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.
My first comment was this:
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.
And another answer:
Of course, there is a cap in parallel to the inductor.
As a consequence, we have a tuned circuit, which determines the oscillator frequency.
And a third reply:
The two capacitors are across the inductor with the battery in series.
I made some comments about turning the transistor ON and OFF and get this reply:
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.
