### Author Topic: Calculating inductor value from resonant frequency with parallel capacitor.  (Read 767 times)

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#### OLderDan

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##### Calculating inductor value from resonant frequency with parallel capacitor.
« on: May 27, 2024, 04:26:52 am »
So I am trying to understand how to calculate the value of an inductor in a LC circuit with a known value capacitor and I would like to know if one or both of my current systems are flawed.
The method to my madness currently is 1/ get the tank (3rd pic) to resonate with a pulse and then count the frequency of the oscillation (1st pic) or, 2/ vary the frequency of a sine wave till I hit the highest amplitude of the tank and measure the frequency (2nd pic) which then gives me enough to calculate the inductor value.

Edited because I can't load pictures in order...
« Last Edit: May 27, 2024, 04:34:44 am by OLderDan »

#### Kim Christensen

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #1 on: May 27, 2024, 04:38:13 am »
Both methods will work. Looks like the inductor is 150uH or so... (33pf Capacitor)

#### T3sl4co1l

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #2 on: May 27, 2024, 05:48:20 am »
You may find this helpful: https://www.seventransistorlabs.com/Calc/RLC.html#frq

Note that the probe has some capacitance, roughly 10pF in 10x mode.  This will load the circuit; you can use another capacitor in parallel (see the next calculator below the highlighted one) to measure the total by difference.

Tim
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#### OLderDan

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #3 on: May 27, 2024, 06:23:14 am »
You may find this helpful: https://www.seventransistorlabs.com/Calc/RLC.html#frq

Note that the probe has some capacitance, roughly 10pF in 10x mode.  This will load the circuit; you can use another capacitor in parallel (see the next calculator below the highlighted one) to measure the total by difference.

Tim

I have wondered how to account for not only the probe capacitance, but also the 10K resistor I have in series between the signal generator output and the tank, thanks for bringing that extra question up! I will digest the link you provided.

#### MrAl

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #4 on: May 27, 2024, 08:23:10 am »
So I am trying to understand how to calculate the value of an inductor in a LC circuit with a known value capacitor and I would like to know if one or both of my current systems are flawed.
The method to my madness currently is 1/ get the tank (3rd pic) to resonate with a pulse and then count the frequency of the oscillation (1st pic) or, 2/ vary the frequency of a sine wave till I hit the highest amplitude of the tank and measure the frequency (2nd pic) which then gives me enough to calculate the inductor value.

Edited because I can't load pictures in order...

Hi there,

[Note this has been edited for content]

Actually the two methods will yield two different values for the resonate frequency.  That might sound strange but that's life.
The difference may not be that great, but in general that's the way it is.

Also, to actually have resonance in the first place the values of the external resistance and the inductor series resistance have to be of certain values or else it's just a damped response.  That does not mean you cannot calculate the inductance that way though, but the values will be different.  How much different depends on the resistances and the L and C values.
The nice thing is, the pulse (stepped) wave technique can show if there is resonance by observing the wave as it looks like you did.  If there is no apparent sinusoidal part then it's not resonating.

The test circuit would be comprised of an inductor L with some series resistance RL (present in all real inductors we normally see), and the generator series resistance we can call Rs, and we can also add more resistance to Rs by placing a resistor in series with Rs, but then the Rs in the formulas below is the total series resistance not just the resistance of the generator.

Ok, so in the time domain with a stepped input wave the 'resonant' frequency w (if there is a real one) will measure:
w=sqrt(-Rs^2*C^2*RL^2+2*Rs*C*L*RL-L^2+4*Rs^2*C*L)/(2*Rs*C*L)
and that part under the radical must be greater than zero or there is no real resonance.

In the frequency domain (frequency sweep) the 'resonant' frequency w (if there is one) will be:
w=sqrt(sqrt(2*Rs^2*C*L*RL^2+2*Rs*L^2*RL+Rs^2*L^2)/(Rs*C*L^2)-RL^2/L^2)
and also this expression illustrates the departure from the ideal LC resonance which is w=sqrt(1/(L*C)).

Note that I put the word 'resonant' in quotes.  That is because there are three different forms of resonance, so in general I like to call this the 'resonant peak' and the 'resonant peak frequency' we often refer to as w0.  That's the frequency where we observe the maximum peak when we apply a variable frequency sine wave for testing.

Note that often the two w's above are not that far apart so an estimate either way MAY get you there.  There could be times when they do not match well though so beware of that.

The nice thing about the pulse method is it seems to show if the circuit can resonate to begin with.

Also note that you can also just use the voltage divider formula to calculate the inductance using:
VL=Vs*w*L/sqrt(w^2*L^2+Rs^2)
where
VL is the voltage across the inductor,
Vs is the source sine wave voltage,
L is the inductance,
Rs is the total resistance of the generator plus any added series resistance for the test.
You will note that there is no capacitor in this circuit so it simplifies a little.
With that expression you solve for L.
Note however that in that expression there is no RL.  If we do include RL the expression becomes more complicated:
V=Vs*sqrt(RL^2+w^2*L^2)/sqrt((RL+Rs)^2+w^2*L^2)
and to solve for L we also have to solve for RL (unless we rely on a low frequency or DC measurement).  To solve for both we would also have to measure the phase and include that into a more advanced formula.

