[SOLVED] Microstrip T Junction (S-Parameter Question)

Hi all,

Theoretically, the S parameter of the equally power divided T-junction with quarterwave, (Z0=70.71 ohms) transformation at the output ports is

( 0 -j/sqrt(2) -j/sqrt(2) )
( -j/sqrt(2) 1/2 -1/2 )
( -j/sqrt(2) -1/2 1/2 )

Here S12=S21, S13=S31, and S22=S33=1/sqrt(2).

This seems to suggest that this can be used as a "one way" lossless divider since S11 = 0 (no mismatch loss), S21=-j/sqrt(2), and S31=-j/sqrt(2) (equal power spit to port 2& 3)

However, with S22 and S33 being a non-zero element, this implies there is mismatch loss.
S12 = S13 = -j/sqrt(2), S23 = S32 = -1/2, and S22 = S33 = 1/2. This means half the power is transmitted to port 1, quarter of the power to either port 2 or 3, and another quarter of the power reflected. (Correct me if I am wrong Please!)

I see no problem using this as a divider. However, when used as a combiner, the power level takes a big hit.

Here is the question.

I still see many design,as recent as 2011, used it as a combiner and the power level did not drop from mismatch loss like I expected. In fact, only conductor and dielectric loss were seen. Can someone please explain what is going on?

Thanks,

Microwave123

It should work as a loss-less combiner as long as Ports 2 and 3 are fed in phase.

At Port 2, there is reflection due to non-zero S22. But you also have coupling from Port 3 because of non-zero S23. Since they are 180-deg out of phase, they are canceled, creating virtual short. At Port 1, you should see a sum of voltages that came from Ports 2 and 3, which are in phase. Assuming that the input voltages at Ports 2 and 3 are unity (1V), the output voltage at Port 1 is 2*j/sqrt(2). The law of conservation of energy holds.