# 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. Dec 17 '15 at 17:08
• I don't know what I'm doing wrong then: screenshot 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 Dec 17 '15 at 17:25
• Re: schematic (please don't sue me, Silicon Chip magazine) i.imgur.com/uyST5lQ.jpg 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