Detecting 16 V AC with a GPIO

Question

My project involves detecting a ringing doorbell. The doorbell circuit is 16 V AC and I want a low-voltage binary signal from it. The obvious way to me is to transform that electrical signal until its properties are useful (read: not harmful) to a digital GPIO. Firstly, by getting a DC signal by feeding it through a rectifier, secondly bringing it to my microcontroller's operating voltage with a voltage regulator. These ought to solve the AC and high voltage problems, but they don't cover current. A breadboard test showed 0.38 A from the regulator output, which I'd expect to be too much for a data GPIO.

Is this a real problem, or does the GPIO not actually need to sink all of that current?

If it is a real problem, what kind of circuit is needed to cut down that ~400 mA by a couple orders of magnitude?

Answer 1

You fail to describe the nature of the AC signal, other than its amplitude, and even that is ambiguous. Is that RMS or peak voltage? This source has two terminals, and all we know about it is that one of the "terminals" of that AC source rises and falls in potential with respect to the other, cycling between 16V higher in potential (or 23V if it's RMS), then 16V (or 23V) lower. We don't know the potential of either individual terminal with respect to the microcontroller's own ground, and without such knowledge it's foolish to connect either directly to your circuit.

I do not doubt this is why you have 0.38A flowing. You don't say where the 0.38A are flowing, but I suspect diode D1 is responsible, because of the unknown potential of the node you call "NEUT" with respect to ground.

When in doubt like this, the only safe solution is to galvanically isolate both terminals of the source of 16V AC from the rest of your circuit:

^{simulate this circuit – Schematic created using CircuitLab}

In the top circuit, I use transformer T1 as the isolating element. It will be bulky, heavy and expensive, and this method is not recommended. In this arrangement, T1 provides me with a "copy" of the potential difference V1, but also permits me to safely "ground" one side of that copy (node G), so that I can be sure that the other terminal (node X) is rising and falling in potential by ±23V with respect to that ground. Only then can I rectify it, with D1 and C1.

Your regulator AZ1117 would probably work as the "limiter" here, but it's not designed for this. It's a complex IC designed for a (relatively speaking) stable input potential, and which provides a stable, lower, output potential. Using it as a level translator here is not a good idea.

The lower circuit employs a 4N25 (or similar) opto-isolator (sometimes called opto-coupler), to galvanically isolate the 16V AC source from the MCU circuit. In this setup the 16V source is used to illuminate an LED, which switches on a transistor. We can safely connect that transistor to our MCU circuitry, bringing the "information" signal to our own ground level, without worrying what voltage shenanigans are happening on the AC side.

Diode D3 is needed to protect the LED, by ensuring that it can never be significantly reverse biased. Do not omit D3.

The above implementation will cause the MCU input to pulse once for every input AC cycle, which is probably not what you want. You can condition the pulses to become a steadier, prolonged "low" (0V) signal when input AC is present, but returning high (+3.3V) 0.1s or so after AC disappears:

^{simulate this circuit}

Answer 2

Several answers have mentioned optocouplers but they have not mentioned that there are AC input optocouplers that have back to back LEDs, for example this one.

With these the phototransistor is illuminated on both halves of the AC cycle, and it eliminates having to add an external diode or rectifying the AC voltage.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

Safest way is to use an optoisolator. Here is a method, showing the response time to a brief buzz of the bell.

The 4 diodes can be a packaged bridge rectifier of just about any kind with an adequate voltage rating.

Answer 4

Your circuit makes a very little sense.

One solid/safe way is to use a standard optocoupler (like old 4N25X or VO615A, page4 )

A 10k series resistor and extra diode on input (in parallel, in opposite direction) will do the job.

Answer 5

Thanks to the many helpful answers, I've learned a lot about the properties of this problem space. It seems like my main problem is that even if I found a "neutral" pole on the incoming AC, I couldn't trust its potential relative to my circuit's GND, which would absolutely fry my components. Because of this, galvanic isolation is needed so to acquire a voltage relative to the device ground. All of the answers suggested an optocoupler because of their size, cheapness, and ease of integration here by pulling the GPIO low.

Here is my revised design. It uses a small capacitor on the output as @Simon Fitch suggested to prevent cut-off at the cycle midpoint, and an AC input optocoupler as @GodJihyo suggested to remove the need for a rectifying diode. This schematic assumes the MCU GPIO (bell_in) has an internal pull-up resistor.