Resonant LC circuit, half cycle modulation?

[Megamox]

Hi Everyone!

Maybe a silly question but here goes. A parallel LC tank which is connected to the power supply and then released shows an oscillation which gradually decays. As the oscillations decay, surprisingly the frequency of the waveform stays the same. Just like a tuning fork when tapped, plays the same note, just the sound gets weaker and weaker. Or a swinging pendulum whose swings get shorter and shorter but keeps the same period.

So anyway, if an LC circuit which is losing energy is still oscillating at its resonant frequency, then to me this implies that an LC circuit being given energy to keep it going should exhibit the exact same frequency too. Again like a pendulum, every time the swing draws back you give it a push to keep it going. This will not make its frequency increase but the amplitude of its motion will be maintained.

My question is then (in principle) how would you frequency modulate a resonant circuit like that, if when left to its own devices will swing back and forth at its own rate. Obviously giving it more energy doesn't seem to change its frequency, so perhaps must you give it energy at different places to change the frequency?

Can pushing a pendulum at different points in its swing make its period change? Can tapping a tuning fork at different times make it vibrate at another frequency? None of these sounds correct to me.

The reason I ask is I've been trying to understand this transmitter circuit and how it modulates:
http://talkingelectronics.com/projects/Wasp/images/The-Wasp_Circuit.gif

Seems like the signal modulates Vbe which varies the point at which the transistor turns on and off (via the feedback capacitor). So this modulation scheme seems to rely on moving the WHERE the transistor is turned on and off, like me analogy of WHERE you push the pendulum. But again, can you really change the period of a pendulum, or the frequency of this oscillator circuit, by giving it energy in different places? If it works like this wouldnt the waveform by asymmetric, with one half of the wave being forced to complete faster because we're supplying energy to it. The other half of the waveform taking its usual resonant speed of time to complete.

If I can put my question in simplest terms: This transmitter circuit obviously works which means varying the point in the cycle where you energise the tank circuit does make it oscillate faster or slower. If this is true, then the only conclusion I can see is the output waveform must not be a sinewave but asymmetric. Is this true, or is something wrong with my logic? In fact, if the transistor is in control of energising the waveform only during negative cycles then the time taken for positive cycles when the transistor is off will never change at all. The transistor has no control over them. Frequency modulation is only achieved by varying the width of the negative half cycles!

Thanks in advance for any help!

MX

 

[alec_t]

There isn't any modulating source in the oscillator schematic
If the transistor is operated over a non-linear region of its characteristic curve harmonics of the fundamental frequency will be generated.
If you replaced Ct with a varactor you could frequency-modulate the oscillator easily.

 

[Megamox]

Sorry wrong schematic
http://talkingelectronics.com/projects/Wasp/images/The-Wasp_Circuit.gif

Apologies

MX

 

[JimB]

The obvious lumped capacitances are not the only ones in the circuit.

Internally within the transistor, there are capacitances between the junctions.
These capcitances contribute to the frequency at which the circuit oscillates.

Varying the voltages and currents in the transistor will change the junction capacitances and hence the oscillation frequency.

JimB

 

[Megamox]

So even when the transistor is off, it still represents a double PN junction capacitance? If the input signal modulates Vbe which modulates this capacitance, how would this affect the frequency of the waveform in the tank though? The only thing I can see modulating the frequency of the a tank circuit is a change in either the parallel L or C.

Any ideas?

Thanks,
MX

 

[JimB]

What make you think that the transistor is ever "off" ?

JimB

 

[Megamox]

The feedback capacitor feeds current back into the emitter resistor, raising and dipping its voltage slightly (small cap, small charge movement, and small emitter resistor, so small voltage wobble). With the emitter wobbling a tiny bit up and down and the base kept steady with the 1nf coupling to ground (assuming no modulating signal) then when Vbe junction drops below 0.7v (ie emitter voltage wobbled up slightly), so transistor behaves like an open circuit, ie off during positive cycles?

MX

 

[JimB]

Using the test jig which formed the basis for discussion in this thread:
https://www.electro-tech-online.com/threads/common-base-amplifiers-and-oscillators.127836/
The emitter did indeed "wobble a tiny bit up and down", but at no time did the transistor get anywhere near to being cut off, it was conducting all the time.

JimB

 

[Megamox]

Thanks for your help. Strange, I would have thought that the transistor being on all the time would drain the LC tank and kill the oscillation, especially when the coil eventually became a short, it'd sink a constant current through the transistor and it's field would never collapse which is needed for half of the oscillation. Also I'm having trouble trying to include the capacitance of the transistor in the LC tank, I cant see how this would affect the tank as it's not in parallel in it. The only thing I can think of is the 22nf AC coupling capacitor on the rails, which at AC would look like a short and put the Cbe across the tank? I've been trying to get simulations working of the oscillator but nothing's working on Multisim.

