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Inductance extraction: L reduce with frequency? (Read 29304 times)
neoflash
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Inductance extraction: L reduce with frequency?
Mar 11th, 2008, 6:05am
 
Hi:

In some papers, they reported their extracted inductance. And the inductance decrease with the increase of frequency.

However, according to some other papers, the inductance increased. What's the difference in their used method of extraction?

I knows only attached one which L increase with F.

Thanks.
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pancho_hideboo
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Re: Inductance extraction: L reduce with frequency
Reply #1 - Mar 11th, 2008, 8:43am
 
Assume cylindrical wire.

Self inductance is composed from sum of internal and external inductance.
For high frequency, internal inductance decrease due to skin effect.

So self inductance decreases with frequency increase.
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« Last Edit: Mar 11th, 2008, 9:53am by pancho_hideboo »  
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didac
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Re: Inductance extraction: L reduce with frequency
Reply #2 - Mar 11th, 2008, 11:43am
 
Hi,
pancho's answer is completely correct although it doesn't conflict with the reported measurements where inductance increases with frequency. I think that in fact if you made(or simulate) a wide bandwidth with an EM simulator you end up with 3 zones:
1.In the first one as pancho explains the internal inductance goes down due to skin effect.
2.Then when the internal inductance it's negligible due to skin effect inductance will eventually go up again.
3.Inductance will reach a maximum at self-ressonance frequency and then it will start to decrease again.
This can be seen(except probably first zone because it will be needed a series resistance varying with frequency,modelling the skin effect) with a lumped pi model of the inductor and using the formula you provided: for low frequency the imaginary part(since we cannot differentiate between inductive and capacitive components) increases since it's dominated by w*L, from resonance the capacitive part starts to dominate the imaginary part so it changes the slope and it starts to decrease.
I suppose the best measurement will involve a measurement of V and I and calculate L from v=Ldi/dt but with Y,Z or S parameters you cannot obtain this directly.
Hope it helps,
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email_gz
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Re: Inductance extraction: L reduce with frequency
Reply #3 - Mar 27th, 2008, 10:45pm
 
yes,just as didac siad, they are not conflict with each other.The key point is which kind of inductor is you care for, effective induct or physical inductor?
I think ,the effective inductance  will increase with frequency because C//L,but in fact physical inductor deincrease with frequency because skin and eddy effct.
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neoflash
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Re: Inductance extraction: L reduce with frequency
Reply #4 - Mar 30th, 2008, 3:57am
 
Thanks for the replies.

I understand that inductance will seek minimum "impedance" path, and reduces its inductance in higher frequency. But, how people could extract the real inductance from measured S-parameters?
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pancho_hideboo
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Re: Inductance extraction: L reduce with frequency
Reply #5 - Mar 30th, 2008, 5:52am
 
neoflash wrote on Mar 30th, 2008, 3:57am:
Thanks for the replies.
I understand that inductance will seek minimum "impedance" path, and reduces its inductance in higher frequency. But, how people could extract the real inductance from measured S-parameters?

From point of view of electrical magnetic wave, there are no separate inductance and capacitance for solid object.
Just there exists electrical and magnetic energy arround it.

From Foster's reactance theorem, a derivative of reactance of lossless object regarding frequency is always positive.
So if we map reactance X to only L like 2*pi*freq*L, L must be propotional to freq^n, here n>-1.
-1 < n <0  ----->> L decrease with frequency.
0  < n       ----->> L increase with frequency.
If there exists loss, a derivative could be negative. For example, assume second resonance point of simple dipole antenna.

Extraction's issue results in what lumped model you apply for a target solid object around frequency you are considering.

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« Last Edit: Mar 30th, 2008, 8:16am by pancho_hideboo »  
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didac
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Re: Inductance extraction: L reduce with frequency
Reply #6 - Mar 30th, 2008, 8:32am
 
Hi,
As pancho said  we are measuring a black box(an inductance is something teorical in the end), and as email_gz said why do you want physical inductance instead of apparent inductance?For example look at the following attachment of what they provide to you as an inductor:
First one is a discrete inductor from Murata, second one it's a family of inductors from Coilcraft and last it's a capture from a presentation of UMC about how they made their inductors available at: http://www.ansoft.com/deliveringperformance/UMC2.pdf(all this material it's freely available at the web so no NDA problems here).
So in fact what every manufacturer provides it's a table or graph with the apparent inductance and Q factor at different frequencies because it's what you care about when making a circuit.
Hope it helps,
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Re: Inductance extraction: L reduce with frequency
Reply #7 - Mar 30th, 2008, 6:54pm
 
"neoflash wrote on Mar 30th, 2008, 3:57am:
Thanks for the replies.

