Skin Effect Calculation
Michael Mirmak of Intel Corp writes, with a question about skin effect calculations in HighSpeed Digital Design: A Handbook of Black Magic.
On page 155 of your book, a formula is given for the "AC resistance" (skin depth impedance) of a pcb trace with rectangular crosssection:
Where:
 w = trace width, inches
 d = trace height, inches
 f = frequency, Hz
 ρ_{R} = relative resistivity, compared to copper = 1.00
 R_{AC} = AC resistance, ohms/inch
Quad tools use a version of this formula in matrix form.
However, their units are considerably different
Where:
 R_{S} is the resistive coefficient of skin effect impedance, in units of (ohms)(nsec)^{1/2} per inch, and
Where:
 w (frequency) must be specified in units of (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.
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:
This wellknown expression corroborates with the literature at least as far back as Frederick Terman, "Radio Engineers’ Handbook", McGrawHill, 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 ρ
, the bulk resistivity of copper, expressed in units of ohminch, and μ
, the magnetic permeability of free space, in units of weber/ampinch, we get the following:
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 voltsec) in more general terms. Here is equation [4.44a], showing the reduction of units.
As you can see, all the volt, amp, and sec terms cancel out, leaving the inch terms to reinforce. Taking the square root, we can pull out units of just inches.
Now we are in a position to fill in the values for ρ and μ.The value for ρ, the bulk resistivity of copper, is found on page 411. This value is already in units of ohmin (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 μ, 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.
With these values for ρ and μ, and choosing an operating frequency of, say, 1 GHz, you should get a skin depth of 0.000082 inches (8.2E05 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 ρ is like this:
In the case of a skineffectlimited 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 π
D, where D is the diameter of the conductor. The total effective are of the conducting ring is:
When the terms Skin_Depth and D are both expressed in units of inches, the resulting area is in units of inches^{2} (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].
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.
Simplifying common terms, we arrive at the following, which comes out in units of plain old ohms:
Now plug in the numbers above for ρ and μ, and you get:
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 ρ into ρ_{0}, the bulk resistivity of copper, and ρ_{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 nonround 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 twodimensional E&M field analysis. Below are plotted the results of a twodimensional field analysis of a microstrip trace.
This trace is 0.006 inches wide, and 0.0015 inches thick (1ounce 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 75ohm 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 75ohm 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 2.61E07, instead of 2.16E07.

In the denominator, use 2(w+d), where w and d are in inches (you had wd 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 frequencybased 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 0.006inch 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 .006in at 1 GHz.
Hope this helps (and sorry for the typo in the book)!
Sincerely,
Dr. Howard Johnson