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?


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\$ – Fizz Jan 11 '15 at 23:47
  • \$\begingroup\$ Okay but how is the DC supply removed to allow oscillation to occur, or how does oscillation start. \$\endgroup\$ – NanyBany96 Jan 12 '15 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\$ – Fizz Jan 12 '15 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\$ – Fizz Jan 12 '15 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 Jan 12 '15 at 0:21

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.

| improve this answer | |

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".

| improve this answer | |
  • \$\begingroup\$ holy cow the original question changed dramatically. This response is likely void. \$\endgroup\$ – lm317 Jan 20 '15 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.