Printed circuit boards for high speed logic.

High speed (1 nsec.) logic systems require voltage planes to separate
successive layers of signal lines (Fig.52) in order to keep signal cross coupling
(noise) down to a reasonable level (say 10% of signal amplitude).At a frequency
of 1 GHz, skin depth in copper is
cm. (=0.00008 in.) (Ref.20). At 10 GHz, skin depth drops to
cm. (=0.00003 in.).The thinnest practicable laminated copper is about
0.001 in., which is more than ten times the skin depth. This means that
in practice, a signal travelling down a signal line and returning by the immediately
adjacent voltage plane(s) will not penetrate beyond the plane(s), and each voltage
plane will screen signals above it from signals below it with negligible crosstalk
through voltage plane. So long as a signal is transmitted down between a signal
line and the voltage plane(s) immediately above and/or below it, the only crosstalk
(noise) of significance will be between (adjacent) signal lines in the same
plane[1].
Care must be taken that at its source and destination, the signal is referenced
to the correct voltage plane(s), and this will now be discussed.
Calculation.
As a practical example, consider (Fig.49) a surface passive line with potentially active lines on each side of it, with w=0.010" and d=0.010".
FX from one adjacent line is 11%, so from the two active lines it is 22%. This
is rather high unless the lines are short. We might therefore reduce h by
from 0.008" to 0.006". The effect is about the same as increasing
d by
to 0.013", which reduces FX to an acceptable 8%, or 16% for the two active
lines[2].
DX is given in one curve only in Figure 49, and this curve relates to w=0.010". With w=0.010" and d=0.010" the velocity difference F is 15%, or 0.15. It can be shown that as a % of the active signal,
where F is the % velocity difference,
Assume; the length of the passive line
; the velocity of propagation
per nsec in an epoxy-glass PCB; and the active signal rise time
For w=0.010" and d=0.010, Figure 49 shows that F=15%, or 0.15 . Thus,
Decoupling by Voltage Planes.

In order to understand the nature of the decoupling action at a point between
parallel voltage planes, first consider the parallel-plate transmission line
(Fig.24), which has a
. This formula still applies for each small section of a transmission line where
the width a is varying, for example for a wedge-shaped line (Fig.53.).

Over a distance
the above formula becomes
.
Now if
, we are considering a complete plane. The signal travels out radially between
the planes (Ref.21), and we get
.
Now we know that, in a medium of permittivity e and permeability m, the outwards velocity of the signal through the epoxy-glass dielectric is
.
Therefore
,
where t is the time since the signal was introduced at the centre.
So using this last equation for the distance r we get
where a is in metres.
A reflection related to
arrives back at the centre at time 2t. If
is small when
, (the risetime of the output of the logic gate,) then natural decoupling between
planes is satisfactory.
Calculation.
As a practical example, if 2t=1ns, d=0.5mm.,
, then
This calculation shows that natural decoupling between voltage planes is satisfactory
for a single switching load of high speed logic, or for a number of loads in
the same integrated circuit[3]. We then prevent superposition
of the current transients into a number of integrated circuits switching at
the same instant from generating unacceptably large voltage transients between
planes by adding extra discrete (
) decoupling capacitors distributed at intervals of a few centimetres. The
proportion of the surface area occupied by the miniature tantalum capacitors
will be insignificant.
[1]
One argument which can be used to dismiss the feared
additional effect of a further active line next to the nearest active line
is as follows. In Figure 49,
if line width w and line spacing d are both 0.010", then maximum FX
is 11%. However, the next parallel active line is at a spacing d of 0.030",
causing it to add a negligible further FX of 2.5%. The reality, of course,
is that the nearer active line shields the further active line from the
passive line.
[2]
There is, however, a price to be paid for reducing h by 3/4 to 0.006"
(equivalent to w=0.013", h=0.008"), because, see Figure 47,
Zo drops from 74 ohms to 64 ohms.
As a result, it takes more current to drive the line, slowing down driving
circuit and increasing power dissipation. Generally, reducing crosstalk
by hiding lines close to planes (a) reduces (impedance and therefore)
speed and (b) increases power dissipation.
[3]
perhaps switching a transient 100ma or 500mw. Think of a 5v power supply
with source impedance of 0.2 ohms driving a 50 ohm load, resulting in a
drop of (0.2/50.2)..5v=20mv across the source and a drop of the effective
supply from 5v to 4.98v.