- 易迪拓培训,专注于微波、射频、天线设计工程师的培养
2-port impedance measurement in ADS
I am wondering about a simple thing. In ADS, I have a lumped network consisting of a series branch (L and R) and a shunt branch (C and R), as shown in the attached figure. If I want to measure the impedance between both terminals of this lumped network, how should I proceed ? Clearly, I cannot ground the terminal at the shunt branch because this will eliminate the shunt branch.
Is there any building block in ADS that allows me to get a 2-terminal impedance like this network ? I need to extract this impedance to be used in a MeaEqn.
It seems that "Zin" block is not for floating impedance like this one, it should be for measuring impedance of a grounded network.
I should figure out how to do, but it seems not clear to me now.
Please suggest
DYL
You mean the impedance between +7 and +8 ?
Just delete the two ports to ground, and connect one port between these two points. Now you have the series impedance for a source that is floating from the circuit ground.
It's very easy.
Zdiff=Z(7,7)+Z(8,8 )-Z(7,8 )-Z(8,7)
This is a formulation based on Z-parameters.
If you prefer formulation based on S-parameters, see The Designer's Guide Community Forum - Inductance extraction: L reduce with frequency?
Hi volker_muehlhaus and pancho_hideboo,
Thank you very much for your replies. I tried both approaches, and they worked very well with the same results.
Thank you for the info again
DYL
---------- Post added at 18:26 ---------- Previous post was at 16:43 ----------
But, by the way, why are the definitions of "differential" and "common-mode" voltages are different from those of current ?
--> Vdiff = V1 - V2
--> Vcom = (V1+V2)/2
--> Idiff = (I1 - I2)2
--> Icom = I1+I2
Is there any specific reason ?
DYL
Consider power.
V1=Vcom+Vdiff/2
V2=Vcom-Vdiff/2
I1=Icom/2+Idiff
I2=Icom/2-Idiff
V1*I1=Vcom*Icom/2+Vdiff*Idiff/2+Vcom*Idiff+Vdiff*Icom/4
V2*I2=Vcom*Icom/2+Vdiff*Idiff/2-Vcom*Idiff-Vdiff*Icom/4
V1*I1+V2*I2=Vcom*Icom+Vdiff*Idiff=Pcom+Pdiff
Pcom=Vcom*Icom
Pdiff=Vdiff*Idiff
Oh, I see. It's about total power delivered to the 2-Port to consist of common-mode power and differential-mode power.
Thank you very much for your explanation, pancho_hideboo,
DYL
申明:网友回复良莠不齐,仅供参考。如需专业帮助,请学习易迪拓培训专家讲授的ADS视频培训课程。
上一篇:How to plot gm in ADS?
下一篇:Please can someone tell me how to design phassed array antenna in ads 2009Please
国内最全面、最专业的Agilent ADS培训课程,可以帮助您从零开始,全面系统学习ADS设计应用【More..】
- Agilent ADS教学培训课程套装
- 两周学会ADS2011、ADS2013视频教程
- ADS2012、ADS2013射频电路设计详解
- ADS高低阻抗线微带滤波器设计培训教程
- ADS混频器仿真分析实例视频培训课程
- ADS Momentum电磁仿真设计视频课程
- ADS射频电路与通信系统设计高级培训
- ADS Layout和电磁仿真设计培训视频
- ADS Workspace and Simulators Training Course
- ADS Circuit Simulation Training Course
- ADS Layout and EM Simulation Training Course
- Agilent ADS 内部原版培训教材合集