Hi Sonaiko,
The transfer ratio for an optocoupler is how well the current in the LED causes the transistor to conduct collector current. With your 4.8k emitter resistor and a 5V supply, the max collector current is about 0.94mA. With its minimum transfer ratio of 100%, the LED current must also be at least 0.94mA. Its forward voltage is about 1.2V, so with 4.5V feeding the current-limiting resistor and LED then 3.3V will be across the resistor and its value will be 3.3/0.94m= 3.5k. :lol: