You cannot take the example of a 1.5v battery because the voltage is very low impedance.
In this circuit the impedance is much higher.
The mere fact that the current is changing, means the "back-voltage" produced by the inductor will change.
Hi again,
You're trying to say that the resistance in series with the inductor is much higher so that limits the current, but i covered that case when i mentioned that if the resistance was really that high then the circuit would not oscillate. There's a point there if we gradually increase the resistance it reaches a point where the circuit no longer can start up, and guess what point that is...it's when the transistor can not be forced out of sat by the inductor current rise.
The resistance is quite low, low enough to allow enough current to flow to force Q1 out of sat.
So you really have to point out exactly what you think makes the inductor current stop rising. If you believe it is something else then you have to point it out exactly if you want to convince anyone else. I doubt this will be possible however because simulations verify what myself and others have been saying about this circuit.
The current through the inductor can charge markedly while maintaining a constant voltage across the inductor terminals. That's the way an inductor behaves and that's just basic theory. The equation was given by skyhawk:
v=L*di/dt (with assumed polarity)
Note in this equation that the current has to be a ramp for the voltage to be constant, or the other way around if the voltage is constant then the current will be a ramp. The current is not exponential it is a ramp. It can be a ramp (or very very close to a ramp) because of the short time period we are talking about here which is only part of the full cycle time.
Note that if di/dt changes the voltage wont be constant anymore, but if i is a ramp and only a ramp the voltage will be constant. Since we apply a constant voltage with relatively low impedance the current approximates a ramp very closely.
Do a simulation and look at the inductor current. Note it is a ramp. Also note that the current continues to rise for longer than 200ns after the transistor starts to turn off. If the inductor current stopped all by itself the transistor would still be on and the current would level off rather than reach a sharp peak and then start to fall.
So in the sequence of operation we see the transistor start to turn off, and then some 200ns later the current starts to fall. And we can also note that the current does not level off it reaches a point and then falls with little horizontal portion. There should be a tiny tiny horizontal portion or rather a very small curve (instead of a sharp point at the top) because the transistor can not turn off instantaneously but takes some finite amount of time even with the snap-off feedback of C2. After the tiny downward curve the current then ramps down.
And when exactly does the current start to fall? It falls exactly when the collector voltage passes through the same voltage as the output voltage. As soon as it falls just a tiny bit lower than the output voltage the current starts to fall.
Here is what the two waveforms will look like. Note that the peak of the inductor current is past the point where the collector voltage starts to fall. The collector voltage was hard to draw because it doesnt fall instantaneously but takes some time. The point in time where the peak of the current occurs is the place where the collector voltage passes through 5 volts (the output voltage) on it's way down.
LATER: I added a better diagram below.
Code:
-------
| |<--collector voltage falls here
| |
| | /\
| |/ \ <--inductor current (triangular)
| / \
| /| \
| / | \
| / | \
-- ---------------> time
Note that the peak of the inductor current is PAST the collector voltage fall point.
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