Zero cross detector output

Question

I have realized what I thought was a classic zero cross detector circuit using an H11AA1 chip: https://www.digikey.com/product-detail/en/on-semiconductor/H11AA1SR2M/H11AA1SR2MCT-ND/2094174

My circuit is the following:

Note, I am using a H11AA1SR2MCT chip in my physical circuit (linked above) which should give me all zero-crossings.

I am now probing with an oscilloscope; I did not attach this circuit directly to 110VAC, I am using a toroidal transformer that gives a ~42V peak to peak isolated output. In the following image, the yellow trace is the transformer output, the green one is the signal at "PAD3" in the above circuit referenced to the local GND. I am using a 3.3V power supply.

I was planning to use that signal to generate an interrupt on a microcontroller very close to zero crossing, but so far it's simply unusable, voltage only goes down to 2.1V instead of zero and the timing characteristics are very far from the actual zero-crossing, even if they stay in sync with the wave. Is there an obvious mistake I am not seeing? Thank you

Answer 1

Either increase R3 or decrease R9/R10, that optocoupler only has a current transfer ratio of 20%,

42V / 94K = 447uA

447uA * 0.2 = 89.3uA

10K * 89.3uA = 0.893V, so everything lines up perfectly,

Answer 2

• You have no reverse polarity protection on the LED. It is likely to fail. Generally it's a bad idea to apply more than 5 V in reverse across an LED.
• With almost 100 kΩ of current limiting on the LED current you are getting 42 / 100k = 0.42 mA RMS through the LED. This is a bit mean and the LED is barely lighting.
• According to the datasheet the CTR (current transfer ratio) is 50% under specified conditions. You need to take that into consideration in designing the transistor side of the isolator. As it stands the pull-down isn't strong enough.

Note on schematic: It's generally a good idea to have the current flow from top to bottom. Invert the left side of the opto-isolator and move V_CC to the top of the page and the emitter and ground towards the bottom.

Answer 3

the timing characteristics are very far from the actual zero-crossing

It looks pretty close to me - see the red lines I've added. In other words the middle of the peak of the yellow waveform corresponds entirely with the centre-points of the yellow waveform's rising and falling edges.

I did not attach this circuit directly to 110VAC, I am using a toroidal transformer that gives a ~42V peak to peak isolated output.

Well, that doesn't really help you. 110 volt AC has a peak to peak amplitude of 311 volts and you are testing your circuit with a signal that is 7.4 times smaller. This means that the output waveform will look a bit sloppy.

You would much improve things if you either reduced your input resistors by 7.4 times and tested on 42 volts p-p. And, what you'll find is that the green waveform will become rectangular and the positive peak will be "thinner" and much more indicative of the AC input's zero crossing points.

As a side note - don't expect perfection from circuits like these - they do need some signal processing to get to the information you want.