How does a tank circuit work with a dc power supply? I understand that if you charge a capacitor that's in parallel with an inductor and then remove the power supply then the capacitor and inductor will exchange energy back and fourth and then slowly die out. But how does the oscillation begin with a constant DC power supply?


[ed: probable source: LC Oscillator Tutorial and Tuned LC Oscillator Basics | Electronics Tutorials ]

Can someone please examine to me how exactly this circuit works? Example, I think what's happening here is, the top plate of the capacitor has a positive charge and bottom negative, and when fully charged no current flows on the capacitor branch; with current flowing through the inductor L a electromagnetic field is building up, now with a field building up it would induce a current in L2 effectively turning on the transistor allowing the current from the inductor branch to flow through the transistor via collector-emmiter. Now with L fully charged the field is no longer changing and No current is induced in L2 so the transistor turns off. NOW, with the transistor off the current can still flow through L to the output line with the sine wave so L's field never changes and the capacitor never discharges because of the constant DC, now the only way I see it is to add a capacitor on the output line with the sine wave to block the DC and allowing the oscillations to occur. But in the diagram no capacitor has been added, so can someone explain if my explanation is right and Im missing something or give a thorough explanation like mine, of what's going on here.

  • \$\begingroup\$ Oh, that's not simply an LC tank. It applies feedback to the transistor/amp's input via the transformer (magnetic coupling), which produces a 180-degree phase shift... which maintains the oscillation. I honestly had prepared a pretty answer with an LTspice simulation for your initial question, but I not gonna post it as answer anymore, since what you really want[ed] is something else. If you are somewhat interested in the question that you textually asked intially see imgur.com/IDfDUzP \$\endgroup\$ Commented Jan 11, 2015 at 23:47
  • \$\begingroup\$ Okay but how is the DC supply removed to allow oscillation to occur, or how does oscillation start. \$\endgroup\$
    – NanyBany96
    Commented Jan 12, 2015 at 0:03
  • \$\begingroup\$ Well, the power supply does have to come ON at one point doesn't it? That's exactly what sets in motion the oscillation. The point from my simulation image is that there's no absolute DC in the real world... for all time. You do have a turn-on pulse, which you can see can set off oscillations. In your circuit, all you have to do is maintain them thereafter. This is where the transistor and feedback loop comes in. \$\endgroup\$ Commented Jan 12, 2015 at 0:05
  • \$\begingroup\$ The point of my little simulation is that while the typical introductory DC experiment is shown by removing the power supply (i.e. by stepping the voltage from E to 0), the phenomenon is actually symmetrical: you also initiate oscillations when you plug [back or just initially] the power supply (step from 0 to E). \$\endgroup\$ Commented Jan 12, 2015 at 0:16
  • \$\begingroup\$ So I would have to manually remove the power supply in the circuit above to set oscillation into motion? There's no way that the DC power supply can remain connected for this oscillator to work? \$\endgroup\$
    – NanyBany96
    Commented Jan 12, 2015 at 0:21

3 Answers 3


Providing positive feedback increases the gain of amplifier. Even noises in the atmosphere will be significantly amplified and fed back.

The circuit given by OP is amplifier + positive feedback. But the gain and phase shift provided by the feedback network depends on the frequency of the signal.

At some frequency, the loop gain becomes unity and phase shift around the loop becomes 360 deg and the circuits starts oscillating (see Barkhausen Criteria).

Conclusion: The input comes from noise. DC supply is for biasing the amplifier. The value of L and C decides the feedback factor and hence the frequency of operation.


I scanned through the video and found no mention of DC power supply. In fact he mentions the driving source being a frequency(signal) generator. Furthermore the schematic he shows shows an AC input signal.

This would explain your confusion. This video is not trying to show a circuit that generates an oscillation. Instead it shows the response to an input oscillation. Another way of looking at it is he is filtering the AC input using the LC circuit and showing how the output changes as he changes the input frequency.

If you are interested in knowing how to generate an AC signal from a DC source, then you should look at videos concerning electronic oscillators. They usually involve some 'active' component equivalent to "removing the power supply".

  • \$\begingroup\$ holy cow the original question changed dramatically. This response is likely void. \$\endgroup\$
    – lm317
    Commented Jan 20, 2015 at 14:33

HOW THE TANK CIRCUIT WORKS An "in-depth" description of how the Tank Circuit delivers the energy from the capacitor to the coil (inductor) and then back to the capacitor. The "secret" of its operation has never been described before and all discussions have glossed-over "how and when and why" the capacitor gets fully discharged before the cycle starts again.

