This website explains how an electronic ballast of a compact fluorescent lamp works. I understand the first half cycle. I don't understand what will happen after the magnetic core saturates and TR1 cuts off. Would you answer the following two questions,Please? They will make me understand it better.

1. When the core saturates and TR1 cuts off, Why current reverse its direction? What is the motivation?
2. In the second half cycle, The current through L1 is opposite to the mains current. Is it ok? Does the current of the mains stop in this half cycle?
Here is the link of the website: http://kakopa.com/CFL_electronic_ballast/index.html

Thank you very much,

 

The 4 diodes to the left will rectify the AC mains power and C1 will smooth the rectified DC. The top and bottom rails are now DC and essentially isolated from the mains frequency.

I think the whole circuit kicks off with the path through R1, giving power the TR1's collector and to R3 (currnet flows into the base of TR1). Then look at the capacitors that do not allow continuous current flow - they are everywhere. As current slows in one leg of the circuit, the flow of current to transistor bases slows and the other half of the circuit is energized to cause the high frequency AC that a fluorescent tube needs.

So, you are going from 50/60Hz AC mains, to DC, to the inverter on the right half of the circuit that creates, with the help of the coupled coils, high voltage, high frequency AC.

 

1. When the core saturates and TR1 cuts off, Why current reverse its direction?

When TR1 is cutoff, the magnetic field built up around T1 collapses, or reverses direction, which causes the current direction to reverse.

2. In the second half cycle, The current through L1 is opposite to the mains current. Is it ok? Does the current of the mains stop in this half cycle?

It flows opposite the rectified DC current. This is OK, because you have an AC current imposed over a DC current.