Calculated result for inductance seems wrong?

I'm building an FM radio transmitter kit which includes an LC tank, where C is a tuner capacitor of ~30pF and the resonant frequency is ~100Mhz. The value of the PCB trace inductor, however, is not specified and I want to know what it is.

If I plug:
$100000000=\dfrac{1}{2π\sqrt{L3\times10^{-11})}}$

into an equation solver the result is apparently 52771500 Henries. That seems... a little off? Considering most of the FM transmitter examples I've looked at on the net are like, 0.5uH?

How do I work this out?

• You seem to be doing a mathematical error. I'm getting $84.43 nH$ when I do the plugging in, which seems to be a reasonable number. – deadude Dec 17 '15 at 17:08
• I don't know what I'm doing wrong then: screenshot – Ashlyn Black Dec 17 '15 at 17:14
• Is the tuner capacitor the only capacitor in the LC tank or is there a fixed one in parallel? Please show the circuit diagramm. – Curd Dec 17 '15 at 17:18
• @AshlynBlack: note how your link at the end claims that there is no solution. after their "switch sides" step they do crappy things to resolve that formula – PlasmaHH Dec 17 '15 at 17:25
• Re: schematic (please don't sue me, Silicon Chip magazine) i.imgur.com/uyST5lQ.jpg – Ashlyn Black Dec 17 '15 at 17:31

I think this is a mathematical error as I specified in a comment. To clarify further, rearrange for L as follows.

$$L = \frac{(1e8*2\pi)^{-2}}{3e-11} = \frac{1}{12\pi^2} \cdot \frac{1e-16}{1e-11} = 8.443e-8 H = 84.43 nH$$

The equation holds and seems to yield a reasonable value.

Normally, the resonant frequency is calculated according to the following equation:

$$fr=\frac{1}{2\pi\sqrt{LC}}$$

Solving fo L you have

$$L=\frac{1}{4\pi^2Cfr^2}$$

Seting your values into it I found L= 84nH