# tuned (tank) circuit, oscillator, Q

Discussion in 'General Electronics Discussion' started by pgib8, Jan 25, 2016.

1. ### pgib8

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Jul 26, 2015
Hi!
I've been wrapping my head around the beautiful and magical workings of tuned circuits and everything about inductance and capacitance.

For a tuned circuit I understand (and btw. i'm specifically referring to parallel LC), that any combination of capacitor and inductor will have one and only one natural frequency. It is where the capacitive and inductive reactances (both measured in Ohm) for both L and C become equal.

So let's talk about an oscillator. I can select any inductor and capacitor and bam I get a nice sine wave at the appropriate frequency?

Let's take a "relatively" big capacitor of 0.1uF (0.0000001 F) with a tiny inductor of 0.0062 uH (0.0000000062 H), should I get a nice oscillator at 6.4 MHz?

Somehow I'm not so sure about this and very few sources on the web mention that the inductor and capacitor shall be matched so that the respective amount of energy they can each absorb is close to equal.
For example I don't think the wikipedia article talks about this (anybody want to expand on this?) https://en.wikipedia.org/wiki/LC_circuit

This finally brings me to the Q (quality) factor. Most websites refer to this in regards to how much energy is lost in the circuit (dampening). Very few sites actually talk about it referring to the bandwidth of the circuit.
So which is it or how can it be both?
In my case, the losses are negligible/irrelevant so I'm liking Q to describe the bandwidth.

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?

Now, how do I achieve a high Q in my parallel LC oscillator?

Are we talking about 2 different kinds of Q (losses versus bandwidth)?

Is matching the energy of capacitor and inductor conducive to high Q or does that not have anything to do with it? In general, what actually happens when the energy of C and L are significantly unmatched? In which way does that affect the signal which is being generated?

As you can see I have a lot of questions and I would really like to understand this subject all the way.

2. ### Y2KEDDIE

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Sep 23, 2012
A specific C and L theoretically resonates at a specific frequency. In the real world there is always some resistance within the components. This resistance broadens the response, thus the tuned circuit resonates a little higher and lower. Resistance in a tuned circuit changes the Q. Q can be good or bad. You want high Q for selectivity and efficiency,and you want low Q to broaden the tuned circuit as in a high fidelity music receiver. High Q, high selectivity, narrow band, rejects interference. Lower Q broadens the tuning ( wider band width) allowing better fidelity,

Construction of components dictate size, L, C, and R. The more turns on a coil will add resistance . Different dielectrics of various capacitors have more leakage (resistance). different ratio's of L and C affect Q.

I hope this helps. I'm sure others can elaborate on this.

3. ### duke37

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Jan 9, 2011
The energy will be the same in the capacitor and inductor, the energy passes from one to the other twice per cycle at the resonant frequency.

Your example of a large capacitor and small inductor will not work as you expect, the capacitor will have sufficient inductance to affect the frequency. Better to make the capacitor resonant frequency well above the working frequency.

The Q is defined as the energy lost per cycle, so a Q of 100 will imply 1% of energy lost per cycle. For an oscillator you will need a Q as high as possible. The loss in the inductor will be much higher than in a good capacitor. The oscillator puts a load across the tuned circuit to drive the transistor and the drive will also add a load. To get oscillation, the power put in by the transistor must exeed the power lost.

In the case of filters, the Q values can be chosen to get the correct filter response. High Q will give a narrow filter such as needed for reception and low Q would be needed for wide band FM reception.

4. ### Ratch

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335
Mar 10, 2013

There is nothing "magical" about them. They all follow the principles of physics.

For a series resonant circuit, there is only one resonant frequency. For a parallel resonant circuit, some textbooks and folks define three frequencies. The first is where the capacitance and inductive reactance are equal. The second is where the impedance across the parallel circuit is at a maximum. The third is where the phase shift is zero across the parallel circuit. If the resistance is small in each branch or alternately, if the Q is high, then the three frequencies are quite close to the same value. I personally don't think definitions two and three are valid definitions of resonance.

