Inductance of a coil using parameters

Hello,
I have created a core coil and I need to simulate it to find the inductance using HFSS. I have simulated it and got some parameters (S, Y and Z) can any one help me how can I calculate Inductance of the coil using the parameters. Thank you in advance.

The impedance of an coil inductor depends on the frequency it encounters, but the inductance value is ideally independent of frequency. This is not always true because some core materials have permeability that can be a function of frequency, and some coil structures may have stray capacitance that makes the model a little more complex.

That all said - if the coil was substantially in air, and you have simulated it in HFSS, and come up with a Z-value, or Y-value, then you are nearly there.

Z = |jωL| = where ω=2*pi*frequency, L is the inductance and j=SQRT(-1)

So we have Z=2*pi*frequency*L
Then you can get L = Z/(2*pi*frequency)

You may find that if you place markers on a few Z values at different frequencies as displayed on either a rectangular or maybe a Smith Chart display, and try them in a column in a spreadsheet, with the frequencies in a column next to them, and then generate the third column of calculated inductance values, you will see how constant is your inductance. Sorry - I have lost the little spreadsheet example I used, but it is easy to re-create.

Y-values can also be used, because Y = 1/Z = 1/(2*pi*frequency*L)
Then L = 1/(2*pi*frequency*Y)
Be careful with the units. frequency is in Hz. If you use GHz in the Z formula, you have to remember the 10^9 multiplier.

Hello Darktrax,
thank you for the response, what if there is a core material having some relative permeability inside the coil. Then how can we calculate the inductance????

The role of permeability is reasonably easy - a constant for the material.
Inductance, even though related to the number of turns in a coil, and the physical dimensions of the thing, is actually defined by the voltages that will appear across a (coil, post, structure, whatever) because of the rate of change of current through it.

This is the v=L*dI/dT, though you will find it needs a minus sign to get it to work by Lenz's Law. But that does not show you the dependence on permeability..

To really get to inductance (say for a solenoid -type coil) L = μ*area*(N^2)/length..
..where N is the number of turns and L is the inductance, and μ is the permeability.
The explanation of how this is derived, and why it depends on the square of the number of turns is given in Wikipedia about 2/3 through under "Inductance of a Solenoid".

--> https://en.wikipedia.org/wiki/Inductance

Permeability uses symbol "μ", the Greek "mu", and is made up of two components..
μ =μ0*μr where μ0 = 4*pi*10^-7 called permeability of free space.
and μr is the relative permeability, value 1 in vacuum, 2000 for mild steel, 7000 for silicon iron, 200,000 for purified iron, 1E+6 for super-magnetic alloy.

You can see that ferrite toroid for example is a magnetic circuit that is all ferrite, and the flux would get to very high values easily, maybe even saturate it. In a pot-core with a tiny air gap, even fractions of a millimeter, the air gap part would have permeability near 4*pi*1.0E-7, so preventing saturation.

If you Google-search "inductance", and maybe "inductance calculator" you will get way more information than ever you wanted. Many are very elaborate, aimed at special structures, like multi-layer, or air-cored helix.

--> https://hamwaves.com/antennas/inductance.html
-->https://www.g3ynh.info/zdocs/magnetics/part_1.html
--> https://www.qsl.net/in3otd/indcalc.html

Hi Darktrax
if the coil was substantially in air, and the stray capacitance has to be considered, means Z=|jωL+1/jωC|, then how can I get L by HFSS？

In addition, are there any methods to get the Q of the coil? in any frequancy? just by the view of Q=(power of EM near-field)/(power of dissipation)?


Thank you in advance.

Hi everyone, can you help me?

I have a helical coil with open end and short end as showed below.

For helical coil, we need to take into account the stray capacitance of coil.

So, how can I have the value of L and C, separately, using HFSS

While I am sure there are many elegant ways to do this, my mind always just thinks up a "quick and dirty"!
My first thought was ..
You can model it at a particular frequency, and find the feed point impedance. The problem is, you don't know how much of the reactive (ie. complex number) part is made up of the (negative) C contribution, and how much is made up of the positive L part.

The Smith chart goes around in circles? OK - I know you can just pick two points at different frequencies off the chart and play with complex numbers, but I like an easier life. Put the marker on the resonance and find the Z.

So OK - you can find the resonance by sweeping over the range of frequencies.
At resonance, you know at least that the capacitive part equals the inductive part.

jωL = -1/jωC or.. jωL = j/ωC the well known resonance condition..

