It's said that an oscillator's frequency will drift with a changing voltage.

[Andy gw0jxm]

Hi guys, if this is true that an oscillators frequency changes with a changing voltage, which part of 'one over two pi root LC' is responsible for the change?
I can understand thermal drift in an L or a C but how does a change in voltage lead to a change in frequency?
Kind Regards ... Andy

Heres one, changing volts changes the gain and hence the Miller capacitance. There are more!

[Herald1360]

Because it can cause changes in gain of the amplifier used plus most capacitors vary in capacitance to some extent with applied voltage, semiconductor junction capacitances ditto even resistors can change in value with applied volts. In addition, increased volts usually means increased power dissipation in the circuit, hence temperature rise etc etc.

Literally 'one over two pi root LC' is not responsible but plug in the value changes in components characteristics and it will help to tell you what the frequency shift might be.

[Andy gw0jxm]

Originally Posted by merlinmaxwell

Heres one, changing volts changes the gain and hence the Miller capacitance. There are more!

So the implication is that an external series component influences a parallel tank circuits resonant frequency. Is that correct?
Kind Regards ... Andy

[Andy gw0jxm]

Originally Posted by Herald1360

Because it can cause changes in gain of the amplifier used plus most capacitors vary in capacitance to some extent with applied voltage, semiconductor junction capacitances ditto even resistors can change in value with applied volts. In addition, increased volts usually means increased power dissipation in the circuit, hence temperature rise etc etc.

Literally 'one over two pi root LC' is not responsible but plug in the value changes in components characteristics and it will help to tell you what the frequency shift might be.

So what you’re implying is that a parallel resonant tanks frequency is independent of voltage change but in an oscillatory circuit (with feedback) it’s the active component (valve, transistor etc) that is susceptible to Miller effect, phase shift etc?
And that's fine as that makes sense.
Kind Regards ... Andy

[Andy gw0jxm]

Originally Posted by Herald1360

Because it can cause changes in gain of the amplifier used plus most capacitors vary in capacitance to some extent with applied voltage, semiconductor junction capacitances ditto even resistors can change in value with applied volts. In addition, increased volts usually means increased power dissipation in the circuit, hence temperature rise etc etc.

Literally 'one over two pi root LC' is not responsible but plug in the value changes in components characteristics and it will help to tell you what the frequency shift might be.

Just one point. In the case of a simple FM transmitter with a parallel tuned tank circuit with no taps or feedback dose the actual tank frequency remain constant as this is where I'm managing to confuse myself. I've always taken it that a tank's f is set by 1/root 2pi LC but I've also manage to convince myself that the tank in the simple FM transmitter is fluctuating as a result of the changing voltage, say on the collector of feeding transistor. Am I right then to say that FM is merely the addition/subtraction of the two frequencies not a changing of the tank frequency. If that's right I'll be able to sleep tonight hi hi.
Kind Regards ... Andy

[Herald1360]

FM is effectively a constant amplitude carrier whose frequency is deviated from nominal by the instantaneous amplitude of the modulating signal. The transmitter tank circuit will be resonant at the centre carrier frequency but its Q will be low enough relative to the bandwidth of the signal that it won't have too significant an effect on the amplitude of the FM signal. The FM signal itself is generated at low level and simply amplified to the required output level. Since it's constant amplitude, class C is good enough. It sounds a bit like you are confusing high level amplitude modulation where the amplitude of the carrier is effectively varied by modulating the dc supply to the final stage with how an FM transmitter is modulated.

Any varying series component in line between a parallel tuned circuit and its associated driving circuit will have some effect on the resonant frequency, though it may be fairly minor.

[Andy gw0jxm]

Originally Posted by Herald1360

... Any varying series component in line between a parallel tuned circuit and its associated driving circuit will have some effect on the resonant frequency, though it may be fairly minor.