Provided you drive it at its resonant frequency.

Why not? Your calculations are correct. However, why \$pend on a large coil or capacitor that can absorb many times the energy of its opposite component. The frequency won't change, but the L or C will be underutilized.

They are two different things. The energy refers to the definition of "Q". The bandwidth refers to effect of Q.

Definition and effect, they are two different things.

High Q or low Q, once the resonant frequency is established, it will not change or "move around".

What is a "solid" frequency? A single frequency? It is possible in a parallel circuit to adjust the L,C and the two resistances in each branch to be resonant at all frequencies.

Keep the resistance to a minimum.

One is a definition, the other is the effect. Question does not make sense.

It doesn't have anything to do with it.

It just costs more money and takes up more space.

It doesn't have anything to do with it.

Keep'em coming.

Ratch

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5. ### Colin Mitchell

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Aug 31, 2014
"Somehow I'm not so sure about this and very few sources on the web mention that the inductor and capacitor shall be matched so that the respective amount of energy they can each absorb is close to equal.""

That's the very thing I say in my discussion.
To get an even width (similar width) for the part of the cycle that involves the charging of the capacitor to the time taken to fully energise the inductor, the two components must be matched.

6. ### Colin Mitchell

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Aug 31, 2014
Basically the "Q" of a circuit is the voltage across a coil when energy (current) is being delivered to it, compared to the voltage it produces when the current is turned off.
In most cases for a tuned circuit, the Q is 2 because the voltage applied to the coil (inductor) is approximately equal to rail voltage and when the current is turned off, the magnetic flux collapses and creates a voltage in the turns that charges the capacitor in the reverse direction. By the time all the magnetic flux has been converted, you will ind the capacitor has only charged to a potential about the same as rail voltage (but in the opposite direction). This gives a total of twice rail voltage for the TANK CIRCUIT.

7. ### Ratch

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Mar 10, 2013
I think that is more complicated than it has to be. As long as the LC product is the same, the resonant frequency will be the same. The value of the circulating current at resonance is the exciting voltage across the tank circuit divided by either the inductive reactance or the capacitive reactance (both reactances are the same at resonance). If the L is high compared to the C, the impedance across the tank will be high and the circulating current low. Likewise, if the L is low, the tank impedance will be lower and the circulating current will be higher. There is no reason to have a higher circulating current than necessary when it can be adjusted to a lower value if necessary.

Ratch

8. ### Colin Mitchell

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Aug 31, 2014
Ratch

I don't know what you are talking about, and neither do you.

9. ### Ratch

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Mar 10, 2013
What is it I said that you do not understand? You have to understand what I am saying before you can reach a conclusion.

Ratch

10. ### Colin Mitchell

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Aug 31, 2014
The tuned circuit is "topped up" on each cycle by the circuit (transistor or IC) and the current passes from the inductor to the capacitor and is then topped up again via the circuit.
You imagery of a "circulating current" does not exist.

"There is no reason to have a higher circulating current than necessary when it can be adjusted to a lower value if necessary."

Where did you get this guff from?????
You obviously don't know what you are talking about.
I have been describing TANK CIRCUITS for 40 years an NO-ONE has ever come up with line of nonsense.

11. ### Ratch

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Mar 10, 2013
You have been wrong about that for 40 years. A tank circuit circulates the current at the resonant frequency. In fact, once the circulation starts, the exciting voltage can be disconnected and it will circulate the current forever provided no resistance is present. Of course, it is impossible to completely eliminate all resistance. The L and C swap energy back and forth at twice the resonant frequency.

Applied electrical principles and physics.

It is analogous to a mass tied to a flexible spring hanging from a ceiling. The mass will oscillate up and down until mechanical friction slows and stops it.

Which is surprising. I would think you would discern the operation of a tank circuit by now.