If you feel like leaving the cap in the circuit but still employ the voltage divider method, you can use:
V=Vs*sqrt(w^2*C^2*RL^2+(1-w^2*C*L)^2)/sqrt((w*C*RL+Rs*w*C)^2+(1-w^2*C*L)^2)
but of course that just complicates things.

Hey, let us know how you make out

« Last Edit: May 28, 2024, 06:34:41 am by MrAl »

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#### MrAl

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #5 on: May 28, 2024, 04:50:05 am »
Hello again,

Here is a little example plot of the two ways used to calculate resonance.
This is using the little circuit with generator resistance Rs=22 Ohms and letting the inductor series resistance vary from 0 to 1.5 Ohms.
Also, C=1F and L=1H.

Note there are two places where the measured frequencies are exactly the same: one at RL=0 and another near RL=about 0.96 or so.
Only the sine frequency sweep method shows the actual frequency where the peak amplitude appears though while the other is only the same at those two points.  There could be some cases where they are never the same.
« Last Edit: May 29, 2024, 04:27:14 am by MrAl »

#### ledtester

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #6 on: May 28, 2024, 05:49:33 am »
I just translated the formulas to MathJax... please let me know if I made any mistakes.

Update: The corrections and suggestions mentioned in later replies have been made.

...
Ok, so in the time domain with a stepped input wave the 'resonant' frequency w (if there is a real one) will measure:
$$w=\frac{\sqrt{-{R_S}^2C^2{R_L}^2+2R_S C L R_L-L^2+4{R_S}^2 C L}}{2 R_S C L}$$
and that part under the radical must be greater than zero or there is no real resonance.

In the frequency domain (frequency sweep) the 'resonant' frequency w (if there is one) will be:
$$w = \sqrt{ \frac{ \sqrt{2 {R_S}^2 C L {R_L}^2 + 2 R_S L^2 R_L + {R_S}^2 L^2 } } { R_S C L^2 } - \frac{{R_L}^2}{L^2} }$$

and also this expression illustrates the departure from the ideal LC resonance which is $w=\frac{1}{\sqrt{LC}}$.

Note that I put the word 'resonant' in quotes.  That is because there are three different forms of resonance, so in general I like to call this the 'resonant peak' and the 'resonant peak frequency' we often refer to as w0.  That's the frequency where we observe the maximum peak when we apply a variable frequency sine wave for testing.

Note that often the two w's above are not that far apart so an estimate either way MAY get you there.  There could be times when they do not match well though so beware of that.

The nice thing about the pulse method is it seems to show if the circuit can resonate to begin with.

Also note that you can also just use the voltage divider formula to calculate the inductance using:
$$\frac{V_L}{V_S} = \frac{w L}{\sqrt{w^2 L^2 + {R_S}^2}}$$
where
VL is the voltage across the inductor,
Vs is the source sine wave voltage,
L is the inductance,
Rs is the total resistance of the generator plus any added series resistance for the test.
You will note that there is no capacitor in this circuit so it simplifies a little.
With that expression you solve for L.
Note however that in that expression there is no RL.  If we do include RL the expression becomes more complicated:
$$\frac{V}{V_S} = \sqrt{ \frac {{{R_L}^2+w^2 L^2}} {{(R_L+R_S)^2+w^2 L^2}} }$$
and to solve for L we also have to solve for RL (unless we rely on a low frequency or DC measurement).  To solve for both we would also have to measure the phase and include that into a more advanced formula.

If you feel like leaving the cap in the circuit but still employ the voltage divider method, you can use:
$$\frac{V}{V_S} = \sqrt{\frac{{w^2 C^2 {R_L}^2+(1-w^2 C L)^2}} {{w^2 C^2 (R_L+R_S)^2+(1-w^2 C L)^2}} }$$
but of course that just complicates things.
...
« Last Edit: May 29, 2024, 03:39:16 am by ledtester »

#### MrAl

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #7 on: May 28, 2024, 06:32:40 am »
I just translated the formulas to MathJax... please let me know if I made any mistakes.

...
Ok, so in the time domain with a stepped input wave the 'resonant' frequency w (if there is a real one) will measure:
$$w=\frac{\sqrt{-R_S^2C^2R_L^2+2R_S C L R_L-L^2+4R_S^2 C L}}{2 R_S C L}$$
and that part under the radical must be greater than zero or there is no real resonance.

In the frequency domain (frequency sweep) the 'resonant' frequency w (if there is one) will be:
$$w=\sqrt{\frac{ \sqrt{2C_1 L_1 R_1^2 R_2^2 + L_1^2 R_2^2 + 2L_1^2 R_1 R_2} } {C_1 L_1^2 R_2} -R_1^2/L_1^2}$$
and also this expression illustrates the departure from the ideal LC resonance which is w=sqrt(1/(L*C)).

