Quick EQ Question...Math Yet Again...

What I would like to know is where I can find the needed formlae to calculate the following. I know the freq., I know the capacitance and resistance and would like to find out the inductance. The paticular circuit in question has 1 inductor in series with a cap and then a resistor to Gnd. Any help appreciated.

if the known frequency is the resonance frequency then:

L=square(1/(2 x PI x f)) /C

hope this makes sense, it´s 2:20 AM here....

steff

For more background info about Steffen's formula see this pdf, which will also tell you how the series resistor affects the results, amongst other things.

or rewritten:

L=1/(4∏^2 x ?^2 x C)

or if your lazy like me:

http://www.opamplabs.com/cfl.htm

the resistor will not affect the resonant frequency, but maybe the Q.

Thanks guys. I'll investigate further from here.

After several calculations and use of the op-amp labs calculator I seem to come to different results with the same criteria. That is, between the calculator and Steffen's formula. Calculator gives me 54mH and the math gives me 40mH. Close but not really. Which should I trust more??

Measure it in-circuit - that is the only way to be sure.

Remember that inductance change when the inductor core is "activated" - at high signal levels the inductance will go down..

I use this nomograph a lot for that type of circuits:

http://www.testecvw.com/carl/images/ImpedanceNomograph.pdf

Jakob E.

> Calculator gives me 54mH and the math gives me 40mH.

That is reasonable if one way includes the resistor and the other way ignores it.

How about posting actual values and letting everybody try to give an answer? Majority wins. :grin:

PRR, you´re a real democrat.

How about posting actual values and letting everybody try to give an answer? Majority wins.

Click to expand...

Do you want to become president? Your country needs you (no shit). :thumb: :thumb: :thumb:
Jens :wink:

Resistor value was not in the equation in either instance (unless the calculator has some hidden basic value it applies without showing). Actual values...

Freq. 100Hz
C 47uf
R 1k8

I get 53.89 mH using your values and my formula.

L=square(1/(2 x PI x f)) /C is not good to read...

this is better?:




steff

F = 1/(2π*√(L*C)) {"2π" is two-pi}

Freq. 100Hz
C 47uf
R 1k8

Reactance chart eye-balls 100Hz and 47uFd as about 35Ω and a bit over 0.050H.

If this is a series-tuned tank with a 1K8 load resistor, output across the 1K8 resistor, the Q is about 35/1800 or 0.02, the response is a VERY broad "peak". I doubt you can even say it has a resonance. Looks to me like it is flat from 2Hz to 5KHz.

Assuming your 40mH answer is right:

F = 1/(2π*√(L*C))
F = 1/(6.28*√(0.040*47E-6))
F = 116Hz

Assuming your 54mH answer is right:

F = 1/(2π*√(L*C))
F = 1/(6.28*√(0.054*47E-6))
F = 100Hz

Steff, I was able to read your formula okay and I too came to the same answer (rounded off to 54).

PRR, I don't doubt the Q to be very wide...it's supposed to be a characteristic of this particular piece although I'd be suprised if it was that large. So all in all, trying to figue out L value using math is more of a general area kinda thing. At least this is what I gather from your posts. Cool. From this I can at least get going at what I need to do and be quite close without an actual measurement of the inductor.

Thanks a lot guys!