula in matrix form.
However, their units are considerably different:
Rs = resistive coefficient of skin effect
impedance
= (ohms)(nsec)^0.5 per inch
where [Zskin] = [Rs](2*j*omega)^0.5 in (Grad/sec)
My math education leads me to the obvious solution of converting units and getting a
result from [4.45] in terms of resistance, length and frequency in Grads/sec; however, this solution does
not in any way resemble the order of magnitude of results given in the book for the same type of board.
Is there a more complete version of formula [4.45] in the proper units which I might use? Or is there a
better "breakdown" of the components used to derive [4.45], so that I can get the result in the
right form?
Thanks!
- Michael Mirmak, Intel Corp.
mmirmak@pcocd2.intel.com
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Well, Mike, as far as I can tell, you are
making pretty good use of the formula, with a couple of minor
exceptions, neither of which is an "order of magnitude"
effect. For the benefit of you, and others who have written to me
asking about the units for this and other related equations in
the book, I’ll try to list out the whole derivation. Then we
can discuss some of the finer points of doing skin effect
calculations on rectangular (as opposed to round) traces. Also, I
can point out a typographical error in the book, which will be
corrected in the next printing.
The formula that gets everything started
appears on page 153 in the text:
 [4.44]
This well-known expression corroborates with
the literature at least as far back as Frederick Terman,
"Radio Engineers’ Handbook", McGraw-Hill, 1943,
and is also is reproduced in Clayton Paul’s excellent (and
somewhat newer) book, "Introduction to Electromagnetic
Compatibility", Wiley InterScience, 1992. A highly
theoretical, although almost impossibly dense, derivation of this
formula may be found in (among many other places) the book by
Seshadri, "Fundamentals of Transmission Lines and
Electromagnetic Fields", Addison Wesley, 1971. That’s
the book I slogged through to learn this stuff in the first
place.
Anyway, let’s do an example. Before we do,
I am going to rewrite the equation to express the operating
frequency, f, in units of hertz. Then, with
r
, the bulk resistivity of copper, expressed in units of ohm-inch, and
m
, the magnetic permeability of free space, in units of weber/amp-inch, we get the following:
 [4.44a]
Manipulating the units expressions, we see that
skin depth computed in this way comes out in units of inches. To
do these manipulations we need to express ohms (1 ohm = 1
volt/amp) and webers (1 weber = 1 volt-sec) in more general
terms. Here is equation [4.44a], showing the reduction of units.
 [4.44b]
As you can see, all the volt, amp,
and sec terms cancel out, leaving only the in terms
to reinforce. Taking the square root, we are left with just in (inches).
 [4.44c]
Now we are in a position to fill in the values for
r
and
m
.The value for
r
, the bulk resistivity of copper, is found on page 411. This
value is already in units of ohm-in (although it doesn’t
say it in the text, you can infer this from, among other places,
its usage on page 412 in the formula RROUND). Sorry I didn’t
say anything about the units on page 411. The value for
m, the magnetic permeability of free space, is listed on page 410. Make sure you
use the value calculated for units of inches, as shown here.
 [4.44d]
 [4.44e]
With these values for
r
and
m
, and choosing an operating frequency of, say, 1 GHz, you should get a skin depth of 0.000082
inches (8.2E-05 inches, or 0.082 mil). This is the same value
calculated by Terman, Paul, and Seshadri.
Between here and equation [4.45], on page 155,
there are a couple of steps I skipped. Perhaps it would be
helpful if I showed them. First, in a general sense, the way to
use the resistivity term
r
is like this:
 [4.45a]
In the case of a skin-effect-limited conductor,
the current flows only in a thin ring around the outer
circumference of the conductor. The effective average ring
thickness is called the skin depth. The ring circumference is
just
p
D, where D is the diameter of the conductor. The total
effective are of the conducting ring is:
 [4.45b]
When the terms Skin_Depth and D
are both expressed in units of inches, the resulting area is in units of inches2 (that’s what we want).
Substituting the expressing [4.45b] for A in [4.45a] we are on the
way toward deriving equation [4.45].
 [4.45c]
Note that I’ve chosen to compute the total
AC resistance of a wire in [4.45c], not the resistance per unit
length. I like this form because, after a few more steps, you can
readily see how to evaluate this AC resistance in various
different systems of units. Let’s next substitute in
expression [4.44c] for skin depth, and expand all terms to show
the units.
