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I made a basic LC parallel circuit.

  • 2.9 µH inductance (measured with a cheap meter)
  • 560 nF capacitor in parallel
  • following f = 1/((2pi)sqrt(LC)) it should resonate at 124.89 kHz, but was manually found to resonate at 174 kHz.

From what I saw online (I could be wrong) it should have ~infinite impedance, act as a band stop filter if a load is in series (after the coil?), and a band pass filter if the load is in parallel.

It was all going well until I set up the band stop filter. I attached an LED in series at the end, it was very bright, and moving away from the resonate frequency hardly did anything. It did make a neat wave in my scope.

series scope

At resonance, 174 kHz, and 15 Vpk-pk sine wave, the red LED had 8.88 Vpk-pk. Not at resonance the best I could get was 9.8 V across the led. First question: Is this the "band stop", a drop of one volt? What about the ~infinite impedance?

For my parallel set up, the LED only turned on at resonance, which felt like a band pass. The voltage across it dropped smoothly as I altered the frequency. Not the sharp subtle rise in voltage from its series counterpart.

parallel scope 2

Parallel (174 kHz and 15 Vpk-pk wave) the LED had 4.3 V across it. That is less voltage than the band stop set up, so should my LED brighter in the band stop set up?

Last question (sorry it may seem unrelated). How can I measure the current in the coil? I could buy an expensive current probe :( I could also measure the voltage in a series resistor. Would measuring the voltage across a resistor be adequate in estimating my current? Would the current be zero if there is ~infinite impedance?

Labeled the parallel high res picture labeled pic

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  • \$\begingroup\$ I did just notice something. I'm using a 15 Vpk-pk sine wave. In series, the LED had ~10 volts. In parallel it had ~5 volts. 10+5 = 15? \$\endgroup\$
    – cj91
    Commented Aug 12, 2023 at 16:18
  • \$\begingroup\$ Schematics of how you connected the LED please. \$\endgroup\$
    – Andy aka
    Commented Aug 12, 2023 at 16:25
  • \$\begingroup\$ just added it! I think it's correct \$\endgroup\$
    – cj91
    Commented Aug 12, 2023 at 16:42
  • \$\begingroup\$ The left side schematic has the LCD shorted out. Even the signal source is shorted. right side schematic has the LC shorted out. It has no function. But at least the LED should be getting current. \$\endgroup\$
    – user338146
    Commented Aug 12, 2023 at 16:44
  • \$\begingroup\$ the left side schematic had the 4.3 volt drop across the LED, if I'm showing everything correctly \$\endgroup\$
    – cj91
    Commented Aug 12, 2023 at 16:50

2 Answers 2

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In circuit shown in the schematic below, during about one half of a cycle, the LED is not conducting, because it is reverse biased, or insufficiently forward biased. As a result, much of the supplied voltage will be across the LED rather than across the parallel LC tank circuit. This in turn means that the voltage supplied to the LC tank circuit is not a pure sine wave, but a composite of many different frequency components. Even if the LC circuit were ideal, it would have "infinite" impedance only at the resonant frequency. All other frequencies of voltage would cause current to flow, to one degree or another. The resulting current would light up the LED.

schematic

simulate this circuit – Schematic created using CircuitLab

Here is a simulation:

enter image description here enter image description here

In the circuit shown in the left hand schematic, one terminal of the LED is connected to one terminal of the AC supply, and the other LED terminal is connected to the other terminal of the AC supply. As a result, the voltage across the LED will be exactly the voltage across the AC supply.

The LED will light while it is forward biased ... until it burns out. The left hand schematic lacks any kind of current limiting. LEDs don't like that.


Both circuits shown as schematics have the LC circuit shorted out. The L and C will have no effect on the circuit.

In the left hand circuit, the LED is also shorted out. The result is that the LED will never light.

In the right hand circuit, the LED is in series with the AC voltage supply, and not shorted out. Therefore, it will light ... until it burns out. You have no current limiting resistor shown in the schematic, and unless the LED is really an LED module, with its own internal current limiting, it will burn out if there is no current limiting.

the left side schematic had the 4.3 volt drop across the LED, if I'm showing everything correctly

You are not showing it correctly, if there is a voltage drop. The LED in the schematic is in parallel with a wire, i.e. it is shorted out.


What about the ~infinite impedance?

An ideal parallel LC circuit will have infinite impedance at exactly the resonant frequency. However, no real LC circuit is ideal. They all have parasitic impedances, the most important in the case of most LC circuits is the resistance of the inductor and the effective series resistance of the capacitor. It may seem paradoxical, but the added resistance in the LC circuit makes the impedance of that circuit at resonant frequency less than infinite.


Is there a set up where I can visualize, in my scope, band stop behavior?

Yes. Remove the LED and add a resistor like this:

schematic

simulate this circuit

Attach your probe to Vout, and your ground to ground. If your input frequency is far from the resonant frequency, you should see a sine wave. If you adjust the frequency to the resonant frequency, the amplitude of the sine wave should become smaller.

Here is a Bode plot of the above circuit.

enter image description here enter image description here

Note, the width of the "notch" will depend upon resistances within the LC tank circuit.

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  • \$\begingroup\$ thanks for your input and helping fix the schematic. Just did my same tests with a pure resistor. in parallel 5 volts across. in series, 15 volts across. Is there a set up where I can visualize, in my scope, band stop behavior? \$\endgroup\$
    – cj91
    Commented Aug 12, 2023 at 17:43
  • \$\begingroup\$ @Krits I have added a schematic which you can use tovisualize the band stop behavior with your scope. Attach your probe to Vout, and your ground to ground. If your input frequency is far from the resonant frequency, you should see a sign wave. If you adjust the frequency to the resonant frequency, the amplitude of the sign wave should become very small. \$\endgroup\$ Commented Aug 12, 2023 at 18:04
  • \$\begingroup\$ appreciate it. Your simulation does show what I expected. My set up cant seem to replicate it. I always get ~ 15 volts across my resistor. Maybe my set up is not ideal enough but I thought I would at least see the behavior a little. In the parallel set up, it is very clear. I would add a picture to show you but i cant :| \$\endgroup\$
    – cj91
    Commented Aug 12, 2023 at 18:17
  • \$\begingroup\$ what value resistor are you using? \$\endgroup\$ Commented Aug 12, 2023 at 18:18
  • \$\begingroup\$ 47 kΩ ±5%. Should I try a 100 ohm ? \$\endgroup\$
    – cj91
    Commented Aug 12, 2023 at 18:20
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Try arranging your schematics to be like this: -

enter image description here

But, don't overdrive the circuits i.e. put a 1 kΩ resistor in series with the signal source.

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