digital-logicmicrocontrollerpcbschematicsswitches
Here is an example of circuit using VIshay Optocoupler 4N25
5V to 3.3V GPIO Output (Inverted)
schematic http://www.savagecircuits.com/attachment.php?attachmentid=294&d=1430452949
Logic
+5V Input —> 0.0V output GPIO LOW
+0V Input —> 3.3V output GPIO High
What happens to the output if my input is -5V???
For the logic that you descibed you need a 32 input OR gate followed by an inverter. You do not want the XOR that you show in your diagram.
If you cannot get an OR gate with 32 inputs then work with some smaller sizes. For example, if all you can get are 4-input OR gates then it will take 8 of these gates to accept the 32 total inputs. Next use a second bank of two more of the 4-input gates to pickup the 8 outputs of the first stage. Finally use a 3rd stage of one final gate to accept the two outputs of the 2nd stage. The two remaining inputs of that 3rd stage gate can be tied to GND. The output of that 3rd stage gate connects to your inverter to produce the final output.
Not trying to be offensive, but are you 100% sure your simulation agrees with the board you're debugging? Is this a production-released board that works, or someone's design concept that hasn't been proven yet?
The _MCLR connection definitely isn't normal-looking to me. Being an active low signal it needs a pull-up (as you've discovered). I also am not totally confident that the voltage divider created with that 1.9k and your 10k pullup is going to work over a large population of parts.
You are also right in that the 4N25 biasing current is way off with the original resistor. The Vishay datasheet only specs CTR at a test condition of 10mA, so your expectation of 20% CTR with 33k is not valid. I generally stick with current that matches the test condition and avoid going higher than that if possible, to avoid premature aging / CTR degradation over time.
Best Answer
As was mentioned in the comments, what you stated is actually opposite of what the outcome actually is.
When you have an input of 5V, the LED is off, which means no photons to turn on the phototransistor, so the transistor is off. If the transistor is off, then your your output is 3.3V
When you apply 0V to your input, you have current flow through the LED, and photons turn on the transistor, which basically brings it down to ground, so your output is 0V. '
In order to get the optocoupler to behave the way you specified in your description, you need to change where the input is applied. If you make the anode of the LED your input node, and ground the right side of R1, you can get the circuit to work as you described in text.
So what happens when you have -5V as your input ? The thing to know is that when the LED lights up, is when the transistor turns on, and in order for the LED to light up, you need to have current flow through the LED. In this case, you do because the input is at a lower potential than the anode.
However you also have more current going through that branch now and if the resistor is not sized appropriately, you will blow your LED rendering your optocoupler dead.
But what happens if you used the alternate wiring scheme where the input is applied to the anode instead ?
When the input goes -5V, the LED is reversed biased, since the anode is at -5V and the cathode is at 0V. IF the LED can withstand -5V reverse voltage, you are ok. Otherwise, you risk breaking the LED and you now have a dead optocoupler (again).
Looking at the datasheet, the absolute max reverse voltage is 5V, so you are really pushing boundaries, so your circuit may work always, work until it suddenly dies, or just die right off the bat.