Wilkinson Divider Current Density

Status
Not open for further replies.
G

ge

Full Member level 5
Joined
Jun 23, 2006
Messages
250
Helped
29
Reputation
58
Reaction score
29
Trophy points
1,308
Location
Pennsylvania, USA
Activity points
2,772
If I simulate a Wilkinson Power Divider current density without the isolation resistors, will I get current density minimums at the locations where the isolation resistors belong? I'm running AWR/Microwave Office.

Tnx,
GE
 

Where is your feed? If you feed from the common port, the structure is symmetric and there is no current through the resistor anyway.
 
Reactions: ge

    G

    ge

    Points: 2
    Helpful Answer Positive Rating
The feed could be from either direction depending on how the divider/combiner is used. If I feed from the divider ports (not common port), I guess that I will need to excite both divider ports in-phase.
 

ge said:
The feed could be from either direction depending on how the divider/combiner is used. If I feed from the divider ports (not common port), I guess that I will need to excite both divider ports in-phase.
Click to expand...

Yes, the resistor is only needed for isolation (asymmetric currents).
 

volker@muehlhaus said:
Yes, the resistor is only needed for isolation (asymmetric currents).
Click to expand...

Yes, the resistor is only needed for isolation (asymmetric currents)[/I][/B].....
Not only, The resistor is also needed to provide the matching to Z0 system of the output ports.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…