# Getting resonant frequency in LC circuit with L as air core coil

im trying to find inductance of the air core coil,i read that in LC parallel circuit with known capacitor value we can find unknown inductance value by the formula

so need to find 'f' resonant frequency.to find L

Im getting that resonant frequency in normal inductor (resistor like) but not in air core coil. check the scope output with normal inductor and air core coil below.

1).with normal inductor 42uh getting resonant frequency

2). this is the output of air core coil,i didnt get the waveform like normal inductor,not able to find resonant frequency.

3) the coil i used ,68mh , i checked that inductance value with LCR meter

why the resonant frequency is not coming in air core coil?,how do i find inductance value of air core coil.?

• Use a sine wave and keep increasing the frequency until the output amplitude hits a peak - that is resonance. – Andy aka Oct 8 '15 at 11:36
• The "unknown" inductor has a large value, it needs some time to "charge" (build up a magnetic field). I think the 1 kHz and the 1 kohm resistor do not allow it much time for this. So there might be a resonance but the amplitude is so small you don't see it. – Bimpelrekkie Oct 8 '15 at 11:50

Let's say the ferrite-cored inductor is 42 uH with an internal self-resistance of 10 ohms and the air core inductor is also 42 uH but, due to it needing (say) ten times as many turns it's self-resistance is 100 ohms: -

Does that simulation help?

2) the inductance of air coil very smaller than the inductor you have used.

You can simply calculate the inductance of air coil. The high quality of resonant tank is also needed, for this you will need that Xl=Xc, jwL=-1/jwC, or wC=1/wL -> C=1/w^2L, low ESR capacitor matched with inductor inductance. Therefore first calculate L, choose the right cap and then measure.

It may be that you're having a hard time with the air-cored coil because its resistance is spoiling its Q.

The surest way to find the inductance of the coil in-circuit is to sweep the input of the circuit with a sine wave and look for a peak across the LC. That peak will occur at resonance, when the reactances of the cap and the coil will be equal and the impedance of the tank will become resistive and climb to its maximum.

Then, knowing f and C, you can calculate Xc, set Xl = Xc, and calculate L. It'll also help if you replace the 1k resistor with 10k since that'll make the peak much more visible.

You also may be able to see the circuit ring with the rig you have set up if you speed up the horizontal sweep so that you get a better look at the signal across the LC.

If you go here and enter L = 42${\mu H}$, D = 6mm, l = 6mm, and a wire size of 24AWG, you'll get a nice little coil with a resistance of about 0.26 ohm, and if you simulate your circuit:

with LTspice, you'll get a plot that looks like this:

I've assumed your signal source has a 50 ohm output impedance, and the plot shows a 10 volt square wave offset by 3 volts, DC. The offset is artificial and I added it just to make the plot clearer.

If you'd like to play around with the circuit, here's the LTspice circuit list:

Version 4
SHEET 1 880 680
WIRE -32 96 -96 96
WIRE 64 96 -32 96
WIRE 192 96 144 96
WIRE 288 96 192 96
WIRE 336 96 288 96
WIRE -32 128 -32 96
WIRE 288 128 288 96
WIRE 192 192 192 96
WIRE -32 240 -32 208
WIRE 288 240 288 208
WIRE -32 368 -32 320
WIRE 192 368 192 256
WIRE 192 368 -32 368
WIRE 288 368 288 320
WIRE 288 368 192 368
WIRE -32 416 -32 368
FLAG -32 416 0
FLAG 336 96 Vout
FLAG -96 96 Vin
SYMBOL ind 272 112 R0
SYMATTR InstName L1
SYMATTR Value 42µ
SYMBOL cap 176 192 R0
SYMATTR InstName C1
SYMATTR Value 1n
SYMBOL res 160 80 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 1000
SYMBOL voltage -32 224 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(0 10 1u 100n 100n 25u 50u)
SYMBOL res 272 224 R0
SYMATTR InstName R2
SYMATTR Value .26
SYMBOL res -48 112 R0
SYMATTR InstName R3
SYMATTR Value 50
TEXT -18 392 Left 2 !.tran 100u

• @VigneshVicky: I don't see why not, but I'm not really interested enough to do the legwork required to find out. That's your job. – EM Fields Oct 8 '15 at 12:51
• No. Take a look at your horizontal sweep speed control and you'll see that it's calibrated in terms of time. If you're triggering the horizontal sweep, which you appear to be doing, then speeding up the sweep means decreasing the sweep time so you get fewer microseconds per box displayed; in effect, spreading out the displayed signal so you can examine it more closely. – EM Fields Oct 8 '15 at 13:36