Formula Transformation...

[Relayer]

Hey guys,
This is a tad embarrassing, but I need to know the following:

How do I transform the following resonant frequency formula to find L, then C from
the equation below:

I need to have L where F is, and C where F is. Two separate formulas.

I realize this request is a fairly basic question, and having done an electronics apprenticeship,
I should know how to do this, but its been over 40 years since I did formula transformations
in high school.
I've tried looking on the internet for those formula variations, but I couldn't find
anything relevant, just calculators. I need a hard copy of the formulas themselves.
Any help with the above would be most appreciated. :smile:
Thanx in advance.
Regards,
Relayer

Edit: I just realized I've posted in the wrong section... Sorry about that :bang:

 

[Aussie Susan]

(Not sure how to use any of the fancy display capabilities this forum has!)
Starting with your equation (in the form I'll use for the rest of the post):
F = 1/(2.pi.sqrt(LC))
you can rearrange to :
F.2.pi.sqrt(LC) = 1
Next:
sqrt(LC) = 1/(2.F.pi)
(and yes I know you can go directly to that from the start)
Now square both sides
LC=1/(4.F.F.pi.pi)
Finally take whichever L or C you want to the other side:
L = 1/(C.4.F.F.pi.pi)
C = 1/(L.4.F.F.pi.pi)
You may also find the '2.pi.f' written as 'w' (that should be the Greek letter lowercase omega) as this is a very commonly used alternative rather than writing the '2.pi.' constants all the time. There is a technical reason that is not important here but it may be the reason you could not find the equation you were looking for.
By the way, you can't use this formula to "find L, then C". It is a case of chose one and then calculate the other. Depending on your requirements (e.g. if this is in a tune-able radio front-end or a fixed frequency filter etc.) and the frequency involved then it is normal to select the hardest-to-match component first, using whatever value you can actually get hold of, and then calculate the other. I often start with the capacitor as these come in set values or cover a variable range for a radio tuning capacitor) and then make the inductor (it is often easier to make one with a few fewer or extra windings to get the value you want.)
Susan (from the other side of Melbourne)

 

[BradtheRad]

For a guideline you can consider what is a suitable L:C ratio. Many applications commonly use between 100:1 and 100000:1.

As another factor consider the amount of Amperes going through your LC network.
High A, high C, low L.
Low A, low C, high L.

Formulas for reactive impedance are useful.
XL = 2*Pi*f*L.
Etc.

 

[FvM]

Or more easily Z = √(L/C)

 

[c_mitra]

Or more easily Z = √(L/C)...

Z is commonly used for Impedance and I do not understand how you get that equation...

In absence of any resistance, Z will be simply j*(XL-XC) where j is the square root of -1.

 

[Relayer]

Hey guys,
Thank you very much for your replies, its very much appreciated. :grin:
I hope my reply works this time. :-(
I've had trouble getting into eda due to slow loading or timing out. Plus my PC has been randomly
freezing to the point that I need to reset my whole system. :-?

@Susan
I realize I need to nominate either L or C in the equation/s. :wink:
I take it that L = 1/(C.4.F.F.pi.pi) and C = 1/(L.4.F.F.pi.pi), where the two "F"'s and two "pi"'s means
they are squared. i.e. L=1/C*4*F squared* and pi squared?
Thank you. :grin:
I see the limitations of the Forum in regards to characters available.
Good to know a fellow Aussie, and one not far from me, is in the electronics field and on this Forum. :grin:

@BradtheRad
Yes, those impedance formula's will help as well. Though I wasn't aware of those ratios. But good to know.
Thank you. :grin:

Thanx once again guys. It just shows how invaluable edaboard Forum really is. There are some really clued
up people on here and so friendly and helpful to boot. I'm very happy to be a part of this.
Regards,
Relayer

P.S. Yay, I was able to post without any hassles. :smile:

 

[CataM]

Z is commonly used for Impedance and I do not understand how you get that equation...

I think he was referring to the characteristic impedance. All in all, I do not see why use it in this context.

- - - Updated - - -

I see the limitations of the Forum in regards to characters available.

I see no limitation. You need to go to the "advanced reply". L=1/(C·4·f²·Π²)

 

[c_mitra]

You need to go to the "advanced reply"....

But I do not know "how"... I do not see any button labelled "advanced reply"...

 

[CataM]

But I do not know "how"... I do not see any button labelled "advanced reply"...

I think if you are using a phone, you can not go there.
Right beneath the window where you type your "Quick Replay", there are 4 buttons:
-Post Quick Replay
-Go Advanced <--- THIS IS THE "Advanced Replay" option.
-Add an Image
-Cancel

 

[Aussie Susan]

I see no limitation. You need to go to the "advanced reply".

Thank you - I had never seen that or (before this) needed anything other than basic text.
Susan

 

[BradtheRad]

If we use values of 1 in the basic electrical formulas, it makes for convenience toward the aim of grasping how they interact. (That is, 1 Farad, 1 Henry, 1 ohm, 1 second, etc.)
So there could be something enlightening if we use 1H & 1F in the resonance formula. Frequency simplifies to 1/(2*Pi) second per cycle, or cps=2*Pi.