# 180 degree hybrid coupler

180 degree hybrid coupler step by step tutorials.

The second reciprocal directional coupler we will discuss is the 180° hybrid. As the name implies, the outputs from such a device can be 180° out of phase. There are two primary objectives for this lecture. The first is to show that the S matrix of the 180° hybrid is with reference to the port definitions in The second primary objective is to illustrate the three common ways to operate this device. These are:
1. In-phase power splitter: With input at port 1 and using column 1 of [S], we can deduce that port 1 is matched, the outputs are ports 2 and 3 (which are in phase) and port 4 is the isolation port.
2. Out-of-phase power splitter: With input at port 4 and using column 4 of [S], we can deduce that port 4 is matched, the outputs are ports 2 and 3 (which are completely out of phase) and port 1 is the isolation port.
3. Power combiner: With inputs at ports 2 and 3 and using columns 2 and 3 of [S], we can deduce that both ports 2 and 3 are matched, port 1 will provide the sum of the two input signals and port 4 will provide the difference.
Because of this, ports 1 and 4 are sometimes called the sum and difference ports, respectively. there are different ways to physically implement a 180° hybrid, as shown in We’ll focus on the ring hybrid and specifically consider the first two pplications described above.
Ring Hybrid
The ring hybrid (aka the rat race) is shown in We’ll analyze this structure using the same even-odd mode approach we applied to the Wilkinson power divider and the branch line coupler in the previous two lectures. In the present case, the physical symmetry plane bisects ports 1 and 2 from 3 and 4 in the figure above.
1. In-phase power splitter. Assume a unit amplitude voltage wave incident on port 1: As in Lecture 26, proper symmetric and anti-symmetric excitations of this device are required to produce the even and odd mode problems, as shown in  As we derived in Lecture 26,
B1=(1/2)te+(1/2)t0
B2 =(1/2)Te +(1/2)T0
B3 =(1/2)te-1/2)t0
B4=(1/2)Te-(1/2)T0
Each of the even and odd solutions for Bi (i =1,…,4) can be found by cascading ABCD matrices, then converting to S parameters. Since the ports are terminated by matched loads, we can directly determine te0 and Te0 from these S parameters. As given in the text,
te =- t0 =(-j/ Ö 2)
Te =T0 =(-j/ Ö 2)
Using these values in (1)-(4) produces
B1 =B4 =0
B2 =B3 = -j/Ö2
These results in (8) and (9) form the first column of [S] in (1). They indicate that with an input at port 1 and all output ports terminated by matched TLs and loads, the signal is equally divided in phase at ports 2 and 3, while none is delivered to port 4.
2. Out-of-phase power splitter. Assume a unit amplitude voltage wave is incident on port 4. To generate symmetric and anti-symmetric problems, we’ll excite the circuit at ports 2 and 4, as shown in  These two excitations sum to +1 at port 4 and 0 at port 2, as required. From the even mode problem is From this figure (and the even symmetry), we can write Be1 =(1/2)Te =Be3 and
Be2 =(1/2)te =Be4
From the odd mode problem is From this figure (and the odd symmetry), we can writ
B01 =(-1/2)T0 =-B03 and
B02 =(-1/2)t0 =-B04
Summing (10)-(13), we find t
B1 =Be1 +B01 =(1/2)Te -(1/2)T0
B2 =Be2 +B02 =(1/2)te -(1/2)t0
B3 =Be3 +B03 =(1/2)Te +(1/2)T0
B4 =Be4 +B04 =(1/2)te +(1/2)t0
Cascading ABCD matrices and converting to S parameters, the text shows that
B1 =B4 =0
B2 =-B3 =j/Ö2
These values form the fourth column of [S] in (1). They indicate that with excitation at port 4 and all output ports terminated by matched TLs and loads, port 1 is isolated and the signal is equally split between output ports 2 and 3 with a 180° phase shift between them.
Design of 180° Hybrid
The ring hybrid is extremely easy to design. First compute the effective permittivity for the chosen substrate from which the strip widths can be determined for the Z0 and ° 2Z0 sections of the device. Then after choosing a center frequency, the lengths l /4 and 3l /4 can be calculated again using the effective permittivity. That’s basically it.Typical |S1j| results for this device are shown in Can you interpret the meaning of these results? How do you expect |S14| and |S32| to behave