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1.47726744996646E-18 is this too big of a number for a variable capacitor as I am trying to do a project on the computer related to tuning "when I enter a frequency like (133,855,000)

The formula would be \$C = \dfrac{1}{(2\pi \cdot freq)^2 \cdot L}\$

I end up with the above number.

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    \$\begingroup\$ That's waaaay small.... 0.0015 femto farads -ish \$\endgroup\$
    – Spoon
    Feb 3, 2015 at 13:00
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    \$\begingroup\$ Do you realize that the number you posted is shown in scientific notation and that the number before E is being divided by 10 followed by 18 zeroes? So the actual number is 0.00000000000000000147726744996646, which is practically zero. \$\endgroup\$
    – Ricardo
    Feb 3, 2015 at 13:05
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    \$\begingroup\$ Why do you want to use a 1H inductor? Choose a smaller L and you will be fine. \$\endgroup\$
    – GT.
    Feb 3, 2015 at 13:08
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    \$\begingroup\$ Presumably you've used an inductor value of no practical use for that frequency. Maybe you can add details of what you're actually trying to do for a more practical answer. \$\endgroup\$
    – PeterJ
    Feb 3, 2015 at 13:08
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    \$\begingroup\$ Using 15 significant digits to express the capacitance you are asking about it absurd, and wrong in a engineering context. \$\endgroup\$ Feb 3, 2015 at 14:18

3 Answers 3

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1.4137449e-18, as others have said, is incredibly small.

The equation given allows you to calculate the capacitor required for an LC (inductor-capacitor) circuit resonant at the given frequency. The number you calculated is for an L of 1 Henry. Looking at the equation it's clear that as the inductance (L) decreases, the capacitance increases. Inductors can be found in a range of inductances from perhaps 1nH and larger. Given that a nanoHenry is \$0.000000001\$ Henries (\$1 \times 10^{-9}\$), you should be able to see that the capacitance required could be much more sensible.

Picking an inductor value of \$100\$nH (or \$0.0000001\$ Henries) gives \$1.4137 \times 10^{-11}\$ Farads, or \$14\$ pico Farads of capacitance. By reducing the inductor size, you can increase the capacitor value further - an inductor of \$10\$nH would give a capacitance of \$141\$pF and so on.

If you don't understand scientific notation it's probably worth finding a maths textbook which explains it, because you can't really do any engineering or electronics design without having a good grasp of maths.

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Variable (hand turned) capacitors generally have a range of for instance around 10%..100%, where 100% is 50 .. 500 pF. Trimmer capacitors (screwdriver turned) can have a lower maximum, down to ~ 10 pF.

Twisted pair wire has a capacity in the order of 1 pF /cm.

1 pF = \$10^{-12}\$ F

Your 1 * \$10^{-18}\$ F is 106 times smaller than 1 pF, which makes it totally impractical because it would be drowned by the stray capacitances in your circuit.

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For a discrete-component tank circuit with 134MHz resonance your inductor should likely be less than 1uH, so the capacitor will be some pF, which is quite practical.

One limitation is the self-resonant frequency of the inductor- you can look up datasheets of real inductors to see what is practical to manufacture. For example, a TDK 330nH inductor with 0.75 ohm resistance has a SRF of 230MHz, and the capacitor would be around 4pF, which is fine.

Example:

inductor: TDK MLF1608DR33KTA00 330nH 0603

variable capacitor: Murata TZC3Z060A110B00 2-6pF

Capacitance for resonance at 134MHz- about 4.3pF

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