# tuned (tank) circuit, oscillator, Q

#### Ratch

Mar 10, 2013
1,098
Colin to answer your question, I do not really have a reason, I'm just curious about it. Maybe one reason would be that it would be (a little) harder (for me) to match them.
Is it bad if my capacitor is 100 times larger than it needs to be (ignoring losses), when we're talking about a part that costs less than a cent...
Or is there another reason to match the parts; that's maybe where i'm getting at.

To be fair, too, I'm not actually a total beginner and I do understand inductors and capacitors fairly well I would say.
I do see too what Ratch is saying about circulating currents.
I feel that we got a little hung up on this technicality, but I might as well add my two cents:

The current does not technically oscillate, of course the energy does.
Inductor: Likes to keep current flowing through it constant at all cause and will happily inject a voltage to keep current going.
Capacitor: Likes to keep the voltage constant at all cause and will happily inject a current to keep the voltage the same.

This btw. is what I meant with "magic". The mirror like features between capacitor/electric field and inductor/magnetic field. It's like ying and yang right?

During the oscillation, all the energy exists in the form of an electric field in the capacitor (voltage). The voltage is driving a current through the inductor. The capacitor wants to keep the voltage up and is injecting more current, which flows through the inductor. Eventually the capacitor is depleted and there is a point when there is no voltage on the circuit and now all the energy is stored in the form of the magnetic field in the inductor (current). This must be the point where voltage polarity switches. Now that the capacitor can no longer push any more current, the inductor is creating a voltage, trying to keep the current going (while eating up it's magnetic field), and this in turn charges the capacitor again.
So yes, energy definitely goes back and forth, switching between voltage and current; and yes, there is always some resistance and other losses (including the radiation of energy) which dampens this oscillation; and of course in any practical example we also need some type of driver to keep topping off the circuit (just like a swing set) to keep things going.

One thing I would add. The energy, voltage, and current all oscillate. I will modify what I said about mismatching the L and C. As I said before, as long as the product of L and C are the same, the resonant frequency will be the same. If you increase L and lower C, the impedance of the tank circuit will increase and the circulating current will be less. The opposite will be true is L is made smaller and C increased. Therefore you can adjust the circulating oscillating current by fiddling with the L and C ratios as long as you keep the product the same.

Ratch

#### duke37

Jan 9, 2011
5,364
A valve transmitter pi match circuit could have a loaded Q of 12 but the coil and hence the circuit should have an unloaded Q of over 100 to keep losses down. No tuned circuit worthy of the name will have a Q of 2.

At any frequency there will be an optimum L:/C ratio to minimise losses, depending on the construction of the components. An air cored coil can get a Q of over 100 on radio frequencies but will need a ferrite core for audio frequencies. Radiation losses can be minimised with approriate screening and a helical filter or coaxial filter (silver plated) can provide a Q of several hundred.

I am reminded of the optimum voltage and current in electrical machines. I was told that 1Ω motors were about the best. So, a small motor would use 1A and 1V whereas a bigger one would use 1000V and 1000A. Of course, deviation from this is possible and a car starter motor may use 10V and take 300A, efficiency in this case is not too critical since it only runs for a short time. I have not looked at the optimum reactance for tuned circuits but suspect that it will not be too far from 1k.

The optimum place in the cycle for adding energy was investigated hundreds of years ago by the clock makers, hence the variation in escapements.

#### Merlin3189

Aug 4, 2011
250
There is nothing "magical" about them. They all follow the principles of physics.
This is obviously true, but we are getting into areas where there are lots of competing issues and it becomes a bit of an art rather than a precise science. The physics tends to be based on ideal components. We can take account of the non-ideal characteristics of real components, but it gets ever more complex. So experience and awareness of good practice is often a more useful guide to a realistic starting point, than taking an ideal circuit then trying to refine it by taking account of all the non-ideal bits.

I've no idea what people are on about with balancing the energy content of L and C. I agree with Ratch, current oscillates around the tank. At some point in each cycle there is no energy in the C, it's all in L, and 90deg later it's all in C and none in L. So they each carry the same amount of peak energy.

Anyhow, back to the original quest. Yes there are an infinite number of pairs of values for L and C for any particular freq.(that's physics.) So why choose one rather than another? Say you choose a very large L (maybe to get a high Q) then you need a very small C. Then you have to worry a lot more about the small stray and parasitic capacitances, particularly varying ones, like someones hand touching the control knob, or the variation of C with temperature (in the capacitor itself, the inductor and in the transistor.) So pick a large C and small L, but then you get a lower Q. As Q lowers there is less attenuation of noise and harmonics -not a lot, but you really want these to be as small as possible if you are going to amplify this signal a million times in a transmitter, or mix it with a microvolt level received signal. With the large C you don't have to worry so much about stray capacitances, but your big capacitor itself needs to be very stable. If you want a variable capacitor, there is not so much choice and physical size goes up with capacitance. So even if the physics is clear, the choices here become much more murkey.

That's why I was asking about the Q to see if it may have something to do with it. Also interestingly I saw RF energy at 40 MHz as well as harmonics at 80, 120, and 160.
Not at all surprising. In an oscillator the amplitude increases by +ve feedback until something limits it. If it is unloaded, the limit will be hitting the saturation and/or the cut-off of the transistor. This squashes the nice sine wave a bit, generating harmonics (that's the 120, 200, 280..) and making it assymmetric (that's the 80, 160, 240...) Yes, higher Q helps discriminate against these and lower gain in the amplifier also reduces their generation.

Good luck and have fun with your experiments. People are so accustomed to radio devices these days that most just take it for granted. I was lucky enough to grow up when the only radio device in the house was a battery powered valve radio (which taught me a lot about electricity before I ever met it at school) and got me interested in this fascinating and "magical" bit of physics.