We are allowed to ignore the lossy resistive part here. It only alters the Q.
ω=2*pi*frequency. ω^2 = 1/(L*C) ... we have a relationship between L and C for one frequency
L=1/(4*pi^2*freq1^2)*C

At this point, there are probably a whole bunch of things one can do. Maybe even there is some cute way to get HFSS to drag it out for you. If you don't mind manipulating complex numbers, then just get the impedance at some new frequency.

For me, I would use the same trick as is normal in HF frequency Q-Meters. Simply temporarily add some artificial (perfect) Cx across the coil at the feed points, and sweep again to find the new lower frequency resonance.

You now have two equations for two unknowns which will solve easily.
L=1/(4*pi^2*freq2)*(C+Cx)

Set them equal to each other, several things cancel..
I (think) we end up with C = Cx*(freq2^2)/[(freq1-freq2)*(freq1+freq2)]
Easy! Two resonant frequencies, and you know everything on the right hand side.

This C will be the equivalent C for an inter-turn capacitance distributed between all the turns.
Maybe get the complete model by simply (temporarily) using two turns??

A really truthful model might be a little more complex, with some capacitance being from between turns, and some being the C part of stored energy in the near-field space. There will be some L that is similarly stored. An open coil like that is to some extent a radiating antenna object. Bringing your hand, or a screen, or anything else near it will affect it. I don't really want to go there.

or.. maybe we can use one of the many on-line calculators - very tempting..
https://www.deepfriedneon.com/tesla_f_calchelix.html
https://hamwaves.com/antennas/inductance.html

How can we find the frequency resonance of the inductance and its quality factor by HFSS?

@Darktrax

Thank for your long discuss. Your methods look good. I will try.

I have another idea. For a series LC tank: Z= j(wL-1/w*C). Two variables are L & C.

So, we simulate this coil in HFSS with a sweep of frequency. Take Imaginary part of Z(1,1) at two different frequency and solve for L and C.

For example: at w1=10 MHz, Z1= Im(Z(1,1))=w1*L - 1/(w1*C)

w2=15 MHz, Z2= Im(Z(1,1))=w2*L - 1/(w2*C).

So we can have: L = (Z1*w1-Z2*w2)/(w1^2-w2^2), and then C = 1/(w1^2*L-Z1*w1).

What do you think about this method?

However, when I applied this method at 10 MHz and 15 MHz, solutions are L = 2.71E-4 H and C = 3.35e -11 F.

At different pair of frequencies w1 and w2, values of L, C is different!

What do you think about this result?

The purpose of my coils is to perform Wireless Power Transfer System at 13.56 MHz. And these coils are sefl - resonance.

Below is the graph of Imaginary part of Z(1,1)

@MrElec: You just make a frequency sweep and simulate in HFSS to find resonance frequency. To find Q, Q = wL/R.

So you need to calculate L and R of inductor, coils... I am also studying about it.

Many thanks

@nvt088 thanks for the response. Does the frequency of resonnace of your inductor situated @15,6 MHz?. Do you have an idea how to get the quality factor from HFSS? Some are using this formulae: Q=-Im(Y(1,1))/re(Y(1,1)) (why Y parameter and not the Z one?). However, this relation do not consider the capacitance of the coil and for the resistance it is in relation with re(Y(1,1)).

@MrElec

My helical coil resonated at 13.2 MHz.

Actually, Y(1,1) = 1/Z(1,1) right? But ImY(1,1) different from 1/ImZ(1,1). The same for Real part.

I think in this case, using Z parameters is better.

In this formulae, it's just right if capacitance = 0. For example: One loop.

In Helical coil or spiral coil, we can not do this formulae. So we need to calculate R, L, C separately and use general formula Q=wL/R.

You need to find a way to calculate them.

nvt088 said:

Actually, Y(1,1) = 1/Z(1,1) right?

Click to expand...

Only for 1-port data.
For multiport data Z11 is the input impedance with all other ports open, and 1/Y11 is the input impedance with all other ports shorted.

nvt088 said:

But ImY(1,1) different from 1/ImZ(1,1). The same for Real part.

Click to expand...

Both must be different.
I think you wanted to compare Im{Y11} and Im{1/Z11}
Im{1/Z11} is not the same as 1/Im{Z11}.

@nvt088

From the image of Im(Z(1,1)), i think that your self resonant frequency (SRF) of the helical coil is about 15 MHz. Am i wrong?

volker@muehlhaus said:

Only for 1-port data.
For multiport data Z11 is the input impedance with all other ports open, and 1/Y11 is the input impedance with all other ports shorted.

