Help with using 4N25 optoisolator
You have two things that need changing:

1: A 4N25 only has what is called a Current Transfer Ratio of as low as 20% (4N25 datasheet). This means that the current out of the photo-transistor may be as little as 1/5 of the current the is injected into the LED. You have less than 2&nbsp;mA being put into the LED, but require about 10&nbsp;mA from the 4N25 phototransistor.

Also from the data sheet the absolute maximum current you are permitted to inject into the LED is 60&nbsp;mA. You may be able to inject enough to get the output current you require, but it would probably be better to put amplification after the optocoupler.

The LM339 is only guaranteed to be able to sink 6&nbsp;mA (typically 16&nbsp;mA), so it will have difficult time directly driving the 4N25 with a high current.

2: You have the output LED on the ground side of the photo-coupler. This will prevent the output (Out) going below about 2&nbsp;V (LED voltage plus the voltage across the photo-coupler). Again if you put a transistor after the photo-coupler to amplify the current it should be no problem to meet your requirements.

You could also do something like using two couplers with the LEDs in series driven by about 40&nbsp;mA, but put their outputs in parallel. Drive the output LED in parallel with R8, it could be operated with only 1&nbsp;mA or so as modern LEDs can be bright with only a small current.