#### LvW

Apr 12, 2014
604
I imagine if one were to use this to build an oscillator, one would want to have a very high Q as it would restrict the natural frequency as opposed to a low q (and high bandwidth), which would let the oscillator move around in frequency by much more.
So for the oscillator high Q means solid frequency, is that true?
.
I do not intend to enter the discussion regarding the current (circulating yes/no).
However, a short answer to the Q value of the LC tank.
Yes - in general, we prefer a high Q value for the following reason:
The phase slope of the tank is proportional to the Q value - and we prefer a high phase slope because , in this case, a phase error of the amplifier circuit (to be compensated by the phase of LC tank in order to ensure a loop phase of zero deg) will result in a small oscillation frequency error only. This is the main reason for a high Q value.

#### Ratch

Mar 10, 2013
1,098
To All:

All I am saying is you can adjust the value of the oscillating circulating current of the tank circuit by using different L vs C ratios as long as the correct LC product is maintained. I can elaborate on this if anybody does not understand some aspect of what I said. In most cases, you want to minimize the resistance so as to keep the Q as high as possible. Hopefully that restriction will not get in the way of the L or C selection. It doesn't make any sense to install a large capacitor in order have a relatively large circulating current if such a current is not needed.

Ratch

#### Ratch

Mar 10, 2013
1,098
This is obviously true, but we are getting into areas where there are lots of competing issues and it becomes a bit of an art rather than a precise science. The physics tends to be based on ideal components. We can take account of the non-ideal characteristics of real components, but it gets ever more complex. So experience and awareness of good practice is often a more useful guide to a realistic starting point, than taking an ideal circuit then trying to refine it by taking account of all the non-ideal bits.

True, but we have to start from somewhere.

I've no idea what people are on about with balancing the energy content of L and C. I agree with Ratch, current oscillates around the tank. At some point in each cycle there is no energy in the C, it's all in L, and 90deg later it's all in C and none in L. So they each carry the same amount of peak energy.

I am afraid I propagated that concept. You can make the L and C ratios anything you like as long as it doesn't get ridiculous, like the distributed capacitance of the inductor coils approaching the value of C. What I should have said is that the L vs C ratios determine the value of the oscillating circulating current of the tank circuit. By the way, the switch of energy between the L and C components occurs at 180°, not 90°.

Anyhow, back to the original quest. Yes there are an infinite number of pairs of values for L and C for any particular freq.(that's physics.) So why choose one rather than another?

As stated above, in order to adjust the circulating current of the tank circuit.

Say you choose a very large L (maybe to get a high Q) then you need a very small C. Then you have to worry a lot more about the small stray and parasitic capacitances, particularly varying ones, like someones hand touching the control knob, or the variation of C with temperature (in the capacitor itself, the inductor and in the transistor.) So pick a large C and small L, but then you get a lower Q. As Q lowers there is less attenuation of noise and harmonics -not a lot, but you really want these to be as small as possible if you are going to amplify this signal a million times in a transmitter, or mix it with a microvolt level received signal. With the large C you don't have to worry so much about stray capacitances, but your big capacitor itself needs to be very stable. If you want a variable capacitor, there is not so much choice and physical size goes up with capacitance. So even if the physics is clear, the choices here become much more murkey.Not at all surprising. In an oscillator the amplitude increases by +ve feedback until something limits it. If it is unloaded, the limit will be hitting the saturation and/or the cut-off of the transistor. This squashes the nice sine wave a bit, generating harmonics (that's the 120, 200, 280..) and making it assymmetric (that's the 80, 160, 240...) Yes, higher Q helps discriminate against these and lower gain in the amplifier also reduces their generation.

Good luck and have fun with your experiments. People are so accustomed to radio devices these days that most just take it for granted. I was lucky enough to grow up when the only radio device in the house was a battery powered valve radio (which taught me a lot about electricity before I ever met it at school) and got me interested in this fascinating and "magical" bit of physics.

#### Merlin3189

Aug 4, 2011
250
... By the way, the switch of energy between the L and C components occurs at 180°, not 90°.
Probably not worth reopening this, but I still think the switch is every 90deg.
Twice per cycle the current is 0 and twice per cycle the voltage across the cap is 0.
Since the voltage is 90deg shifted from the current, that gives 4 energy switches per cycle..

So say we start with a fully charged capacitor at 0 deg in the cycle. Then
at 0 deg, Energy is in the capacitor and current is 0 with no energy in the inductor,
at 90 deg the capacitor is discharged, current is max and the energy is in the inductor,
at 180 deg the cap is fully charged in the opposite polarity, current is 0 and again no energy in L,
at 270 deg C is again discharged, I is max in the opposite sense with energy in L,
and at 360 deg we are back to the start.

#### Ratch

Mar 10, 2013
1,098
Probably not worth reopening this, but I still think the switch is every 90deg.
Twice per cycle the current is 0 and twice per cycle the voltage across the cap is 0.
Since the voltage is 90deg shifted from the current, that gives 4 energy switches per cycle..

So say we start with a fully charged capacitor at 0 deg in the cycle. Then
at 0 deg, Energy is in the capacitor and current is 0 with no energy in the inductor,
at 90 deg the capacitor is discharged, current is max and the energy is in the inductor,
at 180 deg the cap is fully charged in the opposite polarity, current is 0 and again no energy in L,
at 270 deg C is again discharged, I is max in the opposite sense with energy in L,
and at 360 deg we are back to the start.
Yes, you are correct. I was thinking of how often the energy in L and C considered separately reach a peak or zero, which is twice per hertz. However, those points are 90° apart for L and C.

Ratch

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