integrated-circuitpower
Looking at the datasheet of H11L1 Datasheet, for the Total Device, the power dissipation is 250 mW, does this mean that assuming I put 5V (Arduino) on it, it uses 250 mW / 5 V = 50 mA ?
I think it's quite a lot (assuming I need 3 or 4 in my enclosure), that alone is already 150-200 mA.
(Also I'm wondering if very basic components like diodes, capacitors between GND/5V of ICs etc), pullup/down resistors also uses a lot of power?
A simple battery backup IC will do what you want.
Essentially, a comparator samples two voltage inputs, the main power, and a backup power supply (Batteries). Based on that, two cascading inverters paired with two p-channel fets power the circuit at Vo, while two n-channel fets are used as open-drain outputs. Pull them up with a resistor, and they are the perfect interrupt source for your arduino.
The specific one shown is the ICL7673, but it's maximum current is 38mA. It's designed for rtc and ram with a coincell, but an extra pair of P-channels can be used to increase the current.
There are other battery backup or power switches which will do higher current builtin.
You can use ideal diode circuit LTC4357/4359. The 12VDC has to have a slightly higher voltage then your "transformer" if you want 12VDC to take over.
Best Answer
This 250 mW is an absolute maximum rating which means that if something goes wrong you can be sure that the device won't blow up if the power dissipation is below this value. This is far above the normal operational conditions.
What you need is figure 6. from page 3.
A couple of mA that is normally needed.