I'm learning about oscillators, but I can't seem to get the tank to oscillate. I have built a very simple circuit using the 555 timer and a LC tank (sorry for the bad circuit layout, this was my first drawing):

enter image description here

My voltage source is 5v, not 2v, I made a mistake in the drawing. According to the formula, I'm getting a 4 hertz signal at 66% duty cycle: 138ms - high, and 69ms low.

I'm trying to produce the following pattern on my oscilloscope every time I get a signal from the 555:. I do not have a specific goal for the frequency/voltage or any of the other factors, I just want to generate this pattern.

enter image description here

If I remove the inductor, the wave becomes as expected:

enter image description here

But the problem is - if I put back the inductor in the tank circuit, my binary signal becomes a flat line with randomized distortions, I can't even make out where the signal starts or ends, it looks very similar to this:

enter image description here

I thought maybe the inductor is too high of a value, so I tried to use 10uh inductor instead of 100uh, then I can see again my binary wave with slight distortions, but nothing close to oscillation.

I believe my values are totally off for the tank circuit or the frequency coming from the 555 is too high and the 100uh inductor can't work 'fast enought', but I know nothing about eletronics, so it's just a wild guess.

Any hints on where did I go off the track?

  • \$\begingroup\$ Have you considered the output impedance of the output pin of the 555? Is it too low perhaps? Hint: Ask your self what is the ideal source impedance needed to run a parallel tank circuit with minimum damping? \$\endgroup\$ – Adil Malik Jan 19 '17 at 13:35
  • \$\begingroup\$ @AdilMalik thanks! Never heard of these terms before, will have to learn about them first, thanks! \$\endgroup\$ – 0x29a Jan 19 '17 at 13:43
  • \$\begingroup\$ The way you do this in your schematic is that the 555 "dictates" what happens in the LC tank. You don't want this, you want to see the tank's behavior, not what the 555 is doing to it. As in mentioned in the answers you need to add a series resistor or a small capacitor between the 555's output and the tank. Then the 555 and the tank are more separated and you can see the tank's own behavior. \$\endgroup\$ – Bimpelrekkie Jan 19 '17 at 14:19

You need a high impedance source to drive a parallel LC tank circuit with minimal damping. Think about what this tank circuit wants to do when it oscillates, it wants to slosh the energy between the capacitor and inductor and not interact with the "outside" world.

Thus ideally, when this resonates it will appear as a high value resistor. Now if you ring this tank circuit with a source such as the 555 which will have a relatively low output impedance, it is effectively going to dominate it and thus completely squash the natural behaviour of the tank circuit. Look up quality factor of resonators, and how to maximise it for a parallel tank circuit.

As an analogy imagine a child's swing, what your circuit is doing is giving that swing a push but at the same time holding it firmly thus not letting it oscillate!

The solution is simply to have a relatively large resistor perhaps something like 10K between the output pin and the tank circuit!

EDIT: Longer but informal explanation

As you seen unfamiliar with some terms, lets makes things less technical:

Lets get something straight. Lets assume for simplicity that your 555 output pin behaves like a perfect voltage source, i.e that no matter how much current you draw in/out it will always stay at the voltage it wants to stay at. You must follow this point carefully.

Now, if you follow, you must agree that the output pin will always stay at the voltage the 555 wants it to be at. Now consider the tank circuit which is ALSO connected to the output node, what does the tank circuit want to do to the output pin? It wants to creates a nice sine wave at the frequency of the tank circuit, ie RING.

So we just made two contradictory statements: The 555 perfect voltage source wants to keep the output pin stuck at whatever it feels like (firm grip on the swing analogy), but at the same time the tank circuit wants to make that very node oscillate at its own frequency. So who wins i ask? The answer in the ideal case is always going to be the voltage source by definition. So you will get not ringing by simply connecting the tank circuit to a perfect voltage source.

So, now the solution is to increase the output resistance (impedance) of the voltage source, ie we are trying to make it less ideal (so it loosens its grip on the swing after giving it the first push!) You can do that by adding some resistance between the tank and the voltage source.

This is a very informal explanation, hopefully this encourages you to read on and possibly dig into the math!

  • \$\begingroup\$ Makes sense, thanks for the good explanation. So my circuit is holding the swing, not because of the too high of a frequency from 555 (that it was not able to do the first swing before the next out was already made), but the problem is that when the inductor draws the power the voltage drops? or that when it discharges it feeds back the power into 555? \$\endgroup\$ – 0x29a Jan 19 '17 at 13:54
  • \$\begingroup\$ Read the edit in the answer \$\endgroup\$ – Adil Malik Jan 19 '17 at 14:08
  • \$\begingroup\$ Thanks a lot, now it's much more clearer, never encountered the term Independence before ! \$\endgroup\$ – 0x29a Jan 19 '17 at 15:02

Ideally you need to feed the tank circuit from a current source and not a voltage source. The output from your 555 is a voltage source so try feeding the tank through a 1 kohm resistor (a compromise between a voltage and current source). You could even feed the tank via a small capacitor (such as 10 nF) to get voltage amplification but the Q of this circuit may be too high for what you want.

enter image description here

The inverter on the left can be your 555 output.

You could also use a transistor to connect the tank to a (the) DC supply then quickly disconnect it using the transistor. The time period that the transistor is on should be equivalent to half the time period of the resonant frequency ideally.

  • \$\begingroup\$ Thanks! So the whole problem is that the 555 timer gives me out 5v, but very little amps? or did I misunderstood the problem? The bench supply jumps to 3amps when I get a signal from 555. Can you please point me to few keywords where I could learn more about this, as this is the first time I hear about the voltage vs current source (as I though power source always provides both), I only can think of that the 555 has current limit, but then again I got the 3amps from the bench power supply. Is this the case? Thanks! \$\endgroup\$ – 0x29a Jan 19 '17 at 13:47
  • \$\begingroup\$ "try feeding the tank through a 1 kohm resistor (kind of simulates a current source)" could you please point me where I could learn more about this as atm I don't understand how the resistor makes it into a current source, I'm totally lost sorry \$\endgroup\$ – 0x29a Jan 19 '17 at 13:49
  • \$\begingroup\$ I'm not sure what key words to use other than don't drive a tank with a voltage source because all that you will see is the voltage from the voltage source. If you drive it with a current source then you will freely produce tank oscillations and, using a 1 k resistor is just something that works about half way between a voltage and current source. \$\endgroup\$ – Andy aka Jan 19 '17 at 13:54
  • \$\begingroup\$ Given that the 555 tries to drive with a DC bias (average value of the output signal) the parallel inductor imposes itself on this dc voltage and basically shorts it out so, there are multiple problems and issues. \$\endgroup\$ – Andy aka Jan 19 '17 at 14:01
  • \$\begingroup\$ Try this pdf file: mlg.eng.cam.ac.uk/mchutchon/ResonantCircuits.pdf and note section 3 about driving a parallel RLC with a current source rather than a voltage source used for driving a series RLC circuit. Don't ignore "R", it is always there unless the coil is a superconductor (don't go there). \$\endgroup\$ – Andy aka Jan 19 '17 at 14:13

The picture you are using of an LC-tank ringing down:

LC-tank ring down

comes from the web page:

Measuring the Q-factor of a resonator with the ring-down method

and in the section Measuring LC tank circuits it shows three ways to couple a function generator:

coupling methods

as well as how to connect it to an oscilloscope:

oscilloscope connections

There's a lot of other practical advice for generating ring-down patterns on that page as well.


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.