How to get the Q value of a spiral in HFSS

[llooll0001]

Hi guys,

I'm working on a Wireless Power Transfer (WPT) project, and the Q value is important for my study. Before making a real spiral I hope to do some simulation by HFSS.

The problem is how to get the Q value in HFSS when the spiral is in self-resonance. I have already get its self-resonant freq by sweeping the S para. But how to calculate the Q value? Hope to get some details here

Thanks

 

[drkirkby]

Hi guys,

I'm working on a Wireless Power Transfer (WPT) project, and the Q value is important for my study. Before making a real spiral I hope to do some simulation by HFSS.

The problem is how to get the Q value in HFSS when the spiral is in self-resonance. I have already get its self-resonant freq by sweeping the S para. But how to calculate the Q value? Hope to get some details here

Thanks

I know there's an example in the Examples directory of finding the Q of a resonator using the eigenmode solver. That might help you.

Dave

 

[tallface65]

the inductance L = 1 / (2*pi*Freq*im(Y(1,1)))
the Q = re(Y(1,1))/im(Y(1,1))

Have Fun

 

[snaildr]

If you model it as a 2-port element, make sure you have a good return path (the two port need to share ground).

To talface65, I somehow found using Z-parameter generates more accurate result than Y-parameters.

 

[tallface65]

If using a 2 port element, you should use the Admittance as it assumes all other ports in the network are shorted except the port of interest. For a single port element, Impedance and Admittance will give same results. For a multi-port network, you will get very different answers as the Z matrix determines all other impedances by exciting 1 amp of current and measuring the voltage over all other ports, hence the other ports are open (or in the case of an inductor, parasitic capacitance due to poor return path ). The Y matrix determines all other Admittances by exciting 1 volt and measuring the current at all other ports, hence the ports are short and a return path is offered.

I may be wrong, but it has worked for me

Have Fun

 

[snaildr]

Thanks for the explanation.

I was doing a one-port simulation the other port of the inductor is simply short by a PEC boundary condition. I don't quite understand in this case why Y and Z-parameter give slightly different result.