Observe if terminals on a PCB became connected. Know nothing about the PCB, except it is powered with the same power supply transformer
The PCB is powered by a 24 VAC transformer, the same powers my uC. I have build a rectifier for my uC, so they have different ones.


You have to be very careful here. If the PCB has a full-wave (bridge) rectifier and your microcontroller has one then the PCB ground and the micro ground are not directly connected. Opto or relay isolation would be a very good idea.


  On the terminals, I measure 34 V when open.


Unusual, but OK.


  But for observing the terminals, I for cannot use an optocoupler, as surely it's not possible to drive a diode with them, right? They must have some high ohm resistors beside them, so when connected, there isn't a short.








simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. (a) What you've got. (b) A simple solution if ...

Switch your multimeter to mA and measure the current by connecting 60 to 61 or 62. If the current is between, say 5 to 20 mA then you can use the circuit of Figure 1b. This allows you to monitor the status of IC1 and IC2 with complete isolation.

If the current is not between 5 and 20 mA then update your question with the additional information.


  I believe there are two possible situations: 1) the rectifiers are in sync (I could connect the terminal I observe 0 V relative to my uC ground, or will I even observe that?), or 2) they are shifted in phase by π, so I would just switch the wires that lead to my rectifier.


Try to avoid connecting the systems together since you don't know the rectifier arrangement.



From the comments:


  I measured 3.85 mA directly between 70 and 72 contacts. 


That should be enough for an opto-isolator.


  I would prefer, if the optocoupler was parallel to the switches. ... If the switch was disconnected, more current would flow through the optocoupler circuit, than the switch circuit (none), charging a capacitor and then letting out pulses for the LED.








simulate this circuit

Figure 2. Your idea won't work. There will be almost no difference in the current to your PCB if the switch is open or closed.

The voltage will decrease by about 1.4 V (because of the infrared LED) so the current could be expected to drop by 1.4/34 = 4%. The PCB will think the switches are always closed.