 [4.45d]
Simplifying common terms, we arrive at the
following, which comes out in units of plain old ohms:
 [4.45e]
Now plug in the numbers above for
r
and
m
, and you get:
 [4.45f]
A side benefit of expression [4.45e] is that it
clearly shows that as long as L and D are in the same units, and
f is in Hertz, the result is unchanged.
When we convert into the exact form of equation
[4.45], a similar property holds. That is, if D is in inches,
then RAC comes out in ohms per inch. If D is in cm, then RAC
comes out in ohms per cm, etc.
OK! So there was the derivation. You may have
noticed that the constant is 2.61, not 2.16 as shown in the book
(oops! We found a typo.). I’ve submitted the typo to
Prentice Hall and they assure me that it will be corrected in the
next printing. Also, you may have noticed that I did not split
out
r
into
r
0, the bulk resistivity of copper, and
r
r, the relative resistivity of whatever material you are using, as was done for
equation [4.45]. No big deal.
When we started I promised to go into the
effects of non-round conductors, so let’s get into that
next. There are two effects I’d like to discuss, the
proximity effect, and the power dissipated in the returning
signal path.
On printed circuit boards with a solid
reference plane (ground or Vcc) near the trace, the presence of
currents in the returning signal current distorts the flow of
current in the conducting trace. Basically, it pulls more current
onto the bottom side of the trace (the side nearest the reference
plane). In a stripline configuration (reference planes above and
below a trace) the current is going to be pulled over toward
whichever reference plane is closest. This effect is called the
proximity effect. It can be accurately calculated using
two-dimensional E&M field analysis. Below are plotted the
results of a two-dimensional field analysis of a microstrip
trace.
This trace is 0.006 inches wide, and 0.0015
inches thick (1-ounce copper). Current density flowing in the
surface of the conductor is proportional to the magnetic field
strength tangent to the surface at that point. As you can see,
the magnetic field strength on the top side of the conductor
(side away from the reference plane) is less intense than the
field strength on the bottom side (side near the reference
plane). In addition, there is a small local peaking of currents
in the corners of the conductor. If there had been no reference
plane nearby, the current would have distributed itself more
uniformly around the conductor. This distortion of the current,
this proximity effect, increases the apparent resistance of the
conductor. The skin effect squeezes the current into a thin band
around the circumference, the proximity effect pulls more of that
current onto the side facing the opposing flow of returning
signal current. (At the same time, for you physics buffs out
there, Ampere’s forces tend to push the trace up, off the
board.) The apparent increase in resistance of the signal
conductor in this example is about 33%. That’s typical for
75-ohm PCB traces.
The drawing above clearly shows a substantial
magnetic field near the reference plane. Near this field flows
the returning signal current. The returning signal current at any
one point is, just like in the signal conductor, proportional to
the magnetic field tangent to the surface at that point. From
this principle we can work out the distribution of returning
signal current in the reference plane, and from that we may
compute the power dissipated in the reference plane. The result
of this additional dissipation is a further reduction in the
received signal size.
You see, from the point of view of transmission
line theory, it doesn’t matter whether we dissipate power in
the reference plane or in the signal conductor, the net effect is
pretty much the same: the signal at the end of the wire shrinks.
The reference plane dissipation has the effect, in this
case, of a 36% increase in the apparent resistance of the signal
conductor. That’s a typical value for a 75-ohm PCB trace.
So, Mike, my overall recommendation for you is
to use the formula pretty much as you show it, with a couple of
exceptions:
- Use the constant 2.61E-07, instead of
2.16E-07.
- In the denominator, use 2*(w+d),
where w and d are in inches (you had w*d in your original email to me,
which could be part of your problem). The term w
in this expression should be the trace width and d
the trace thickness.
- After you are done, multiply the result by
a fudge factor of 1.69 = 1 + 0.33 + 0.36, to account for
the proximity effect and extra dissipation in the ground
plane.
- When you do frequency-based simulations,
you should multiply the result by an additional factor of
(1+j) = sqrt(2j), to account for the phase terms. So far,
all we have talked about is the magnitude of the
resistive component of skin effect. There is an inductive
component, too, with the same magnitude. Multiplying by
the term (1+j) will include in both R and L
terms. Your simulator probably already does this.
When you evaluate the AC resistance of the
0.006-inch trace in my example, at a frequency of 1 GHz, you
should come out to a value of 0.93 ohms per inch. As a rule of
thumb, I assume one ohm per inch for a .006-in at 1 GHz.
Hope this helps (and sorry for the typo in the
book)!
Sincerely,
Dr. Howard Johnson
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