Take a look at this link and also the rest of the tutorial. This should give you a better understanding.
http://www.electronics-tutorials.ws/electromagnetism/magnetic-hysteresis.html

 

@Roderick Young I consider your last explanation every words. That's clear and simple. I appreciate. However, I did this calculation because I just want to develop the idea that you've proposed before to be able to draw a waveform.

So, I did some calculation (a 2nd order problem) about when Q301 is ON. As shown in the picture, the waveform is the voltage across the primary winding of XFMR2. The winding is shown in the attached picture as an inductor.

Is my following understanding correct?
In the view of frequency, due to a large capacitance value, the period of free-oscillated inductor's voltage is very long compared to the ON-time of Q301, so the actual voltage across the inductor is trimmed by the two transistors. (In the manner that Vcc is connected to this L-C circuit for a very very short time before disconnecting it) What's remained is the red waveform. If the capacitor is large enough, the voltage across an inductor at Q301's ON-time will be almost constant near Vcc, which corresponds to the idea that the large capacitor shorts the Vcc to the primary winding when Q301 is on.

I'm not exactly sure I understand all your words, but disregarding any resistance, the equation for voltage looks good for the excitation being a step function. Exactly as you mentioned, for large C, just about the entire Vcc is applied across the transformer winding for the short time that the transistor is on, and you can consider the current to be a linear ramp at that time.

 

Sir, this solution image file might clarify what I want to say . This is the differential equation solving when Q301 is "ON" and Q302 is "OFF".

The final result is the inductor's voltage. How long should we keep Q301 staying "ON" and Q302 staying "OFF"? If the phase of operation is too long, the voltage across the inductor (primary winding) will swing up and down and up again as the result showed. But if this phase is so short, the voltage across the inductor will only be positive.

Thank you for keeping tracking my curiosity
BlackMelon

 

There's the math question, and then there's the practical application. Unfortunately, I have no interest in the math question. We learned these things in college, enough to understand how things work (and write a simulator if necessary) then moved on. You can address the differential equation to Yahoo Answers - there are occasional mathematicians who visit.

In the practical application, how long the pulse transformer is driven is dictated by the needs of the switching converter. After figuring out what the maximum pulse width must be, the capacitor value can be chosen to be "large". Possibly the application notes give a suggested value. I don't think the choice of capacitor value is too critical. If too large a value is chosen, it would serve as a short across that diode, and we might have to wait too long for the primary to reset its current to 0 (you can read that link on Hysteresis that someone provided earlier for why the inductor current should be reset.)

 

@recklessrog Thank you very much for the link. I know now that the diode is actually carrying a reversed current. And this current will make the magnetic flux zero, preventing it from rising too far to saturation. That helps me much!!