During Period A to B, Q1 is supplying current to the Load via the L1 inductor.
This is due to the forward Vbe voltage on Q2 , relative to Vout.
At time B, Vout rises so that Q2 stops conducting, as it does so positive (negative) feedback via C2 drives it off hard and fast.
At time B the current flowing thru the L1 inductor is suddenly cut off, this causes the magnetic field around the inductor to collapse back into the inductor , this results in a current thru the D1 diode , L1 and the Load. So the collapsing field of the inductor is the source of this current.
At time C, the Vout (actually the voltage produced by the inductor - not Vout) has fallen enough so that Q2 starts to conduct, as it does so positive feedback via C2 drives it hard and fast on.
So the major current sources for the load during time A to B is Q1 and from time B to C is the current loop around the L1, Load and D1 due to the collapsing magnetic field of the inductor.
The function of C2 is positive feedback at the switching times of Q2, so that Q2 and Q1 switch hard and fast.
You are entirely incorrect ericgibbs.
We are NOT talking about positive or negative feedback as applied to an oscillator.
We are talking about positive or negative voltage on the base of one of the transistors.
From what I have read of your posted descriptions, how you can claim this is what I have been saying all along is amusing
If you read my full description on my website, you will find that the description in #22 is what I have been saying all along.
C2 provides a negative voltage (a reduced voltage ) on the base of Q2 to turn it OFF and a positive voltage to turn it on harder.
You can also call these FEEDBACK VOLTAGES.
Your above reply displays your complete lack of understanding regarding electronics fundamentals.
I can see no further point in trying to explain this topic to you, as are not prepared to listen and learn.
Just wondering if you would care to share your analysis and simulation you are basing your textual description of the circuit from.
No. You have got it wrong.
The overall feedback is called POSITIVE FEEDBACK as it keeps the circuit oscillating. But we are now talking about the individual positive and negative part of the feedback signal (actually the CONTROL SIGNAL - because it is not directly connected from the output to the input) and its effect on turning the transistor ON or OFF.
because it is not directly connected from the output to the input)
hi Roman,
LTSpice simulations of your circuit.
...
Finally the tripover point is reached where the base drive in Q1 and the inductor current forces Q1 to start to turn off. This small change causes the collector voltage of Q1 to fall slightly too because of the load from L1. This small change is immediately felt at the base of Q2 because of the coupling from C2, which pulls the base down slightly. Thus some of the energy from C1 is pulled through C2 now and the current goes through the inductor to the output because the inductor provides the path to the output load to ground. Thus some of the energy from C1 gets to the output during this short time when Q1 starts to turn off.
The above needs to expanded.
Finally the tripover point is reached because the current through the inductor is not increasing and thus the flux is not expanding. Thus the voltage produced across it decreases and this reduced voltage is passed through C2 to lower the voltage on the base of Q2.
This action charges C2 but the energy from C2 does not get passed to the load though the inductor.
The inductor has a much-greater effect on charging the capacitor during this time. The capacitors C1/C2 have no “strength” or “ability” to pass their energy to the load.
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