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In many cases, insertion of a 50-ohm attenuator in the sig-
nal path will provide an approximate impedance match
and will absorb most reflections. It should be noted, how-
ever, that the attenuation factor will not be the same as
it would be if the impedances were the same on both sides.
Fig. 2-7. Simple resistive impedance-matching network providing
minimum attenuation.
Fig. 2-7 illustrates a simple resistive impedance-matching
network that provides minimum attenuation. To match imped-
ances with the network, the following conditions must exist:
(
)
+
R
Z
R
must equal Z
1
2
2
(
)
+
+
R
Z
R
1
2
2
must equal Z
.
2
Therefore:
R
Z
= Z
Z
;
and R
Z
1
2
1
2
1
or R
=
1
and R
=
2
As an example, to match a 50-ohm system to a 125-ohm
system:
Z
= 50 ohms; and Z
= 125 ohms.
1
2
Therefore:
R
=
= 96.8 ohms
−
125
(
125
50
)
1
125
and R
=
= 64.6 ohms
50
2
−
125
50
Z
R
; and
+
R
1
2
1
1
+
Z
R
1
2
= R
(Z
-Z
)
1
2
2
1
−
Z
(
Z
Z
)
;
2
2
1
Z
Z
2
1
−
Z
Z
2
1
Operating Instructions – Type 1S1
When constructing such a device, the environment sur-
rounding the components should also be designed to provide
a transition between the impedances. Keep in mind that the
characteristic impedance in a coaxial system is determined
by the ratio between the outside diameter of the inner
conductor and the inside diameter of the outer conductor
(Z
= 138 log
D
/D
)
0
10
1
2
Though the network in Fig. 2-7 provides minimum atten-
uation for a purely resistive impedance-matching device, the
attenuation as seen from one end does not equal that seen
from the other end. A signal applied from the lower imped-
ance source (Z
) encounters a voltage attenuation (A
1
may be determined as follows:
Since:
i
= i
;
R1
Z2
E
R
Therefore:
=
=
A
1
1
1
E
Z
2
2
A signal applied from the higher impedance source (Z
will encounter a greater voltage attenuation [A
be determined similarly:
Since:
i
= i
+ i
R1
R2
Therefore: A
=
2
2
Z
<
A <
1 (
2
)
2
Z
1
In the example of matching 50 ohms to 125 ohms,
96
8 .
=
+
=
A
1
. 1
77
;
1
64
6 .
96
8 .
96
8 .
and
=
+
+
=
A
1
2
64
6 .
50
Note that if the 50-ohm source were used for pulsing a
high-impedance load, R
1
impedance of the load (high R) and R
equal the 50 ohms of the pulse source. In this situation, volt-
age attenuation would be about 2.
If a low-impedance load (<50 ohms) were to be en-
countered, the 50-ohm pulse source would be the Z
If the load impedance were to approach 0 ohms, the value
of R
would then approach the load impedance (low R).
1
Voltage attenuation in this case would become quite signifi-
cant:
2
Z
Attenuation =
=
2
Z
L
The illustrated network can be modified to provide dif-
ferent attenuation ratios by adding another resistor (<R
in series between Z
and the junction of R
1
Probes
For relatively high-impedance measurements of nanosec-
ond signals, special passive or cathode-follower signal probes
are available for use with the Type 1S1 Sampling Unit. Pas-
sive probes may also be built into or onto the circuits to be
monitored, to minimize changes in loading.
E −
E
E
=
2
1
2
R
Z
1
2
+
<
<
; 1
1 (
A
) 2
1
) that may
2
E −
E
E
E
;
=
+
1
1
2
1
z1
R
Z
R
1
2
1
E
R
R
=
+
2
1
E
R
Z
1
2
. 4
44
would approximately equal the
would approximately
1
2
100
(very high)
Z
L
and R
.
1
2
) that
1
)
2
;
+
1
1
1
source.
)
1
2-11
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