Click to expand...

What is the different between short and open input impedance. Can you explain more clearly? Because when I analyze Z-matrix, I see that it's OK to use Z(1,1).

@MrElec

The self-resonant of this coils is about over 12 MHz when Im(Z(1,1))=0.

At 15MHz, Im(Z(1,1)) is large, I think it called "antiresonance". Right?

nvt088 said:

@MrElec

The self-resonant of this coils is about over 12 MHz when Im(Z(1,1))=0.

At 15MHz, Im(Z(1,1)) is large, I think it called "antiresonance". Right?

Click to expand...

@nvt088, yes you are right, i'm sorry i've confused it with the anti-resonant.

nvt088 said:

What is the different between short and open input impedance. Can you explain more clearly? Because when I analyze Z-matrix, I see that it's OK to use Z(1,1).

Click to expand...

If your circuit has only one port, then Z11 = 1/Y11.
If your circuit has more than one port, the input impedance into port 1 also depends on the load at the other ports.

By definition, Z11 is the input impedance with current i=0 at the other ports => Z11 is the input impedance with all other ports open (i=0A)

By definition, Y11 is the input admittance impedance with voltage u=0 at the other ports => 1/Y11 is the input impedance with all other ports shorted (u=0V).

But as you found, for circuits with only one port we have Z11=1/Y11.

The self-resonant of this coils is about over 12 MHz when Im(Z(1,1))=0.

At 15MHz, Im(Z(1,1)) is large, I think it called "antiresonance". Right?

Click to expand...

The two resonance frequencies can be only explained by a combination of series and parallel resonance. You have more parameters than a single L and C.

volker@muehlhaus said:

If your circuit has only one port, then Z11 = 1/Y11.
If your circuit has more than one port, the input impedance into port 1 also depends on the load at the other ports.

By definition, Z11 is the input impedance with current i=0 at the other ports => Z11 is the input impedance with all other ports open (i=0A)

By definition, Y11 is the input admittance impedance with voltage u=0 at the other ports => 1/Y11 is the input impedance with all other ports shorted (u=0V).

But as you found, for circuits with only one port we have Z11=1/Y11.

Click to expand...

Thank for your explain. I understood that.

But, can you explain for multi-port system, when calculating L value, should we use L=Im(Z(1,1))/(2*pi*Freq) or use L=Im(1/Y(1,1))/(2*pi*Freq) or other formulae? and why?

I have read in this forum, many people like to use the second formular. In my case, I think that the first formular also OK?

Many thanks

nvt088 said:

But, can you explain for multi-port system, when calculating L value, should we use L=Im(Z(1,1))/(2*pi*Freq) or use L=Im(1/Y(1,1))/(2*pi*Freq) or other formulae? and why?

Click to expand...

If you simulate the inductor with one port, where (+) and (-) terminals of the port are connected between both ends of the coil, both equations are equivalent.

Often, inductor are simulated with two ports, where port 1 (+) is on one end of the coil, and port 2 (+) is on the other end the coil, and the port references 1(-) and 2(-) are connected to some common ground conductor. In that case, you can't use Z11 because this implies that one end of the inductor is open (floating). No current can flow. You will see an open circuit if you look into port 1, with port 2 open.

In this case, you use 1/Y11 instead, which implies that port 2 is shorted to the common ground. Now, current will flow through the conductor and inductance can be measured.

So the simple reason for using 1/Y11 is that it works for both cases: it works for 1-port and 2-port models.

volker@muehlhaus said:

If you simulate the inductor with one port, where (+) and (-) terminals of the port are connected between both ends of the coil, both equations are equivalent.

Often, inductor are simulated with two ports, where port 1 (+) is on one end of the coil, and port 2 (+) is on the other end the coil, and the port references 1(-) and 2(-) are connected to some common ground conductor. In that case, you can't use Z11 because this implies that one end of the inductor is open (floating). No current can flow. You will see an open circuit if you look into port 1, with port 2 open.

In this case, you use 1/Y11 instead, which implies that port 2 is shorted to the common ground. Now, current will flow through the conductor and inductance can be measured.

So the simple reason for using 1/Y11 is that it works for both cases: it works for 1-port and 2-port models.

Click to expand...

Thank you very much for your explanation.
I will check your method by my simulation and post my results at here later.
Many thanks!