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I'm very new to oscillators but very keen to learn more so I connected this very simple circuit:

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where the voltage source is my signal generators and I have an oscilloscope connected across the capacitor.

Here is a picture:

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The formula to calculate the resonant frequency is:

$${\displaystyle f_{0}={\frac {\omega _{0}}{2\pi }}={\frac {1}{2\pi {\sqrt {LC}}}}\,.}$$

My inductor has a value of 100uH, and the capacitors is 100uF, so the formula gives the resonant frequency of around 1.6kHz. For a parallel LC circuit, this should give the highest impedance for the circuit, so therefore the voltage reading on the oscilloscope should be the highest at this frequency.

However, this is not what I'm observing. The highest voltage seems to be at around 4.7kHz:

enter image description here

It has about 440mV volt peak to peak (the signal generator is set to 2V) while around the calculated resonant frequency:

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The voltage is around 340mV peak to peak, much lower value. For higher voltages than 4.7kHz, the reading is smaller again.

So what am I misunderstanding/doing wrong? I understand that a basic LC circuit is not an ideal filter, but how is the measured resonant frequency so different from the calculated one?

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    \$\begingroup\$ You have a capable 'scope - try a much smaller-value capacitor, like 0.01uF. That should give you resonance @ 159 kHz. One more thing: feed this parallel resonant circuit from your function generator through a resistor of about 10k ohm so that the resonator has decently-high quality factor. \$\endgroup\$
    – glen_geek
    Dec 12, 2018 at 18:19
  • \$\begingroup\$ You should measure the actual values of L and C...the tolerance on these components can be extremely broad. Also, remember that neither of the these components is ideal...they both have some series resistance. \$\endgroup\$
    – Elliot Alderson
    Dec 12, 2018 at 18:25
  • \$\begingroup\$ @glen_geek Alright, I did. You were right - I put a 10nF cap and the resonance seemed to be at 160kHz. But why did this change things? Why is the smaller cap different is this case, why does the circuit follow the calculated resonance with a smaller cap not with a larger one? \$\endgroup\$
    – S. Rotos
    Dec 12, 2018 at 18:46
  • \$\begingroup\$ At 1.59 kHz, your circuit is not resonant. Its Q is too low - I'm guessing you're driving it with a function generator having a low source resistance (perhaps 600 ohms? or 50 ohms?). At this low frequency, the inductor reactance is only one ohm. The inductor's internal resistance will be considerably higher than one ohm - it is wound with very small-diameter wire with many turns. \$\endgroup\$
    – glen_geek
    Dec 12, 2018 at 18:58
  • \$\begingroup\$ Ah, I understand. So the inductor, at the low frequency essentially behaves as a resistor instead of inductor and so the circuit is not really a resonant LC? But you also mentioned the function generator resistance, what does that have to do with this? How does the function generators resistance affect Q, and how does Q affect the resonant frequency? I thought Q is the one set by resonant frequency and bandwidth. \$\endgroup\$
    – S. Rotos
    Dec 12, 2018 at 19:16

1 Answer 1

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Another consideration is that electrolytic capacitors are completely inappropriate for this application. Their capacitance changes significantly with applied voltage. They are leaky, with high ESR and very low linearity.

Ceramic capacitor is much more appropriate for this application. Film capacitors are better yet.

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