Exceeding _Enable input voltage_ of optocoupler

I have microcontoller board with following optocoupler interface

which is obviously made for input up to 5V. Based on datasheet I found, Vf = 1.5 V and If = 15 mA (whereby recommended range is from 0.3 mA up to 10 mA). On page 10, they say that

Enable input voltage (Not to Exceed Vcc by more then 500mV): Max: Vcc + 0.5 V

My question is, if I replace R17 with higher resistance resistor (i.e. 2.4 kOhms) I am still not allowed to connect 24V on OCVCC? With higher resistor, I can limit current through optocoupler, but wouldn't it still exceed Max. enable input voltage? Or will such high resistor cause voltage drop?

BTW, I am amateur doing microcontoller just for fun. If I am asking stupid question, I apologize in advance!

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Commented Jan 1, 2016 at 16:09

On the first page of the datasheet, you can see the following diagram. Enable pin has nothing to do with the LED. You can have any voltage on the connectors, as long as you don't exceed the maximum LED current.

At 24V and with a 2.4k resistor, you would have If = (24-1.5)/2400 = 9.375mA. I wouldn't set it so high for extended periods of time, but it is within the limits.

Don't change the resistor, add a new one in series, because the parallel one will burn. THe 10k resistor is called bleeding resistor it could be placed also in parallel with diode, but they decided in such way, OK. A rough calc gives you the forward current (10k resistor is ommited from calc), as you said Vf is 1.5V (I will believe you, but most optos have 1.2V) so when connecting 5V the current is I = (5V-1.5V) / 390 ohm = 8.9mA. If you want the exact current with 24V , then R = (24-1.5) / 8.9mA = 2.5k total, now subtracting the existing one 390 comes to 2.1k (again the 10k is ommited).
Let we take into consideration also the bleeding resitor, at 5V the current is I=5/10k = 0.5mA and the total current is 0.5 + 8.9 = 9.4mA, now let's compute again the input resitor R= (24V-5V) / 9.4mA = 2.02k - not really big differnce as before calc.