Thanks,
MX

 

[JimB]

I would have thought that the transistor being on all the time would drain the LC tank and kill the oscillation,

no, it is the transistor which maintains the oscillation.

especially when the coil eventually became a short, it'd sink a constant current through the transistor and it's field would never collapse which is needed for half of the oscillation.

The coil IS a short circuit at DC, the DC resistance is (or should be) low.
At the resonant frequency of the (parallel) tuned circuit, the impedance is high.
So for the collector current of the transistor the DC component passes through the coil with very little volt drop, the AC component of the collector current sees the high impedance of the circuit and develops a large voltage at the resonant frequency.

Some of this voltage is fed back to the emitter of the transistor by the capacitor connected from collector to emitter in order to maintain the oscillation.

Also I'm having trouble trying to include the capacitance of the transistor in the LC tank, I cant see how this would affect the tank as it's not in parallel in it. The only thing I can think of is the 22nf AC coupling capacitor on the rails, which at AC would look like a short and put the Cbe across the tank?

Consider the base-collector capacitance of the transistor, it is in series with the base decoupling capacitor and the supply decoupling capacitor, both of which are very large compared with the
base-collector capacitance, and so the effective capacitance is the collector-base capacitance.
As far as AC is concerned these three series connected capacitors are effectively in parallel with the LC circuit in the collector.

I've been trying to get simulations working of the oscillator but nothing's working on Multisim.

I am a bit of a sceptic (dinasaur?) where simulators are concerned, I cannot help you there.

JimB

 

[Megamox]

Thank's alot for your really informative reply, I think I've got a better understanding now. So one of my original questions which has been solved is:

Question: How can a signal modulate the LC oscillation, if all it does is vary Vbe?
Answer: Varying Vbe varies the PN junction capacitance in the transistor but the rest of the components of the circuit must be considered and not just the 3-4 components that make the oscillator part. This way, you can see that the varying Junction capacitance of the transistor can be put in parallel with the LC tuned circuit through the base capacitor and the supply decoupling capacitor and therefore effect modulation. I don't think I would have seen this at all.

What I was hoping to ask was about the transistor being on all the time. It sounds like the transistor must be giving energy to the tank during both cycles of the waveform, thereby keeping the ouput maintained and symmetric, is this right? If so couldnt we just get rid of the feedback capacitor and bias the BJT as a constant current sink? The tank would be happily oscillating, with a constant current source pulling current down through it as you describe? The only niggling thing is that when I learned about tank circuits, I remember seeing how resonance was maintained with bursts of energy, on, off, on, off like a square wave into a tank at resonance could produce a sinewave. However here it sounds like a constant source of energy (the transistor always being on) creates resonance too? Bit confused.

Thanks,
MX

 

[JimB]

It sounds like the transistor must be giving energy to the tank during both cycles of the waveform, thereby keeping the ouput maintained and symmetric, is this right?

Yes.

If so couldnt we just get rid of the feedback capacitor and bias the BJT as a constant current sink?

No.

The tank would be happily oscillating, with a constant current source pulling current down through it as you describe?

LC circuits dont just "happily oscillate" when a DC currrent flows through them.
Otherwise tuned amplifier circuits, as used in 99.9% of radio receivers would not amplify but oscillate.

The only niggling thing is that when I learned about tank circuits, I remember seeing how resonance was maintained with bursts of energy, on, off, on, off like a square wave into a tank at resonance could produce a sinewave. However here it sounds like a constant source of energy (the transistor always being on) creates resonance too? Bit confused.

I think that you are being confused by terminology.
Resonance is not the same as oscillation.

Resonance is a property of a tuned circuit, whether it is a series tuned circuit (the L and the C in series) or a parallel tuned circuit (where the L and C are in parallel).
At resonance the inductive reactance and the capacitive reactance are equal and of opposite phase, whereby they cancel each other out.

A parallel LC circuit just sitting on the workbench has a resonant frequency, but it is not oscillating.
If there is something nearby which is radiating a magnetic field at the resonant frequency of the LC circuit, there will be a current flowing around that LC circuit.
When the magnetic field goes away the current oscillating in the LC circuit will die away as energy is dissipated in the resistance of the coil (and the capacitor).

Oscillation is the property of an electronic circuit.
Where an amplifying device such as the transistor in this case has some feedback from the output back to the input so that the gain around the feed back loop is greater than 1 and the phase shift of the feed back is 0 degrees, then the circuit will oscillate.