Note that I put the word 'resonant' in quotes.  That is because there are three different forms of resonance, so in general I like to call this the 'resonant peak' and the 'resonant peak frequency' we often refer to as w0.  That's the frequency where we observe the maximum peak when we apply a variable frequency sine wave for testing.

Note that often the two w's above are not that far apart so an estimate either way MAY get you there.  There could be times when they do not match well though so beware of that.

The nice thing about the pulse method is it seems to show if the circuit can resonate to begin with.

Also note that you can also just use the voltage divider formula to calculate the inductance using:
$$\frac{V_L}{V_S} = \frac{w L}{w^2 L^2 + R_S^2}$$
where
VL is the voltage across the inductor,
Vs is the source sine wave voltage,
L is the inductance,
Rs is the total resistance of the generator plus any added series resistance for the test.
You will note that there is no capacitor in this circuit so it simplifies a little.
With that expression you solve for L.
Note however that in that expression there is no RL.  If we do include RL the expression becomes more complicated:
$$\frac{V}{V_S} = \sqrt{ \frac {{R_L^2+w^2 L^2}} {{(R_L+R_S)^2+w^2 L^2}} }$$
and to solve for L we also have to solve for RL (unless we rely on a low frequency or DC measurement).  To solve for both we would also have to measure the phase and include that into a more advanced formula.

If you feel like leaving the cap in the circuit but still employ the voltage divider method, you can use:
$$\frac{V}{V_S} = \sqrt{\frac{{w^2 C^2 R_L^2+(1-w^2 C L)^2}} {{w^2 C^2 (R_L+R_S)^2+(1-w^2 C L)^2}} }$$
but of course that just complicates things.
...

Hi,

Oh that's nice.

I don't think you made any mistakes but I did.  The formula with w using variables R1, R2, L1, C1, i used different variable names.  It's the same thing but to keep in line with the previous information it should read:
w=sqrt(sqrt(2*Rs^2*C*L*RL^2+2*Rs*L^2*RL+Rs^2*L^2)/(Rs*C*L^2)-RL^2/L^2)

Note these formulas use Rs, RL, L, and C, not R1, R2, L1, and C1.

Also, the shorter formula for VL was missing the sqrt() in the denominator.  It should read:
VL=Vs*w*L/sqrt(w^2*L^2+Rs^2)

Those are the only two corrections.

« Last Edit: May 28, 2024, 06:35:48 am by MrAl »

#### ledtester

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #8 on: May 28, 2024, 07:29:11 am »
...
Those are the only two corrections.

Ok - I made those changes to reply #6.

#### MrAl

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #9 on: May 28, 2024, 08:43:53 am »
...
Those are the only two corrections.

Ok - I made those changes to reply #6.

Hello again,

Oh yes that looks better.
Just one little thing that may make that second formula down from the top look a little better.   The part with -RL^2/L^2 appears just like that:
-RL^2/L^2
all on one 'line'.
Is there any way using the math rendering software to make it appear with RL^2 actually on top and L^2 on the actual bottom?
Something like:
Code: [Select]
     RL^2-  ------        L^2
How are you getting the math expressions to appear like that ?

$$\frac{{RL}^{2}}{{L}^{2}}$$

Oh ok.
« Last Edit: May 28, 2024, 08:47:51 am by MrAl »

#### ledtester

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #10 on: May 28, 2024, 01:07:10 pm »

Is there any way using the math rendering software to make it appear with RL^2 actually on top and L^2 on the actual bottom?
Something like:
Code: [Select]
     RL^2-  ------        L^2

#### T3sl4co1l

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #11 on: May 28, 2024, 05:02:48 pm »
FYI, you may want to brace the subscripts to make the superscripts show better:

$$\frac{{R_L}^2}{L^2}$$

Maybe more of a preference thing, or depends what the scripts are meaning in context, but I prefer it this way.

Also sub/sup-scripts are single-entity commands, so if you have multiple characters, put them in braces.  Words should probably be written in Roman style:

$$R_{L1} \quad R_\text{load}$$

Also consider changing w (ASCII shorthand) to lowercase omega $\omega$

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!

#### MrAl

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##### Re: Calculating inductor value from resonant frequency with parallel capacitor.
« Reply #12 on: May 29, 2024, 03:32:37 am »

Is there any way using the math rendering software to make it appear with RL^2 actually on top and L^2 on the actual bottom?
Something like:
Code: [Select]
     RL^2-  ------        L^2

Hi,

Yes that makes it look very nice.

I almost never rely on the kind of code to get that to show up like that because over the years I found that the rendering software sometimes does not work and then the reader is left with a post with completely unreadable formulas in it.
If I have to post formulas like that for some reason, I always use an image attachment.  Even that is not perfect.
Also, if someone wants to use the formulas they can not copy and past that into their math software they have to type it all over again, and possibly make mistakes typing it in.  That's why I almost always use regular text like y=1/(2*x^2+x+1) instead of forcing the web site software to render it into a more human readable form.

« Last Edit: May 29, 2024, 03:41:57 am by MrAl »

Smf