The way it works is that the on/standby button goes to a circuit on the motherboard, and that circuit drives the enable signal to the power supply. Because we don't know exactly what that motherboard circuit is, the safest thing is to connect the output of an optocoupler in parallel with the switch.
220Vac > blackout detector > optocoupler
The circuit must hold the switch closed for at least several seconds. After that, continuing to hold it closed until the circuit's energy supply runs out is not a problem. For applications like this, that energy supply usually is a large capacitor; but you can get away with a smaller one. Here is a plan that is a bit more complicated but will give better results:
220Vac > optocoupler > blackout detection > 5 second delay > output optocoupler > chassis on/standby switch
The 5 second delay prevents the computer from shutting down during a quick power bounce. This can be lengthened to any time period you are comfortable with. Since you can use the computer's own +12 V supply to run the circuit, the circuit down to two separate optocouplers, 1 transistor, and some D's, R's and C's.
Blackout detector - either a relay with a 220Vac coil or an optocoupler (with a 10 W resistor in series with the input diode). This keeps a timing capacitor discharged.
Delay - when power goes out, the timing capacitor starts charging. After 5 seconds it has enough voltage across it to turn on a small MOSFET.
Output - the FET drives an optocoupler continuously until the computer's +12 V supply goes away, indicating the power suppl has shut down. The optocoupler's output transistor is in parallel with the on/standby switch contacts. You will need to measure the switch contacts with a voltmeter to determine which way to connect the opto.
I can whip up a schematic, but first - How does that sound?
ak