JimB

 

[MrAl]

Hello guys,


There is still one small detail that has not been mentioned yet. It has to do with the resonance of an LC tank circuit that is not made with a pure L and a pure C but has resistance R also in the circuit either parasitic or intentionally added.

A pure LC circuit resonates at the frequency given by:
[LATEX]f=\frac{1}{2\pi\sqrt{LC}}[/LATEX]
and im sure we've all seen that a million times. That's the resonate frequency of an LC circuit but it must be stressed that this is only exact when there is absolutely no resistance whatsoever that is also connected to the circuit in some way. In terms of the angular frequency w we have the equivalent:

[LATEX]w=\frac{1}{\sqrt{LC}}[/LATEX]

and slightly changing the form of this equation we have:

[LATEX]w=\sqrt{\frac{1}{LC}}[/LATEX]

We will use this last equation to compare to circuits that actually contain resistance so we can see at a glance how the resistance changes the resonant frequency.

You will however find books that mention that the extra resistance does not matter for the resonant frequency and this is partly true because usually the resistance only has a small effect on the resonant frequency so it is often ignored. In the circuit under consideration here however we must pay attention to that detail because the very small change in resonant frequency due to some resistance in the circuit could have a profound effect, and that is because for this particular circuit we only need to change the resonant frequency by less than 0.1 percent (that's one tenth of one percent) in order to obtain frequency modulation. That's a very small change and it just so happens that it is so small that the resistances of the circuit have a big impact on the operation of the circuit.

To see how this works all we have to do is replace the transistor with a resistor from collector to emitter and excite the DC power supply by simply energizing the circuit. The damped oscillations (damped with no real transistor now) will occur at a frequency that is partly determined by the LC of the circuit and partly by the resistance of the transistor (we replaced) and the resistance of the 470 ohm resistor.

But before we look at that lets look at a simpler LC circuit with only one resistor in series with the inductor. This is a common scenario because real life inductors always have some series resistance even though small.

With the resistance equal to zero we have the equation shown above. That equation only contains L and C. But when that resistance is non zero, we end up with this instead:

[LATEX]w=\frac{\sqrt{4LC-C^2R^2}}{2LC}[/LATEX]

Here again we see that we are calculating the resonant frequency but this time it includes R. That means that R is partly responsible for the resonant frequency.

Changing the form of this equation we end up with something we can compare to the equation with pure L and C above:

[LATEX]w=\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}[/LATEX]

where we can see the difference is the subtraction of a term with R and L in it and that is what alters the resonant frequency away from the pure LC resonant frequency.

With the actual circuit with the transistor replaced by resistor R1 and R2 is the 470 ohm resistor we get a more complicated equation but again the resistance has to be included in the resonant frequency formula:

[LATEX]w=\frac{\sqrt{4LCR_2^2+8LCR_1R_2+4LCR_1^2-L^2}}{2LCR_2+2LCR_1}[/LATEX]

It's not immediately apparent but if both resistors are the same value R we get something we can also easily compare to the pure L and C resonant frequency:

[LATEX]w=\sqrt{\frac{1}{LC}-\frac{1}{16C^2R^2}}[/LATEX]

and here we can see that again we have a subtractive term under the radical that alters the resonant frequency and again it contains resistance (and this time the capacitance also).



So just how much does R1 (the transistor) have to change in order to achieve around one tenth of one percent change in resonant frequency?

If R1 changes from 100 to 500 ohms we can see a change in resonant frequency of 75kHz which is enough to provide frequency modulation. Are there other effects? Most likely yes, but again the required change is so small that if you breath on the circuit you can probably change the frequency by one tenth of one percent

But the most important fact to be considered is the resonant frequency of a tank circuit that also contains resistance.

 

[alec_t]

Nice explanation MrAl. +1

 

[Megamox]

Thanks for all the really informative answers! I guess for a circuit like this, you'll make the layout as compact and as tight as possible to reduce the parasitic effects. But if im not mistaken, a circuit with traces closer together represents more of a problem with it comes to cross capacitance and inductance, so you're trading less resistance and skin effect for more in circuit capacitance and inductance. If you can tune the Tank to take this into account and cancel it, you've at least got a point of reference before things like temperature begin to skew things again. This is far beyond my level though, I'm sure it's much more complicated. Maybe with some sort of negative feedback you could keep the oscillator going at one steady frequency. I think that's how PLL's do it anyway.

But I suppose the best tips for building a transmitter like this is, tight layout, close circuitry. Anything else? Position of the LC tank, 1nf, 22nf and transistor should be close as theyre all one parallel tank anyway?

Thanks,
MX