HP 8753D - Page 359

The
line standard
must meet
specic frequency
related
criteria,
in
conjunction
with
the
length
used
by the
thru
standard.
In
particular
,
the
insertion
phase
of
the
line
must not
be the
same
as
the thru.
The
optimal
line
length
is
1/4
wavelength
(90
degrees)
relative
to
azero
length
thru
at the
center
frequency
of
interest,
and
between
20
and
160
degrees
of
phase
dierence
over
the frequency
range
of
interest.
(Note:
these
phase
values
can
be
6
N
2
180
degrees
where
Nis
an
integer
.)
If
two
lines
are
used
(LRL),
the
dierence
in
electrical
length of
the
two
lines
should
meet
these
optimal
conditions
.
Measurement
uncertainty
will
increase
signicantly
when
the
insertion
phase
nears
zero
or
is
an
integer
multiple
of
180
degrees,
and
this
condition
is
not
recommended.
F
or
a
transmission
media
that
exhibits
linear
phase
over
the
frequency
range
of
interest,
the
following
expression
can
be
used
to
determine
a
suitable
line
length
of
one-quarter
wavelength
at
the
center
frequency
(which
equals
the
sum
of
the
start
frequency
and stop
frequency
divided
by
2):
El
ectrical
leng
th
(
cm
)
=
(
LI
N
E
0
0
l
eng
th
T
H
R
U )
E
l
ectr
ical l
ength
( cm
)=
(15000
2
V
F
)
f
1(
M
H
z
)
+
f
2(
M
H
z
)
let:
f1
=
1000
MHz
f2
=
2000
MHz
VF
=
V
elocity
F
actor
=
1
(for
this
example)
Thus
,
the
length
to
initially check
is 5
cm.
Next,
use
the
following
to
verify
the
insertion phase
at
f1
and
f2:
P
hase
(
deg
r
ees
)
=
(360
2
f
2
l
)
v
where:
f
=
frequency
l
=
length
of
line
v
=
velocity
=
speed
of light
2 velocity
factor
which
can be
reduced to
the
following
using
frequencies
in
MHz
and
length
in
centimeters:
P
hase (
deg r
ees )
appr ox
=
0
:
012 2
f (
MH
z )
2
l
(
cm
)
V
F
So
for an
air line
(velocity factor
approximately 1)
at
1000
MHz,
the
insertion
phase
is
60 degrees for a 5 cm line; it is
120 degrees at 2000 MHz. This line would be a suitable line
standard.
For microstrip and other fabricated standards
, the velocity factor is signicant. In those cases
,
the phase calculation must be divided by that factor
.F
or example
, if the dielectric constant for
a substrate is 10, and the corresponding \eective" dielectric constant for microstrip is 6.5,
then the \eective" velocity factor equals 0.39 (1
4 square root of 6.5).
Using the rst equation with a velocity factor of 0.39, the initial length to test would be 1
.95 cm. This length provides an insertion phase at 1000 MHz of 60 degrees; at 2000 MHz,
120 degrees (the insertion phase should be the same as the air line because the velocity factor
was accounted for when using the rst equation).
6-92 Application and Operation Concepts