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As Vout increases it becomes high enough to start reverse biassing Q2 , via the emitter connection. This causes the voltage across D1 to fall, this fall is negative going and is coupled back to the Q2 base via C2 , which makes Q2 switch off faster.
That's exactly what happens.That statement suggests that the inductor current somehow magically stops increasing just before the transistor turns off
Because the two transistors are finally fully turned ON.What i'd like to see is how Collin can explain why he thinks the inductor current magically stops increasing.
That's exactly what happens.
The current gets to a maximum due to the two transistors being fully turned ON. It is only the EXPANING FLUX that produces the "back voltage" that separates the high voltage on the emitter of Q1 and the 5v on the output of the circuit.
As soon as this expanding flux cannot be maintained, the voltage on the left terminal of the inductor reduces and this is passed to the base of Q2 via the 1n to turn the transistor OFF
I expect at this point somebody may want to close this thread.
I hope that does not happen because the topic under debate is electronics and more then a few of us are finding it interesting.
The problem is you cannot "see" the circuit working.
I have been investigating circuits for the past 40 years and can "see" how they work.
I wrote up the operation of the circuit weeks ago, long before ericgibbs produced a simulation and an explanation.
However ericgibbs and Mr Al does not understand the intricacies of the operation of the circuit and glosses over the points such as the vital fact that starts the “turn-off” portion of the cycle.
” It's basic inductor circuit operation where we have a (relatively) constant voltage across an inductor and that causes a continual increase (ramp) in current until something else occurs to stop that increase.”
This is entirely incorrect.
How can we have a constant voltage across an inductor and an increasing current through it???
How can we have a constant voltage across an inductor and an increasing current through it"
You cannot take the example of a 1.5v battery because the voltage is very low impedance.For example, connect a 1.5v battery across an inductor.
The mere fact that the current is changing, means the "back-voltage" produced by the inductor will change.How can we have a constant voltage across an inductor and an increasing current through it"
What do you mean?
3v0 said:I expect at this point somebody may want to close this thread.
I hope that does not happen because the topic under debate is electronics and more then a few of us are finding it interesting.
3v0 said:We may have other people here with EE degrees that have not yet looked at this. Perhaps they can try their hand at it.
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.
Oh well, here we go....again.
Shakes head
This Forum is going NOWHERE unless we all pull together. It is not hard to do.
Please stop arguing amongst yourself Members. Common.
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