hfss port impedance
While HFSS doing the computations for one lumped port, what are the properties of the other ports? Do they become open circuits, or perfect-conductor planes, or something else?
when you define a lumped port you define an impedance for each port that represent properation of the other port
B16rahmati I believe is tellying you that it looks like whatever impedance you set the line to so if you have 3 ports 1 = 25ohms, 2 = 50ohms, and 3 = 100ohms. Then when you analyze port 1 it loos like port 2 and 3 are 25ohms and 100ohms
hope this helps,
Babbage
The replies seem reasonable but did not help. So here's a more detailed explanation of my problem.
Basically, I'm using HFSS to simulate a CMOS transformer, like in the picture below. When I use two lumped ports as differential inputs, as in the top diagram, I get correct results, based on Z parameters.
However I also want to simulate results with single-ended input. And I want to add a ground tap to one of the coils. I tried to do this by creating a ground plane, and making four lumped ports between the ground plane and the coils. But once I did this, the results became screwed up. At high frequencies the values were OK, but at low frequencies the Z parameters were wrong. Eventually I figured out that the *amount* of Z parameter inaccuracy was roughly equal to the effects of a parasitic capacitor of about 1pF. But I have no idea why adding ports should create a whole new parasitic capacitance! (I once saw a similar effect in ADS Momentum, but it magically disappeared before I could figure out what caused it :D )
I could try going back to one port for each coil - but then my "ground connection" becomes just another conducting path with inductance of its own, which would make the results inaccurate again.
I also tried using waveports at each input, but I could never get the Z parameters to come out correctly. In all the inductor examples, people always use lumped ports - as far as I can tell waveports just don't work for an inductor - but why should that be the case?
What am I missing here?
I was having the same problem, when I used two lumped ports in my structure. I never figured out why the Z results were wrong. Which lumped port of the two was used as an excitation port, what is the property of the other one was difficult to say. To solve the problem, I used only one lumped port and the other one was substituted by a resistor. I think you can do the same.
I am not really sure I can give you any reasonable explanation. But the way the setup the two simulations are totally different.
In setup #1 you used lumped port as differential excitation which leaves your ground plane floating, no common-mode field. This means your impedance is mainly DC impedance i.e. resistance.
Where as in setup #2, you used single-ended excitation withe the ground plane as reference. A low frequencies, the ground plane is included in your impedance calculation.
At high frequency, however, the current starts to take the path with least impedance -> lowest inductance hence smallest loop area.
It is better to use waveport for this kind of problem. You need a certain distance between a waveport and discontinuity. Extend you feeding line and de-embed your waveport as much.
wlcsp
hi
this a replay from EVAN , it can help you
h??p://w?w.edaboard.com/ftopic293420.html
Lumped ports:
Lumped ports are similar to traditional wave ports, but can be located internally and have a complex user-defined impedance. Lumped ports compute S-parameters directly at the port.The complex impedance Zs defined for a lumped port serves as the reference impedance of the S-matrix on the lumped port. The impedance Zs has the characteristics of a wave impedance; it is used to determine the strength of a source, such as the modal voltage V and modal current I, through complex power normalization. (The magnitude of the complex power is normalized to 1.) In either case, you would get an identical S-matrix by solving a problem using a complex impedance for a lumped Zs or renormalizing an existing solution to the same complex impedance.
By default, the interface between all 3D objects and the background is a perfect E boundary through which no energy may enter or exit. Wave ports are typically placed on this interface to provide a window that couples the model device to the external world.
Wave ports:
HFSS assumes that each wave port you define is connected to a semi-infinitely long waveguide that has the same cross-section and material properties as the port. When solving for the S-parameters, HFSS assumes that the structure is excited by the natural field patterns (modes) associated with these cross-sections. The 2D field solutions generated for each wave port serve as boundary conditions at those ports for the 3D problem. The final field solution computed must match the 2D field pattern at each port.
HFSS generates a solution by exciting each wave port individually. Each mode incident on a port contains one watt of time-averaged power. Port 1 is excited by a signal of one watt, and the other ports are set to zero watts. After a solution is generated, port 2 is set to one watt, and the other ports to zero watts and so forth.
With lumped ports you should know the characteristic impedance of the connected feeding line for calculating S-matrix, while with wave ports, if correctly sized, portZ0 defines the reference impedance for calculating S parameters, and it automatically takes the value of the Zo impedance of feeding line. If I were you, I would prefer wave ports, always if you are not obliged of defining an internal port.
Regards,
Ivan