Suppose the capacitor is charged and is placed across the inductor. Current will flow into the inductor and produce magnetic lines of force in the core that will cut all the other turns and produce a voltage in these turns that is opposite to the incoming voltage. This means the incoming voltage will see a voltage produced by the inductor that will be as high as 99% of the incoming voltage. This means the incoming voltage will appear as a very small voltage and it will increase the flux lines very slowly. The capacitor will keep supplying current but since the voltage across it is reducing, the current will be reducing and thus the flux will be expanding at a reduced rate. The back voltage produced by the expanding flux depends on the rate of expansion and since this expansion is getting less, the back voltage is reducing. The amazing thing is this: as the voltage of the capacitor decreases, the back voltage decreases and the current increases. I can explain it this way. Suppose you put a 9v battery across the coil, after a short time the flux will be a maximum but it will not be expanding flux and inductor will produce the maximum flux and take the maximum current. When the capacitor is almost fully discharged, the current will be a maximum and because the flux is not expanding, there will be no back voltage. So a point comes when the capacitor has no voltage across it and the inductor produces no voltage. This is the secret to how the oscillator works. Because the inductor has a very small resistance, it only takes a very small voltage to deliver a very high current and produce a very large amount of magnetic flux. But eventually this small voltage cannot maintain the flux and all the voltage and current-capability is taken from the capacitor. At this point in the cycle, the flux cannot be maintained and it starts to collapse. As it collapses, it can only produce a certain amount of current and this current charges the capacitor. In other words the capacitor controls the rate of collapse of the inductor and the voltage across the capacitor gradually increases. In actual fact, the inductor "can and will" produce a very large voltage during a collapse if nothing is connected to it and this is called a fly-back voltage. But since a capacitor is connected, the voltage can only rise as the capacitor allows it to rise. So it rises until the flux has almost fully collapsed and even at this point the collapsing flux is able to produce a voltage much higher than the voltage across the capacitor and that's why it can keep charging the capacitor right up to the point when the flux has almost completely collapsed. That's why the capacitor gets charged to almost the original voltage. Even the tiniest amount of flux will produce a charging voltage. But eventually the flux is zero and the voltage across the capacitor sees the inductor as a very small resistance and it starts to deliver a current. This current produces magnetic flux in all the turns of the winding and each turn produces a back voltage so that the actual magnetizing voltage is very small and thus only a very small current flows to create the second cycle.

THE SECRET Here's the reason why the capacitor is able to deliver all its energy to the coil: As the voltage across the capacitor decreases, the coil can only produce a back voltage that is slightly less than the capacitor voltage. That's why the energy keeps flowing from the capacitor to the inductor. It is only when the capacitor cannot deliver any more current, that the circuit starts to change direction. Just before this occurs, the voltage of the capacitor can be very small because the resistance of the inductor comes into play since the back-voltage is very small and it is the back-voltage that turns the resistance of the coil into an inductance. Now we have a very small capacitance voltage being able to deliver a high current into a small resistance to maintain the magnetic field. Only when this voltage finally reduces to almost zero, does the circuit start to change direction. Now, going back in the other direction, why is the inductor able to keep charging the capacitor when it is nearly out of magnetic flux? The reason is this. If the capacitor was not connected, the inductor would be able to produce a very high voltage when the magnetic field is collapsing because the size of the back-voltage depends on the speed of the collapsing field. Even when the inductor is almost out of flux, it can produce a very high voltage when nothing is connected to it. That is: when no capacitor is connected, it will collapse very fast and produce a very high voltage. So, it is the capacitor that is controlling this voltage, BUT it is always slightly higher than the voltage across the capacitor so the charging keeps occurring until the inductor is finally out of flux. Don't forget, when the magnetic field of the inductor is collapsing, the voltage it is producing is in the opposite direction to the original voltage. This means the capacitor gets charged in the opposite direction. In the diagram above, the top rail is the supply rail and the bottom rail is connected to a transistor. If we connect a multimeter or digital CRO to the transistor, we will see the voltage reduce lower than rail voltage during half the cycle and then become higher than rail voltage during the second half of the cycle. This means the effective voltage at this point is TWICE RAIL VOLTAGE. The Tank Circuit can double the supply voltage !!

  • \$\begingroup\$ Also posted as a comment at the bottom of this post and apparently somewhere on this site as well. \$\endgroup\$ Commented Apr 27 at 15:27

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