# oscillator help

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#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
In a circuit one can freely decided the direction of current (or voltage),if done consistently it has no effect on the circuit analysis /behavior .
Hey! I was going to say that but you headed me off at the pass!

Chris

#### Arouse1973

Dec 18, 2013
5,178
Here is another example of your miss-information and actually very basic misunderstanding !
And you brag about teaching teachers...

Current(of electrons) does infect flow from negative to positive.
it is only by convention that we prefer to describe it flowing the other way around from positive to negative.

In a circuit one can freely decided the direction of current (or voltage),if done consistently it has no effect on the circuit analysis /behavior .

Yes and in an A.C circuit they dont go anywhere, they just vibrate back and forth.

#### Arouse1973

Dec 18, 2013
5,178
I have added this to my website under SPOT THE MISTAKE Page 26:

98MHz OSCILLATOR
This discussion has been taken from an electronics forum where a student has asked for component values for the following 98MHz FM oscillator.

You really need a cap across the coil to make a reliable osc.

The answer I got from a technician was:
The effective capacitor across the coil is the one from collector to emitter.

Of course, there is a cap in parallel to the inductor.
As a consequence, we have a tuned circuit, which determines the oscillator frequency.

The two capacitors are across the inductor with the battery in series.

I think, it is not appropriate for this circuit to think in "ON" and "OFF" terms. The circuit produces a sinusoidal output!

Then we have another comment:
The tuned circuit will have a Q value of maybe 100, in other words, the circulating current will be much higher than the input current, this will be built up over several cycles.

Let's look at these comments and see how incorrect they are.

Firstly we need to look at the circuit and see if it oscillates.

The circuit above oscillates. It has a coil and capacitor connected in parallel to make a circuit called a TUNED CIRCUIT.
This type of circuit produces a sinewave when connected to a power supply and then instantly removed.
It does not need any other components and it does not need a transistor to produce this effect.
So, what do all the surrounding components do?
They connect the TUNED CIRCUIT to the supply and quickly remove it.
It must be quickly removed, otherwise the voltage produced by the TANK CIRCUIT will be reduced. In other words the surrounding circuitry will put a load on the TANK CIRCUIT.
That means the transistor must be turned on and then turned OFF very quickly.
This clears up the faulty thinking of one the replies above.
Now we come to the amplitude of the waveform produced by the TANK CIRCUIT.
We are going to remove all the surrounding components and talk about the TANK CIRCUIT.
For a correctly designed tank circuit, the energy stored in the coil is equal to the energy stored in the capacitor.
This is necessary because the TANK CIRCUIT produces a waveform consisting of half a cycle that is below the power rail and half a cycle that is higher than the power rail.
For both these half-cycles to be identical, the component values must match.
The full amplitude (called the peak-to-peak value) will be the addition of these two values.
Q-values have nothing to do with current. They are a voltage-determined value.
You can now see the first circuit does not have a capacitor across the coil and even though some of the cycle may be produced by the surrounding components, the equal parts of the waveform cannot be produced.
We have not described how the TANK CIRCUIT produces a waveform above AND below the power rail.
The easiest way is to remove all the surrounding components and tap the TANK CIRCUIT across the battery.
This will put energy into the uncharged capacitor and it will be charged to 3v. During this time the voltage will appear across the coil and it will produce a small amount of flux that will oppose the voltage and thus very little current will flow into the coil.
We are delivering energy to the circuit for a very short period of time and this is too short for the coil.
Now the supply is removed and the energy from the capacitor is slowly fed to the coil and it produces magnetic flux. The coil does not accept energy any faster than a certain rate because the "applied voltage" - the voltage on the capacitor, produces magnetic flux called EXPANDING FLUX and this cuts the turns of the coil to produce a voltage in the opposite direction to OPPOSE the incoming voltage and that's why the capacitor discharges slowly.
So far we are producing the lower part of the waveform because the bottom plate of the capacitor is at 0v and the capacitor is gradually discharging.
The capacitor continues to deliver current until a point is reached where the coil is producing a "back voltage" equal to the capacitor and suddenly the capacitor cannot deliver any current.
The magnetic flux in the air surrounding the coil cannot be maintained and it collapses. This produces a voltage in the turns of the coil that is in the OPPOSITE DIRECTION.
We not have a situation where the voltage produced by the coil is OPPOSITE to the previous voltage and as the magnetic flux collapses it delivers a voltage to charge the capacitor in the opposite direction.
Since the two components are equally matched in "energy storage" capability, the capacitor is charged to a voltage EQUAL to the original voltage (but in the opposite direction).
Now you can see how the waveform rises to the voltage of the supply rail when the capacitor has no charge and then rises ABOVE rail voltage as the capacitor charges in the opposite direction.
This exchange could continue FOREVER but there are some losses with the magnetic flux and each cycle becomes smaller and smaller.
If we have a transistor that delivers a small amount of energy at exactly the right time during each cycle, the full waveform will be maintained. That is what the transistor and surrounding components do.
For a 3v supply, the peak-to-peak value of the waveform will be 6v.
The "Q" of the circuit is 2. Not 100.

