Page 2–2
rF hybrids System Manual
A basic balanced hybrid can be used to illustrate
how it isolates two inputs from one another and
matches impedances as shown in Figure 2–1. The
hybrid acts as a balanced bridge network and when
the output’s load resistance is exactly twice the
center tap resistor value you get infinite loss, theo-
retically, between the 2 inputs. The hybrid, in this
case, is made up of a resistor of 25 ohms, and a
transformer with a center tap on the primary. The
transformer turns ratio is √2/1 with the √2 turns on
the center tapped primary.
Let’s assume the secondary of the transformer is
terminated with a 50 ohm resistor and a voltage (V)
is applied to input port #1. The 50 ohm load will be
reflected in the primary of the transformer as a 25
ohm quantity from point (a) to the center tap (ct).
This is because there is 1 turn on the primary, (a) to
(ct), for every √2 turns on the secondary. The
impedance will be transferred as the square of the
turns ratio, which in this case is 2 to 1. The voltage
V will divide equally between the 25 ohm resistor
and the 25 ohm reflected load into the top half of
the primary. Thus each voltage has a value of V/2,
and in the direction as shown. Since the center
tapped primary of the transformer will act as an
autotransformer, a voltage V/2 will also appear on
the other half of the primary between point (ct) and
(b). The voltage appearing across input port #2, due
to the voltage V at input port #1, is the sum of the
voltages around the loop from (g) to (y). This
resultant voltage is 0 volts. And as shown in Figure
2–1, the hybrid isolates the voltage at one input port
from the other input port. A price must be paid for
this isolation and that is in the loss from the inputs
to the output. One half the power is dissipated for
each input in the center tap balance resistor causing
a 3 dB minimum loss in the power going to the
output from each input.
Figure 2–1.
resistive hybrid.
Input
Port
#1
Input
Port
#2
V
(x)
V/2
V/2
V/2
(g)
(y)
(a)
(b)
(ct)
0
50
W
2:1
2
V
25
W
