Lower the current and/or lower the voltage drop over the zeners, (change from 2 to 3 zeners)Pelle
What I can see from your layout you have the indication LED in parallell with the optocoupler.Connect the indication LED in series with the optocoupler and the current are half of what is now.One less component (2,2K?)Pelle
Your revised schematics need amending. If you are going to connect the indicator LED in series with the opto isolator, then the 1N4148 diode should be in parallel with the series combination of the indicator and the opto isolator. Otherwise, if the 80V input is reversed, you will blow the LED.
Another issue with both the original and the revised circuit is that if you reverse the 80V input, then almost the full 80V appears across the 2.2k resistor, and it will dissipate about 3W. This is a consequence of adding the zener diodes and reducing the series resistor.
You haven't told us what the forward voltage of your indicator LEDs is (or even what colour they are), so I can't calculate the exact current flowing through the circuit.
The full design is here, as you can see there are already 2 x 1n400x at the begining as protection. however, previously I was told to put also this other diode across the optocoupler as it was a faster diode. Is this correct? should I leave it or remove it?
Let me go through your calculations, CRT is new to me. After I read about it I will post again.