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 FastHenry2
 Simulation of a current shunt (kelvin connection)
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maba1971

Germany
4 Posts

Posted - Apr 08 2019 :  16:16:39  Show Profile  Reply with Quote
I try to simulate a current shunt with four-terminal kelvin connection.

I specified two external ports:
...
.external nout nin current_path
.external nsense2 nsense1 voltage_sense

But I can't understand the impedance matrix in Zc.mat:
Row 2: nsense2 to nsense1, port name: voltage_sense
Row 1: nout to nin, port name: current_path
Impedance matrix for frequency = 10 2 x 2
0.0894106 +9.63507e-07j 0.0187796 +2.67151e-07j
0.0187796 +2.67391e-07j 0.118482 +5.27858e-07j
Impedance matrix for frequency = 100000 2 x 2
0.0896302 +0.00770253j 0.0188284 +0.00188867j
0.0188287 +0.00189023j 0.118695 +0.00389998j

I fact I have a current path nout <-> nin and need the resistance (voltage drop) between nsense2 nsense1.
Can anybody help to interpret the impedance matrix?
Thanks in advance!

Mario

mb

Enrico

410 Posts

Posted - Apr 08 2019 :  19:09:10  Show Profile  Reply with Quote
Hi Mario,

the resistance is the real part of the impedance matrix. At your frequencies vs. your structure dimensions, it seems it does not change with the frequency. Now your port 'voltage_sense' is between nsense2 and nsense1, so I would say that you are interested in the resistance of this part, i.e. 0.119 ohm. However I am a bit puzzled by the fact that you use two frequencies, as your statement is 'need the resistance (voltage drop)'. In AC the voltage drop is determined by the overall impedance, not only by the resistance; but as you talk about current shunt with 4 terminal kelvin connection, I expect you are interested in DC current.

If the resistance result makes sense for you, ok; otherwise, please provide a small drawing of the structure as you defined and simulated it.

Ciao,
Enrico
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maba1971

Germany
4 Posts

Posted - Apr 08 2019 :  21:04:00  Show Profile  Reply with Quote
Hey Enrico,
thanks for your fast answer and sorry for the inaccuracy of my question!

I analyze Foil-Shunts for the use at higher frequencies and compare the "DC" impedance (at 10 Hz) with the impedance at 100 kHz. In addition to the simulation I also do some measurements and in my measure setup it is more easy to use 10 Hz than DC, the impedance should be almost the same as at DC.

I am interesed mainly in the real-part of the impedance due to skineffect, current distribution and other effects.
Of course the voltage drop rises due to the inductance too, but this can be eliminated in my setup -- the problem is the increase of the real part.

I did not find a possibility to upload the postscript file generated by fasthenry and zbuf, so I did a small ascii-drawing of the foil structure:


.units mm
 ------------------------------
|     thick=0.075              |
|     sigma=758                |
|     ---   --------   ---     |
|    |   | |        | |   |    |
|    |   | |        | |   |    |
|    |   ---        ---   |    |
|    | nsense1    nsense2 |    |
------                    ------
 nin                       nout


The material of the complete foil structure is NiCr.
Nearby the foil is the base plate made of aluminum with eddy currents.

I think the real part of the impedance from nin to nout is: 89 mohms.
What I need is the real part of the impedance which is responsible for the sense voltage between nsense1 and nsense2, so maybe: 89 mohms - 2*18 mohms??


Mario

Edited by - maba1971 on Apr 08 2019 21:12:39
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Enrico

410 Posts

Posted - Apr 09 2019 :  16:41:22  Show Profile  Reply with Quote
Not really. If I read correctly your drawing, this is a single conductive structure with a common part and the 4 contacts (as per Kelvin, but including the 'measured' part in the simulation).

Now from a black block perspective you have a two-port network (as you defined two ports) so your matrix structure is:


|V1| = |Z11 Z12| |I1|
|V2|   |Z21 Z22| |I2|


(where the bars indicate matrices and vectors. Note also that in your case V2 is Vsense and V1 is Vcurrent_path).

But internally your network is this:



 I1 ->  --R1----R2---   <- I2
              |
   ^          R         ^
V1 |          |         | V2
        --R1----R2---


where R1 and R2 are the small segment resistances, and R is the central part resistance (I'm using symmetry considerations here, and using only the real part)

Now from simple circuit theory you have:

Z11 = 2*R1+R
Z12 = Z21 = R
Z22 = 2*R2+R

So this gives you

R = 0.0188 ohm
R2 = (Z22-R)/2 = (Z22-Z12)/2 = (0.119-0.019)/2 = 0.05 ohm
R1 = (Z11-R)/2 = (Z11-Z12)/2 = (0.0896-0.019)/2 = 0.035 ohm

If nsense1 and nsense2 contacts are narrow, it is reasonable that their resistance is higher.

Does this make sense for you?

BTW why do you not use FastModel for visualization or, even better, FreeCAD E.M. workbench for modelling the FastHenry 3D structures? They make your life easier. BTW If you want to attach pictures, please send me via email (see contacts page). As users are not verified on the Forum, because of spam and virus uploading images and/or posting links is limited. We may move to user email verification in the future to avoid this issue.

Best Regards,
Enrico

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maba1971

Germany
4 Posts

Posted - Apr 09 2019 :  18:21:09  Show Profile  Reply with Quote
quote:
Originally posted by Enrico
... Does this make sense for you?


Yes, your interpretation of my small drawing was correct. I understood the meaning of the impedance matrix completely wrong -- thanks for your explanation!
I expected 50 mohms for the central part resistance R.
And less than 2*35 mohms for the current path pins R1.
I think I have to check and correct errors in the structure of the NiCr part!
quote:
BTW why do you not use FastModel for visualization


because I love my Linux so much ;-)
quote:
or, even better, FreeCAD E.M. workbench for modelling the FastHenry 3D structures? They make your life easier."


I think I will try this, if the structure in the next project is maybe more complex ...

Mario
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Enrico

410 Posts

Posted - Apr 09 2019 :  20:34:37  Show Profile  Reply with Quote
quote:
because I love my Linux so much ;-)


..then definitely you need to use FreeCAD ;)

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