Detect 220 VAC with a Raspberry Pi

Question

I want to build a simple device to check if the light is on or off. I found some schematics but I need to be sure that it is safe.

Source : How to detect 220 VAC voltage using an opto-isolator

My problem is that I do not understand this very well. I found that PS2501 have a chance of working up to 80V so how can I safely connect it to 220 VAC? Which resistors make a limit for the voltage here? Why do we need to add a capacitor and diode (D1,C1)?

I'm a beginner in this field so please be patient.

Answer 1

If safety is a primary concern, rather than build that circuit up, you could consider purchasing a standard AC-input module for about $10-$15. They are UL, CSA, CE, and TÜV safety certified (it's still possible to go wrong and create a dangerous situation, especially if the wiring is sloppy, but less likely).

Best to have someone knowledgeable look it over before you power it up in any case.

The circuit inside the housing is somewhat similar to the one you show, however it will respond much faster.

---

With regard to the circuit you show the PS2501 has a short-term isolation voltage rating of 5000V RMS. That is adequate to safely withstand the 240VAC mains and most transients that might appear on it.

For safety you need to keep the creepage (surface leakage) distance between input and output leads (of the opto) to at least 8mm and make sure it can never get wet or otherwise contaminated with conductive materials.

The circuit on the low voltage side should be earthed and fuses or other current limiting used so a failure of the opto cannot cause a hazardous condition. R1 and R4 may or may not be acceptable depending on the type. They can certainly burn up under some conditions.

Without an earth connection on the isolated side you are depending on a few mils of plastic inside that opto for safety.

As far as your other questions, the 90V rating is only the output transistor- in operation, it sees only 5V and the 240V input is reduced to the 1.2V the LED needs through the components to the left of the isolator.

All the latter parts are electrically "hot" (including that side of the opto) and need to be well protected against accidental contact.

The capacitor C1 MUST be an X2 type which is a safety certification for cross-mains use. R1 and R2 must be capable of withstanding mains voltage and transients. Vishay VR25, 35 etc. series is appropriately rated.

C1 is what really does the work of dropping the 240VAC down to 1.2V. Most of the mains voltage appears across it. On positive half-cycles the current flows through the LED in the optocoupler, on negative half-cycles the current flows through the 1N4007. The resistors are there mainly to limit the current if powered up when the voltage is not zero.

Answer 2

The cheapest and simplest solution is just to buy a tiny mains-powered USB charger and wire up the USB +5 and ground lines to detect when the charger is powered by AC.

Answer 3

I strongly suggest to reconsider using your circuit in favor of detection without direct connection to mains voltage. Some examples are:

• Photodiode to detect light
• Coil-based current transformer to detect current
• Current transformer with Hall effect sensor to detect current

All of the above have the benefit that they do not trigger if your light bulb is burnt out (if that is what you want).

• Non-contact voltage sensor (NCV) circuit to detect AC voltage

The latter can trigger when the switch is turned ON even if the bulb is burnt out.

Not only the options above are much safer, their circuits are usually simpler than the monstrosity in your question.