Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

About the T-juntion power divider

Status
Not open for further replies.

CQCQ

Member level 3
Joined
Nov 26, 2006
Messages
61
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,721
Hi!I got a simple but awkward problem about the T-juntion power divider.I am using a T-juntion power divider to dividing a signal from port1(50ohm) to port2 and port3 and the two port's are 25ohm load(pic1).While from the simulating result(pic2) I see the port1 is matched to 50ohm ,but the port2,3's impedence are 75ohm not matched to 25ohm.So I have some questions....
1.I see the port 1 is matched to 50ohm but port2,3 are not matched.So I think it's have refelction loss in the circuit but why the IL is also ideal 3 dB?
2.What is the real impedence of port2,3? I'm using two-section 1/4 line to transform the 25ohm to 50ohm I think that is right way ,but why the port2,3's impedence are 75ohm not 25ohm?

 

Basically u r trying to simulate Wilkinson div with out 100ohm,
Q1 ans: reflections are not happening to source as it is matched. so loss is ideal.
reflections are between 2 and 3 causing poor isolation.
Q2 ans: having 100 ohm between port 2 and 3 leads 25 ohms if 50 ohm load present at port 2 and 3.
this 100 ohm sees lambda/2 distance means potentials are opposite to each other and cause signal absorption at all times and improves the isolation between port 2 and 3.

hope this explanation makes you better understand.
 
  • Like
Reactions: CQCQ

    CQCQ

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top