[SI-LIST] Why 50 ohms? The history.
- From: agathon <hreidmarkailen@xxxxxxxxx>
- To: cygnul@xxxxxxxxx
- Date: Wed, 20 Dec 2006 18:30:27 -0800
Aw, what's a few ohms between friends?
http://www.edn.com/contents/images/47191.pdf
Why 50 Ohms?
*(Originally published in EDN Magazine <http://www.ednmag.com/>, September
2000)*
*Q: **Why do most engineers use 50W pc-board transmission lines (sometimes
to the extent of this value becoming a default for pc-board layout)? Why not
60 or 70W ??Tim Canales*
*A: *Given a fixed trace width, three factors heavily influence
pc-board-trace impedance decisions. First, the near-field EMI from a
pc-board trace is proportional to the height of the trace above the nearest
reference plane; less height means less radiation. Second, crosstalk varies
dramatically with trace height; cutting the height in half reduces crosstalk
by a factor of almost four. Third, lower heights generate lower impedances,
which are less susceptible to capacitive loading.
All three factors reward designers who place their traces as close as
possible to the nearest reference plane. What stops you from pressing the
trace height all the way down to zero is the fact that most chips cannot
comfortably drive impedances less than about 50W. (Exceptions to this rule
include Rambus, which drives 27W, and the old National BTL family, which
drives 17W).
It is not always best to use 50W. For example, an old NMOS 8080 processor
operating at 100 kHz doesn't have EMI, crosstalk, or capacitive-loading
problems, and it can't drive 50W anyway. For this processor, because very
high-impedance lines minimize the operating power, you should use the
thinnest, highest-impedance lines you can make.
Purely mechanical considerations also apply. For example, in dense,
multilayer boards with highly compressed interlayer spaces, the tiny
lithography that 70W traces require becomes difficult to fabricate. In such
cases, you might have to go with 50W traces, which permit a wider trace
width, to get a manufacturable board.
What about coaxial-cable impedances? In the RF world, the considerations are
unlike the pc-board problem, yet the RF industry has converged on a similar
range of impedances for coaxial cables. According to IEC publication 78
(1967), 75W is a popular coaxial impedance standard because you can easily
match it to several popular antenna configurations. It also defines a solid
polyethylene-based 50W cable because, given a fixed outer-shield diameter
and a fixed dielectric constant of about 2.2 (the value for solid
polyethylene), 50W minimizes the skin-effect losses.
You can prove the optimality of 50W coaxial cable from basic physics. The
skin-effect loss, *L* , (in decibels per unit length) of the cable is
proportional to the total skin-effect resistance, *R* , (per unit length)
divided by the characteristic impedance, *Z* 0 , of the cable. The total
skin-effect resistance, *R* , is the sum of the shield resistance and center
conductor resistances. The series skin-effect resistance of the coaxial
shield, at high frequencies, varies inversely with its diameter* d* 2 . The
series skin-effect resistance of the coaxial inner conductor, at high
frequencies, varies inversely with its diameter *d* 1 . The total series
resistance, *R* , therefore varies proportionally to (1/*d* 2 +1/*d* 1 ).
Combining these facts and given fixed values of *d* 2 and the relative
electric permittivity of the dielectric insulation, *E* R , you can minimize
the skin- effect loss, *L* , starting with the following equation:
(see pdf) L is proportional to ((1+*d* 2 /*d* 1 )/ln(*d* 2 /*d*
1))/Zo ---
eq.1
Z0 = (60/sqrt(Er))*ln(d2/d1)
In any elementary textbook on electromagnetic fields and waves, you can find
the following formula for *Z* 0 as a function of *d* 2 , *d* 1 , and *E* R :
Substituting *Equation 2* into *Equation 1* , multiplying numerator and
denominator by *d* 2 , and rearranging terms:
[ see pdf link above ]
*Equation 3* separates out the constant terms /60)´(1/*d* 2 )) from the
operative terms ((1+*d* 2 /*d* 1 )/ln(*d* 2 /*d* 1 )) that control the
position of the minimum. Close examination of *Equation 3* reveals that the
position of the minima is a function only of the ratio *d* 2 /*d* 1 and not
of either *E* R or the absolute diameter *d* 2 .
A plot of the operative terms from *L* , as a function of the argument *d* 2/
*d* 1 , shows a minimum at *d* 2 /*d* 1 =3.5911. Assuming a solid
polyethelene insulation with a dielectric constant of 2.25 corresponding to
a relative speed of 66% of the speed of light, the value *d* 2 /*d* 1 =
3.5911 used in *Equation 2* gives you a characteristic impedance of 51.1W. A
long time ago, radio engineers decided to simply round off this optimal
value of coaxial-cable impedance to a more convenient value of 50W. It turns
out that the minimum in *L* is fairly broad and flat, so as long as you stay
near 50W, it doesn't much matter which impedance value you use. For example,
if you produce a 75W cable with the same outer-shield diameter and
dielectric, the skin-effect loss increases by only about 12%. Different
dielectrics used with the optimal *d* 2 /*d* 1 ratio generate slightly
different optimal impedances.
On 12/19/06, Sam Pete <cygnul@xxxxxxxxx> wrote:
>
> Hi Friends,
>
> Whenever we refer to SI books or application notes, the impedance of the
> transmission line is mentioned as 50-ohms or 60-ohms. What is the speciffic
> reason behind this value? why can't it be something different than this? OR
> Why this 50-ohms conceptualized from the driver's perspective? (evolution of
> transmission line theory for Digital Design)
> Please flood me some info on this.
>
> Regards,
> Sam
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