I understand that inductance will seek minimum "impedance" path, and reduces its inductance in higher frequency. But, how people could extract the real inductance from measured S-parameters?
"

From the two port theory, first we get s11 s12 s21 s22, then to z11,z12 ,z21,z22. finally,we get L1=IM(z11)/(2pi*f),
L2=IM(z22)/(2pi*f),M12=IM(z12)/(2pi*f),M21=IM(z21)/(2pi*f),k=sqrt(M12*M21)/sqrt(
L1*L2),Q1=IM(z11)/Re(z11),Q2=IM(z22)/Re(z22)
for accurate measure ,maybe you need de-embed : two more dummy pad with open and short.  
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pancho_hideboo
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Re: Inductance extraction: L reduce with frequency
Reply #8 - Mar 31st, 2008, 8:39am
 
See http://www.designers-guide.org/Forum/YaBB.pl?num=1133858081/1#1

This is for floating inductor.

port1 >---Inductor ----< port2

L1 is inductance evaluated with port2=gnd.
L2 is inductance evaluated with port1=gnd.

L1 = imag(1/y11)/(2*pi*xval(y11))
L2 = imag(1/y22)/(2*pi*xval(y22))  

Q1 = imag(1/y11)/real(1/y11)  
Q2 = imag(1/y22)/real(1/y22)
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pancho_hideboo
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Re: Inductance extraction: L reduce with frequency
Reply #9 - Apr 1st, 2008, 12:07am
 
This is differential inductor. Ldiff and Qdiff are calculate by the following.

s11 = getData("s11")
s12 = getData("s12")
s21 = getData("s21")
s22 = getData("s22")

R0=50
Sdiff = (s11-s12-s21+s22)/2.0
Zdiff = 2*R0*(1+Sdiff)/(1-Sdiff)

freq = xval(s11)
Ldiff = imag(Zdiff)/(2*pi*freq)
Qdiff = imag(Zdiff)/real(Zdiff)
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Diff_Ind2.jpg
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pancho_hideboo
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Re: Inductance extraction: L reduce with frequency
Reply #10 - Apr 1st, 2008, 12:08am
 
This is Qdiff, Ldiff and Zdiff=Rdiff+j*Xdiff.
Here Ldiff increase with frequency except for first resonance point.
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Diff_Ind1.jpg
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Re: Inductance extraction: L reduce with frequency
Reply #11 - Apr 2nd, 2008, 1:16am
 
pancho_hideboo wrote on Apr 1st, 2008, 12:07am:
This is differential inductor. Ldiff and Qdiff are calculate by the following.

s11 = getData("s11")
s12 = getData("s12")
s21 = getData("s21")
s22 = getData("s22")

R0=50
Sdiff = (s11-s12-s21+s22)/2.0
Zdiff = 2*R0*(1+Sdiff)/(1-Sdiff)

freq = xval(s11)
Ldiff = imag(Zdiff)/(2*pi*freq)
Qdiff = imag(Zdiff)/real(Zdiff)

。。。。。nice reply with detail, thank you very much!
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Re: Inductance extraction: L reduce with frequency
Reply #12 - Jan 31st, 2012, 10:05pm
 
pancho_hideboo wrote on Apr 1st, 2008, 12:07am:
This is differential inductor. Ldiff and Qdiff are calculate by the following.

s11 = getData("s11")
s12 = getData("s12")
s21 = getData("s21")
s22 = getData("s22")

R0=50
Sdiff = (s11-s12-s21+s22)/2.0
Zdiff = 2*R0*(1+Sdiff)/(1-Sdiff)

freq = xval(s11)
Ldiff = imag(Zdiff)/(2*pi*freq)
Qdiff = imag(Zdiff)/real(Zdiff)


Hi, thanks for the detailed information.
Would you kindly provide any reference to the equations used for calculating L using S parameters (both floating and differential )?
thanks.
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