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2-port impedance measurement in ADS

时间:04-07 整理:3721RD 点击:
Hi all,

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

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