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I should be able to crank this out easily! Unfortunately not. So need some basic help here...

I'm trying to determine parasitic inductance in an unknown inductor based on change of ringing frequency. By inserting a capacitance across the low-side switch (nChannel FET) when the ringing frequency is cut in half, the parasitic capacitance is equal to 1/3 of the capacitance value.

To determine the inductance I have the formula: $$ \ f = 0.5 \pi \sqrt LC $$

...which I believe is a derivation of: $$ \ Fr = 1 / (2\pi \sqrt LC) $$

To solve for L I believe it's first necessary to remove the square root by squaring both sides? But what of $$ 0.5 \pi $$

My known terms are f=9.3MHz, C=1000pf. How to solve for L?

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    $\begingroup$ Solve for $\sqrt{L}$ first, then square it. $\endgroup$ – dxiv Nov 28 '16 at 18:47
  • $\begingroup$ You probably want the square root to cover both the $L$ and the $C$. To get that, put them in braces, so \sqrt{LC} gives $\sqrt{LC}$ $\endgroup$ – Ross Millikan Nov 28 '16 at 19:22
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Isolate $\sqrt L$: $$\sqrt L = \dfrac f{.5\pi \cdot C} = \frac{2f}{\pi C}\tag {$\frac 1{\frac 12} = 2$}$$

Now square each side square each side of the equation to get

$$(\sqrt L)^2= \left(\dfrac{2f}{\pi C}\right)^2$$

$$L = \left(\frac {4f^2}{(\pi C)^2}\right)$$

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Assuming both $L$ and $C$ are supposed to be under the square root sign, just square the equation. The $0.5\pi$ when squared gives $0.25\pi^2$, so $$f=0.5\pi \sqrt {LC}\\f^2=0.25\pi^2LC\\L=\frac {f^2}{0.25\pi^2C}=\frac {4f^2}{\pi^2C}$$

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  • $\begingroup$ Yes, both L and C should be under the square root sign! Sorry for my mistake in latex. What I don't follow is how 0.5pi sqrt LC becomes 0,25pi sq * LC. To remove sqrt, I thought the next step would become 0.5*((LC)squared). $\endgroup$ – Mark Richards Nov 28 '16 at 19:51
  • $\begingroup$ You square the whole equation because $(\sqrt{LC})^2=LC$. The other terms need to be squared as well. We have $0.5^2=0.25$ and the $\pi$ and $f$ are squared as well. One of the laws of exponents is $(ab)^2=a^2b^2$ You apply that here. $\endgroup$ – Ross Millikan Nov 28 '16 at 20:55
  • $\begingroup$ Excellent! Thank you very much for this guidance! $\endgroup$ – Mark Richards Nov 28 '16 at 21:04
  • $\begingroup$ ... I learned how to put more than one term within the square root using Latex. Put terms in curly braces as \sqrt{LC}: $$ \sqrt {LC} $$ $\endgroup$ – Mark Richards Nov 29 '16 at 1:08

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