OK on the FM modulation but can you qualify the last sentence as this is the bit I can't come to terms with, that is that a series component can influence a parallel tank. I can understand that you could do something physical ether capacitively or inductively to pull the frequency but I can't otherwise see how a series comp can pull the frequency, even a little bit. I've worked with tank circuits for 30 years but all reasonably high power stuff 1 to 350 KW at 3KHz to several hundred KHz for induction heating applications were a tiny variation in frequency would have gone unnoticed.
... Andy

[julie_m]

Depends on what kind of oscillator it is. If it uses an R-C circuit for timing, quite possible. If it's L-C or crystal based, it should be stable (barring effects of voltage on parasitic capacitance of valves and semiconductors; only really likely to be an issue at RF) as long as there's enough gain.

[Andy gw0jxm]

Originally Posted by ajs_derby

Depends on what kind of oscillator it is. If it uses an R-C circuit for timing, quite possible. If it's L-C or crystal based, it should be stable (barring effects of voltage on parasitic capacitance of valves and semiconductors; only really likely to be an issue at RF) as long as there's enough gain.

Hi there, I think that I have my answer now. My question related specifically to a parallel tank circuit of an FM transmitter.
I think the fact of the matter is that the frequency of a parallel tank is essentially immune from voltage fluctuation. Any voltage induced variation in frequency in an oscillator (as opposed to a fixed tank) is associated with the other components in the circuit. I must admit I was trying to convince myself that the tank frequency was 'pulled' by the audio component where as in reality the change in frequency as seen at the antenna is the addition/subtraction of the AF and RF frequencies.
Kind Regards ... Andy

[kalee20]

Not with FM!

Consider a 100MHz signal, modulated by a pure 1kHz sine wave. The AF is 1kHz and the RF is 100MHz. If it were AM then the only frequencies present would be 99.999MHz, 100MHz, and 100.001MHz, and a perfect spectrum analyser would confirm this. However, if it were FM then the frequency at the tank could typically vary continuously from 99.925MHz to 100.075MHz and back again, at a rate of 1kHz. The total variation in frequency, which I've taken here as 75kHz, is dependent on the amplitude of the 1kHz modulating signal, not its frequency.

[G8HQP Dave]

In all oscilators the tank, whether series or parallel, forms just one part of the total circuit. In a well-designed circuit the tank will dominate the amplitude and phase response but it it will never be the only factor. For the tank to do anything you have to couple to it. As soon as you do that the circuit will load it. I don't think you really understand what you mean by 'the frequency' of the tank. For a start, the simple 1/2 pi sqrt(LC) assumes no damping but a real tank will always have losses.

Try to find and read a good textbook on oscillator theory. You will discover that it is more complicated than the naive stuff you tend to see in amateur books.

[kalee20]

Originally Posted by G8HQP Dave

You will discover that it is more complicated than the naive stuff you tend to see in amateur books.

Definitely.

[Andy gw0jxm]

Originally Posted by G8HQP Dave

...Try to find and read a good textbook on oscillator theory. You will discover that it is more complicated than the naive stuff you tend to see in amateur books.

Hi there, thank you,
Someone, who I respect, suggested that the fundamental frequency of a tank circuit, not an oscillator you understand, but a tank circuit, had a voltage dependency. To me that's like saying that a tuning fork or a bell will vary its frequency depending up on how hard you hit it. We know that if you 'over strike' a bell for too long that would damp the oscillation but would it change the frequency? I don’t think so.
I think where I went wrong was in the way I asked the question as I mentioned an 'oscillator' I should have said 'a tank', so maybe I should rephrase the question and ask ...
In isolation, can the magnitude of the exciting voltage, in anyway, influence the fundamental frequency of a simple parallel LC tank circuit given that the exciting voltage is not applied for such a period that would damp the oscillation? So in other words, if I hit a tank with a single pulse can the voltage of that pulse influence the resonate frequency, say beyond the first cycle? Personally I think the answer is no. I know if the tank is part of a loop then other components in the loop will be voltage, phase and temperature dependant and in some circuits I've built even having an R in the month seem to influence them.
As I say I think the answer to the rephrased question is 'no' but I've been around long enough to realise that a lot of what we know (or think we know) seems to be based on half truths but knowing I'd get an expert opinion (or two) from the vintage-radio net I thought I'd pose the question, it's just a pity I didn't ask the right question to start with. Sorry about that.
Kind Regards ... Andy

[julie_m]

Originally Posted by Andy gw0jxm

In isolation, can the magnitude of the exciting voltage, in anyway, influence the fundamental frequency of a simple parallel LC tank circuit given that the exciting voltage is not applied for such a period that would damp the oscillation? So in other words, if I hit a tank with a single pulse can the voltage of that pulse influence the resonate frequency, say beyond the first cycle?

