...it is flyback smps (as power fractor corrector).
Anyway, the opto (4N35) has a max Current Transfer Ratio of 1 (min is 0.5).
So there is no way that the opto transistor can have any more current through it than the opto diode...since CTR is 1 at maximum. ?
...in figure 16, (page 17) the 5k1 resistor above the opto diode would have 15 - 1.2 - 2.5 / 5100 = 2.2mA.....so the opto diode has 2.2mA through it when its on.
However, the opto transistor in figure 16 feeds into a 2k4 resistor..from a 14Volt rail...this would mean the opto transistor passing 14/2400 = 5.8mA.
...This is more than the 2.2mA in the opto diode so how can it be possible?
(since the CTR is 1 maximum for 4N35 opto-coupler)
No, the CTR is a maximum of 1. That means the transistor cannot have more than 2.2mA through it. And that simply means the voltage across the 2.4k resistor will never be more than 2.2*2.4=5.28V. In other words, the transistor will never saturate with that resistor.