There, the AC phase is monitored with a H11AA1 optocoupler. The 2 15k resistors seem to make sense, with 240V RMS, they bring the current to around 10mA, as required by the IC. The only problem is: 10mA at 240V gives more than 2W power dissipation.

Look again. That article is written for 120vac /60 Hz (USA)

The AC signal (in the US anyway) is 60 Hz. What this means is that the AC signal crosses zero, reaches peak positive voltage, crosses zero, reaches peak negative voltage and returns to zero 60 times each second. The period (length of time this takes) is 1/60 or 0.01667 seconds (16.67 milliseconds). A half cycle (the time between two zero-crossings) occurs in 8.33 milliseconds. This is t3 in the figure above.

The H11AA1 forward voltage is 1.2V @ 10 mA

With a 15k resistor in series with the LINE and another in series with the Neutral, the total series resistance is 30 k ohms. The peak voltage is SQRT(2) * Vrms = 1.414* 120 V = 169.68 (170 V)

R _{CL} =(V_{peak}-V_{f})/I_{f} = (170V-1.2 V)/0.010 A = 168.8/0.01=**16880 ohms**

If you calculate for Vrms it comes to 120vac/0.010 A = 12000 ohms.

P_{RCL} = I^{2} * R_{CL} = (0.010 A)^{2}*(15000+15000) =

**3 W**

I really don’t need a device that blows out more than 2W of power for the monitoring alone. Am I overlooking anything? Are there more efficient ways to monitor AC phase?

It is what it is…

You might be able to use an op amp. The input current is in the order of nA. I don’t see what the issue is with “blowing out 3W”. Is the electricity expensive where you are ?