[TowerTalk] S21 to Rp & Cp - Equivalent Circuit

Jim Brown jim at audiosystemsgroup.com
Thu Jan 11 13:18:50 EST 2018


You're over-Excelling the problem, Maximo. Best to go back to 
fundamentals. A simple parallel resonant circuit looks like an inductor 
below resonance and a capacitor above resonance. Go far enough away from 
resonance and the other terms become increasing insignificant.

To find Lp, take one or more measured Z values well down in amplitude 
and frequency from the resonant peak (say 1/20 of both) and use the 
simple equation for X = 2 pi F L   Far below resonance, Z becomes 
approximately equal to XL, so use the measured value for Z at some low 
frequency and solve that equation for L.  Do that at 2-3 frequencies, 
and you should get values for Lp that are quite close to each other.

73, Jim K9YC

On 1/11/2018 10:01 AM, Máximo EA1DDO_HK1H wrote:
> Hi Jim, Steve, all,
>
>
> Playing with those formulas I´ve got an inconsistency.
>
>
> To get R and jX from LogMag and Phase I can use two different formulas, one from Steve and one from Jim, but the results are not same, so I don´t know where the mistake is.
>
>
> Example from Excel;
>
>
> LogMag -29.601925
>
> Phase  -77.10851
>
>
> Steve´s formula:
>
> Rx = (100*10^(-LogMag/20))/(SQRT(1+(TAN(Phase/57.29578))^2))-100  =  573.916
>
> X = -(Rx+100)*TAN(Phase/57.29578)  =  2944.484
>
>
>
> Jim´s formula:
>
>
> R = COMPLEX(10^((LogMag/20)*(COS((Phase*PI())/180))),10^((LogMag/20)*(SIN((Phase*PI())/180))))
>
> then Rx+jX = IMPRODUCT(50,IMDIV(IMSUM(1,R),IMSUB(1,R)))  =  -49.930  +3.606
>
>
> As you can see, 573 is not same as -49 OHm
>
> And 2944 is not 3.606
>
>
> Anyone know where the error is?
>
>
> Thank you
>
>
> 73, Maximo
>
>
> ________________________________
> De: TowerTalk <towertalk-bounces at contesting.com> en nombre de Jim Brown <jim at audiosystemsgroup.com>
> Enviado: martes, 9 de enero de 2018 17:53
> Para: towertalk at contesting.com
> Asunto: Re: [TowerTalk] S21 to Rp & Cp - Equivalent Circuit
>
> This, of course, is classic circuit analysis that we learned in our
> first EE year. If, however, you have a plot of data vs frequency that
> you can identify as a parallel resonant circuit, there's a relatively
> easy way to find Rp, Lp, and Cp. Rp is the peak value of Z at resonance.
> Lp is the value that yields Z far below resonance, and Cp is the value
> that resonates with Lp.  Graphical techniques like this, by the way,
> were taught in another course in that first year of EE.
>
> When I was doing the point by measurements on the first ferrite chokes
> using a signal generator and voltmeter, I put the raw data into a
> Quattro Pro spreadsheet, computed Z magnitude, and plotted it. To find
> the equivalent circuit, I used the method above, plotted the result on
> the same axes as the raw data, and tweaked the three values so that the
> plot of the equivalent circuit matched the measured data. These tweaks
> are necessary because mu' and mu'' vary with frequency. Mu' is the value
> that defines the series inductance, mu'' is the value that defines the
> series equivalent resistance.
>
> 73, Jim K9YC
>
> On 1/7/2018 2:01 PM, jimlux wrote:
>> On 1/7/18 12:59 PM, Máximo EA1DDO_HK1H wrote:
>>> Hi all,
>>>
>>>
>>> I know there is a formula to get Rp and Cp from S21 or S11 data
>>> (LogMag and Phase, or R & I).
>> Convert S11 from logmag, angle to complex
>>
>> Z = Z0 * ((1+gamma)/(1-gamma))
>> in Excel
>> gamma =
>> imcomplex(10^(S11mag/20)*cos(s11phase*pi()/180),10^(S11mag/20)*sin(s11phase*pi()/180))
>>
>> Z= immult(50,imdivide(imsum(1,gamma),imsub(1,gamma))
>>
>> that gives you R+jX series form R and C/L
>>
>> You then have to convert to the parallel form
>>
>> Rp = Xs*(Rs/Xs+Xs/Rs)
>> Xp = Rs*(Rs/Xs+Xs/Rs)
>>
>> The Rs/Xs+Xs/Rs is common to both, so in excel, adding a column for
>> that makes it easier.
>>
>>
>>
>>> Anyone know that formula?
>>>
>>> It is to use it in Excel and get Rp and Cp from the VNA´s exported data.
>>>
>>>
>>> Thank you very much
>>>
>>>
>>> 73, Maximo
>>> _______________________________________________
>>>
>>>
>>>
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>>> TowerTalk at contesting.com
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