# Coil resonance on oscilloscope

Hello All,

I have been playing around with this digital oscilloscope and function generator that I am using at school. I first connected the function generator directly to the oscilloscope to get the hang of adjusting it, then I took a small coil I had laying around and connected it in parallel with the function generator and scope to see what the voltage waveform over the coil looked like.

I ran through a wide range of frequencies and I noticed that some frequencies caused the voltage to rise a lot higher than the majority of other frequencies. Is this a good method to figure out what the natural resonant frequency of the coil is for use in an oscillator circuit, or would the coil itself need to be connected to a capacitor in an LC setup to find the resonant frequency?

I'm trying to build some oscillator circuits but I wanted to pick the capacitor value based on whatever the natural frequency of the coil happens to be (for now). But can I use my oscilloscope and function generator in the fashion I described above to find that natural frequency of my coils?

Thanks,
Jason O

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berkeman
Mentor
The output impedance of your signal generator is likely 50 Ohms (or maybe 300 Ohms), and that is going to limit your ability to measure the natural resonant frequency of the coil (where it resonates with its own parasitic parallel capacitance, which is usually small). If you had a high-impedance signal source, then you might be able to do it. And adding an explicit parallel capacitor will lower the resonant frequency to make it easier to see in your setup.

Try this -- come out of your signal generator, through a series 10kOhm resistor, and into your coil with an explicit capacitor connected in parallel. Write the equation for the Z of the parallel LC part, and solve for the gain from Vin of the signal generator to the Vlc across the LC part. Then drive your signal generator around near the resonant frequency that you got by maximizing the gain equation above, and see if your measurements agree with the equations.

Then calculate what you would get with only a few 10s of picofarads of parasitic capacitance of the coil itself. What happens to the gain and resonant frequency (still using the 10kOhm drive isolation resistance)?

Hi Berkeman,

Thank you for your help. I will try that setup this evening as soon as I can. This brings me to another question I've been wondering about. If I setup a simple LC circuit driven with my signal generator in series with it, how do I properly hook my scope to the circuit to monitor it without affecting the circuits operation?

Going back to your previous explanation, what is the purpose of the 10 kOhm resister connected to the signal generator? Is that used to limit the current coming out or something? Also, for oscillator circuits, if I am driving the LC circuit with my signal generator, is it better to put it in series or parallel with the setup?

Thanks again for your help, I'm learning this all from scratch

Thanks,
Jason O

berkeman
Mentor
The 10kOhm resistor is meant to isolate the low output impedance of your signal generator from the LC "tank" resonant circuit, so that the LC is more free to resonate. If you drive it directly with the 50 Ohm sig-gen, that low output impedance will spoil the Q of the tank.

Go through the calculations that I mentioned, and try different values for the source resistance. You'll see how that affects the Q.

Your oscilloscope probe most likely will look like 10MOhm to ground in parallel with about 5pF (it will say usually on the probe). That's if it's a 10x probe. If you need to make a less invasive measurement of an oscillator circuit, you can use a 100x probe, which has higher input impedance and less capacitance, or you can even do a non-contact measurement with a sensitive spectrum analyzer.

Thanks berkeman. I'll check out those tutorials. I really don't yet have a solid understanding of how reactance works in LC circuits (especially when you throw in resisters and other factors from the instruments) so these tutorials will help .

Thanks,
Jason O