# Easier frequency calculations – g3jkx

One of the best short cuts that I know is a simpler way to calculate the frequency of a tuned circuit involving a coil and a capacitance.

The formula usual formula is f = _____1_____

2 pi root LC

( this formula will be given to you if there is a question on this in the Advanced exam)

Ahaa, but L has to be in Henrys and C in Farads for this formula! Mmmm!

This means lots of ten to the minus goodness knows what for both L and C; a daunting sum.

Even for an experienced calculator operator this can betroublesome, so
here is an easy sure fire method. Being amateurs, our questions will be
about short waves, involving Megahertz, coils in microhenrys (µH) and
capacitors in picofarads (pF) So we can rearrange the formula and it
becomes really easy….

f 2 = 25,330 f in MHz, L in microHenries, C in picoFarads

L C **Yes, this really works.**

First, multiply L in µH by C in pF.

Divide 25,330 by this number and then, because f is squared, we must
find the square root of it to get the frequency in MHz! Let’s take an
example.

We are told that a variable capacitor has a range of 5pf to 100pf.

This is in parallel with a coil of 10µH. What frequency range will be covered?

First work out the L x C using 10µH and 5pF. L x C = 10 x 5 = 50

Divide this into 25,330 = 506·6 The square root of this is f = 22·5 (MHz)

Now to work out the frequency for 100pf. L x C = 10 x 100 = 1000

Divide this in to 25,330 = 25·33 The square root of this is f = 5 MHz.(+ a tiny bit)

So the frequency range with a 5 to100pf capacitor and a 10µH coil is about 5 to 22·5 MHz

Another question might be: Using a 5microH coil,
what capacitance will be needed to bring the circuit to resonance at 10
MHz? This means we have to juggle the formula a bit.

We know that f2 = 25,330 We need to know C, so that has to be on the Left Hand Side

LC and the f, which we know, on the RHS.

As with all formulae, if you do the same thing to the left and the right hand sides…..

*This make no difference to the formula!*

So, by multiplying both sides by C and we get f2 C = 25,330 x C

LC

Now C divided by C = 1, so we get f2 C = 25,330 x 1 (25,330 times 1 is 25, 330)

L x 1 (L times 1 is L)

If we now divide both sides by f2 we get f2 C = 25,330

f2 f2 L

f2 divided by f2 = 1 so we end up with C = 25,330

f2 L

Calculating f2 x L = 10 x 10 x 5 = 500. Divided into 25,330 = C = 50·66pFWe
can do a similar sum if we are given the C and don’t know the L. Easy
really, when you know how. Practice makes perfect. So try out your own
values until you can do these calculations easily using your calculator.
There WILL be a question on this for sure on the advancedexam.