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I’m trying to reproduce a simple oscillator conductor / inductor circuit as shown here. enter image description here

I replicated exactly the same scheme on a breadboard and check all the connections using a led enter image description here

However when I try to get a measure of voltage it seems to stay constant even though according to the simulation on here (with exactly the same parameters) I’m supposed to have an oscillation.

Link to simulation

Here’s my wiring where I connected my osciometer

enter image description here

What am I doing wrong and how can I have voltage oscillations back and forth like in the simulation?

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    \$\begingroup\$ The simulation is reasonably accurate. Notice that the oscillation comes to a rest after only a few cycles after power is applied. Your circuit will do the same. For continuos oscillation you need some kind of negative feedback, which can be as simple as switching power on and off at regular intervals. \$\endgroup\$
    – JimmyB
    Jan 18, 2019 at 14:29
  • \$\begingroup\$ Do you know how I could increase its longevity? Like installing a resistor with less resistance or maybe adding something into this circuit, apart from the switch? \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 14:32
  • \$\begingroup\$ See e.g. en.wikipedia.org/wiki/Electronic_oscillator, en.wikipedia.org/wiki/Clapp_oscillator, en.wikipedia.org/wiki/Colpitts_oscillator \$\endgroup\$
    – JimmyB
    Jan 18, 2019 at 14:39
  • \$\begingroup\$ The resistor damps the oscillations (resistor dissipates power), therefore the oscillations will decay. Short circuit the resistor and what remains is the resistance of the source, so the oscillations will last a little longer. \$\endgroup\$
    – Chu
    Jan 18, 2019 at 14:59
  • \$\begingroup\$ @Chu how to short circuit the resistor? \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 15:37

4 Answers 4

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A couple of points: that's not an Arduino board, it's a breadboard (although it may have come with your Arduino which is the PCB with the microcontroller on it).

You missed some details in your simulation.

enter image description here

Figure 1. Screengrab from the OP's simulation.

The initial oscillation is caused by the simulator starting up. In real life it will be caused by connecting the battery.

  1. Note that the oscillation is decaying in very few cycles. On the second cycle it's already below half the initial amplitude.
  2. Note the time. This is just 60 ms into your simulation.

The result is that the oscillation has decayed before you even have time to read it.

You'll need to add some active electronics to make it a self-sustaining oscillator.

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  • \$\begingroup\$ Thank you! Do you know what I could add to sustain the oscillation for longer? \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 14:34
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    \$\begingroup\$ @DmitryParanyushkin You need a transistor at the minimum (though I guess the rarely-seen tunnel diode or gunn diode could work too). Look up the terms "colpitts oscillator" and "hartley oscillator" for some simple one-transistor designs. \$\endgroup\$
    – Hearth
    Jan 18, 2019 at 14:36
  • \$\begingroup\$ @Hearth and if I don't want to use transistors is there any way to do it? \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 14:45
  • \$\begingroup\$ @DmitryParanyushkin You could use a vacuum tube or a tunnel diode..? You need some form of amplification for oscillation to be possible. \$\endgroup\$
    – Hearth
    Jan 18, 2019 at 14:46
  • \$\begingroup\$ Note that it is hard to get series oscillations using LC parts with these values since R as the resonant reactance is < 1 Ohm. So until the OP gets familiar with RLC practical values, he has an over-damped circuit with a little glitch. \$\endgroup\$
    – Tony Stewart EE75
    Jan 18, 2019 at 17:37
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The RLC circuit continously oscilates after applying pulsed voltages, not DC. The simulation runs because of power supply first shot.

Try with squarewave voltage power source.

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  • \$\begingroup\$ Yes, but then in this case i'm just going to see that AC voltage oscillating, no? \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 14:33
  • \$\begingroup\$ @DmitryParanyushkin That's right. The 'trick' is that in actual oscillators the AC voltage is generated and amplified from the oscillator's output. What you have now is only the frequency-determining part of an oscillator, which, when given the appropriate feedback, will result in an oscillator of the desired frequency. \$\endgroup\$
    – JimmyB
    Jan 18, 2019 at 14:37
  • \$\begingroup\$ Not exactly, the RLC will modify the waveform. But yes, if you put your scope on power supply, you will see the generator voltage only. If you want to see some shift phase, measure the capacitor voltage etc. \$\endgroup\$
    – user208862
    Jan 18, 2019 at 14:46
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The oscillation decay time is OVERDAMPED so there is none if you had R=10, L=100uH, C= 100uF so the voltage rise time or current decay time T=RC here 10*100uf= 1ms with voltage at 64% of target or steady state. enter image description here

I added realistic ESR of the cap , choke and voltage source which must always be considered in any analog design when considering Q values for Q=1/ζ with damping factor ζ.

Graphical Solution using RLC Nomograph expects Zo=1 Ohm and with series R=10 to have Q=0.1overdamped and RC= 1ms pulse decay time.

enter image description here Now if you short the 10 Ohms what decay you get of the oscillation envelope is your ESR total in the loop. ( we call it DCR for chokes)

Read the graph see the resonant frequency from the intersection of 3 variables to get the 4th (resonant frequency) You can choose any 3 of 4 variables or just 3 as in RC,f-3dB.

Or add your expected ESR's and simulate

But be aware that short circuiting the electrolytic or e-cap can result in oscillations that momentarily reverse the voltage more than it's 10% tolerance perhaps to short of time to cause permanent damage if done only a few times.

e.g. with ESR values reduced.

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Without an active device such as a BJT or FET acting as an amplifier with feedback the oscillations will die down after a few cycles, as in your simulation, and you will be unable to see them with a scope. You need to construct a suitable oscillator circuit using a Hartley or Colpitts oscillator.

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  • \$\begingroup\$ I just want a very simple construction, with the most basic components, so I could explain them through simple physical metaphors... \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 14:35
  • \$\begingroup\$ Won't work! You need an oscillator. \$\endgroup\$
    – Leon Heller
    Jan 18, 2019 at 14:49
  • \$\begingroup\$ But I want to create one! \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 15:35
  • \$\begingroup\$ @DmitryParanyushkin And I want to create an infinite energy device, but neither of us are going to manage to, because those pesky laws of physics keep getting in the way. \$\endgroup\$
    – Hearth
    Jan 18, 2019 at 15:52
  • \$\begingroup\$ Of course but I mean this above one works just for a short time. So I want to extend its lifetime to a few seconds. \$\endgroup\$
    – Aerodynamika
    Jan 18, 2019 at 15:53

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