Ratch

12. ### Colin Mitchell

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Aug 31, 2014
"In fact, once the circulation starts, the exciting voltage can be disconnected and it will circulate the current forever"

Yes, the current does circulate AFTER the applied voltage is removed. But when it is oscillating, the energy is topped up during each cycle. The energy is added during each cycle and passed from the capacitor to the inductor and then back to the capacitor and topped up.
This is not a "circulation" because the top-up is considerable.

13. ### Ratch

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Mar 10, 2013
The top off is dependent on the energy loss due to resistance, so you cannot say that the top off is considerable unless you know the energy losses. If no losses, then no top off is needed. Are you telling me that the current in each branch is not alternating, and the phase of the current in each branch is not 180° apart? If the above is not true, then that appears to be an oscillating circulation to me.

Ratch

14. ### Colin Mitchell

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Aug 31, 2014
The mere fact that you can detect radiation and get a signal some metres from the circuit proves the losses to be considerable.

15. ### Ratch

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Mar 10, 2013
That in itself proves nothing. If something is taking energy from the tank circuit, whether a resistor or antenna, a damped sine wave will result. The tank circuit will still circulate the current with decreasing amplitude after each cycle unless replenished by the exciting voltage. Do you still believe that the current does not circulate?

Ratch

16. ### pgib8

107
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Jul 26, 2015
I should probably have mentioned that I had a transistor in there too to top it off. I thought it would be fun to build a little FM transmitter for the purpose of learning more about tuned circuits.

I wind my own air-core RF coil, I have one with 5 turns and I made another one with 9 turns. This is just to experiment with. The schematic calls for a very small capacitor but just for fun I've been using 0.1uF.

I don't know if the losses are considerable or not. I only figured in my case they weren't because it's practically a few inches of solid copper wire (coiled up) and soldered directly to a through-hole ceramic capacitor. That's the only reason why I was ignoring losses.

I was using one of those SDR (NooElec) USB dongles, which are pretty fun in their own right. I was seeing the RF energy being radiated from my oscillator and it wasn't a clean sine wave, but had lot's of other noise in it. 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.

There is no need to debug my circuit because it's just for me to play around and learn new things. It approximately looks like this:
http://www.talkingelectronics.com/projects/Spy Circuits/imagesP2/Page2-Cct0.gif

Except I omitted anything to the left of the 47k.

Anyways, I think that many people understand the tuned circuit at the surface but once you get really into the details it becomes quite complicated.

I did not really get any information from the first two answers, so I really appreciate Ratch and Colin's input. Even though you two may disagree, from my perspective it looks like both of you have a deep understanding of the tank circuit.

Colin you said:
To get an even width (similar width) for the part of the cycle that involves the charging of the capacitor to the time taken to fully energise the inductor, the two components must be matched.

Are you referring to the same thing too that Ratch pointed out, that if the L and C are not matched, you are essentially just "wasting" extra inductance/capacitance but that otherwise it does not have any real effect on the oscillation (losses ignored)?

Colin do you agree that L and C do not need to be matched, provided that material is not a matter and provided we can ignore losses?

17. ### Colin Mitchell

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Aug 31, 2014
What I am trying to emphasis is this.

You are talking to beginners and talking about "circulating current" infers that current is circulating around and around the the two components.
Current (energy) is added to the circuit on each cycle and is then passed to the inductor and back to the capacitor and topped up by the "driver."
You are giving the wrong impression to a beginner by not explaining things clearly.

18. ### Colin Mitchell

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Aug 31, 2014
Colin do you agree that L and C do not need to be matched, provided that material is not a matter and provided we can ignore losses?

What is your reason for not matching the two components?

19. ### Ratch

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Mar 10, 2013
I disagree. I explained very clearly what is going on. It is you who first denied that the current was circulating. Whether extra energy is supplied by the driver is irrelevant, the current is still circulating back and forth.

Ratch

20. ### pgib8

107
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Jul 26, 2015
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.

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