What's the purity of the output in your circuit versus the other? Whar's the level of harmonic attenuation versus the other circuit as this may be important in some cases.

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Colin, you said that you posted your monologue on your website. Got the balls to provide us with a link?

Chris

#### GPG

Sep 18, 2015
452
"For a 3v supply, the peak-to-peak value of the waveform will be 6v.
The "Q" of the circuit is 2. Not 100".
Not an accurate statement.

#### LvW

Apr 12, 2014
604
Just to complete my previous comment:
Excerpt from "talking Electronics":
Suppose the transistor gets warm on a hot day and looses gain.
The collector voltage will rise and the current through the load resistor will decrease. The current through Re will also decrease and the voltage across it will decrease. This means the voltage between the base and emitter will increase and the transistor will turn ON more to counteract the voltage-rise on the collector.

..."looses gain" for rising temperatures? Which gain? Current gain?
"The current through Re will also decrease" for rising temperatures ("on a hot day")?

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Since Colin isn't going to do it will someone please post the link to where Colin posted this on his site?

Chris

#### dorke

Jun 20, 2015
2,342
Since Colin isn't going to do it will someone please post the link to where Colin posted this on his site?

Chris
What site ?

May 8, 2012
4,960

Dec 18, 2013
5,178

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Adam, it's not working for me. I can't see anything past page 15. The website is laid out much like the man's mind... erratically!

Did you see a link to our discussion there?

Thanks,
Chris

#### Arouse1973

Dec 18, 2013
5,178
Adam, it's not working for me. I can't see anything past page 15. The website is laid out much like the man's mind... erratically!

Did you see a link to our discussion there?

Thanks,
Chris

Yes it's there, click on the page 15 and more page numbers should appear. It took me a while to work it out.

May 8, 2012
4,960

#### duke37

Jan 9, 2011
5,364
My first disagreement with Colin was when he said that inductors cost nothing.

I found the Colin's comments on this thread by clicking on 'spot mistakes', then P24 then P25.
I gave up on this thread, there seems to be no point in arguing when the action of a tuned circuit is not understood.
In any tuned circuit including a pendulum or watch balance wheel, there are two methods of storing energy and the energy switches between these twice per cycle. In order to get a constant frequency the loss of energy per cycle needs to be low. This is quoted as the Q of the circuit. A Q of 100 means that 1% of the energy is lost per cycle. A Q of 2 means that the circuit is almost critically damped.

The circuit which Colin advocates would be easier to set up since the feedback and tuning are separated but this was not the circuit we were asked to comment on. I would prefer a Clapp circuit since one side of the tuning capacitor can be connected to earth so easier to adjust but we were not asked for what circuit we thought best.

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Adam, I thought I deleted that image-less screen shot. Actually it wasn't a screen shot but that's what I should have done. No matter though because I've had it up to my keester with him, his website and anything he posts. All I want is for him to stay the hell out of my way! He has his own private sandbox and I think it would be best for all concerned that he confine his playtime (where he rules) there!

The following missing rant was deleted because I like it here at EP and I don't want to be banned!

Chris

#### Martaine2005

May 12, 2015
4,903
Oh dear Uncle Chris,
I told you not to hit the Martini....Especially plurals
Relax with some red grapes (crushed) and bottled.. Works wonders.xx

Marty

#### CDRIVE

##### Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Oh dear Uncle Chris,
I told you not to hit the Martini....Especially plurals
Relax with some red grapes (crushed) and bottled.. Works wonders.xx

Marty
Last night that didn't work at all. In fact I'm amazed that I had the wherewithal to delete the majority of my last post!

If this thread continues I don't know if Colin is going to hang back in the shadows or not but I do know that whether it's here or in another topic he is going to continue to monologue all his arguments on his own website.

Chris

#### Arouse1973

Dec 18, 2013
5,178
Colin does it on purpose to get this sort of reaction I am sure. You only have to look at his website full of capitalised words, and he loves the saying "You obviously know nothing about electronics" This is the sort of thing I am on about.

"A diagram without any values does not give you any idea of how it works. A circuit of this nature obviously has not been produced by a competent technical person"

I used to work with diagrams that had no values many years ago. There are some very good reasons why some companies will do this. But instead of talking about all the possible scenarios he just point blank says it's wrong and you are stupid if you do this.

I think if he insists on being the way he is, people should just ignore him. It's a shame as he has obviously put in a huge amount of work in collating all those designs on his website and could offer much more to this forum.

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