It shouldn't.

Consider a swinging pendulum. Energy is constantly changing from potential to kinetic and back. At each end, when it changes direction, KE is a minimum and PE is a maximum; mid-travel, at its lowest point, KE is a maximum and PE is a minimum. Ideally all the PE lost on the way down (= m * g * h) would end up as KE (= .5 * m * v ** 2) and all of that would be converted back to PE on the way up, but friction in the pivot robs some of it every swing.

Note also that no matter how hard you push the pendulum initially, its period remains the same for as long as it continues to swing; T =2 * π * (L / g) ** .5.

Now in a tank circuit, energy is being transferred from the electric field in the capacitor (E = .5 * C * V ** 2) to the magnetic field in the coil (E = .5 * L * I ** 2) and vice versa (and instead of friction in the pivot, we have resistance in the wires). The principle is exactly the same as the pendulum, though, and the period is not influenced by the initial excitation.

[Andy gw0jxm]

[QUOTE=ajs_derby;540772]

Originally Posted by Andy gw0jxm

...The principle is exactly the same as the pendulum, though, and the period is not influenced by the initial excitation.

OK that’ how I see it but I just wanted to ask the question in case there was some subtle point I’ve been missing for the last 50 years, like ‘i’ before ‘e’ ... (except in certain circumstances) . Thanks for the input. Kind regards ... Andy

[kalee20]

Originally Posted by ajs_derby

Note also that no matter how hard you push the pendulum initially, its period remains the same for as long as it continues to swing; T =2 * π * (L / g) ** .5

I'm afraid that isn't a good example! A pendulum does change its period according to how far it is swinging - the time of a swing decreases with increasing amplitude. The formula you give is valid for small swings only.

It is possible to make the period constant (the so-called isochronous pendulum) by arranging the string to wind around a curve at the fixed end. Alternatively, pendulum clocks were designed to keep the amplitude as constant as possible, as the spring or weight ran down.

The reasons for a pendulum's swing period being amplitude dependent are really OT but basically it's because the restoring force is not proportional to the angular displacement, but to the sine of the angular displacement.

For an electronic oscillator, the resonant frequency of the tank should be truly amplitude-independent. But to the nth degree, the values of L and C are to some extent voltage-dependent (a large voltage on an air-spaced capacitor will case some electrostatic attraction between plates, for instance, which drawing them together will minutely change capacitance).

Far more significant, is the effect of the external circuit. Valves or transistors will add voltage-dependent external damping, and external capacitance. And the resonant frequency of the circuit is the circuit as a whole, including wiring capacitance and valve capacitance. Pull out the valve, and the oscillations will die away - but the time between zero crossings will be very slightly higher than with the valve in place.

the time of a swing decreases with increasing amplitude. The formula you give is valid for small swings only.

Not quite, it goes up with large swings, take the limiting case a pendulum sticking up vertically (and perfectly balanced) it would take forever to swing down. Also the 'isochronous cheeks' used shorten the pendulums length at larger swings, making it go faster.

[Andy gw0jxm]

Originally Posted by kalee20

...Far more significant, is the effect of the external circuit. Valves or transistors will add voltage-dependent external damping, and external capacitance. And the resonant frequency of the circuit is the circuit as a whole, including wiring capacitance and valve capacitance. Pull out the valve, and the oscillations will die away - but the time between zero crossings will be very slightly higher than with the valve in place.

Hi there, did you see that I rephrased the question (see #14) initially I’d confused the issue by referring to an oscillator where as I really wanted the question to relate to a tank in isolation, i.e. without feedback.
Kind Regards ... Andy

[kalee20]

Originally Posted by merlinmaxwell

Not quite, it goes up with large swings, ... Also the 'isochronous cheeks' used shorten the pendulums length at larger swings, making it go faster.

Right on both counts!