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.

Want help in designing 1-2 wilkinson power divider in HFSS

Status
Not open for further replies.

sarslanali

Junior Member level 2
Joined
Aug 12, 2011
Messages
24
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,505
I designed a 50 ohm 1-2 wilkinson power divider for for 2.4 GHz. I don't know why but my S11 is not correct. It is -40db at 2.8 GHz instead of 2.4 GHz. My S22, S33 and S23 are also not correct please help me. This is my HFSS design. I used Appcad for initial line width measurement and then used HFSS seperately for tuning the lines to requirement. Please help.
 

Attachments

  • power divider.rar
    130.7 KB · Views: 263

I haven't had time to simulate the design. However I don't understand why you have copper layers that are almost half a millimetre thick.That might be the reason for the inaccuracy.
 

I haven't had time to simulate the design. However I don't understand why you have copper layers that are almost half a millimetre thick.That might be the reason for the inaccuracy.

Actually I used AppCad to calculate 100 ohm transmission line and optimized it using HFSS (0.35 mm t-line). I think that's the mistake. When I simulated the same design in ADS and used 100 ohm resistor in place of transmission line, the result was satisfactory but I don't understand how can I design 100 ohm resistor in HFSS. Do I have to use some other material?

---------- Post added at 17:11 ---------- Previous post was at 17:10 ----------

I haven't had time to simulate the design. However I don't understand why you have copper layers that are almost half a millimetre thick.That might be the reason for the inaccuracy.

Actually I used AppCad to calculate 100 ohm transmission line and optimized it using HFSS (0.35 mm t-line). I think that's the mistake. When I simulated the same design in ADS and used 100 ohm resistor in place of transmission line, the result was satisfactory but I don't understand how can I design 100 ohm resistor in HFSS. Do I have to use some other material?
 

you need to right click on an object an assign boundary>impedance OR Lumped RLC OR Finite impedance. I have never used these options but I am sure you can find the best one for yourself with a quick google search.
 

you need to right click on an object an assign boundary>impedance OR Lumped RLC OR Finite impedance. I have never used these options but I am sure you can find the best one for yourself with a quick google search.

Thanks for reply it really helped me a lot. I searched edaboard and found many threads on same issue. Thanks.
 

Actually I used AppCad to calculate 100 ohm transmission line and optimized it using HFSS (0.35 mm t-line). I think that's the mistake. When I simulated the same design in ADS and used 100 ohm resistor in place of transmission line, the result was satisfactory but I don't understand how can I design 100 ohm resistor in HFSS. Do I have to use some other material?

---------- Post added at 17:11 ---------- Previous post was at 17:10 ----------



Actually I used AppCad to calculate 100 ohm transmission line and optimized it using HFSS (0.35 mm t-line). I think that's the mistake. When I simulated the same design in ADS and used 100 ohm resistor in place of transmission line, the result was satisfactory but I don't understand how can I design 100 ohm resistor in HFSS. Do I have to use some other material?

I'm not familiar w/ HFSS, but in ADS Layout, you can use thin-film resistor layers to make nice loads for circuits like this. You might give that a look. Define the sheet resistance to be 50 ohms/square, then make your resistor have a 1:2 aspect ratio. Each "square" of thin-film material will look like a 50 ohm resistor. If you have a long, skinny strip, they will add up in "series"... so a 1:2 aspect ratio will give you 100 ohms (R = Rs*L/W.... 50*2/1 = 100 ohms). Conversely, a short, wide thin-film resistor will act like "square" 50 ohm resistors in parallel, so a 4:1 aspect ratio would yield 50*1/4 = 12.5 ohms.

In all cases, make sure that your resistor is electrically short (< 5 degrees), or you'll mess up the reflected signal phases, which will reduce the voltage-wave cancellation and hurt your port-to-port isolation.

For these frequencies, you could use a thin-film R that is similar in dimensions to your final packaged part (0201, 0402, 0603, 0805, etc), then simply install the real part when/if you fabricate it.
 
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top